Patent Publication Number: US-2012043968-A1

Title: Variable equalizer circuit

Description:
CROSS REFERENCE TO RELATED APPLICATIONS  
     This application is the U.S. National Stage of International Patent Application No. PCT/JP2010/002357 filed on Mar. 31, 2010, the disclosure of which is hereby incorporated by reference in its entirety. 
    
    
     BACKGROUND OF THE INVENTION  
     1. Field of the Invention 
     The present invention relates to an equalizing circuit configured to equalize a signal. 
     2. Description of the Related Art 
     In order to test whether or not a semiconductor device operates normally after the semiconductor device is manufactured, a semiconductor test apparatus (which will also be referred to simply as the “test apparatus” hereafter) is employed. The test apparatus receives a signal (signal under test) output from a DUT (device under test), and compares the signal under test with an expected value, so as to judge the quality (Pass or Fail) of the DUT, or so as to measure the amplitude margin or the timing margin of the signal under test. 
     RELATED ART DOCUMENTS  
     Patent Documents 
     
         
         [patent document 1] 
       
    
     U.S. Pat. No. 6,937,054 B2 Specification
     [patent document 2]   

     U.S. Pat. No. 7,394,331 B2 Specification 
     In general, a receiver circuit included in the test apparatus and the DUT are electrically connected to each other via a transmission line and a connector. The characteristic impedance Zo (e.g., 50Ω) of the transmission line or the connector is designed so as to provide impedance matching with a circuit block to be connected. Ideally, such an arrangement causes no waveform distortion due to signal transmission via the transmission line or the connector. However, in reality, it is impossible to provide such impedance matching over the entire pass band. Accordingly, such a transmission line or the like functions as an undesirable filter which causes waveform distortion. That is to say, the receiver circuit of the test apparatus receives a distorted waveform even if the waveform output from the DUT is satisfactory. This prevents the performance of the DUT itself from being measured. 
     By providing an equalizer circuit configured to compensate for the distortion of the signal under test as a component upstream of the receiver circuit (e.g., comparator) of the test apparatus, such an arrangement is capable of improving the waveform distortion of the signal under test due to the transmission line or the like. For example, Patent document 1 discloses an equalizer circuit monolithically integrated together with a differential amplifier. Also, Patent document 2 discloses a passive equalizer employing an LRC. 
     SUMMARY OF THE INVENTION 
     The present invention has been made in view of such a situation. Accordingly, it is an exemplary purpose of an embodiment of the present invention to provide a variable equalizer circuit which is capable of adjusting the equalization level using a new approach that differs from conventional approaches. 
     An embodiment of the present invention relates to a variable equalizer circuit configured to equalize a signal received via a transmission line from a device which is a communication partner device. The variable equalizer circuit comprises: an input terminal connected to the transmission line; an output terminal; a first resistor arranged between the output terminal and a fixed voltage terminal, and configured to have a variable resistance; a first capacitor arranged between the output terminal and the fixed voltage terminal, arranged in parallel with the first resistor, and configured to have a variable capacitance; a second resistor arranged between the input terminal and the output terminal; a second capacitor arranged between the input terminal and the output terminal, and arranged in parallel with the second resistor; and a shunt resistor arranged on a path including the first capacitor and the second capacitor between the input terminal and the fixed voltage terminal. 
     Another embodiment of the present invention also relates to a variable equalizer circuit configured to equalize a signal received via a transmission line from a device which is a communication partner device. The variable equalizer circuit comprises: an input terminal connected to the transmission line; an output terminal; a first capacitor arranged between the output terminal and a fixed voltage terminal, and configured to have a variable capacitance; a second resistor arranged between the input terminal and the output terminal; a second capacitor arranged between the input terminal and the output terminal, and arranged in parallel with the second resistor; a shunt resistor arranged on a path comprising the first capacitor and the second capacitor between the input terminal and the fixed voltage terminal; and a level shifter configured to shift the voltage level of the output terminal, and to have a variable resistance between the output terminal and the fixed voltage terminal. 
     Such an equalizing circuit according to any one of the aforementioned embodiments functions as a high-frequency emphasis filter configured to emphasize the high-frequency component of the input signal, and has an advantage of being capable of adjusting the amount of boost and the time constant. Furthermore, such an equalizing circuit can be integrated on a semiconductor chip. Such an arrangement uses no inductor, thereby providing an advantage of a small circuit area, and an advantage of involving no unintended oscillation. 
     Yet another embodiment of the present invention relates to a test apparatus configured to receive a signal from a device under test via a transmission line, and to test the device under test. The test apparatus comprises: a variable equalizer circuit according to any one of the aforementioned embodiments, configured to equalize a signal received from the device under test; and a receiver circuit configured to receive an output signal from the variable equalizer circuit. 
     Such an embodiment is capable of testing a signal output from the device under test after it corrects signal distortion that occurs due to the transmission line or the like. 
     It should be noted that any combination of the aforementioned components may be made, and any component of the present invention or any manifestation thereof may be mutually substituted between a method, apparatus, and so forth, which are effective as an embodiment of the present invention. 
    
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
       Embodiments will now be described, by way of example only, with reference to the accompanying drawings which are meant to be exemplary, not limiting, and wherein like elements are numbered alike in several Figures, in which: 
         FIG. 1  is a circuit diagram which shows a configuration of a test apparatus including a variable equalizer circuit according to an embodiment; 
         FIGS. 2A through 2C  are circuit diagrams showing example configurations of a variable resistor and a variable capacitor; 
         FIGS. 3A through 3C  are circuit diagrams each showing an example configuration of a level shifter; 
         FIG. 4  is a circuit diagram which shows a configuration of a variable equalizer circuit according to a comparison technique; 
         FIG. 5  is a circuit diagram which shows a simplified configuration of the variable equalizer circuit shown in  FIG. 1 ; 
         FIG. 6  is an equivalent circuit diagram which shows a variable equalizer circuit in a static state; 
         FIGS. 7A and 7B  are simulation waveform diagrams for the variable equalizer circuit shown in  FIG. 1 ; 
         FIG. 8  is a circuit diagram which shows a configuration of a variable equalizer circuit according to a first modification; 
         FIG. 9  is a circuit diagram which shows a configuration of a variable equalizer circuit according to a second modification; and 
         FIG. 10  is a circuit diagram which shows a configuration of a variable equalizer circuit according to a fourth modification. 
     
    
    
     DETAILED DESCRIPTION OF THE INVENTION 
     Description will be made below regarding preferred embodiments according to the present invention with reference to the drawings. The same or similar components, members, and processes are denoted by the same reference numerals, and redundant description thereof will be omitted as appropriate. The embodiments have been described for exemplary purposes only, and are by no means intended to restrict the present invention. Also, it is not necessarily essential for the present invention that all the features or a combination thereof be provided as described in the embodiments. 
     In the present specification, a state represented by the phrase “the member A is connected to the member B” includes a state in which the member A is indirectly connected to the member B via another member that does not affect the electric connection therebetween, in addition to a state in which the member A is physically and directly connected to the member B. Similarly, a state represented by the phrase “the member C is provided between the member A and the member B” includes a state in which the member A is indirectly connected to the member C, or the member B is indirectly connected to the member C via another member that does not affect the electric connection therebetween, in addition to a state in which the member A is directly connected to the member C, or the member B is directly connected to the member C. 
       FIG. 1  is a circuit diagram which shows a test apparatus  2  including a variable equalizer circuit  100  according to an embodiment. 
     The test apparatus  2  is connected to a DUT  1  via a transmission line  3 . The test apparatus  2  judges the quality of the DUT  1  or identifies the defective portions based upon a signal output from the DUT  1 . The DUT  1  includes a driver Dr and an output resistor Ru. The driver Dr 1  applies a signal under test Vu to one terminal of the transmission line  3 . 
     A terminator  6  includes a terminal driver Dr 2  and a terminal resistor Rd. The terminal driver Dr 2  applies a terminal voltage Vd to the other terminal of the transmission line  3  via the terminal resistor Rd. The terminator  6  may function as a transmitter circuit (driver) configured to output a signal to the DUT  1 . 
     A receiver circuit  8  receives a signal under test Vu output from the DUT  1 . For example, the receiver circuit is configured as a comparator or a buffer. The test apparatus  2  compares the signal under test thus received by the receiver circuit  8  with an expected value so as to judge the quality of the DUT  1 . Alternatively, the test apparatus  2  measures the amplitude margin or the timing margin of the signal under test. 
     With such a test system, waveform distortion occurs in the signal under test output from the DUT  1  when it passes through the transmission line  3 , an unshown connector, or the like (which will be referred to as the “transmission line” or the like hereafter). In order to compensate for such waveform distortion, the test apparatus  2  includes a variable equalizer circuit  100  arranged as an upstream component of the receiver circuit  8 . 
     Description will be made regarding a specific configuration of the variable equalizer circuit  100 . 
     The variable equalizer circuit  100  equalizes a signal Va input via an input terminal P 1  thereof from the DUT  1  which is a communication partner device, at the same time attenuates this signal, and outputs the signal thus processed to the receiver circuit  8  via an output terminal P 2 . 
     The variable equalizer circuit  100  includes an equalizing unit  10  and a level shifter  20 . 
     The equalizing unit  10  includes a first resistor R 1 , a second resistor R 2 , a first capacitor C 1 , a second capacitor C 2 , and at least one shunt resistor Rs. 
     The first resistor R 1  is configured as a variable resistor, the resistance value of which is changeable. The first resistor R 1  is arranged between the output terminal P 2  and a fixed voltage terminal (ground terminal). The first capacitor C 1  is configured as a variable capacitor, the capacitance of which is changeable. The first capacitor C 1  is arranged between the output terminal P 2  and the ground terminal, in parallel with the first resistor R 1 . The second resistor R 2  is arranged between the input terminal P 1  and the output terminal P 2 . The second capacitor C 2  is arranged between the input terminal P 1  and the output terminal P 2 , in parallel with the second resistor R 2 . 
     At least one shunt resistor Rs is arranged on a path including the first capacitor C 1  and the second capacitor C 2  between the input terminal P 1  and the ground terminal.  FIG. 1  shows an arrangement in which the third resistor R 3  and the fourth resistor Rc each function as a shunt resistor Rs. 
     The third resistor R 3  is arranged between the input terminal P 1  and a connection node (N 1 ) that connects one terminal of the second resistor R 2  and one terminal of the second capacitor C 2 . The third resistor R 3  has resistance that is sufficiently greater than the characteristic impedance (50Ω) of the transmission line  3 . For example, the third resistor R 3  is preferably configured to have a resistance on the order of five to ten times the characteristic impedance of the transmission line  3 . By setting the resistance of the third resistor R 3  to be greater than the characteristic impedance of the transmission line  3 , such an arrangement reduces the effects of the variable equalizer circuit  100  on the impedance matching between the terminator  6  and the DUT  1 . 
     The fourth resistor Rc is arranged on a path in parallel with the first resistor R 1 , and in series with the first capacitor C 1 . 
       FIGS. 2A through 2C  are circuit diagrams showing example configurations of the variable resistor and the variable capacitor.  FIG. 2A  shows an example configuration of the first resistor R 1 . The first resistor R 1  includes a first terminal P 11 , a second terminal P 12 , multiple resistors R 1   1  through R 1   6  arranged in series between the first terminal P 11  and the second terminal P 12 , and multiple switches SW 1   1  through SW 1   5  each arranged between a connection node (tap) of the adjacent resistor and the second terminal P 12 . By switching the states of the multiple switches SW 1   1  through SW 1   5  between the on state and the off state, such an arrangement is capable of switching the resistance of a path between the first terminal P 11  and the second terminal P 12 . It should be noted that the switches SW 1   1  through SW 1   5  are arranged on the fixed voltage terminal (ground terminal) side. It should be noted that the number of resistors R 1  can be determined as desired. 
       FIG. 2B  shows an example configuration of a first capacitor C 1 . The first capacitor C 1  includes multiple capacitors C 1   1  through C 1   4  arranged in parallel between a first terminal P 21  and a second terminal P 22 . Multiple switches SW 2   1  through SW 2   4  are respectively arranged in series with the multiple capacitors C 1   1  through C 1   4 . By switching the states of the multiple switches SW 21  through SW 24 , such an arrangement is capable of switching the capacitance that develops between the first terminal P 21  and the second terminal P 22 . The switches SW 21  through SW 24  are preferably arranged on the fixed voltage terminal (ground terminal) side. It should be noted that the number of multiple capacitors C 1   1  through C 1   4  can be determined as desired. 
       FIG. 2C  is a circuit diagram which shows an example configuration of the switches SW 1  and SW 2  employed in  FIGS. 2A and 2B . The switch SW is a so-called transfer gate, and includes a first transistor M 1  configured as an N-channel MOSFET and a second transistor M 2  configured as a P-channel MOSFET arranged in parallel between a first terminal P 31  and a second terminal P 32 . A control signal S 1  is input to the gate of the first transistor M 1 . A control signal #S 1  obtained by inverting the control signal S 1  by the inverter  32  is input to the gate of the second transistor M 2 . The state of a path between the first terminal P 31  and the second terminal P 32  is switched between the connection state and the disconnection state according to the control signal S 1 . 
     It should be noted that N-channel MOSFETs only or P-channel MOSFETs only may be employed, depending upon the relation between the electric potential at the first terminal P 31  and the electric potential at the second terminal P 32 . 
     It should be noted that the configurations of the variable resistor and the variable capacitor are not restricted to such arrangements shown in  FIGS. 2A through 2C . Rather, the topology thereof should be designed based upon the required resistance and capacitance. 
     Returning to  FIG. 1 , the level shifter  20  shifts the voltage level at the output terminal P 2 . In a case in which the receiver circuit  8  is configured as a comparator or a differential amplifier, such an arrangement has a limited input voltage range. Thus, by shifting the electric potential at the output terminal P 2  by means of the level shifter  20  such that it matches the input voltage range of such a comparator or the like, such an arrangement can be expected to provide high-speed or high-precision operation. 
       FIGS. 3A through 3C  are circuit diagrams each showing an example configuration of the level shifter  20 . The level shifter  20  shown in  FIG. 3A  includes a voltage source  22  configured to generate a first voltage V SH  and a fifth resistor R SH  arranged between the voltage source  22  and the output terminal P 2 . The level shifter  20  is capable of adjusting a level shifting amount by varying the first voltage V SH . 
       FIG. 3B  is a circuit diagram which shows another example configuration of the level shifter. A level shifter  20   a  includes a first fixed voltage terminal (power supply terminal) Pvdd to which a first fixed voltage (power supply voltage vdd) is to be applied, a second fixed voltage terminal (ground terminal) Pvss to which a second fixed voltage (ground voltage vss) that differs from the first fixed voltage (power supply voltage vdd) is to be applied, a first variable resistor R SH1  arranged between the first fixed voltage terminal Pvdd and the output terminal P 2 , and a second variable resistor R SH2  arranged between the second fixed voltage terminal Pvss and the output terminal P 2 . 
     Assuming that the level shifter shown in  FIG. 3B  is equivalent to the level shifter shown in  FIG. 3A , the following Expression A1 holds true. 
     
       
      
       R 
       SH 
       =R 
       SH1 
       //R 
       SH2  
      
     
         V   SH =(vdd· R   SH2 +vss· R   SH1 )/( R   SH1   +R   SH2 )   (A1)
 
     Here, “R 1 //R 2 ” represents an operator that expresses the combined impedance of the resistors R 1  and R 2  arranged in parallel. 
     By solving Expression (A1) for R SH1  and R SH2 , the following Expression (A2) is obtained. 
         R   SH1   =R   SH ·(vdd−vss)/( V   SH −vss)
 
         R   SH2   =R   SH ·(vdd−vss)/(Vdd− V   SH )
 
       FIG. 3C  is a circuit diagram which shows a more specific configuration of the level shifter  20   a  shown in  FIG. 3B . In the level shifter  20   a  shown in  FIG. 3C , the variable resistor shown in  FIG. 2A  is employed for each of the first variable resistor R SH1 and  the second variable resistor R SH2 . 
     The first variable resistor R SH1  and the second variable resistor R SH2  preferably have a configuration in which the multiple switches SW are arranged on one fixed voltage terminal Pvdd side and a configuration in which the multiple switches SW are arranged on the other fixed voltage terminal Pvss side, respectively. Each switch SW has a parasitic capacitance (not shown). However, by arranging the switches SW on the fixed voltage terminal side, such an arrangement reduces the parasitic capacitance that occurs at the output terminal P 2 . As a result, such an arrangement reduces the effects of such parasitic capacitance on a signal transmitted via a node to which the output terminal P 2  is connected. 
     The above is the configuration of the variable equalizer circuit  100 . Next, returning to  FIG. 1 , description will be made regarding the operation thereof. 
     First, the DUT  1  outputs a signal under test to the test apparatus  2 , and the signal under test thus output is input to the input terminal P 1  of the variable equalizer circuit  100  shown in  FIG. 1 . 
     A combination of the second resistor R 2  and the second capacitor C 2  functions as a peaking filter for the signal Va input to the input terminal P 1 . The capacitance C 2  of the second capacitor C 2  is determined so as to provide overcompensation. 
     Furthermore, the first resistor R 1  is configured as a variable resistor and the first capacitor C 1  is configured as a variable capacitor. By adjusting the first resistor R 1  and the first capacitor C 1 , such an arrangement has a function of adjusting the overall characteristics of the variable equalizer circuit  100 . Specifically, the first capacitor C 1  having a capacitance C 1  suppresses overcompensation provided by the second capacitor C 2 . Here, the capacitances of the first capacitor C 1  and the second capacitor C 2  are set such that the relation C 2 &gt;C 1  holds true. Furthermore, the amount of boost by the equalizer can be controlled using the resistance of the first resistor R 1 . 
     With such a test system shown in  FIG. 1 , before the test operation, the user of the test apparatus can measure or calculate the amount of distortion or the frequency characteristics of such distortion that occurs in the signal output from the DUT  1  due to the effects of the transmission line  3  or the like. Accordingly, the user can determine the circuit constants of the first resistor R 1  and the first capacitor C 1  so as to cancel out distortion that occurs due to the transmission line  3  or the like. 
     The equalizing unit  10  equalizes the signal input to the input terminal P 1 , and at the same time attenuates this signal. The level shifter  20  shifts the level of the output signal of the equalizing unit  10 , and outputs the resulting signal to the receiver circuit  8 . 
     The above is the operation of the variable equalizer circuit  100 . The advantage of the variable equalizer circuit  100  can be clearly understood in comparison with conventional techniques.  FIG. 4  is a circuit diagram which shows a variable equalizer circuit  300  according to a conventional technique. The variable equalizer  300  includes an equalizing unit  310  and a level shifter  320 . The equalizing unit  310  includes a third resistor R 3 , a second resistor R 2  configured as a variable resistor, and a second capacitor C 2  configured as a variable capacitor. 
     With such a variable equalizer circuit  300  shown in  FIG. 4 , if the second resistor R 2  is configured as a variable resistor having a configuration as shown in  FIG. 2A , the parasitic capacitance C R2  of each switch is connected between the signal line and the ground terminal. Similarly, if the second capacitor C 2  is configured as a variable capacitor as shown in  FIG. 2B , the parasitic capacitance C C2  of the switch is connected between the signal line and the ground terminal. Such parasitic capacitances C R2  and C C2  lead to a dulled signal being input to the receiver circuit  8 . 
     That is to say, such parasitic capacitances counteract the desired functions of the equalizer circuit. This means that there is a reduction in the response speed of the circuit. 
     In contrast, with the variable equalizer circuit  100  shown in  FIG. 1 , the second resistor R 2  and the second capacitor C 2  are respectively configured as a fixed resistor and a fixed capacitor, and the first resistor R 1  and the first capacitor C 1  are respectively configured as a variable resistor and a variable capacitor. With such an arrangement, the parasitic capacitance C R1  of the first resistor R 1  and the parasitic capacitance C C1  of the first capacitor are not directly connected to a signal line via which a signal is transmitted from the input terminal P 1  to the output terminal P 2 . Thus, such an arrangement provides a circuit with an improved response speed. 
     In addition to such an advantage, the variable equalizer circuit  100  has the following advantages. 
     The variable equalizer circuit  100  is capable of changing the amount of boost and the time constant by adjusting the first capacitor C 1  and the first resistor R 1 . 
     Furthermore, the variable equalizer circuit  100  includes resistors, capacitors, and transistors. This means that the variable equalizer circuit  100  has a configuration suitable for integration on a semiconductor chip. Furthermore, the variable equalizer circuit  100  includes no inductor. Thus, such an arrangement provides an advantage of a reduced circuit area and an advantage that unintended oscillation does not occur. 
     Furthermore, the variable equalizer circuit  100  attenuates the signal at the same time as the equalizing operation. Accordingly, such an arrangement reduces the voltage level to be input to the receiver circuit  8 . Thus, such an arrangement allows the receiver circuit  8  to be configured using high-speed and low-voltage transistors. Thus, such an arrangement is capable of receiving a high-speed signal. 
     Furthermore, by providing the third resistor R 3 , such an arrangement reduces the effects of the variable equalizer circuit  100  on the impedance matching between the terminator  6  and the DUT  1 . Furthermore, by providing the fourth resistor Rc, such an arrangement provides improved bandwidth characteristics. 
     Next, description will be made regarding a qualitative analysis of the variable equalizer circuit  100 . 
     Here it is supposed that impedance matching is provided between the output resistor Ru of the DUT  1 , the terminal resistor Rd of the terminator  6 , and the characteristic impedance Zo of the transmission line  3 . In this case, the impedance at the node N 2  is represented by Zo/2. 
     Furthermore, the resistance of the third resistor R 3  is sufficiently higher than the characteristic impedance Zo as described above. Accordingly, it can be assumed that the effects of the variable equalizer circuit  100  on the impedance matching between the terminator  6  and the DUT  1  are negligible. 
       FIG. 5  is a circuit diagram obtained by simplifying the configuration of the variable equalizer circuit  100  shown in  FIG. 1 . 
     R 1  represents the resistance of the first resistor R 1 , R 2  represents the resistance of the second resistor R 2 , R 3  represents the resistance of the third resistor R 3 , R c  represents the resistance of the fourth resistor Rc, C 1  represents the capacitance of the first capacitor C 1 , and C 2  represents the capacitance of the second capacitor C 2 . 
     First, the following Expression (1) is obtained using Kirchhoff&#39;s current law. 
         i ( t )= i   R2 ( t )+ i   C2 ( t )= i   SH ( t )+ i   R1 ( t )+ i   C1 ( t )   (1)
 
     Each current is obtained as represented by Expressions (2) through (6). Here, G 1 =1/R 1 , G 2 =1/R 2 , G 3 =1/R 3 , and G SH =1/R SH . It should be noted that i ci  is separately calculated. 
     
       
         
           
             
               
                 
                   
                     i 
                      
                     
                       ( 
                       t 
                       ) 
                     
                   
                   = 
                   
                     
                       G 
                       3 
                     
                     · 
                     
                       ( 
                       
                         
                           
                             v 
                             A 
                           
                            
                           
                             ( 
                             t 
                             ) 
                           
                         
                         - 
                         
                           
                             v 
                             B 
                           
                            
                           
                             ( 
                             t 
                             ) 
                           
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   ( 
                   2 
                   ) 
                 
               
             
             
               
                 
                   
                     
                       i 
                       
                         R 
                          
                         
                             
                         
                          
                         2 
                       
                     
                      
                     
                       ( 
                       t 
                       ) 
                     
                   
                   = 
                   
                     
                       G 
                       2 
                     
                     · 
                     
                       ( 
                       
                         
                           
                             v 
                             B 
                           
                            
                           
                             ( 
                             t 
                             ) 
                           
                         
                         - 
                         
                           
                             v 
                             C 
                           
                            
                           
                             ( 
                             t 
                             ) 
                           
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   ( 
                   3 
                   ) 
                 
               
             
             
               
                 
                   
                     
                       i 
                       
                         C 
                          
                         
                             
                         
                          
                         2 
                       
                     
                      
                     
                       ( 
                       t 
                       ) 
                     
                   
                   = 
                   
                     
                       
                         C 
                         2 
                       
                       · 
                       
                          
                         
                            
                           t 
                         
                       
                     
                      
                     
                       ( 
                       
                         
                           
                             v 
                             B 
                           
                            
                           
                             ( 
                             t 
                             ) 
                           
                         
                         - 
                         
                           
                             v 
                             C 
                           
                            
                           
                             ( 
                             t 
                             ) 
                           
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   ( 
                   4 
                   ) 
                 
               
             
             
               
                 
                   
                     
                       i 
                       SH 
                     
                      
                     
                       ( 
                       t 
                       ) 
                     
                   
                   = 
                   
                     
                       G 
                       SH 
                     
                     · 
                     
                       ( 
                       
                         
                           
                             v 
                             C 
                           
                            
                           
                             ( 
                             t 
                             ) 
                           
                         
                         - 
                         
                           V 
                           SH 
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   ( 
                   5 
                   ) 
                 
               
             
             
               
                 
                   
                     
                       i 
                       
                         R 
                          
                         
                             
                         
                          
                         1 
                       
                     
                      
                     
                       ( 
                       t 
                       ) 
                     
                   
                   = 
                   
                     
                       G 
                       1 
                     
                     · 
                     
                       
                         v 
                         C 
                       
                        
                       
                         ( 
                         t 
                         ) 
                       
                     
                   
                 
               
               
                 
                   ( 
                   6 
                   ) 
                 
               
             
           
         
       
     
     By generating the Laplace transform of the Expressions (1) through (6), the following Expressions (1)′ through (6)′ are obtained. 
     
       
         
           
             
               
                 
                   
                     I 
                      
                     
                       ( 
                       s 
                       ) 
                     
                   
                   = 
                   
                     
                       
                         
                           I 
                           
                             R 
                              
                             
                                 
                             
                              
                             2 
                           
                         
                          
                         
                           ( 
                           s 
                           ) 
                         
                       
                       + 
                       
                         
                           I 
                           
                             C 
                              
                             
                                 
                             
                              
                             2 
                           
                         
                          
                         
                           ( 
                           s 
                           ) 
                         
                       
                     
                     = 
                     
                       
                         
                           I 
                           SH 
                         
                          
                         
                           ( 
                           s 
                           ) 
                         
                       
                       + 
                       
                         
                           I 
                           
                             R 
                              
                             
                                 
                             
                              
                             1 
                           
                         
                          
                         
                           ( 
                           s 
                           ) 
                         
                       
                       + 
                       
                         
                           I 
                           
                             C 
                              
                             
                                 
                             
                              
                             1 
                           
                         
                          
                         
                           ( 
                           s 
                           ) 
                         
                       
                     
                   
                 
               
               
                 
                   
                     ( 
                     1 
                     ) 
                   
                   ′ 
                 
               
             
             
               
                 
                   
                     I 
                      
                     
                       ( 
                       s 
                       ) 
                     
                   
                   = 
                   
                     
                       G 
                       3 
                     
                     · 
                     
                       ( 
                       
                         
                           
                             V 
                             A 
                           
                            
                           
                             ( 
                             s 
                             ) 
                           
                         
                         - 
                         
                           
                             V 
                             B 
                           
                            
                           
                             ( 
                             s 
                             ) 
                           
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   
                     ( 
                     2 
                     ) 
                   
                   ′ 
                 
               
             
             
               
                 
                   
                     
                       I 
                       
                         R 
                          
                         
                             
                         
                          
                         2 
                       
                     
                      
                     
                       ( 
                       s 
                       ) 
                     
                   
                   = 
                   
                     
                       G 
                       2 
                     
                     · 
                     
                       ( 
                       
                         
                           
                             V 
                             B 
                           
                            
                           
                             ( 
                             s 
                             ) 
                           
                         
                         - 
                         
                           
                             V 
                             C 
                           
                            
                           
                             ( 
                             s 
                             ) 
                           
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   
                     ( 
                     3 
                     ) 
                   
                   ′ 
                 
               
             
             
               
                 
                   
                     
                       I 
                       
                         C 
                          
                         
                             
                         
                          
                         2 
                       
                     
                      
                     
                       ( 
                       s 
                       ) 
                     
                   
                   = 
                   
                     
                       C 
                       2 
                     
                     · 
                     
                       { 
                       
                         
                           s 
                           · 
                           
                             ( 
                             
                               
                                 
                                   V 
                                   B 
                                 
                                  
                                 
                                   ( 
                                   s 
                                   ) 
                                 
                               
                               - 
                               
                                 
                                   V 
                                   C 
                                 
                                  
                                 
                                   ( 
                                   s 
                                   ) 
                                 
                               
                             
                             ) 
                           
                         
                         - 
                         
                           ( 
                           
                             
                               
                                 v 
                                 B 
                               
                                
                               
                                 ( 
                                 
                                   0 
                                   - 
                                 
                                 ) 
                               
                             
                             - 
                             
                               
                                 v 
                                 C 
                               
                                
                               
                                 ( 
                                 
                                   0 
                                   - 
                                 
                                 ) 
                               
                             
                           
                           ) 
                         
                       
                       } 
                     
                   
                 
               
               
                 
                   ( 
                   
                     4 
                     ′ 
                   
                   ) 
                 
               
             
             
               
                 
                   
                     
                       I 
                       SH 
                     
                      
                     
                       ( 
                       s 
                       ) 
                     
                   
                   = 
                   
                     
                       G 
                       SH 
                     
                     · 
                     
                       ( 
                       
                         
                           
                             V 
                             C 
                           
                            
                           
                             ( 
                             s 
                             ) 
                           
                         
                         - 
                         
                           
                             1 
                             s 
                           
                            
                           
                             V 
                             SH 
                           
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   
                     ( 
                     5 
                     ) 
                   
                   ′ 
                 
               
             
             
               
                 
                   
                     
                       I 
                       
                         R 
                          
                         
                             
                         
                          
                         1 
                       
                     
                      
                     
                       ( 
                       s 
                       ) 
                     
                   
                   = 
                   
                     
                       G 
                       1 
                     
                     · 
                     
                       
                         V 
                         C 
                       
                        
                       
                         ( 
                         s 
                         ) 
                       
                     
                   
                 
               
               
                 
                   
                     ( 
                     6 
                     ) 
                   
                   ′ 
                 
               
             
           
         
       
     
     Next, directing attention to the relation between the i C1 (t) and v c (t), the following Expression (7) is obtained. By generating the Laplace transform of Expression (7), Expression (7)′ is obtained. Furthermore, Expression (7)′ is solved with respect to I C1 (s) by removing V P (s), thereby obtaining the following Expression (8). 
     
       
         
           
             
               
                 
                   
                     
                       i 
                       
                         C 
                          
                         
                             
                         
                          
                         1 
                       
                     
                      
                     
                       ( 
                       t 
                       ) 
                     
                   
                   = 
                   
                     
                       
                         1 
                         
                           R 
                           C 
                         
                       
                        
                       
                         ( 
                         
                           
                             
                               v 
                               C 
                             
                              
                             
                               ( 
                               t 
                               ) 
                             
                           
                           - 
                           
                             
                               v 
                               P 
                             
                              
                             
                               ( 
                               t 
                               ) 
                             
                           
                         
                         ) 
                       
                     
                     = 
                     
                       
                         C 
                         1 
                       
                       · 
                       
                         
                           
                             v 
                             P 
                           
                            
                           
                             ( 
                             t 
                             ) 
                           
                         
                         
                            
                           t 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   7 
                   ) 
                 
               
             
             
               
                 
                   
                     
                       I 
                       
                         C 
                          
                         
                             
                         
                          
                         1 
                       
                     
                      
                     
                       ( 
                       s 
                       ) 
                     
                   
                   = 
                   
                     
                       
                         1 
                         
                           R 
                           C 
                         
                       
                        
                       
                         ( 
                         
                           
                             
                               V 
                               C 
                             
                              
                             
                               ( 
                               s 
                               ) 
                             
                           
                           - 
                           
                             
                               V 
                               P 
                             
                              
                             
                               ( 
                               s 
                               ) 
                             
                           
                         
                         ) 
                       
                     
                     = 
                     
                       
                         C 
                         1 
                       
                       · 
                       
                         ( 
                         
                           
                             s 
                             · 
                             
                               
                                 V 
                                 P 
                               
                                
                               
                                 ( 
                                 s 
                                 ) 
                               
                             
                           
                           - 
                           
                             
                               v 
                               P 
                             
                              
                             
                               ( 
                               
                                 0 
                                 - 
                               
                               ) 
                             
                           
                         
                         ) 
                       
                     
                   
                 
               
               
                 
                   
                     ( 
                     7 
                     ) 
                   
                   ′ 
                 
               
             
             
               
                 
                   
                     
                       I 
                       
                         C 
                          
                         
                             
                         
                          
                         1 
                       
                     
                      
                     
                       ( 
                       s 
                       ) 
                     
                   
                   = 
                   
                     
                       C 
                       1 
                     
                     · 
                     
                       
                         
                           
                             s 
                             · 
                             
                               V 
                               C 
                             
                           
                            
                           
                             ( 
                             s 
                             ) 
                           
                         
                         - 
                         
                           
                             v 
                             P 
                           
                            
                           
                             ( 
                             
                               0 
                               - 
                             
                             ) 
                           
                         
                       
                       
                         
                           s 
                           · 
                           
                             C 
                             1 
                           
                           · 
                           
                             R 
                             C 
                           
                         
                         + 
                         1 
                       
                     
                   
                 
               
               
                 
                   ( 
                   8 
                   ) 
                 
               
             
           
         
       
     
     By substituting Expressions (2)′ through (6)′ and (8) into Expression (1)′, Expression (9) is obtained. 
     
       
         
           
             
               
                 
                   
                       
                   
                    
                   
                     
                       G 
                       3 
                     
                     · 
                     
                       ( 
                       
                         
                           
                             V 
                             A 
                           
                            
                           
                             ( 
                             s 
                             ) 
                           
                         
                         - 
                         
                           
                             V 
                             B 
                           
                            
                           
                             ( 
                             s 
                             ) 
                           
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   
                     ( 
                     9 
                     ) 
                   
                    
                   
                       
                   
                    
                   Left 
                 
               
             
             
               
                 
                   = 
                   
                     
                       
                         ( 
                         
                           
                             G 
                             2 
                           
                           + 
                           
                             s 
                             · 
                             
                               C 
                               2 
                             
                           
                         
                         ) 
                       
                       · 
                       
                         ( 
                         
                           
                             
                               V 
                               B 
                             
                              
                             
                               ( 
                               s 
                               ) 
                             
                           
                           - 
                           
                             
                               V 
                               C 
                             
                              
                             
                               ( 
                               s 
                               ) 
                             
                           
                         
                         ) 
                       
                     
                     - 
                     
                       
                         C 
                         2 
                       
                       · 
                       
                         ( 
                         
                           
                             
                               v 
                               B 
                             
                              
                             
                               ( 
                               
                                 0 
                                 - 
                               
                               ) 
                             
                           
                           - 
                           
                             
                               v 
                               C 
                             
                              
                             
                               ( 
                               
                                 0 
                                 - 
                               
                               ) 
                             
                           
                         
                         ) 
                       
                     
                   
                 
               
               
                 
                   
                     ( 
                     9 
                     ) 
                   
                    
                   
                       
                   
                    
                   Middle 
                 
               
             
             
               
                 
                   = 
                   
                     
                       
                         G 
                         DH 
                       
                       · 
                       
                         ( 
                         
                           
                             
                               V 
                               C 
                             
                              
                             
                               ( 
                               s 
                               ) 
                             
                           
                           - 
                           
                             
                               1 
                               s 
                             
                              
                             
                               V 
                               SH 
                             
                           
                         
                         ) 
                       
                     
                     + 
                     
                       
                         G 
                         1 
                       
                       · 
                       
                         
                           V 
                           C 
                         
                          
                         
                           ( 
                           s 
                           ) 
                         
                       
                     
                     + 
                     
                       
                         C 
                         1 
                       
                       · 
                       s 
                       · 
                       
                         
                           
                             
                               V 
                               C 
                             
                              
                             
                               ( 
                               s 
                               ) 
                             
                           
                           - 
                           
                             
                               v 
                               P 
                             
                              
                             
                               ( 
                               
                                 0 
                                 - 
                               
                               ) 
                             
                           
                         
                         
                           
                             s 
                             · 
                             
                               C 
                               1 
                             
                             · 
                             
                               R 
                               C 
                             
                           
                           + 
                           1 
                         
                       
                     
                   
                 
               
               
                 
                   
                     ( 
                     9 
                     ) 
                   
                    
                   
                       
                   
                    
                   Right 
                 
               
             
           
         
       
     
     The following Expression (10) is obtained from the left side and the middle of Expression (9). 
     
       
         
           
             
               
                 
                   
                     
                       V 
                       B 
                     
                      
                     
                       ( 
                       s 
                       ) 
                     
                   
                   = 
                   
                     
                       
                         
                           
                             V 
                             C 
                           
                            
                           
                             ( 
                             s 
                             ) 
                           
                         
                         · 
                         
                           ( 
                           
                             
                               s 
                               · 
                               
                                 C 
                                 2 
                               
                             
                             + 
                             
                               G 
                               2 
                             
                           
                           ) 
                         
                       
                       + 
                       
                         
                           C 
                           2 
                         
                         · 
                         
                           ( 
                           
                             
                               
                                 v 
                                 B 
                               
                                
                               
                                 ( 
                                 
                                   0 
                                   - 
                                 
                                 ) 
                               
                             
                             - 
                             
                               
                                 v 
                                 C 
                               
                                
                               
                                 ( 
                                 
                                   0 
                                   - 
                                 
                                 ) 
                               
                             
                           
                           ) 
                         
                       
                       + 
                       
                         
                           
                             V 
                             A 
                           
                            
                           
                             ( 
                             s 
                             ) 
                           
                         
                         · 
                         
                           G 
                           3 
                         
                       
                     
                     
                       
                         s 
                         · 
                         
                           C 
                           2 
                         
                       
                       + 
                       
                         G 
                         2 
                       
                       + 
                       
                         G 
                         3 
                       
                     
                   
                 
               
               
                 
                   ( 
                   10 
                   ) 
                 
               
             
           
         
       
     
     Here, defining v A (t) to be a step function represented by Expression (11), the Laplace transform of v A (t) is represented by Expression (12). It should be noted that the value of V A1  does not appear in Expression (12). However, the information with respect to the initial state is included in Vc(0−) in expression (10), and accordingly, this poses no difficulty in the downstream calculation step. Furthermore, assuming that the circuit is static at the time point t&lt;0, the following Expression (13) holds true. 
     
       
         
           
             
               
                 
                   
                     
                       v 
                       A 
                     
                      
                     
                       ( 
                       t 
                       ) 
                     
                   
                   = 
                   
                     { 
                     
                       
                         
                           
                             
                               v 
                               
                                 A 
                                  
                                 
                                     
                                 
                                  
                                 1 
                               
                             
                           
                           
                             … 
                           
                           
                             
                               ( 
                               
                                 t 
                                 &lt; 
                                 0 
                               
                               ) 
                             
                           
                         
                         
                           
                             
                               v 
                               
                                 A 
                                  
                                 
                                     
                                 
                                  
                                 2 
                               
                             
                           
                           
                             … 
                           
                           
                             
                               ( 
                               
                                 0 
                                 ≤ 
                                 t 
                               
                               ) 
                             
                           
                         
                       
                        
                       
                         
 
                       
                       ↓ 
                     
                   
                 
               
               
                 
                   ( 
                   11 
                   ) 
                 
               
             
             
               
                 
                   
                     
                       V 
                       a 
                     
                      
                     
                       ( 
                       s 
                       ) 
                     
                   
                   = 
                   
                     
                       V 
                       
                         A 
                          
                         
                             
                         
                          
                         2 
                       
                     
                     s 
                   
                 
               
               
                 
                   ( 
                   12 
                   ) 
                 
               
             
             
               
                 
                   
                     
                       v 
                       P 
                     
                      
                     
                       ( 
                       
                         0 
                         - 
                       
                       ) 
                     
                   
                   = 
                   
                     
                       v 
                       C 
                     
                      
                     
                       ( 
                       
                         0 
                         - 
                       
                       ) 
                     
                   
                 
               
               
                 
                   ( 
                   13 
                   ) 
                 
               
             
           
         
       
     
     Expression (10) and Expression (12) are substituted into the left side of Expression (9), and Expression (13) is substituted into the right side of Expression (9), thereby obtaining Expression (14). Furthermore, Expression (14) is transformed so as to provide the following Expression (15). 
     
       
         
           
             
               
                 
                   
                     
                       G 
                       3 
                     
                     · 
                     
                       ( 
                       
                         
                           
                             V 
                             
                               A 
                                
                               
                                   
                               
                                
                               2 
                             
                           
                           s 
                         
                         - 
                         
                           
                             
                               
                                 
                                   
                                     
                                       
                                         V 
                                         C 
                                       
                                        
                                       
                                         ( 
                                         s 
                                         ) 
                                       
                                     
                                     · 
                                     
                                       ( 
                                       
                                         
                                           s 
                                           · 
                                           
                                             C 
                                             2 
                                           
                                         
                                         + 
                                         
                                           G 
                                           2 
                                         
                                       
                                       ) 
                                     
                                   
                                   + 
                                   
                                     
                                       C 
                                       2 
                                     
                                     · 
                                   
                                 
                               
                             
                             
                               
                                 
                                   
                                     ( 
                                     
                                       
                                         
                                           v 
                                           B 
                                         
                                          
                                         
                                           ( 
                                           
                                             0 
                                             - 
                                           
                                           ) 
                                         
                                       
                                       - 
                                       
                                         
                                           v 
                                           C 
                                         
                                          
                                         
                                           ( 
                                           
                                             0 
                                             - 
                                           
                                           ) 
                                         
                                       
                                     
                                     ) 
                                   
                                   + 
                                   
                                     
                                       
                                         V 
                                         A 
                                       
                                        
                                       
                                         ( 
                                         s 
                                         ) 
                                       
                                     
                                     · 
                                     
                                       G 
                                       3 
                                     
                                   
                                 
                               
                             
                           
                           
                             
                               s 
                               · 
                               
                                 C 
                                 2 
                               
                             
                             + 
                             
                               G 
                               2 
                             
                             + 
                             
                               G 
                               3 
                             
                           
                         
                       
                       ) 
                     
                   
                   = 
                   
                     
                       
                         G 
                         SH 
                       
                       · 
                       
                         ( 
                         
                           
                             
                               V 
                               C 
                             
                              
                             
                               ( 
                               s 
                               ) 
                             
                           
                           - 
                           
                             
                               1 
                               s 
                             
                              
                             
                               V 
                               SH 
                             
                           
                         
                         ) 
                       
                     
                     + 
                     
                       
                         G 
                         1 
                       
                       · 
                       
                         
                           V 
                           C 
                         
                          
                         
                           ( 
                           s 
                           ) 
                         
                       
                     
                     + 
                     
                       
                         C 
                         1 
                       
                       · 
                       
                         
                           
                             
                               s 
                               · 
                               
                                 V 
                                 C 
                               
                             
                              
                             
                               ( 
                               s 
                               ) 
                             
                           
                           - 
                           
                             
                               v 
                               C 
                             
                              
                             
                               ( 
                               
                                 0 
                                 - 
                               
                               ) 
                             
                           
                         
                         
                           
                             s 
                             · 
                             
                               C 
                               1 
                             
                             · 
                             
                               R 
                               C 
                             
                           
                           + 
                           1 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   14 
                   ) 
                 
               
             
             
               
                 
                   
                       
                   
                    
                   
                     
                       
                         V 
                         C 
                       
                        
                       
                         ( 
                         s 
                         ) 
                       
                     
                     = 
                     
                       
                         1 
                         s 
                       
                       · 
                       
                         
                           
                             A 
                             · 
                             
                               s 
                               2 
                             
                           
                           + 
                           
                             T 
                             · 
                             s 
                           
                           + 
                           U 
                         
                         
                           
                             s 
                             2 
                           
                           + 
                           
                             P 
                             · 
                             s 
                           
                           + 
                           Q 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   15 
                   ) 
                 
               
             
           
         
       
     
     The coefficients A, T, U, P, and Q in Expression (15) are represented by the following Expressions (15-1) through (15-5). 
     
       
         
           
             
               
                 
                   A 
                   = 
                   
                     · 
                     
                       
                         
                           
                             
                               
                                 
                                   R 
                                   C 
                                 
                                 · 
                                 
                                   { 
                                   
                                     
                                       
                                         V 
                                         
                                           A 
                                            
                                           
                                               
                                           
                                            
                                           2 
                                         
                                       
                                       · 
                                       
                                         G 
                                         3 
                                       
                                     
                                     - 
                                     
                                       
                                         
                                           ( 
                                           
                                             
                                               
                                                 v 
                                                 B 
                                               
                                                
                                               
                                                 ( 
                                                 
                                                   0 
                                                   - 
                                                 
                                                 ) 
                                               
                                             
                                             - 
                                             
                                               
                                                 v 
                                                 C 
                                               
                                                
                                               
                                                 ( 
                                                 
                                                   0 
                                                   - 
                                                 
                                                 ) 
                                               
                                             
                                           
                                           ) 
                                         
                                         · 
                                       
                                        
                                       
                                         G 
                                         3 
                                       
                                     
                                     + 
                                     
                                       
                                         V 
                                         SH 
                                       
                                       · 
                                       
                                         G 
                                         SH 
                                       
                                     
                                   
                                   } 
                                 
                               
                               + 
                             
                           
                         
                         
                           
                             
                               
                                 v 
                                 C 
                               
                                
                               
                                 ( 
                                 
                                   0 
                                   - 
                                 
                                 ) 
                               
                             
                           
                         
                       
                       
                         
                           
                             R 
                             C 
                           
                           · 
                           
                             ( 
                             
                               
                                 G 
                                 3 
                               
                               + 
                               
                                 G 
                                 SH 
                               
                               + 
                               
                                 G 
                                 1 
                               
                             
                             ) 
                           
                         
                         + 
                         1 
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     15 
                      
                     
                       - 
                     
                      
                     1 
                   
                   ) 
                 
               
             
             
               
                 
                   
                       
                   
                    
                   
                     T 
                     = 
                     
                       
                         
                           
                             
                               
                                 
                                   
                                     
                                       
                                         V 
                                         
                                           A 
                                            
                                           
                                               
                                           
                                            
                                           2 
                                         
                                       
                                       · 
                                       
                                         G 
                                         3 
                                       
                                       · 
                                       
                                         { 
                                         
                                           
                                             C 
                                             2 
                                           
                                           + 
                                           
                                             
                                               C 
                                               1 
                                             
                                             · 
                                             
                                               R 
                                               C 
                                             
                                             · 
                                             
                                               G 
                                               2 
                                             
                                           
                                         
                                         } 
                                       
                                     
                                     - 
                                   
                                 
                               
                               
                                 
                                   
                                     
                                       
                                         G 
                                         3 
                                       
                                       · 
                                       
                                         C 
                                         2 
                                       
                                       · 
                                       
                                         ( 
                                         
                                           
                                             
                                               v 
                                               B 
                                             
                                              
                                             
                                               ( 
                                               
                                                 0 
                                                 - 
                                               
                                               ) 
                                             
                                           
                                           - 
                                           
                                             
                                               v 
                                               C 
                                             
                                              
                                             
                                               ( 
                                               
                                                 0 
                                                 - 
                                               
                                               ) 
                                             
                                           
                                         
                                         ) 
                                       
                                     
                                     + 
                                   
                                 
                               
                             
                           
                         
                         
                           
                             
                               
                                 
                                   V 
                                   SH 
                                 
                                 · 
                                 
                                   G 
                                   SH 
                                 
                                 · 
                                 
                                   { 
                                   
                                     
                                       C 
                                       2 
                                     
                                     + 
                                     
                                       
                                         C 
                                         1 
                                       
                                       · 
                                       
                                         R 
                                         C 
                                       
                                       · 
                                       
                                         ( 
                                         
                                           
                                             G 
                                             2 
                                           
                                           + 
                                           
                                             G 
                                             3 
                                           
                                         
                                         ) 
                                       
                                     
                                   
                                   } 
                                 
                               
                               + 
                             
                           
                         
                         
                           
                             
                               
                                 
                                   v 
                                   C 
                                 
                                  
                                 
                                   ( 
                                   
                                     0 
                                     - 
                                   
                                   ) 
                                 
                               
                               · 
                               
                                 C 
                                 1 
                               
                               · 
                               
                                 ( 
                                 
                                   
                                     G 
                                     2 
                                   
                                   + 
                                   
                                     G 
                                     3 
                                   
                                 
                                 ) 
                               
                             
                           
                         
                       
                       
                         
                           C 
                           1 
                         
                         · 
                         
                           C 
                           2 
                         
                         · 
                         
                           { 
                           
                             
                               
                                 R 
                                 C 
                               
                               · 
                               
                                 ( 
                                 
                                   
                                     G 
                                     3 
                                   
                                   + 
                                   
                                     G 
                                     SH 
                                   
                                   + 
                                   
                                     G 
                                     1 
                                   
                                 
                                 ) 
                               
                             
                             + 
                             1 
                           
                           } 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     15 
                      
                     
                       - 
                     
                      
                     2 
                   
                   ) 
                 
               
             
             
               
                 
                   
                       
                   
                    
                   
                     U 
                     = 
                     
                       
                         
                           
                             V 
                             
                               A 
                                
                               
                                   
                               
                                
                               2 
                             
                           
                           · 
                           
                             G 
                             3 
                           
                           · 
                           
                             G 
                             2 
                           
                         
                         + 
                         
                           
                             V 
                             SH 
                           
                           · 
                           
                             G 
                             SH 
                           
                           · 
                           
                             ( 
                             
                               
                                 G 
                                 2 
                               
                               + 
                               
                                 G 
                                 3 
                               
                             
                             ) 
                           
                         
                       
                       
                         
                           C 
                           1 
                         
                         · 
                         
                           C 
                           2 
                         
                         · 
                         
                           { 
                           
                             
                               
                                 R 
                                 C 
                               
                               · 
                               
                                 ( 
                                 
                                   
                                     G 
                                     3 
                                   
                                   + 
                                   
                                     G 
                                     SH 
                                   
                                   + 
                                   
                                     G 
                                     1 
                                   
                                 
                                 ) 
                               
                             
                             + 
                             1 
                           
                           } 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     15 
                      
                     
                       - 
                     
                      
                     3 
                   
                   ) 
                 
               
             
             
               
                 
                   
                       
                   
                    
                   
                     P 
                     = 
                     
                       
                         
                           
                             
                               
                                 
                                   G 
                                   3 
                                 
                                 · 
                                 
                                   ( 
                                   
                                     
                                       
                                         C 
                                         1 
                                       
                                       · 
                                       
                                         R 
                                         C 
                                       
                                       · 
                                       
                                         G 
                                         2 
                                       
                                     
                                     + 
                                     
                                       C 
                                       2 
                                     
                                   
                                   ) 
                                 
                               
                               + 
                               
                                 
                                   ( 
                                   
                                     
                                       G 
                                       SH 
                                     
                                     + 
                                     
                                       G 
                                       1 
                                     
                                   
                                   ) 
                                 
                                 · 
                               
                             
                           
                         
                         
                           
                             
                               
                                 { 
                                 
                                   
                                     C 
                                     2 
                                   
                                   + 
                                   
                                     
                                       C 
                                       1 
                                     
                                     · 
                                     
                                       R 
                                       C 
                                     
                                     · 
                                     
                                       ( 
                                       
                                         
                                           G 
                                           2 
                                         
                                         + 
                                         
                                           G 
                                           3 
                                         
                                       
                                       ) 
                                     
                                   
                                 
                                 } 
                               
                               + 
                               
                                 
                                   C 
                                   1 
                                 
                                 · 
                                 
                                   ( 
                                   
                                     
                                       G 
                                       2 
                                     
                                     + 
                                     
                                       G 
                                       3 
                                     
                                   
                                   ) 
                                 
                               
                             
                           
                         
                       
                       
                         
                           C 
                           1 
                         
                         · 
                         
                           C 
                           2 
                         
                         · 
                         
                           { 
                           
                             
                               
                                 R 
                                 C 
                               
                               · 
                               
                                 ( 
                                 
                                   
                                     C 
                                     3 
                                   
                                   + 
                                   
                                     G 
                                     SH 
                                   
                                   + 
                                   
                                     G 
                                     1 
                                   
                                 
                                 ) 
                               
                             
                             + 
                             1 
                           
                           } 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     15 
                      
                     
                       - 
                     
                      
                     4 
                   
                   ) 
                 
               
             
             
               
                 
                   
                       
                   
                    
                   
                     Q 
                     = 
                     
                       
                         
                           
                             G 
                             2 
                           
                           · 
                           
                             G 
                             3 
                           
                         
                         + 
                         
                           
                             ( 
                             
                               
                                 G 
                                 SH 
                               
                               + 
                               
                                 G 
                                 1 
                               
                             
                             ) 
                           
                           · 
                           
                             ( 
                             
                               
                                 G 
                                 2 
                               
                               + 
                               
                                 G 
                                 3 
                               
                             
                             ) 
                           
                         
                       
                       
                         
                           C 
                           1 
                         
                         · 
                         
                           C 
                           2 
                         
                         · 
                         
                           { 
                           
                             
                               
                                 R 
                                 C 
                               
                               · 
                               
                                 ( 
                                 
                                   
                                     G 
                                     3 
                                   
                                   + 
                                   
                                     G 
                                     SH 
                                   
                                   + 
                                   
                                     G 
                                     1 
                                   
                                 
                                 ) 
                               
                             
                             + 
                             1 
                           
                           } 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     15 
                      
                     
                       - 
                     
                      
                     5 
                   
                   ) 
                 
               
             
           
         
       
     
     Assuming that partial fraction decomposition of Expression (15) can be done as represented by Expression (16), α, β, γ, ω 1 , and ω 2  are calculated. If α, β, γ, ω 1 , and ω 2  are all real numbers, the inverse Laplace transform of Expression (16) can be calculated, thereby obtaining the response V c (t) on the time domain. The reason why such a partial fraction decomposition can be done as represented by Expression (16) is that the variable equalizer circuit  100  shown in  FIG. 1  is configured using resistors and capacitors, and accordingly, the response of the circuit does not involve oscillation. Expression (16) is reduced, thereby obtaining the following Expression (17). 
     
       
         
           
             
               
                 
                   
                     
                       V 
                       C 
                     
                      
                     
                       ( 
                       s 
                       ) 
                     
                   
                   = 
                   
                     
                       γ 
                       s 
                     
                     + 
                     
                       α 
                       
                         s 
                         + 
                         
                           ω 
                           1 
                         
                       
                     
                     + 
                     
                       β 
                       
                         s 
                         + 
                         
                           ω 
                           2 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   16 
                   ) 
                 
               
             
             
               
                 
                   
                     
                       V 
                       C 
                     
                      
                     
                       ( 
                       s 
                       ) 
                     
                   
                   = 
                   
                     
                       1 
                       s 
                     
                     · 
                     
                       
                         
                           
                             
                               
                                 
                                   ( 
                                   
                                     γ 
                                     + 
                                     α 
                                     + 
                                     β 
                                   
                                   ) 
                                 
                                 · 
                                 
                                   s 
                                   2 
                                 
                               
                               + 
                               
                                 
                                   ( 
                                   
                                     
                                       γ 
                                       · 
                                       
                                         ( 
                                         
                                           
                                             ω 
                                             1 
                                           
                                           + 
                                           
                                             ω 
                                             2 
                                           
                                         
                                         ) 
                                       
                                     
                                     + 
                                     
                                       α 
                                       · 
                                       
                                         ω 
                                         2 
                                       
                                     
                                     + 
                                     
                                       β 
                                       · 
                                       
                                         ω 
                                         1 
                                       
                                     
                                   
                                   ) 
                                 
                                 · 
                               
                             
                           
                         
                         
                           
                             
                               s 
                               + 
                               
                                 γ 
                                 · 
                                 
                                   ω 
                                   1 
                                 
                                 · 
                                 
                                   ω 
                                   2 
                                 
                               
                             
                           
                         
                       
                       
                         
                           s 
                           2 
                         
                         + 
                         
                           
                             ( 
                             
                               
                                 ω 
                                 1 
                               
                               + 
                               
                                 ω 
                                 2 
                               
                             
                             ) 
                           
                           · 
                           s 
                         
                         + 
                         
                           
                             ω 
                             1 
                           
                           · 
                           
                             ω 
                             2 
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   17 
                   ) 
                 
               
             
           
         
       
     
     Expression (15) must be identically equivalent to Expression (17). Thus, by making a comparison of each term between these Expressions, the following Expressions (18-1) through (18-5) are obtained. 
         A=γ+α+β   (18-1)
 
         T =γ·(ω 1 +ω 2 )+α·ω 2 +β ω 1    (18-2)
 
         U=γ·ω   1 ·ω 2    (18-3)
 
         P =ω 1 +ω 2 ( 18-4)
 
         Q=ω   1 ω 2    (18-5)
 
     By solving the Expressions (18-1) through (18-5), the following Expressions (19-1) through (19-5) are obtained. 
     
       
         
           
             
               
                 
                   
                       
                   
                    
                   
                     γ 
                     = 
                     
                       U 
                       Q 
                     
                   
                 
               
               
                 
                   ( 
                   
                     19 
                      
                     
                       - 
                     
                      
                     1 
                   
                   ) 
                 
               
             
             
               
                 
                   
                       
                   
                    
                   
                     
                       ω 
                       1 
                     
                     = 
                     
                       
                         1 
                         2 
                       
                        
                       
                         ( 
                         
                           P 
                           + 
                           
                             
                               
                                 P 
                                 2 
                               
                               - 
                               
                                 4 
                                 · 
                                 Q 
                               
                             
                           
                         
                         ) 
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     19 
                      
                     
                       - 
                     
                      
                     2 
                   
                   ) 
                 
               
             
             
               
                 
                   
                       
                   
                    
                   
                     
                       ω 
                       2 
                     
                     = 
                     
                       
                         1 
                         2 
                       
                        
                       
                         ( 
                         
                           P 
                           - 
                           
                             
                               
                                 P 
                                 2 
                               
                               - 
                               
                                 4 
                                 · 
                                 Q 
                               
                             
                           
                         
                         ) 
                       
                     
                   
                 
               
               
                 
                   ( 
                   
                     19 
                      
                     
                       - 
                     
                      
                     3 
                   
                   ) 
                 
               
             
             
               
                 
                   α 
                   = 
                   
                     
                       
                         - 
                         1 
                       
                       
                         2 
                         · 
                         
                           
                             
                               P 
                               2 
                             
                             - 
                             
                               4 
                               · 
                               Q 
                             
                           
                         
                       
                     
                     · 
                     
                       { 
                       
                         
                           2 
                           · 
                           T 
                         
                         - 
                         
                           
                             U 
                             Q 
                           
                           · 
                           
                             ( 
                             
                               P 
                               - 
                               
                                 
                                   
                                     P 
                                     2 
                                   
                                   - 
                                   
                                     4 
                                     · 
                                     Q 
                                   
                                 
                               
                             
                             ) 
                           
                         
                         - 
                         
                           A 
                           · 
                           
                             ( 
                             
                               P 
                               + 
                               
                                 
                                   
                                     P 
                                     2 
                                   
                                   - 
                                   
                                     4 
                                     · 
                                     Q 
                                   
                                 
                               
                             
                             ) 
                           
                         
                       
                       } 
                     
                   
                 
               
               
                 
                   ( 
                   
                     19 
                      
                     
                       - 
                     
                      
                     4 
                   
                   ) 
                 
               
             
             
               
                 
                   β 
                   = 
                   
                     
                       1 
                       
                         2 
                         · 
                         
                           
                             
                               P 
                               2 
                             
                             - 
                             
                               4 
                               · 
                               Q 
                             
                           
                         
                       
                     
                     · 
                     
                       { 
                       
                         
                           2 
                           · 
                           T 
                         
                         - 
                         
                           
                             U 
                             Q 
                           
                           · 
                           
                             ( 
                             
                               P 
                               + 
                               
                                 
                                   
                                     P 
                                     2 
                                   
                                   - 
                                   
                                     4 
                                     · 
                                     Q 
                                   
                                 
                               
                             
                             ) 
                           
                         
                         - 
                         
                           A 
                           · 
                           
                             ( 
                             
                               P 
                               - 
                               
                                 
                                   
                                     P 
                                     2 
                                   
                                   - 
                                   
                                     4 
                                     · 
                                     Q 
                                   
                                 
                               
                             
                             ) 
                           
                         
                       
                       } 
                     
                   
                 
               
               
                 
                   ( 
                   
                     19 
                      
                     
                       - 
                     
                      
                     5 
                   
                   ) 
                 
               
             
           
         
       
     
     The inverse Laplace transform of Expression ( 16 ) is calculated, thereby obtaining the following Expression (20). 
     
       
         
           
             
               
                 
                   
                     
                       V 
                       C 
                     
                      
                     
                       ( 
                       s 
                       ) 
                     
                   
                   = 
                   
                     
                       γ 
                       s 
                     
                     + 
                     
                       α 
                       
                         s 
                         + 
                         
                           ω 
                           1 
                         
                       
                     
                     + 
                     
                       
                         β 
                         
                           s 
                           + 
                           
                             ω 
                             2 
                           
                         
                       
                        
                       
                         
 
                       
                       ↓ 
                       
                         
                           - 
                           1 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   16 
                   ) 
                 
               
             
             
               
                 
                   
                     
                       v 
                       C 
                     
                      
                     
                       ( 
                       t 
                       ) 
                     
                   
                   = 
                   
                     γ 
                     + 
                     
                       α 
                       · 
                       
                         exp 
                          
                         
                           ( 
                           
                             
                               - 
                               
                                 ω 
                                 1 
                               
                             
                             · 
                             t 
                           
                           ) 
                         
                       
                     
                     + 
                     
                       β 
                       · 
                       
                         exp 
                          
                         
                           ( 
                           
                             
                               - 
                               
                                 ω 
                                 2 
                               
                             
                             · 
                             t 
                           
                           ) 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   20 
                   ) 
                 
               
             
           
         
       
     
     Expression (20) is defined only in the range 0&lt;t. In the range t&lt;0, assuming that the circuit is in the static state, v c (0−) is calculated.  FIG. 6  is an equivalent circuit diagram of a variable equalizer circuit in the static state. In the static state, each capacitor is regarded as being in an open state. In the range t&lt;0, the circuit shown in  FIG. 5  provides the same voltage state and the same current state as those of the circuit shown in  FIG. 6 . Thus, v c (0−) is calculated based upon the circuit model shown in  FIG. 6 , thereby obtaining the following Expression (21). 
     
       
         
           
             
               
                 
                   
                     
                       v 
                       c 
                     
                      
                     
                       ( 
                       
                         0 
                         - 
                       
                       ) 
                     
                   
                   = 
                   
                     
                       
                         
                           V 
                           
                             A 
                              
                             
                                 
                             
                              
                             1 
                           
                         
                         · 
                         
                           R 
                           SH 
                         
                         · 
                         
                           R 
                           1 
                         
                       
                       + 
                       
                         
                           V 
                           SH 
                         
                         · 
                         
                           R 
                           1 
                         
                         · 
                         
                           ( 
                           
                             
                               R 
                               2 
                             
                             + 
                             
                               R 
                               3 
                             
                           
                           ) 
                         
                       
                     
                     
                       
                         
                           R 
                           SH 
                         
                         · 
                         
                           R 
                           1 
                         
                       
                       + 
                       
                         
                           ( 
                           
                             
                               R 
                               SH 
                             
                             + 
                             
                               R 
                               1 
                             
                           
                           ) 
                         
                         · 
                         
                           ( 
                           
                             
                               R 
                               2 
                             
                             + 
                             
                               R 
                               3 
                             
                           
                           ) 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   21 
                   ) 
                 
               
             
           
         
       
     
     The response waveform represented by Expression (20) is attached to the response waveform represented by Expression (21) at the time point t=0, thereby obtaining the response waveform v c (t) which represents the response that occurs when a step input represented by Expression (11) is applied. 
     Next, the attenuation rate is calculated. 
     Assuming that v A (t) is represented by a step function as represented by Expression (11), the circuit is in the static state at the time point t=∞. Thus, v c (∞) can be calculated as represented by the following Expression (22) based upon the equivalent circuit shown in  FIG. 6 . Furthermore, the attenuation rate ATT is represented by the following Expression (23). 
     
       
         
           
             
               
                 
                   
                     
                       v 
                       c 
                     
                      
                     
                       ( 
                       ∞ 
                       ) 
                     
                   
                   = 
                   
                     
                       
                         
                           V 
                           
                             A 
                              
                             
                                 
                             
                              
                             2 
                           
                         
                         · 
                         
                           R 
                           SH 
                         
                         · 
                         
                           R 
                           1 
                         
                       
                       + 
                       
                         
                           V 
                           SH 
                         
                         · 
                         
                           R 
                           1 
                         
                         · 
                         
                           ( 
                           
                             
                               R 
                               2 
                             
                             + 
                             
                               R 
                               3 
                             
                           
                           ) 
                         
                       
                     
                     
                       
                         
                           R 
                           SH 
                         
                         · 
                         
                           R 
                           1 
                         
                       
                       + 
                       
                         
                           ( 
                           
                             
                               R 
                               SH 
                             
                             + 
                             
                               R 
                               1 
                             
                           
                           ) 
                         
                         · 
                         
                           ( 
                           
                             
                               R 
                               2 
                             
                             + 
                             
                               R 
                               3 
                             
                           
                           ) 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   22 
                   ) 
                 
               
             
             
               
                 
                   
                     A 
                      
                     
                         
                     
                      
                     T 
                      
                     
                         
                     
                      
                     T 
                   
                   = 
                   
                     
                       
                         
                           
                             v 
                             c 
                           
                            
                           
                             ( 
                             ∞ 
                             ) 
                           
                         
                         - 
                         
                           
                             v 
                             c 
                           
                            
                           
                             ( 
                             
                               0 
                               - 
                             
                             ) 
                           
                         
                       
                       
                         
                           V 
                           
                             A 
                              
                             
                                 
                             
                              
                             2 
                           
                         
                         - 
                         
                           V 
                           
                             A 
                              
                             
                                 
                             
                              
                             1 
                           
                         
                       
                     
                     = 
                     
                       1 
                       
                         1 
                         + 
                         
                           
                             ( 
                             
                               
                                 1 
                                 
                                   R 
                                   SH 
                                 
                               
                               + 
                               
                                 1 
                                 
                                   R 
                                   1 
                                 
                               
                             
                             ) 
                           
                           · 
                           
                             ( 
                             
                               
                                 R 
                                 2 
                               
                               + 
                               
                                 R 
                                 3 
                               
                             
                             ) 
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   23 
                   ) 
                 
               
             
           
         
       
     
       FIGS. 7A and 7B  are simulation waveform diagrams for the variable equalizer circuit  100  shown in  FIG. 1 . 
       FIG. 7A  shows waveforms for when the parameter of the resistance value R 1  of the first resistor R 1  is set to 2 kΩ, 4 kΩ, 6 kΩ, 8 kΩ, and 10 kΩ. The other circuit constants are as follows. 
       R 2 =1.75 kΩ
 
       R 3 =250 Ω
 
       R c =2 kΩ
 
       C 1 =60 fF 
       C 2 =300 fF 
     It can be understood that the amount of boost can be controlled mainly by changing the resistance value R 1  of the first resistor R 1 . 
       FIG. 7B  shows waveforms when the parameters of the capacitance value C 1  of the first capacitor C 1  is set to 30 fF, 60 fF, 90 fF, and 120 fF. The resistance value R 1  is 4 kΩ, and the other circuit constants are the same as those in the above description. It can be understood that the time constant can be controlled by changing the capacitance value C 1  of the first capacitor C 1 . 
     The above-described embodiment has been described for exemplary purposes only, and is by no means intended to be interpreted restrictively. Rather, it can be readily conceived by those skilled in this art that various modifications may be made by making various combinations of the aforementioned components or processes, which are also encompassed in the technical scope of the present invention. Description will be made below regarding such modifications. 
     [First Modification] 
       FIG. 8  is a circuit diagram which shows a configuration of a variable equalizer circuit  100   a  according to a first modification. The variable equalizer circuit  100   a  shown in  FIG. 8  has a configuration obtained by eliminating the level shifter  20  from the configuration of the variable equalizer circuit  100  shown in  FIG. 1 . With such an arrangement, with R SH  set to ∞, Expressions (2) through (23) hold true without change. In a case in which the output signal of the variable equalizer circuit  100   a  is within the input voltage range of the receiver circuit  8  without the need to level-shift the output signal of the variable equalizer circuit  100   a,  the level shifter  20  can be eliminated. 
     [Second Modification] 
       FIG. 9  is a circuit diagram which shows a variable equalizer circuit  100   b  according to a second modification. The variable equalizer circuit  100   b  shown in  FIG. 9  has a configuration obtained by eliminating the fourth resistor Rc from the configuration of the variable equalizer circuit  100  shown in  FIG. 1 . With such an arrangement, with Rc as 0, Expressions (2) through (23) holds true without change. 
     [Third Modification] 
     A third modification has a configuration obtained by eliminating the third resistor R 3  from the configuration of the variable equalizer circuit  100  shown in  FIG. 1 . In a case in which the second resistor R 2 , the first resistor R 1 , and the fourth resistor Rc each have a large resistance as compared with the characteristic impedance Zo of the transmission line  3 , the variable equalizer circuit  100  does not affect the impedance matching. Thus, in this case, the third resistor R 3  can be eliminated. 
     [Fourth Modification] 
       FIG. 10  is a circuit diagram which shows a configuration of a variable equalizer circuit  100   c  according to a fourth modification. The variable equalizer circuit  100   c  shown in  FIG. 10  has a configuration obtained by eliminating the first resistor R 1  from the configuration of the variable equalizer circuit  100  shown in  FIG. 1 . Furthermore, with such a configuration, the resistor R SH  of the level shifter  20   c  is configured as a variable resistor instead of including the first resistor R 1 . The level shifter  20   c  is configured to be capable of changing the resistance component R SH  between the output terminal P 2  and the fixed voltage terminal (ground terminal or power supply terminal). With such an arrangement, with R 1  as ∞, Expressions (2) through (23) hold true without change. 
     Some modifications described above may be combined with each other. 
     For example, the first modification may be combined with at least one from the second and third modifications. 
     For example, the second modification may be combined with at least one from the first, third, and fourth modifications. 
     For example, the third modification may be combined with at least one from the first, second, and fourth modifications. 
     For example, the fourth modification may be combined with at least one from the second and third modifications. 
     Various combinations and various modifications may be made without harming the advantages of the present invention, which are readily conceived by those skilled in this art. 
     Description has been made in the aforementioned embodiments regarding an arrangement in which the variable equalizer circuit  100  is employed in the test apparatus  2 . However, an application of the variable equalizer circuit  100  is not restricted to such an arrangement. Rather, such a variable equalizer circuit can be applied to various kinds of devices configured to receive a signal from an external circuit. 
     Description has been made regarding the present invention with reference to the embodiments. However, the above-described embodiments show only the mechanisms and applications of the present invention for exemplary purposes only, and are by no means intended to be interpreted restrictively. Rather, various modifications and various changes in the layout can be made without departing from the spirit and scope of the present invention defined in appended claims.