## 3 Results
### Passive EQ-Lumped Elements, Synthesis  
Passive EQ does not amplify any components of the signal, low or high frequencies that pass through it but rather attenuates the low frequency signal components. A method to determine the filter's elements [14] starts from insertion loss representation applying filters' synthesis theory before applying any EQ scheme. At the cut-off frequency, \(\mathrm{f}_{\mathrm{c}}\), the EQ's frequency, determined from the dB IL of the trace, the value of the attenuation is half of the maximum value. In the frequency domain selected of \(\left[0,\ 40\right]\) GHz, the maximum is \(\left.\mathrm{IL}\right|_{\mathrm{max}}=\left.\mathrm{IL}\right|_{\mathrm{40 \ GHz}}=-0.82\) dB so \(\mathrm{f}_{\mathrm{c}}\cong 20\) GHz. Applying the theory of filters' synthesis, the IL \(\left[\mathrm{dB}\right]\) at a given frequency is offered here without proof [14] by the relation:  
\[\mathrm{IL}(\mathrm{f})=10\log\left[1+\frac{K^{2}-1}{1+\mathrm{K}\left(\frac{t }{\mathrm{f}_{\mathrm{c}}}\right)^{2}}\right], \tag{5}\]  
where factor K determines the filter's IL, or the amount of attenuation provided, measured at frequency \(\mathrm{f}_{\mathrm{c}}\), but is also a measure of the flatness of the transfer function, that should be equal with the inverse transfer function of the trace. From relation (5), it can be deduced that the higher the K, the flatter the frequency response. Typically, \(\mathrm{K}\in(1,5)\) otherwise a value greater than five causes a flatness of the response in frequency domain and the IL will increase [2]. Next, value of the factor is determined for the present case for \(\mathrm{f}=10\) GHz. From the representation of IL, we deduce that \(\mathrm{IL}(10\ \mathrm{GHz})=-0.25\) dB. From this, K is determined to have two values, only one corresponding to a positive value in the interval \((1,\ 5)\), and that is \(\mathrm{K}=1.145\). There are several architectures to be considered so that the EQ performs as a high pass filter [14]. For passive type, EQ variants will include RC derivative shunt of a resistor and a capacitor connected as in Figure 13a or RL series of a resistor and an inductor connected as in Figure 13b or combination of the two, or bridge type EQs as in the below representation from Figure 13.  
Figure 12: Voltage variation with time for 50 ps rise and fall times (**a**) and eye diagram associated at the output of the microstrip line (**b**). Figure 13: Types of passive EQs: (**a**) RC, (**b**) RL, (**c**) and (**d**) composed RC and RL in bridged variants. From K coefficient deduced at the working frequency, the cut-off inductance and capacitance [14] are:  
\[\mathrm{L_{c}}=\frac{\mathrm{R_{0}}}{2\pi\mathrm{f_{c}}}\mathrm{C_{c}}=\frac{1}{2 \pi\mathrm{f_{c}}\mathrm{R_{0}}}, \tag{6}\]  
where \(\mathrm{f_{c}}\) is determined from (7):  
\[\mathrm{f_{c}}=\frac{1}{2\sqrt{\mathrm{L_{c}}\mathrm{C_{c}}}}=\left.\frac{1}{2 }\mathrm{IL}\right|_{\mathrm{[dB]}}, \tag{7}\]  
and \(\mathrm{R_{0}}\) is the EQ impedance, in this case \(\mathrm{R_{0}}=123.65\ \Omega\) determined with TxLine. Computing the values for \(\mathrm{L_{c}}\) and \(\mathrm{C_{c}}\) results 0.983 nH and 0.0643 pF so that \(\mathrm{L_{1}}\) and \(\mathrm{C_{2}}\) for the EQ components are:  
\[\mathrm{L_{2}}=\mathrm{L_{c}}\frac{\sqrt{\mathrm{K}}}{\mathrm{K-1}}=7.25\ \mathrm{nH_{c}}\mathrm{C_{1}}=\mathrm{C_{c}}\frac{\sqrt{\mathrm{K}}}{\mathrm{K -1}}=0.474\ \mathrm{pF}, \tag{8}\]  
and the resistive elements are:  
\[\mathrm{R_{1}}=\mathrm{R_{0}}(\mathrm{K-1})=17.92\ \Omega, \tag{9}\]  
The computed values of the elements for the EQ networks were used in the MWO [6] and results are presented in the next subsections. Another variant as presented in the Figure 13d is the constant input and constant output impedance bridge T. The values of the components are the same deduced in relations (8) and (9). The cost of this EQ is increased due to additional resistive parts. The bridged EQ is also implemented in MWO and presented in the next subsections.