# Handling Fault Diagnosis Problem of Linear-Analogue Circuits with Voltage Phasor Measurement

Authors: Xin Gao; HouJun Wang; Zhen Liu
Journal: Mathematical Problems in Engineering
 (2014)

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## Abstract

This paper proposes a novel method to estimate
the influence of hard-fault in linear-analogue circuit system
based on the measurement of voltage phasor with assistant
branch introduced. Furthermore, a new fault diagnosis strategy
based on the voltage phasor modeling is established, and the
tolerance influence on the corresponding voltage measurement
is also discussed. The actual analogue circuit test shows us
that the proposed method is effective and reliable to locate the
accurate fault signature in voltage measurement for the fault
diagnosis. As a matter of fact, it includes both the amplitude and
phase information in a complex value form when the linear-analogue
circuit is under the AC test. Besides, it can be also
applied to ambiguous groups and the sensitive test-frequencies
determination in the process of fault diagnosis,while the effectiveness
of multifrequencies test has also been testified through
test-frequencies sweeping investigation and the maximum error
evaluation of fault component value in the second circuit example.


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## Body

## 1. Introduction

The fault diagnosis problem has been one of the most critical issues in the test of large-scale industrial or military digital-analogue hybrid circuit. Furthermore, according to statistics, in the mixed-signal (circuit) system, more than80% faults occur in the analogue section [1], and the corresponding test cost accounts for 95% of the total test cost and 30%–50% of manufacturing cost, as well as the test time which is more than 80% of the total test time [2]. Therefore, there still exists a pressing need to locate an effective and simple method handling fault diagnosis in the analogue circuits.Up to now, there have been extensive researches on analogue circuit fault diagnosis, in which scientists developed various fault diagnosis methods through different excitation for the circuit test: (a) DC test: it is the simplest and fastest one; however, it might fail because of energy-storage components; (b) AC test [3–5]: it uses a periodic signal as external stimulus for the circuit under test (CUT) and overcomes the shortcomings of DC test, as well as requiring relatively simple test; (c) aperiodic signals stimulus test [6–9]: it owns the most abundant features about circuit state but needs most complex test requirement.To compromise the test simplicity (e.g., less time or money consumption) and the test effectiveness (e.g., higher fault detection and/or isolation rate), one can recommend the AC test in this paper. Moreover, depending on the number of frequency components in a stimulus, AC tests are further classified into two categories: single-tone and multitone test. In practice, multitone test stimuli can be composed of signals of different frequencies; therefore, in this paper, all concerns should be on the single-tone test: sinusoid signal test.Eventually, this paper proposes a robust voltage phasor-based method for the circuit diagnosis, when the sinusoid signal stimulus is used in analogue circuit such that this method is robust, because of the fault modeling being established according to the rigorous theories. What is more, there are at least 3 more advantages: (a) the corresponding fault signature (voltage measurement) is simple to calculate in the process of circuit diagnosis, compared to the feature used in [10, 11]; (b) the fault diagnosis can be found in toleration circumstance to benefit us in the actual fault diagnosis with given requirement (e.g., the fault detection and/or isolation rate); (c) although the voltage measurement is discussed in a form of complex value for a AC test, it actually can be generalized to the case of real voltage measurement in the DC test.In the end of this section, all of critical aspects of fault diagnosis discussed in this paper are listed as follows: (a) the calculation of voltage phasor value for a faulty linear-analogue circuit with a given AC test, (b) the influence of component tolerance on circuit voltage phasor response, (c) the AC test selection for fault diagnosis based on voltage phasor measurement, and (d) the further discussion about ambiguous groups determination and evaluation of multifrequencies test.The rest of paper is organized as follows. At first, Section2 establishes the theoretical basis for all the discussed contents as mentioned above. Then Section 3 tests the effectiveness of theories of Section 2 in some representative linear circuits. In Section 4, considering an important aspect—ambiguity groups determination was often discussed in linear-analogue circuit diagnosis in the past; one also discusses this topic on the basis of proposed fault modeling in this paper. Besides, the effectiveness of multifrequencies test in the second circuit example is further evaluated. In the end, the corresponding conclusions and prospects are made in Section 5.


## 2. Basic Principle

### 2.1. Voltage Phasor Modeling in Fault Diagnosis

Without loss of generality, one assumes that a linear-analogue circuit in Figure1 is excited by an independent voltage stimuli as shown in (1). Besides, this signal is expressed in a phasor form of (2), according to the linear circuit theory in [12]:
(1)UIi=aisin(ωt+ϕi),
where ai is the amplitude of sinusoidal signal of UI, ω is a known fixed frequency, and ϕi represents the phase of this signal. Therefore, voltage phasor UIi→ is a complex value including the information of amplitude and phase from the stimuli of UI(2)UIi→:ai2eiϕ.Figure 1
The linear circuit network.With the input signalUI, the circuit output response is also a sinusoidal signal Uo=aosin(ωt+ϕo), whose voltage phasor form is Uoi→. Then, on the basis of the superposition theorem [13, 14], one can establish (3) for the voltage phasor Uoi→ at test-node To as follows:
(3)Uoi→=•aiUIi→+•aXUXi→,
where •ai is the complex transmission factor between UIi→ and Uoi→, while •aX is the complex transmission factor from UXi→ to Uoi→.As a matter of fact, the value of•aiUIi→ equals the value of voltage phasor response at To, if the input stimulus UI is the sole incentive and component X is in short-circuit state. In order to be convenient speaking in this paper, one can set it as Uosi→=•aiUIi→.Ulteriorly, on the basis of Thevenin’s theorem in the linear-analogue circuit, one can get the following:(4)UXi→=UOXi→•Zx•Zx+•Zo,
where the UXi→ (UOXi→) represents a voltage phasor through the circuit branch Tx-To, when component X is in normal (open) state. Besides, •Zx is a complex impedance for component X. Meanwhile, •Zo is Thevenin’s equivalent impedance seeing through the circuit ports Tx and To, when component X is open.Due to the formulaUosi→=•aiUIi→ and (3)-(4), this paper finds the following:
(5)Uoi→=Uosi→+•aXUOXi→•Zx•Zx+•Zo.In particular, one assumes that componentX is in open-circuit state; then the voltage phasor at To is Uooi→ as follows:
(6)Uooi→=Uosi→+•aXUOXi→.At last, in accordance with (5) and (6), there exists a formula as follows:
(7)•Zx•Zo=-Uoi→-Uosi→Uoi→-Uooi→.In other words, (8) is formulated as follows:
(8)Uoi→=Γ(•Zo,•Zx,Uosi→,Uooi→)=•ZoUosi→+•ZxUooi→•Zo+•Zx,
where Γ represents an invariable continuous function for a given linear-analogue circuit.In the case of parametric fault ofX, the value of Uosi→ or Uooi→ is considered as constant. At last, “Theorem 1” can be found.Theorem 1.
In analogue circuit of Figure1, the potential faulty component X owns the nominal impedance value •Zxr, and the nominal response is Uori→; then the faulty response UoFi→ caused by parametric fault of X at test-node To can be calculated according to the following equation:
(9)UoFi→=•α•ZxFUooi→-Uosi→•α•ZxF-1,
where the value of •α is a constant determined by •α=(1/•Zxr)((Uori→-Uosi→)/(Uori→-Uooi→)).Proof.
With the statement in (8), it is obvious to find following equations:
(10)•Zxr•Zo=-Uori→-Uosi→Uori→-Uooi→,(11)•ZxF•Zo=-UoFi→-Uosi→UoFi→-Uooi→.
Later, if (10) is divided by (11), •Zo will be eliminated. Then Theorem 1 is obtained.Theorem1 builds up the accurate analytic expression of fault response, because of parametric fault of component X. It also tells us that based on the voltage phasor measurement or estimation in a few fault states (e.g., open-circuit state, short-circuit state, and nominal state), one can induce all the fault responses for any continuous parametric faults.In particular, if componentX is resistance R or capacitance C, which means Zx=Rx or Zx=Xx=1/iωCx, “Theorem 2” is given as follows.Theorem 2.
The faulty response curve of voltage phasorUoi→=UoFi→ in Figure 1 can be determined by (14)-(15), respectively, if the fault component X is capacitance Cx or resistance Rx in linear-analogue circuit.Proof.
Without loss of generality, letZx=Rx+iXx, Zo=Ro+iXo, Uooi→-Uosi→=a+ib, and Uoi→-Uosi→=Ure+iUim, then (12) is given as follows:
(12)Rx+Ro=(A1Ro-A2Xo)(Ure-a)(Ure-a)2+(Uim-b)2+(A2Ro+A1Xo)(Uim-b)(Ure-a)2+(Uim-b)2Xx+Xo=(A2Ro+A1Xo)(Ure-a)(Ure-a)2+(Uim-b)2-(A1Ro-A2Xo)(Uim-b)(Ure-a)2+(Uim-b)2,
where A1=aRo-bXo,A2=bRo+aXo.
Furthermore, the results of (12) lead to (13), when component X is capacitance (Zx=ıXx, Rx=0) or resistance (Zx=Rx, Xx=0) as follows:
(13)(Ure-a)2+(Uim-b)2-B1(Ure-a)+B2(Uim-b)=0,(Ure-a)2+(Uim-b)2-B1′(Ure-a)+B2′(Uim-b)=0,
where B1=(A1Ro-A2Xo)/Ro, B2=(A2Ro+A1Xo)/Ro, B1′=(A2Ro+A1Xo)/Xo, and B2′=(A1Ro-A2Xo)/Xo.It is obvious to find thatUoi→-Uooi→=(Ure-a)+i(Uim-b). Then if one sets Uoi→=Uore+iUoim and Uooi→=Uoore+iUooim, (14)-(15) can be found as follows:
(14)(Uore)2+(Uoim)2-C1Uore-C2Uoim=C,(15)(Uore)2+(Uoim)2-C1′Uore-C2′Uoim=C′,
where C1=2Uoore+B1, C2=2Uooim-B2,C1′=2Uoore+B1′, C2′=2Uooim-B2′, C=-(a2+b2)-B1a+B2b, and C′=-(a2+b2)-B1′a+B2′b.


### 2.2. Voltage Phasor Measurement and Estimation

In the Theorems1 and 2, we know that by separately measuring voltage phasor Uori→ in the nominal circuit state and Uooi→(Uosi→) in open (short)-circuit state, one can calculate the value of the output voltage phasor in all parametric fault states, which benefit us to accomplish the circuit fault modeling.However, the open-circuit and short-circuit faults are the most catastrophic fault states, which means the corresponding analogue circuit in these states has to be destroyed in both physical structure and circuit function; that is, one can not drive the actual circuit into these two dangerous states for constructing the fault modeling. Therefore, there are 2 possible means to solve this tricky thing:(1) faulty responses are simulated in PSpice-modeled circuit, which has been utilized in the SBT method; (2) faulty responses can also be estimated through the voltage phasor measurement and calculation as shown in this section.In Figure2(a), when the circuit is in nominal state, one can set stimulus UIi→ and measure the voltage phasor Uxai→ through component X’s branch as well as locate the value of Uoai→. Furthermore, in Figure 2(b), by adding an auxiliary stimulus with voltage phasor stimuli representation of Uxbi→ in Figure 2(b), in this case, the voltage phasor at To will be Uobi→≠Uoai→, while the value of Uxai→ turns to Uxbi→. After this, Lemma 3 is acquired.Short-circuit response evaluation in linear circuit.
(a)
(b)
(c)
(d)Lemma 3.
The faulty responseUosi→, which is caused by short-circuit fault of component X, is as follows:
(16)Uosi→=Uobi→-UoBi→.Proof.
According to the superposition theorem in linear circuit theories and (3), Uobi→=Uosi→+•axUxbi→. Furthermore, the value of UoBi→ is the CUT output at To when the original input stimulus UIi→ is shortened but the auxiliary stimulus Uxbi→ is preserved; that is, UoBi→=•axUxbi→. As a result, Lemma 3 is founded.Lemma3 means that, with the measurement of voltage phasor in Figures 2(b) and 2(d), one can estimate the short-circuit response at To before the potential faulty component X is involved in short-circuit fault state. The cost is the input stimulus UIi→ should be shortened in the process of voltage measurement. In order not to shorten any stimulus or port for the estimation of Uosi→, Theorem 4 is established.Theorem 4.
The faulty responseUosi→, which is caused by short-circuit fault of component X, is as follows:
(17)Uosi→=Uoai→Uxbi→-Uobi→Uxai→Uxbi→-Uxai→.Proof.
According to the process of deduction of Lemma3, it is apparent to know that UoBi→=•axUxbi→. By substituting it to (6), one obtains the following equations:
(18)Uooi→=Uosi→+UoBi→Uobi→UOXi→.
Furthermore, when circuit is in normal state, the complex impedance ofX is Zxa, and then there is the following equation:
(19)UOXi→=Uxai→(•Zo•Zxa+1).
Based on (6) and (18)-(19), one further knows the following equation:
(20)(Uooi→-Uosi→)(1-UoBi→Uxai→Uxbi→(Uoai→-Uosi→))=0.
In general case,Uooi→-Uosi→≠0; that is, 1-(UoBi→Uxai→/Uxbi→(Uoai→-Uosi→))=0. Hence, with UoBi→=•axUxbi→, one can get the solution of Theorem 2.According to Theorems1 and 2, if one can further obtain the value of Uooi→, the fault modeling with voltage phasor will be established. In fact, the faulty response of Uooi→ can be calculated with Theorem 5.Theorem 5.
If one replaces the auxiliary stimulus in Figure2(b) with an auxiliary branch X′′ owning impedance •Za, then the output response Uobi→=Uobi→. Thus, the value of faulty response Uooi→ at To, which is caused by open-circuit fault of component X, is estimated through (21). Here, •Z//=•Za//•Zxa as follows:
(21)Uooi→=Uobi→+Uoai→-Uobi→1+(•Z///•Zxa)((Uoai→-Uosi→)/(Uobi→-Uosi→)).Proof.
Assume that the faulty complex impedance ofX is •ZxF, while the nominal complex impedance of X is •Zxr; then according to (10)-(11), (22) is as follows:
(22)•ZxF=-•Zxr(UoFi→-Uosi→)(Uori→-Uooi→)(UoFi→-Uooi→)(Uori→-Uosi→).
According to (22), there should be (23), in which UoFi→=Uobi→, Uori→=Uoai→ and •Zxr=•Zxa:
(23)•Z//=•Zxa+(Uobi→-Uosi→)(Uoai→-Uooi→)(Uobi→-Uooi→)(Uoai→-Uosi→).
At last, Theorem5 is established by the result of (23).In short, the paralleled impedance of•Za achieves the impedance transformation in Theorem 5; then the hard-fault state (open-circuit) influence can be estimated with the help of auxiliary branch X′′ and other circuit responses (e.g., Uobi→, Uosi→).


### 2.3. Tolerance Influence on the Voltage Phasor

According to (9) in Theorem 1, Lemma 6 is found.Lemma 6.
Set the faulty value of componentX to •ZxF=ZxrF+iZxiF, and the corresponding circuit output voltage phasor at To is UoFi→=UorF+iUoiF; then within the tolerance circumstance, there are 2 continuous functions shown as follows:
(24)UorF=ħ(•ZxF,UooTori→,UosTori→,UroTori→),UoiF=g(•ZxF,UooTori→,UosTori→,UroTori→),
where UooTori→ (UosTori→) is obtained, when component X is in the open-circuit state (short-circuit state) with all other fault-free components according to component tolerance. And the value of UroTori→ is measured in the case that all components are set around the nominal component values within the tolerance range.Proof.
According to (9), one can assure that there are 2 equations without consideration of tolerance:
(25)UorF=ħ(•ZxF,Uooi→,Uosi→,Uroi→),UoiF=g(•ZxF,Uooi→,Uosi→,Uroi→),
where Uroi→=Uori→. Therefore, Lemma 6 is correct while taking the account of tolerance influence in the analogue circuit.On the basis of Lemma6 and the basic principle of continuous function, one can conclude that if the faulty value of component X varies continuously in a given parametric range as shown in (26), the corresponding value of voltage phasor at the circuit output To should be bounded in (27) as follows:
(26)ZxrF∈[{ZxrF}min⁡,{ZxrF}max⁡],ZxiF∈[{ZxiF}min⁡,{ZxiF}max⁡],(27)UorF∈[{UorF}min⁡,{UorF}max⁡],UoiF∈[{UoiF}min⁡,{UoiF}max⁡].Without loss of generality, if the potential faulty componentX is resistance Rx, whose nominal value is Rx0 and the component tolerance is ±α, then the component value variation within tolerance range is [Rx0(1-α),Rx0(1+α)]. Furthermore, the maximum range of component parameters is [Rx0(1-β),Rx0(1+β)], β>α. Then Lemma 7 and Theorem 8 can be listed.Lemma 7.
Assume that componentR owns the following parametric values: (a) Rx1=((1+β)/(1+α))Rx0 and (b) Rx2=((1-β)/(1-α))Rx0, and these values satisfy: (1)  Rx1(1-α)≤Rx0(1+α); (2)  Rx1(1+α)≥Rx0(1-α)); then the conclusions are made in (28) as follows:
(28)∀Uorβ∈Ar=∪κ=0,1,2[{Uorκ}min⁡,{Uorκ}max⁡],∀Uoiβ∈Ai=∪κ=0,1,2[{Uoiκ}min⁡,{Uoiκ}max⁡],
where the potential faulty component Rx owns the parametric range of [Rxκ(1-α),Rxκ(1+α)], κ=0,1,2, when the corresponding voltage phasor value is {Uoκi→}={Uorκ}+i{Uoiκ}. And Uoβi→=Uorβ+ıUoiβ is the voltage phasor corresponding to the maximum range of component parameters being [Rx0(1-β),Rx0(1+β)], β>α.Proof.
If componentRx can take these values: Rx1=((1+β)/(1+α))Rx0 and Rx2=((1-β)/(1-α))Rx0, which means Rx1(1-α)≤Rx0(1+α) and Rx1(1+α)≥Rx0(1-α), then [Rx0(1-β),Rx0(1+β)]= ∪κ=0,1,2[Rx0(1-α),Rx0(1+α)],β>α.
Therefore, if the value of componentX(Rx) varies within parametric range of [Rx0(1-β),Rx0(1+β)], according to the continuous function in (24), Lemma 7 is established.Theorem 8.
The accurate parametric range ofUorF(UoiF) is [{UorF}min⁡,{UorF}max⁡]([{UoiF}min⁡,{UoiF}max⁡]), which can be determined according to (29), when the faulty component value is Rx0(1-β)≤ZxF=RxF<Rx0(1-α) or Rx0(1+α)<ZxF=RxF≤Rx0(1+β). Consider
(29)[{UorF}min⁡,{UorF}max⁡]=Ar∩[{UrorTor}min⁡,{UrorTor}max⁡]¯,[{UoiF}min⁡,{UoiF}max⁡]=Ai∩[{UroiTor}min⁡,{UroiTor}max⁡]¯,
where Ar=∪κ=0,1,2[{Uorκ}min⁡,{Uorκ}max⁡] and Ai=∪κ=0,1,2[{Uoiκ}min⁡,{Uoiκ}max⁡] are shown in Lemma 7, and Uroi→=Uror+iUroi, which has been given in Lemma 6.Lemma7 and Theorem 8 tell us that in tolerance-influencing circuit, if the potential faulty component is in a continuous faulty parametric range, the corresponding variation range of faulty response at To can be estimated through the simulation or measurement of voltage phasor in some particular discrete faulty parameter cases. Or in other words, the limited simulations or measurements in practice could bring us accurate faulty response range estimation.Furthermore, in the simulation or measurement of voltage phasor in discrete faulty parameter cases, Lemma9 is shown.Lemma 9.
As for a general linear-analogue circuit, it can be considered as a network withb branches and n+1 nodes (one of the nodes is reference node, e.g., ground). Let Yn be the branch complex admittance matrix, en represent the voltage phasor measurement vector, and sn be a complex current vector, then it is obvious to find (30) as follows:
(30)Ynen=sn.
The tolerance and discrete fault values influencing the corresponding branches (components) introduce these increment matrices:ΔYn, Δen, and Δsn as follows:
(31)(Yn+ΔYn)(en+Δen)=sn+Δsn.Without loss of generality, assume that the excitation current is fixed; thenΔsn=0. As a result, Theorem 10 is established.Theorem 10.
Given a discrete faulty parameter for componentX through its branch, the corresponding voltage phasor (real part and image part) follows normal distribution in the tolerance-influencing circuit, if (a) the elements in the matrix ΔY for all fault-free components follow the normal distribution; (b) Δs=0.Proof.
Due to the solution of (30)-(31) and Δs=0, one can get
(32)Δen=-(Yn+ΔYn)-1ΔYnen=-[Yn-1ΔYnen+en].
Therefore, each element ofΔen can be linearly expressed by the elements in matrix ΔYn. And the corresponding voltage phasor (real part and image part) follows normal distribution if the elements in the matrix ΔY follow the normal distribution.Theorem10 states that the complex parametric faults (discrete fault values) lead to the faulty responses at To, whose real part and image part number should conform to the normal distribution.


### 2.4. Fault Diagnosis in Linear-Analogue Circuit

In a linear-analogue circuit under test (CUT), the estimated range of voltage phasor is represented in Table1. Here, the fault states set is F={Fκ}, κ=1,2,…,l. And l is the number of fault states in linear-analogue circuit. Each fault state can represent one of following continuous fault parameter ranges: (a) ZxFκ=(Zxr(1+α),Zxr(1+β)] or ZxFκ=[Zxr(1-β),Zxr(1-α)) for ith component branch in the circuit, where β>α and the tolerance is ±α and (b) ZxFκ=(ZxFx(1-γ),ZxFx(1+γ)] for ith component in the circuit, where the component tolerance ±α (γ≥α) exerts on all other fault-free elements in the circuit and ZxFx is the representative discrete fault value as the central value of parametric range.Table 1
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LConsidering the voltage phasor is dependent on the input stimulus at test-nodeTx, which could variate with different test-frequency of signal; thus, the test-frequency set can be shown in the columns of Table 1: QF={QFm},m=1,2,…,♭, for the best fault diagnosis.The corresponding voltage phasor value range in the rows of Table1 can be separated into two parts: one is from the real part of voltage phasor value (e.g., [{UorFκ}min⁡,{UorFκ}max⁡], 1≤κ≤l), while the other is from the image part of voltage phasor value (e.g., [{UoiFκ}min⁡,{UoiFκ}max⁡], 1≤κ≤l). Therefore, based on the corresponding values in Table 1, the fault diagnosis for a set of parametric faults is categorized into fault detection and isolation: (a) a fault can be detected if the corresponding value of voltage phasor range owns no parametric overlaps with the fault-free state. (b) Meanwhile, a fault can be isolated meaning that this fault state can be distinguished from all other fault states because of no intersection of complex voltages range.To be convenient speaking in the following sections, one sets the representation ofRrκ=[{UorFκ}min⁡,{UorFκ}max⁡] and Riκ=[{UoiFκ}min⁡,{UoiFκ}max⁡]]; then following functions are shown in (33)-(34) as follows:
(33){χdet⁡{Fκ}=1,ifRrκ∩Rr0=∅orRiκ∩Ri0=∅χdet⁡{Fκ}=0,otherwise,
where 1≤κ≤l.
(34){χiso{Fκ1,2}=1,ifRrκ1∩Rrκ2=∅orRiκ1∩Riκ2=∅χiso{Fκ1,2}=0,otherwise,
where 1≤κ1,κ2≤l.Then the result of fault diagnosis is measured through fault detection rate (FDR) and fault isolation rate (FIR) as follows:(a)
Fault detection rate (FDR): the ratio of fault states that can be detected in a given fault states set as follows:(35)FDR=∑κ=1lFκL,
where L=l.(b)
Fault isolation rate (FIR): the ratio of fault states pair can be isolated based on a given fault states pair set as follows:(36)FIR=∑κ=1lFκ1,2N,
where N=l(l-1)/2.


## 2.1. Voltage Phasor Modeling in Fault Diagnosis

Without loss of generality, one assumes that a linear-analogue circuit in Figure1 is excited by an independent voltage stimuli as shown in (1). Besides, this signal is expressed in a phasor form of (2), according to the linear circuit theory in [12]:
(1)UIi=aisin(ωt+ϕi),
where ai is the amplitude of sinusoidal signal of UI, ω is a known fixed frequency, and ϕi represents the phase of this signal. Therefore, voltage phasor UIi→ is a complex value including the information of amplitude and phase from the stimuli of UI(2)UIi→:ai2eiϕ.Figure 1
The linear circuit network.With the input signalUI, the circuit output response is also a sinusoidal signal Uo=aosin(ωt+ϕo), whose voltage phasor form is Uoi→. Then, on the basis of the superposition theorem [13, 14], one can establish (3) for the voltage phasor Uoi→ at test-node To as follows:
(3)Uoi→=•aiUIi→+•aXUXi→,
where •ai is the complex transmission factor between UIi→ and Uoi→, while •aX is the complex transmission factor from UXi→ to Uoi→.As a matter of fact, the value of•aiUIi→ equals the value of voltage phasor response at To, if the input stimulus UI is the sole incentive and component X is in short-circuit state. In order to be convenient speaking in this paper, one can set it as Uosi→=•aiUIi→.Ulteriorly, on the basis of Thevenin’s theorem in the linear-analogue circuit, one can get the following:(4)UXi→=UOXi→•Zx•Zx+•Zo,
where the UXi→ (UOXi→) represents a voltage phasor through the circuit branch Tx-To, when component X is in normal (open) state. Besides, •Zx is a complex impedance for component X. Meanwhile, •Zo is Thevenin’s equivalent impedance seeing through the circuit ports Tx and To, when component X is open.Due to the formulaUosi→=•aiUIi→ and (3)-(4), this paper finds the following:
(5)Uoi→=Uosi→+•aXUOXi→•Zx•Zx+•Zo.In particular, one assumes that componentX is in open-circuit state; then the voltage phasor at To is Uooi→ as follows:
(6)Uooi→=Uosi→+•aXUOXi→.At last, in accordance with (5) and (6), there exists a formula as follows:
(7)•Zx•Zo=-Uoi→-Uosi→Uoi→-Uooi→.In other words, (8) is formulated as follows:
(8)Uoi→=Γ(•Zo,•Zx,Uosi→,Uooi→)=•ZoUosi→+•ZxUooi→•Zo+•Zx,
where Γ represents an invariable continuous function for a given linear-analogue circuit.In the case of parametric fault ofX, the value of Uosi→ or Uooi→ is considered as constant. At last, “Theorem 1” can be found.Theorem 1.
In analogue circuit of Figure1, the potential faulty component X owns the nominal impedance value •Zxr, and the nominal response is Uori→; then the faulty response UoFi→ caused by parametric fault of X at test-node To can be calculated according to the following equation:
(9)UoFi→=•α•ZxFUooi→-Uosi→•α•ZxF-1,
where the value of •α is a constant determined by •α=(1/•Zxr)((Uori→-Uosi→)/(Uori→-Uooi→)).Proof.
With the statement in (8), it is obvious to find following equations:
(10)•Zxr•Zo=-Uori→-Uosi→Uori→-Uooi→,(11)•ZxF•Zo=-UoFi→-Uosi→UoFi→-Uooi→.
Later, if (10) is divided by (11), •Zo will be eliminated. Then Theorem 1 is obtained.Theorem1 builds up the accurate analytic expression of fault response, because of parametric fault of component X. It also tells us that based on the voltage phasor measurement or estimation in a few fault states (e.g., open-circuit state, short-circuit state, and nominal state), one can induce all the fault responses for any continuous parametric faults.In particular, if componentX is resistance R or capacitance C, which means Zx=Rx or Zx=Xx=1/iωCx, “Theorem 2” is given as follows.Theorem 2.
The faulty response curve of voltage phasorUoi→=UoFi→ in Figure 1 can be determined by (14)-(15), respectively, if the fault component X is capacitance Cx or resistance Rx in linear-analogue circuit.Proof.
Without loss of generality, letZx=Rx+iXx, Zo=Ro+iXo, Uooi→-Uosi→=a+ib, and Uoi→-Uosi→=Ure+iUim, then (12) is given as follows:
(12)Rx+Ro=(A1Ro-A2Xo)(Ure-a)(Ure-a)2+(Uim-b)2+(A2Ro+A1Xo)(Uim-b)(Ure-a)2+(Uim-b)2Xx+Xo=(A2Ro+A1Xo)(Ure-a)(Ure-a)2+(Uim-b)2-(A1Ro-A2Xo)(Uim-b)(Ure-a)2+(Uim-b)2,
where A1=aRo-bXo,A2=bRo+aXo.
Furthermore, the results of (12) lead to (13), when component X is capacitance (Zx=ıXx, Rx=0) or resistance (Zx=Rx, Xx=0) as follows:
(13)(Ure-a)2+(Uim-b)2-B1(Ure-a)+B2(Uim-b)=0,(Ure-a)2+(Uim-b)2-B1′(Ure-a)+B2′(Uim-b)=0,
where B1=(A1Ro-A2Xo)/Ro, B2=(A2Ro+A1Xo)/Ro, B1′=(A2Ro+A1Xo)/Xo, and B2′=(A1Ro-A2Xo)/Xo.It is obvious to find thatUoi→-Uooi→=(Ure-a)+i(Uim-b). Then if one sets Uoi→=Uore+iUoim and Uooi→=Uoore+iUooim, (14)-(15) can be found as follows:
(14)(Uore)2+(Uoim)2-C1Uore-C2Uoim=C,(15)(Uore)2+(Uoim)2-C1′Uore-C2′Uoim=C′,
where C1=2Uoore+B1, C2=2Uooim-B2,C1′=2Uoore+B1′, C2′=2Uooim-B2′, C=-(a2+b2)-B1a+B2b, and C′=-(a2+b2)-B1′a+B2′b.


## 2.2. Voltage Phasor Measurement and Estimation

In the Theorems1 and 2, we know that by separately measuring voltage phasor Uori→ in the nominal circuit state and Uooi→(Uosi→) in open (short)-circuit state, one can calculate the value of the output voltage phasor in all parametric fault states, which benefit us to accomplish the circuit fault modeling.However, the open-circuit and short-circuit faults are the most catastrophic fault states, which means the corresponding analogue circuit in these states has to be destroyed in both physical structure and circuit function; that is, one can not drive the actual circuit into these two dangerous states for constructing the fault modeling. Therefore, there are 2 possible means to solve this tricky thing:(1) faulty responses are simulated in PSpice-modeled circuit, which has been utilized in the SBT method; (2) faulty responses can also be estimated through the voltage phasor measurement and calculation as shown in this section.In Figure2(a), when the circuit is in nominal state, one can set stimulus UIi→ and measure the voltage phasor Uxai→ through component X’s branch as well as locate the value of Uoai→. Furthermore, in Figure 2(b), by adding an auxiliary stimulus with voltage phasor stimuli representation of Uxbi→ in Figure 2(b), in this case, the voltage phasor at To will be Uobi→≠Uoai→, while the value of Uxai→ turns to Uxbi→. After this, Lemma 3 is acquired.Short-circuit response evaluation in linear circuit.
(a)
(b)
(c)
(d)Lemma 3.
The faulty responseUosi→, which is caused by short-circuit fault of component X, is as follows:
(16)Uosi→=Uobi→-UoBi→.Proof.
According to the superposition theorem in linear circuit theories and (3), Uobi→=Uosi→+•axUxbi→. Furthermore, the value of UoBi→ is the CUT output at To when the original input stimulus UIi→ is shortened but the auxiliary stimulus Uxbi→ is preserved; that is, UoBi→=•axUxbi→. As a result, Lemma 3 is founded.Lemma3 means that, with the measurement of voltage phasor in Figures 2(b) and 2(d), one can estimate the short-circuit response at To before the potential faulty component X is involved in short-circuit fault state. The cost is the input stimulus UIi→ should be shortened in the process of voltage measurement. In order not to shorten any stimulus or port for the estimation of Uosi→, Theorem 4 is established.Theorem 4.
The faulty responseUosi→, which is caused by short-circuit fault of component X, is as follows:
(17)Uosi→=Uoai→Uxbi→-Uobi→Uxai→Uxbi→-Uxai→.Proof.
According to the process of deduction of Lemma3, it is apparent to know that UoBi→=•axUxbi→. By substituting it to (6), one obtains the following equations:
(18)Uooi→=Uosi→+UoBi→Uobi→UOXi→.
Furthermore, when circuit is in normal state, the complex impedance ofX is Zxa, and then there is the following equation:
(19)UOXi→=Uxai→(•Zo•Zxa+1).
Based on (6) and (18)-(19), one further knows the following equation:
(20)(Uooi→-Uosi→)(1-UoBi→Uxai→Uxbi→(Uoai→-Uosi→))=0.
In general case,Uooi→-Uosi→≠0; that is, 1-(UoBi→Uxai→/Uxbi→(Uoai→-Uosi→))=0. Hence, with UoBi→=•axUxbi→, one can get the solution of Theorem 2.According to Theorems1 and 2, if one can further obtain the value of Uooi→, the fault modeling with voltage phasor will be established. In fact, the faulty response of Uooi→ can be calculated with Theorem 5.Theorem 5.
If one replaces the auxiliary stimulus in Figure2(b) with an auxiliary branch X′′ owning impedance •Za, then the output response Uobi→=Uobi→. Thus, the value of faulty response Uooi→ at To, which is caused by open-circuit fault of component X, is estimated through (21). Here, •Z//=•Za//•Zxa as follows:
(21)Uooi→=Uobi→+Uoai→-Uobi→1+(•Z///•Zxa)((Uoai→-Uosi→)/(Uobi→-Uosi→)).Proof.
Assume that the faulty complex impedance ofX is •ZxF, while the nominal complex impedance of X is •Zxr; then according to (10)-(11), (22) is as follows:
(22)•ZxF=-•Zxr(UoFi→-Uosi→)(Uori→-Uooi→)(UoFi→-Uooi→)(Uori→-Uosi→).
According to (22), there should be (23), in which UoFi→=Uobi→, Uori→=Uoai→ and •Zxr=•Zxa:
(23)•Z//=•Zxa+(Uobi→-Uosi→)(Uoai→-Uooi→)(Uobi→-Uooi→)(Uoai→-Uosi→).
At last, Theorem5 is established by the result of (23).In short, the paralleled impedance of•Za achieves the impedance transformation in Theorem 5; then the hard-fault state (open-circuit) influence can be estimated with the help of auxiliary branch X′′ and other circuit responses (e.g., Uobi→, Uosi→).


## 2.3. Tolerance Influence on the Voltage Phasor

According to (9) in Theorem 1, Lemma 6 is found.Lemma 6.
Set the faulty value of componentX to •ZxF=ZxrF+iZxiF, and the corresponding circuit output voltage phasor at To is UoFi→=UorF+iUoiF; then within the tolerance circumstance, there are 2 continuous functions shown as follows:
(24)UorF=ħ(•ZxF,UooTori→,UosTori→,UroTori→),UoiF=g(•ZxF,UooTori→,UosTori→,UroTori→),
where UooTori→ (UosTori→) is obtained, when component X is in the open-circuit state (short-circuit state) with all other fault-free components according to component tolerance. And the value of UroTori→ is measured in the case that all components are set around the nominal component values within the tolerance range.Proof.
According to (9), one can assure that there are 2 equations without consideration of tolerance:
(25)UorF=ħ(•ZxF,Uooi→,Uosi→,Uroi→),UoiF=g(•ZxF,Uooi→,Uosi→,Uroi→),
where Uroi→=Uori→. Therefore, Lemma 6 is correct while taking the account of tolerance influence in the analogue circuit.On the basis of Lemma6 and the basic principle of continuous function, one can conclude that if the faulty value of component X varies continuously in a given parametric range as shown in (26), the corresponding value of voltage phasor at the circuit output To should be bounded in (27) as follows:
(26)ZxrF∈[{ZxrF}min⁡,{ZxrF}max⁡],ZxiF∈[{ZxiF}min⁡,{ZxiF}max⁡],(27)UorF∈[{UorF}min⁡,{UorF}max⁡],UoiF∈[{UoiF}min⁡,{UoiF}max⁡].Without loss of generality, if the potential faulty componentX is resistance Rx, whose nominal value is Rx0 and the component tolerance is ±α, then the component value variation within tolerance range is [Rx0(1-α),Rx0(1+α)]. Furthermore, the maximum range of component parameters is [Rx0(1-β),Rx0(1+β)], β>α. Then Lemma 7 and Theorem 8 can be listed.Lemma 7.
Assume that componentR owns the following parametric values: (a) Rx1=((1+β)/(1+α))Rx0 and (b) Rx2=((1-β)/(1-α))Rx0, and these values satisfy: (1)  Rx1(1-α)≤Rx0(1+α); (2)  Rx1(1+α)≥Rx0(1-α)); then the conclusions are made in (28) as follows:
(28)∀Uorβ∈Ar=∪κ=0,1,2[{Uorκ}min⁡,{Uorκ}max⁡],∀Uoiβ∈Ai=∪κ=0,1,2[{Uoiκ}min⁡,{Uoiκ}max⁡],
where the potential faulty component Rx owns the parametric range of [Rxκ(1-α),Rxκ(1+α)], κ=0,1,2, when the corresponding voltage phasor value is {Uoκi→}={Uorκ}+i{Uoiκ}. And Uoβi→=Uorβ+ıUoiβ is the voltage phasor corresponding to the maximum range of component parameters being [Rx0(1-β),Rx0(1+β)], β>α.Proof.
If componentRx can take these values: Rx1=((1+β)/(1+α))Rx0 and Rx2=((1-β)/(1-α))Rx0, which means Rx1(1-α)≤Rx0(1+α) and Rx1(1+α)≥Rx0(1-α), then [Rx0(1-β),Rx0(1+β)]= ∪κ=0,1,2[Rx0(1-α),Rx0(1+α)],β>α.
Therefore, if the value of componentX(Rx) varies within parametric range of [Rx0(1-β),Rx0(1+β)], according to the continuous function in (24), Lemma 7 is established.Theorem 8.
The accurate parametric range ofUorF(UoiF) is [{UorF}min⁡,{UorF}max⁡]([{UoiF}min⁡,{UoiF}max⁡]), which can be determined according to (29), when the faulty component value is Rx0(1-β)≤ZxF=RxF<Rx0(1-α) or Rx0(1+α)<ZxF=RxF≤Rx0(1+β). Consider
(29)[{UorF}min⁡,{UorF}max⁡]=Ar∩[{UrorTor}min⁡,{UrorTor}max⁡]¯,[{UoiF}min⁡,{UoiF}max⁡]=Ai∩[{UroiTor}min⁡,{UroiTor}max⁡]¯,
where Ar=∪κ=0,1,2[{Uorκ}min⁡,{Uorκ}max⁡] and Ai=∪κ=0,1,2[{Uoiκ}min⁡,{Uoiκ}max⁡] are shown in Lemma 7, and Uroi→=Uror+iUroi, which has been given in Lemma 6.Lemma7 and Theorem 8 tell us that in tolerance-influencing circuit, if the potential faulty component is in a continuous faulty parametric range, the corresponding variation range of faulty response at To can be estimated through the simulation or measurement of voltage phasor in some particular discrete faulty parameter cases. Or in other words, the limited simulations or measurements in practice could bring us accurate faulty response range estimation.Furthermore, in the simulation or measurement of voltage phasor in discrete faulty parameter cases, Lemma9 is shown.Lemma 9.
As for a general linear-analogue circuit, it can be considered as a network withb branches and n+1 nodes (one of the nodes is reference node, e.g., ground). Let Yn be the branch complex admittance matrix, en represent the voltage phasor measurement vector, and sn be a complex current vector, then it is obvious to find (30) as follows:
(30)Ynen=sn.
The tolerance and discrete fault values influencing the corresponding branches (components) introduce these increment matrices:ΔYn, Δen, and Δsn as follows:
(31)(Yn+ΔYn)(en+Δen)=sn+Δsn.Without loss of generality, assume that the excitation current is fixed; thenΔsn=0. As a result, Theorem 10 is established.Theorem 10.
Given a discrete faulty parameter for componentX through its branch, the corresponding voltage phasor (real part and image part) follows normal distribution in the tolerance-influencing circuit, if (a) the elements in the matrix ΔY for all fault-free components follow the normal distribution; (b) Δs=0.Proof.
Due to the solution of (30)-(31) and Δs=0, one can get
(32)Δen=-(Yn+ΔYn)-1ΔYnen=-[Yn-1ΔYnen+en].
Therefore, each element ofΔen can be linearly expressed by the elements in matrix ΔYn. And the corresponding voltage phasor (real part and image part) follows normal distribution if the elements in the matrix ΔY follow the normal distribution.Theorem10 states that the complex parametric faults (discrete fault values) lead to the faulty responses at To, whose real part and image part number should conform to the normal distribution.


## 2.4. Fault Diagnosis in Linear-Analogue Circuit

In a linear-analogue circuit under test (CUT), the estimated range of voltage phasor is represented in Table1. Here, the fault states set is F={Fκ}, κ=1,2,…,l. And l is the number of fault states in linear-analogue circuit. Each fault state can represent one of following continuous fault parameter ranges: (a) ZxFκ=(Zxr(1+α),Zxr(1+β)] or ZxFκ=[Zxr(1-β),Zxr(1-α)) for ith component branch in the circuit, where β>α and the tolerance is ±α and (b) ZxFκ=(ZxFx(1-γ),ZxFx(1+γ)] for ith component in the circuit, where the component tolerance ±α (γ≥α) exerts on all other fault-free elements in the circuit and ZxFx is the representative discrete fault value as the central value of parametric range.Table 1
Voltage phasor-based fault diagnosis with tolerance.
Q
F
1
Q
F
2
⋯
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F
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L
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LConsidering the voltage phasor is dependent on the input stimulus at test-nodeTx, which could variate with different test-frequency of signal; thus, the test-frequency set can be shown in the columns of Table 1: QF={QFm},m=1,2,…,♭, for the best fault diagnosis.The corresponding voltage phasor value range in the rows of Table1 can be separated into two parts: one is from the real part of voltage phasor value (e.g., [{UorFκ}min⁡,{UorFκ}max⁡], 1≤κ≤l), while the other is from the image part of voltage phasor value (e.g., [{UoiFκ}min⁡,{UoiFκ}max⁡], 1≤κ≤l). Therefore, based on the corresponding values in Table 1, the fault diagnosis for a set of parametric faults is categorized into fault detection and isolation: (a) a fault can be detected if the corresponding value of voltage phasor range owns no parametric overlaps with the fault-free state. (b) Meanwhile, a fault can be isolated meaning that this fault state can be distinguished from all other fault states because of no intersection of complex voltages range.To be convenient speaking in the following sections, one sets the representation ofRrκ=[{UorFκ}min⁡,{UorFκ}max⁡] and Riκ=[{UoiFκ}min⁡,{UoiFκ}max⁡]]; then following functions are shown in (33)-(34) as follows:
(33){χdet⁡{Fκ}=1,ifRrκ∩Rr0=∅orRiκ∩Ri0=∅χdet⁡{Fκ}=0,otherwise,
where 1≤κ≤l.
(34){χiso{Fκ1,2}=1,ifRrκ1∩Rrκ2=∅orRiκ1∩Riκ2=∅χiso{Fκ1,2}=0,otherwise,
where 1≤κ1,κ2≤l.Then the result of fault diagnosis is measured through fault detection rate (FDR) and fault isolation rate (FIR) as follows:(a)
Fault detection rate (FDR): the ratio of fault states that can be detected in a given fault states set as follows:(35)FDR=∑κ=1lFκL,
where L=l.(b)
Fault isolation rate (FIR): the ratio of fault states pair can be isolated based on a given fault states pair set as follows:(36)FIR=∑κ=1lFκ1,2N,
where N=l(l-1)/2.


## 3. Computational Example

In this section, there are 2 representative examples to validate the proposed fault diagnosis based on the voltage phasor analysis in both tolerance-free and tolerance-influencing circuit. To be simplifying the question, in the first example of this section, the proposed method to estimate the parametric fault response and hard-faults influence is concerned with the theories from Theorem1 and Theorems 4–8 and their corresponding lemmas. After this, in the second example, one can organize the complex voltages as what has been shown in Table 1 and select appropriate test-signal frequencies; that is, we can use these selected columns for fault diagnosis of a given fault states set, in order to satisfy the requirement of FIR≥90% and FDR≥90%.


### 3.1. Thomas Filter Example

The first example circuit is shown in Figure3, while the circuit output is at T3. This example is used to demonstrate the critical results of proposed method for fault diagnosis based on complex voltages estimation, and the parametric faults are as follows: (a) discrete fault values in tolerance-free case as shown in Tables 2 and 4; (b) the parameter of faulty components varies from 85% to 115% of its discrete fault values in Table 5. Here, the former one is to testify the accuracy of Theorems 1, 4, and 5, while the latter one is used to demonstrate the process of faulty response estimation with consideration of tolerance and continuous parametric fault pattern diagnosis in the linear-analogue circuit.Table 2
Component values in circuit of Figure3.
Nominal values
Discrete fault values
Parametric faults limit
R
1
1
k
Ω
1.50
k
Ω
[
1.28
k
Ω
,
1.73
k
Ω
]
R
2
1
k
Ω
5.00
k
Ω
[
4.25
k
Ω
,
5.75
k
Ω
]
R
3
1
k
Ω
1.25
k
Ω
[
1.06
k
Ω
,
1.44
k
Ω
]
R
4
1
k
Ω
0.75
k
Ω
[
0.64
k
Ω
,
0.86
k
Ω
]
R
5
1
k
Ω
0.50
k
Ω
[
0.43
k
Ω
,
0.58
k
Ω
]
R
6
1
k
Ω
1.20
k
Ω
[
1.02
k
Ω
,
1.38
k
Ω
]
C
1
100
nF
50.00
nF
[42.5 nF, 57.5 nF]
C
2
100
nF
150.00
nF
[127.5 nF, 172.5 nF]Figure 3
The Thomas filter.In addition, in this circuit example, the stimulus signal (green sketch in Figure4) is as follows: (a) the amplitude is 2 (V); (b) the test-frequency is set as 1.5 kHz; (c) the phase of this signal is zero. Furthermore, the circuit response in nominal circuit state is also given in Figure 4 (red sketch), which satisfies: (a) the amplitude is 9.67 (V); (b) the test-frequency is set as 1.5 kHz; (c) the phase of this signal is -0.69. Therefore, according to the relationship between (1) and (2), the corresponding phasor is 1.00+ı0.00 (V) (-5.23+ı4.41 (V)).Figure 4
The circuit stimulus (green) and response (red) in AC test.In order to estimate the faulty responses because of parametric faults in second column of Table2, one should primarily estimate all hard-faults influence in this circuit. For instance, as for test-node T3, if R3 is assumed to be the potential parametric fault component, the corresponding circuit output voltage phasor at test-node T3 could be estimated as the following procedure in the circuit with nominal parametric values: (a) the circuit is in nominal state (fault-free), and the stimulus is UIi→=1.00+i0.00 (V); (b) the voltage phasor through R3 branch is Uxai→=-4.16-i4.93 (V), and one can locate this complex value through the amplitude and phase of sinusoidal signal at test-node T1; (c) as shown in Figure 3, one adds an auxiliary voltage stimulus Uxbi→=1.00+i0.00 (V), and at the same time, the voltage phasor at test-node T3 is Uobi→=-1.10-i0.94 (V); (d) let all these measurements be in (17), the solution of (17) is Uosi→=0.00+i0.00 (V).The similar measurements and calculations are processed for the other hard-faults influence estimations at test-nodeT3; after that, the corresponding voltage phasor values are shown in Table 3.Table 3
Voltage phasors for hard-faults atT3 in Figure 3.
Fault response (short-circuit)
Fault response (open-circuit)
R
1
-
1.13
+
i
1.06
×
1
0
6
0.00
+
i
0.00
R
2
-
0.00
-
i
0.00
-
8.95
+
i
0.00
R
3
-
1.00
+
i
0.00
0.00
+
i
0.00
R
4
-
1.00
+
i
0.00
0.00
+
i
0.00
R
5
0.00
+
i
0.00
-
1.00
+
i
0.00
R
6
-
0.00
+
i
0.00
1.11
+
i
0.12
C
1
0.00
+
i
0.00
-
0.99
+
i
0.10
C
2
0.00
+
i
0.00
-
1.00
+
i
0.00Table 4
Voltage phasors for parametric faults at test-nodeT3 in Figure 3.
e
{
U
o
F
→
}
U
o
F
→
Diagnosis
R
1
=
1.50
k
Ω
-
3.49
+
i
2.94
-
3.49
+
i
2.94
Y
R
2
=
5.00
k
Ω
2.34
-
i
3.93
-
2.33
+
i
3.93
Y
R
3
=
1.25
k
Ω
4.24
+
i
4.52
4.24
+
i
4.52
Y
R
4
=
0.75
k
Ω
-
2.86
+
i
0.61
-
2.86
+
i
0.61
Y
R
5
=
0.50
k
Ω
1.22
+
i
0.30
1.22
+
i
0.30
Y
R
6
=
1.20
k
Ω
4.61
+
i
7.88
4.61
+
i
7.92
Y
C
1
=
50.00
nF
-
1.75
+
i
0.33
-
1.75
+
i
0.33
Y
C
2
=
150.00
nF
2.55
+
i
1.08
2.55
+
i
1.08
YTable 5
Voltage phasor-based fault diagnosis.
U
o
r
1
U
o
i
1
Diagnosis
No faults
[
-
5.50
,
-
4.98
]
[
4.01,5.38
]
100
%
R
2
[
-
4.01
,
-
2.96
]
[
2.12,3.10
]
100
%
C
1
[
-
2.64
,
-
1.99
]
[
0.24,1.40
]
100
%
C
2
[
2.20,3.64
]
[
0.82,2.36
]
100
%
1The voltage phasor response at T3 is Uo→=Uor+ıUoi.These particular hard-fault responses can be used to evaluate the parametric fault responses through the solution of Theorem1. Here, some parametric fault examples are shown in Table 4, and one gives a specific process of calculating fault response at test-node T3, when R2 is the fault component (e.g., R2=5kΩ): (a) according to data in Table 3, Uosi→=-0.00-i0.00 (V) and Uooi→=-8.95-i0.00 (V); (b) when all the components are in nominal circuit state R2=Z2r=1kΩ, the value of Uoi→ on test-node T3 is -5.23+i4.41 (V); (c) according to (9), •a is (-0.00+i1.19)/104; (d) let all these measurements, estimations, and calculations be in (9); one can find the estimated complex value e{UoFi→}=2.34-i3.93 (V).All the estimated results for the discrete parametric faults in Table2 (column 2) have been calculated based on (9) and been listed in Table 4 (column 2). And the results from actual circuit measurements have also been shown in the third column of Table 4. As a consequence, the corresponding fault diagnosis can be well done because of the consistence of data (e.g., the amplitude and phase reflected in the voltage phasor value) in column 2 and 3 of Table 4.The results in Tables3 and 4 point out that the process of calculating faulty response when the circuit suffered from parametric fault without consideration of tolerance. Furthermore, in order to form the corresponding faulty response data while considering the tolerance influence, the basic solutions of Lemmas 6 and 7 and Theorem 8 are used. Here, one gives a example process to calculate the the voltage phasor value range for R1 at test-node T3. Here, the fault component R1 varies in [0.85R1F,0.95R1F)∪[1.05R1F,1.15R1F], and R1F=1.5kΩ. Then, there exist these values: Rx0= 1.500 kΩ, Rx1= 1.643 kΩ, and Rx2=1.342 kΩ (1.358 kΩ), which lead to the following component value ranges with respective 5% tolerance influence:(a)
[1.425 kΩ, 1.575 kΩ], (b) [1.561 kΩ, 1.725 kΩ],(c)
[1.275 kΩ, 1.409 kΩ], (d) [1.290 kΩ, 1.426 kΩ].These parametric sets unify to cover the whole range of[0.85R1F,0.95R1F)∪(1.05R1F,1.15R1F] for component R1. Therefore, by running 100 simulations around the component value Rx0 or Rx1 or Rx2, with the consideration of tolerance ±5%, the complex response ranges (including real part range r and image part range i for each complex response) at test-node T3 are as follows:(a)
r=[-3.67,-3.25], i=[2.32,3.10];(b)
r=[-3.28,-2.96], i=[2.12,2.57];(c)
r=[-4.01,-3.63], i=[2.55,2.99];(d)
r=[-3.75,-3.64], i=[2.55,2.84].Therefore, according to Lemma7 and Theorem 8, the faulty response range for component R1=[0.85R1F,0.95R1F)∪(1.05R1F,1.15R1F] at the test-node T3 is as follows:(a)
[{UorF}min⁡,{UorF}max⁡]=[2.96,4.01]∩[4.98,5.50]¯=[2.96,4.01];(b)
[{UoiF}min⁡,{UoiF}max⁡]=[2.12,3.10]∩[4.01,5.38]¯=[2.12,3.10].Take some representative parametric faults in Table5 into account; the corresponding fault states are continuous fault component values in a given bounded parametric varying range. And the other fault-free components’ values are randomly selected around their corresponding nominal values, while the tolerance is ±α=±5%. Here, according to the contents in Table 5, it is obvious to find both of the values of FDR and FIR are 100% for these 3 parametric faults.


### 3.2. Sallen-Key Filter Example

The proposed fault diagnosis method will be testified in another representative linear circuit: Sallen-Key filter example. In this section, the nominal component value is as follows:R1= 1 kΩ; R2= 3 kΩ; R3= 2 kΩ; R4= 4 kΩ; R5= 6 kΩ; C1,C2= 5 nF. Different from the discussion in the previous subsection, the test-frequencies of sinusoidal signal are carefully selected according to the fault response curves determined by Theorem 2. And all the candidate test-signal frequencies eventually construct a test-frequencies set Q=QFı, ı=1,2,…,♭.The number of♭ (test-frequencies numbers) is quite dependent on the number of systematic parameter in transfer function. For instance, in this example, the transfer function is in (37). Therefore, there are at most 3 test-frequencies required, in order to determine all the systematic parameter in (37). Consider
(37)ħ(s)=((R5+R4)R1R4C1s)×((R1+R2)C1C2R1R2R3s2+(1C2R3+1C1R1+1C1R3-R5C1R2R4)s00100+(R1+R2)C1C2R1R2R3s2)-1.Note that, in this example, if the faulty component can own all possible parametric fault values (0≤∥ZxF∥<∞), and the corresponding stimulus signal is 2sin(2*π*3000t-0), then the corresponding faulty response curve is shown in Figure 6 for each component. With Theorem 2, these curves are determined according to equations as follows: (a) R1: (Uore)2+(Uoim)2-1.86Uore-1.21Uoim=0. (b) R2: (Uore)2+(Uoim)2-2.99Uore-8.46Uoim=0. (c) R3: (Uore)2+(Uoim)2-1.36Uore+1.93Uoim=0. (d) R4,5: (Uore)2+(Uoim)2+3.00Uore+29.76Uoim=0.0035. (e) C1: (Uore)2+(Uoim)2-1.82Uore-0.96Uoim=0. (f) C2: (Uore)2+(Uoim)2-1.50Uore+1.06Uoim=0.The intersections of these curves indicate the potential ambiguous faults in Table6; for instance, when R1 is open, or R3 is short, the voltage phasor on To owns the same value: 0.00+i0.00 (V). Thus, the fault state R3: short, is one of ambiguous faults to the fault state R1: Open.Table 6
Ambiguous faults in the circuit of Figure4.
Fault state
Ambiguous faults
R
1
Open
R
2,3: Short, C1: Short, C2: Open
R
2
Short
R
1: Open, R3: Short, C1: Short, C2: Open
R
3
Short
R
1: Open, R2: Short, C1: Short, C2: Open
R
4
1
All parametric faults
R
5: All parametric faults
R
5
1
All parametric faults
R
4: All parametric faults
C
1
Short
R
1: Open, R2,3: Short, C2: Open
C
2
Open
R
1: Open, R2,3: Short, C1: Short
1For any given parametric fault of R4(R5),there exists a parametric fault of R5(R4) generating the undistinguishable complex faulty response at test-node To.According to Figure6 and Tables 6 and 7, test-frequency QF1= 30 kHz is a sensitive frequency for the circuit under test (CUT) of Figure 5 in parametric fault diagnosis (Table 7). Furthermore, one can sketch the other response curves with different test-frequencies. Such investigation leads to 3 representative sensitive test-frequencies: QF1= 30 kHz, QF2= 20 kHz, and QF2= 60 kHz, for a given fault states set in Table 7.Table 7
Voltage phasor-based tolerance-free fault diagnosis with test-frequencies setQF={QF1,QF2,QF3}.
Q
F
1
Q
F
2
Q
F
3
No faults
1.63
-
ı
0.26
1.54
+
ı
0.44
0.82
-
ı
0.83
R
1
=
1.50
k
Ω
1.27
-
ı
0.45
1.40
+
ı
0.20
0.49
-
ı
0.68
R
2
=
1.50
k
Ω
2.49
-
ı
0.15
1.68
+
ı
1.17
0.81
-
ı
1.17
R
3
=
3.00
k
Ω
1.84
-
ı
0.74
2.13
+
ı
0.14
0.71
-
ı
1.01
R
4
=
2.00
k
Ω
3.79
-
ı
0.89
3.37
+
ı
1.46
1.20
-
ı
1.83
R
5
=
3.00
k
Ω
0.98
-
ı
0.13
0.94
+
ı
0.23
0.57
-
ı
0.49
C
1
=
2.50
nF
1.93
+
ı
0.65
1.47
+
ı
0.88
1.64
-
ı
0.77
C
2
=
2.50
nF
1.18
+
ı
0.28
0.80
+
ı
0.60
0.93
-
ı
0.55Figure 5
The Sallen-Key filter circuit.Figure 6
Complex faulty response curve for each component, whereR-axis represents Uore, whileI-axis represents Uoim.These test-frequencies can be testified as sensitive frequencies, according to the frequency sweeping results in Figure7: (a) each response curve corresponds to a component fault in Table 7; (b) actually, in Figure 6, the test-frequencies QF1= 30 kHz and QF2= 20 kHz are around the center frequency of Sallen-Key filter.Figure 7
The Sallen-Key filter circuit response in frequency domain.As a matter of fact, the sensitive test-frequencies investigated in Figures6 and 7 are also the recommendations in the circuit diagnosis for all potential single parametric faults with tolerance consideration (i.e., component tolerance for fault-free component is 2%, while the parametric range for faulty component is [95%Xc,105%Xc] and Xc is the value in Table 7 for each potential faulty component). For instance, as for the fault states set in Table 8, if one lists the corresponding voltage phasor varying ranges, which can be estimated through the similar manner as what the Thomas filter example does, it is simple to locate the value of FIR=100% and FDR=100%. Therefore, all faults in Table 8 can be 100% diagnosis with given test-signals set.Table 8
Voltage phasor-based tolerance-influencing fault diagnosis with test-frequencies setQF={QF1,QF2,QF3}.
Q
F
1
Q
F
2
Q
F
3
No faults
[
1.37,1.78
]
[
1.32,1.63
]
[
0.73,0.95
]
[
-
0.27
,
-
0.12
]
[
0.35,0.56
]
[
-
0.90
,
-
0.70
]
R
1: [1.43Ω,1.58kΩ]
[
1.09,1.41
]
[
1.15,1.51
]
[
0.42,0.58
]
[
-
0.46
,
-
0.29
]
[
0.14,0.32
]
[
-
0.75
,
-
0.56
]
R
2: [1.43Ω,1.58kΩ]
[
1.91,2.73
]
[
1.27,1.74
]
[
0.73,1.03
]
[
-
0.14,0.15
]
[
0.87,1.35
]
[
-
1.29
,
-
0.94
]
R
3: [2.85kΩ,3.15kΩ]
[
1.66,2.01
]
[
1.76,2.26
]
[
0.61,0.85
]
[
-
0.74
,
-
0.51
]
[
0.10,0.33
]
[
-
1.11
,
-
0.86
]
R
4: [1.90kΩ,2.10kΩ]
[
2.92,4.24
]
[
2.89,3.55
]
[
1.07,1.52
]
[
-
0.89
,
-
0.27
]
[
1.00,1.73
]
[
-
2.07
,
-
1.45
]
R
5: [2.85kΩ,3.15kΩ]
[
0.86,1.05
]
[
0.83,0.98
]
[
0.51,0.65
]
[
-
0.13
,
-
0.05
]
[
0.19,0.27
]
[
-
0.52
,
-
0.45
]
C
1: [2.34nf,2.63nf]
[
1.74,2.05
]
[
1.19,1.54
]
[
1.19,1.54
]
[
0.30,0.46
]
[
0.73,0.96
]
[
0.73,0.96
]
C
2: [2.34nf,2.63nf]
[
0.97,1.26
]
[
0.65,0.83
]
[
0.78,1.04
]
[
0.21,0.37
]
[
0.49,0.66
]
[
-
0.57
,
-
0.42
]


## 3.1. Thomas Filter Example

The first example circuit is shown in Figure3, while the circuit output is at T3. This example is used to demonstrate the critical results of proposed method for fault diagnosis based on complex voltages estimation, and the parametric faults are as follows: (a) discrete fault values in tolerance-free case as shown in Tables 2 and 4; (b) the parameter of faulty components varies from 85% to 115% of its discrete fault values in Table 5. Here, the former one is to testify the accuracy of Theorems 1, 4, and 5, while the latter one is used to demonstrate the process of faulty response estimation with consideration of tolerance and continuous parametric fault pattern diagnosis in the linear-analogue circuit.Table 2
Component values in circuit of Figure3.
Nominal values
Discrete fault values
Parametric faults limit
R
1
1
k
Ω
1.50
k
Ω
[
1.28
k
Ω
,
1.73
k
Ω
]
R
2
1
k
Ω
5.00
k
Ω
[
4.25
k
Ω
,
5.75
k
Ω
]
R
3
1
k
Ω
1.25
k
Ω
[
1.06
k
Ω
,
1.44
k
Ω
]
R
4
1
k
Ω
0.75
k
Ω
[
0.64
k
Ω
,
0.86
k
Ω
]
R
5
1
k
Ω
0.50
k
Ω
[
0.43
k
Ω
,
0.58
k
Ω
]
R
6
1
k
Ω
1.20
k
Ω
[
1.02
k
Ω
,
1.38
k
Ω
]
C
1
100
nF
50.00
nF
[42.5 nF, 57.5 nF]
C
2
100
nF
150.00
nF
[127.5 nF, 172.5 nF]Figure 3
The Thomas filter.In addition, in this circuit example, the stimulus signal (green sketch in Figure4) is as follows: (a) the amplitude is 2 (V); (b) the test-frequency is set as 1.5 kHz; (c) the phase of this signal is zero. Furthermore, the circuit response in nominal circuit state is also given in Figure 4 (red sketch), which satisfies: (a) the amplitude is 9.67 (V); (b) the test-frequency is set as 1.5 kHz; (c) the phase of this signal is -0.69. Therefore, according to the relationship between (1) and (2), the corresponding phasor is 1.00+ı0.00 (V) (-5.23+ı4.41 (V)).Figure 4
The circuit stimulus (green) and response (red) in AC test.In order to estimate the faulty responses because of parametric faults in second column of Table2, one should primarily estimate all hard-faults influence in this circuit. For instance, as for test-node T3, if R3 is assumed to be the potential parametric fault component, the corresponding circuit output voltage phasor at test-node T3 could be estimated as the following procedure in the circuit with nominal parametric values: (a) the circuit is in nominal state (fault-free), and the stimulus is UIi→=1.00+i0.00 (V); (b) the voltage phasor through R3 branch is Uxai→=-4.16-i4.93 (V), and one can locate this complex value through the amplitude and phase of sinusoidal signal at test-node T1; (c) as shown in Figure 3, one adds an auxiliary voltage stimulus Uxbi→=1.00+i0.00 (V), and at the same time, the voltage phasor at test-node T3 is Uobi→=-1.10-i0.94 (V); (d) let all these measurements be in (17), the solution of (17) is Uosi→=0.00+i0.00 (V).The similar measurements and calculations are processed for the other hard-faults influence estimations at test-nodeT3; after that, the corresponding voltage phasor values are shown in Table 3.Table 3
Voltage phasors for hard-faults atT3 in Figure 3.
Fault response (short-circuit)
Fault response (open-circuit)
R
1
-
1.13
+
i
1.06
×
1
0
6
0.00
+
i
0.00
R
2
-
0.00
-
i
0.00
-
8.95
+
i
0.00
R
3
-
1.00
+
i
0.00
0.00
+
i
0.00
R
4
-
1.00
+
i
0.00
0.00
+
i
0.00
R
5
0.00
+
i
0.00
-
1.00
+
i
0.00
R
6
-
0.00
+
i
0.00
1.11
+
i
0.12
C
1
0.00
+
i
0.00
-
0.99
+
i
0.10
C
2
0.00
+
i
0.00
-
1.00
+
i
0.00Table 4
Voltage phasors for parametric faults at test-nodeT3 in Figure 3.
e
{
U
o
F
→
}
U
o
F
→
Diagnosis
R
1
=
1.50
k
Ω
-
3.49
+
i
2.94
-
3.49
+
i
2.94
Y
R
2
=
5.00
k
Ω
2.34
-
i
3.93
-
2.33
+
i
3.93
Y
R
3
=
1.25
k
Ω
4.24
+
i
4.52
4.24
+
i
4.52
Y
R
4
=
0.75
k
Ω
-
2.86
+
i
0.61
-
2.86
+
i
0.61
Y
R
5
=
0.50
k
Ω
1.22
+
i
0.30
1.22
+
i
0.30
Y
R
6
=
1.20
k
Ω
4.61
+
i
7.88
4.61
+
i
7.92
Y
C
1
=
50.00
nF
-
1.75
+
i
0.33
-
1.75
+
i
0.33
Y
C
2
=
150.00
nF
2.55
+
i
1.08
2.55
+
i
1.08
YTable 5
Voltage phasor-based fault diagnosis.
U
o
r
1
U
o
i
1
Diagnosis
No faults
[
-
5.50
,
-
4.98
]
[
4.01,5.38
]
100
%
R
2
[
-
4.01
,
-
2.96
]
[
2.12,3.10
]
100
%
C
1
[
-
2.64
,
-
1.99
]
[
0.24,1.40
]
100
%
C
2
[
2.20,3.64
]
[
0.82,2.36
]
100
%
1The voltage phasor response at T3 is Uo→=Uor+ıUoi.These particular hard-fault responses can be used to evaluate the parametric fault responses through the solution of Theorem1. Here, some parametric fault examples are shown in Table 4, and one gives a specific process of calculating fault response at test-node T3, when R2 is the fault component (e.g., R2=5kΩ): (a) according to data in Table 3, Uosi→=-0.00-i0.00 (V) and Uooi→=-8.95-i0.00 (V); (b) when all the components are in nominal circuit state R2=Z2r=1kΩ, the value of Uoi→ on test-node T3 is -5.23+i4.41 (V); (c) according to (9), •a is (-0.00+i1.19)/104; (d) let all these measurements, estimations, and calculations be in (9); one can find the estimated complex value e{UoFi→}=2.34-i3.93 (V).All the estimated results for the discrete parametric faults in Table2 (column 2) have been calculated based on (9) and been listed in Table 4 (column 2). And the results from actual circuit measurements have also been shown in the third column of Table 4. As a consequence, the corresponding fault diagnosis can be well done because of the consistence of data (e.g., the amplitude and phase reflected in the voltage phasor value) in column 2 and 3 of Table 4.The results in Tables3 and 4 point out that the process of calculating faulty response when the circuit suffered from parametric fault without consideration of tolerance. Furthermore, in order to form the corresponding faulty response data while considering the tolerance influence, the basic solutions of Lemmas 6 and 7 and Theorem 8 are used. Here, one gives a example process to calculate the the voltage phasor value range for R1 at test-node T3. Here, the fault component R1 varies in [0.85R1F,0.95R1F)∪[1.05R1F,1.15R1F], and R1F=1.5kΩ. Then, there exist these values: Rx0= 1.500 kΩ, Rx1= 1.643 kΩ, and Rx2=1.342 kΩ (1.358 kΩ), which lead to the following component value ranges with respective 5% tolerance influence:(a)
[1.425 kΩ, 1.575 kΩ], (b) [1.561 kΩ, 1.725 kΩ],(c)
[1.275 kΩ, 1.409 kΩ], (d) [1.290 kΩ, 1.426 kΩ].These parametric sets unify to cover the whole range of[0.85R1F,0.95R1F)∪(1.05R1F,1.15R1F] for component R1. Therefore, by running 100 simulations around the component value Rx0 or Rx1 or Rx2, with the consideration of tolerance ±5%, the complex response ranges (including real part range r and image part range i for each complex response) at test-node T3 are as follows:(a)
r=[-3.67,-3.25], i=[2.32,3.10];(b)
r=[-3.28,-2.96], i=[2.12,2.57];(c)
r=[-4.01,-3.63], i=[2.55,2.99];(d)
r=[-3.75,-3.64], i=[2.55,2.84].Therefore, according to Lemma7 and Theorem 8, the faulty response range for component R1=[0.85R1F,0.95R1F)∪(1.05R1F,1.15R1F] at the test-node T3 is as follows:(a)
[{UorF}min⁡,{UorF}max⁡]=[2.96,4.01]∩[4.98,5.50]¯=[2.96,4.01];(b)
[{UoiF}min⁡,{UoiF}max⁡]=[2.12,3.10]∩[4.01,5.38]¯=[2.12,3.10].Take some representative parametric faults in Table5 into account; the corresponding fault states are continuous fault component values in a given bounded parametric varying range. And the other fault-free components’ values are randomly selected around their corresponding nominal values, while the tolerance is ±α=±5%. Here, according to the contents in Table 5, it is obvious to find both of the values of FDR and FIR are 100% for these 3 parametric faults.


## 3.2. Sallen-Key Filter Example

The proposed fault diagnosis method will be testified in another representative linear circuit: Sallen-Key filter example. In this section, the nominal component value is as follows:R1= 1 kΩ; R2= 3 kΩ; R3= 2 kΩ; R4= 4 kΩ; R5= 6 kΩ; C1,C2= 5 nF. Different from the discussion in the previous subsection, the test-frequencies of sinusoidal signal are carefully selected according to the fault response curves determined by Theorem 2. And all the candidate test-signal frequencies eventually construct a test-frequencies set Q=QFı, ı=1,2,…,♭.The number of♭ (test-frequencies numbers) is quite dependent on the number of systematic parameter in transfer function. For instance, in this example, the transfer function is in (37). Therefore, there are at most 3 test-frequencies required, in order to determine all the systematic parameter in (37). Consider
(37)ħ(s)=((R5+R4)R1R4C1s)×((R1+R2)C1C2R1R2R3s2+(1C2R3+1C1R1+1C1R3-R5C1R2R4)s00100+(R1+R2)C1C2R1R2R3s2)-1.Note that, in this example, if the faulty component can own all possible parametric fault values (0≤∥ZxF∥<∞), and the corresponding stimulus signal is 2sin(2*π*3000t-0), then the corresponding faulty response curve is shown in Figure 6 for each component. With Theorem 2, these curves are determined according to equations as follows: (a) R1: (Uore)2+(Uoim)2-1.86Uore-1.21Uoim=0. (b) R2: (Uore)2+(Uoim)2-2.99Uore-8.46Uoim=0. (c) R3: (Uore)2+(Uoim)2-1.36Uore+1.93Uoim=0. (d) R4,5: (Uore)2+(Uoim)2+3.00Uore+29.76Uoim=0.0035. (e) C1: (Uore)2+(Uoim)2-1.82Uore-0.96Uoim=0. (f) C2: (Uore)2+(Uoim)2-1.50Uore+1.06Uoim=0.The intersections of these curves indicate the potential ambiguous faults in Table6; for instance, when R1 is open, or R3 is short, the voltage phasor on To owns the same value: 0.00+i0.00 (V). Thus, the fault state R3: short, is one of ambiguous faults to the fault state R1: Open.Table 6
Ambiguous faults in the circuit of Figure4.
Fault state
Ambiguous faults
R
1
Open
R
2,3: Short, C1: Short, C2: Open
R
2
Short
R
1: Open, R3: Short, C1: Short, C2: Open
R
3
Short
R
1: Open, R2: Short, C1: Short, C2: Open
R
4
1
All parametric faults
R
5: All parametric faults
R
5
1
All parametric faults
R
4: All parametric faults
C
1
Short
R
1: Open, R2,3: Short, C2: Open
C
2
Open
R
1: Open, R2,3: Short, C1: Short
1For any given parametric fault of R4(R5),there exists a parametric fault of R5(R4) generating the undistinguishable complex faulty response at test-node To.According to Figure6 and Tables 6 and 7, test-frequency QF1= 30 kHz is a sensitive frequency for the circuit under test (CUT) of Figure 5 in parametric fault diagnosis (Table 7). Furthermore, one can sketch the other response curves with different test-frequencies. Such investigation leads to 3 representative sensitive test-frequencies: QF1= 30 kHz, QF2= 20 kHz, and QF2= 60 kHz, for a given fault states set in Table 7.Table 7
Voltage phasor-based tolerance-free fault diagnosis with test-frequencies setQF={QF1,QF2,QF3}.
Q
F
1
Q
F
2
Q
F
3
No faults
1.63
-
ı
0.26
1.54
+
ı
0.44
0.82
-
ı
0.83
R
1
=
1.50
k
Ω
1.27
-
ı
0.45
1.40
+
ı
0.20
0.49
-
ı
0.68
R
2
=
1.50
k
Ω
2.49
-
ı
0.15
1.68
+
ı
1.17
0.81
-
ı
1.17
R
3
=
3.00
k
Ω
1.84
-
ı
0.74
2.13
+
ı
0.14
0.71
-
ı
1.01
R
4
=
2.00
k
Ω
3.79
-
ı
0.89
3.37
+
ı
1.46
1.20
-
ı
1.83
R
5
=
3.00
k
Ω
0.98
-
ı
0.13
0.94
+
ı
0.23
0.57
-
ı
0.49
C
1
=
2.50
nF
1.93
+
ı
0.65
1.47
+
ı
0.88
1.64
-
ı
0.77
C
2
=
2.50
nF
1.18
+
ı
0.28
0.80
+
ı
0.60
0.93
-
ı
0.55Figure 5
The Sallen-Key filter circuit.Figure 6
Complex faulty response curve for each component, whereR-axis represents Uore, whileI-axis represents Uoim.These test-frequencies can be testified as sensitive frequencies, according to the frequency sweeping results in Figure7: (a) each response curve corresponds to a component fault in Table 7; (b) actually, in Figure 6, the test-frequencies QF1= 30 kHz and QF2= 20 kHz are around the center frequency of Sallen-Key filter.Figure 7
The Sallen-Key filter circuit response in frequency domain.As a matter of fact, the sensitive test-frequencies investigated in Figures6 and 7 are also the recommendations in the circuit diagnosis for all potential single parametric faults with tolerance consideration (i.e., component tolerance for fault-free component is 2%, while the parametric range for faulty component is [95%Xc,105%Xc] and Xc is the value in Table 7 for each potential faulty component). For instance, as for the fault states set in Table 8, if one lists the corresponding voltage phasor varying ranges, which can be estimated through the similar manner as what the Thomas filter example does, it is simple to locate the value of FIR=100% and FDR=100%. Therefore, all faults in Table 8 can be 100% diagnosis with given test-signals set.Table 8
Voltage phasor-based tolerance-influencing fault diagnosis with test-frequencies setQF={QF1,QF2,QF3}.
Q
F
1
Q
F
2
Q
F
3
No faults
[
1.37,1.78
]
[
1.32,1.63
]
[
0.73,0.95
]
[
-
0.27
,
-
0.12
]
[
0.35,0.56
]
[
-
0.90
,
-
0.70
]
R
1: [1.43Ω,1.58kΩ]
[
1.09,1.41
]
[
1.15,1.51
]
[
0.42,0.58
]
[
-
0.46
,
-
0.29
]
[
0.14,0.32
]
[
-
0.75
,
-
0.56
]
R
2: [1.43Ω,1.58kΩ]
[
1.91,2.73
]
[
1.27,1.74
]
[
0.73,1.03
]
[
-
0.14,0.15
]
[
0.87,1.35
]
[
-
1.29
,
-
0.94
]
R
3: [2.85kΩ,3.15kΩ]
[
1.66,2.01
]
[
1.76,2.26
]
[
0.61,0.85
]
[
-
0.74
,
-
0.51
]
[
0.10,0.33
]
[
-
1.11
,
-
0.86
]
R
4: [1.90kΩ,2.10kΩ]
[
2.92,4.24
]
[
2.89,3.55
]
[
1.07,1.52
]
[
-
0.89
,
-
0.27
]
[
1.00,1.73
]
[
-
2.07
,
-
1.45
]
R
5: [2.85kΩ,3.15kΩ]
[
0.86,1.05
]
[
0.83,0.98
]
[
0.51,0.65
]
[
-
0.13
,
-
0.05
]
[
0.19,0.27
]
[
-
0.52
,
-
0.45
]
C
1: [2.34nf,2.63nf]
[
1.74,2.05
]
[
1.19,1.54
]
[
1.19,1.54
]
[
0.30,0.46
]
[
0.73,0.96
]
[
0.73,0.96
]
C
2: [2.34nf,2.63nf]
[
0.97,1.26
]
[
0.65,0.83
]
[
0.78,1.04
]
[
0.21,0.37
]
[
0.49,0.66
]
[
-
0.57
,
-
0.42
]


## 4. Further Discussion

So far, this paper has presented a novel parametric fault modeling to the fault diagnosis of linear analog circuits via voltage phasor measurements. Particularly, it has discussed the tolerance and multifrequency measurements in the fault diagnosis as follows.(a)
The tolerance often introduces a variable parameter for the circuit components which leads to the variation of circuit response from one circuit board to another. In this paper, its influence is estimated according to the principles from Lemmas6–9 and Theorems 8 and 10. At last, the corresponding applications have been shown in the illustrative examples of previous section.(b)
The analytic parametric fault modeling (see, (9), (14)-(15)) and tolerance estimation in this paper give us a simple rule to select test-frequency for fault diagnosis in a given fault states set, which means the following:  (1) the maximum number of test-frequencies is determined by the number of systematic parameter in transfer function (i.e., (37)); (2) the final test-frequencies set should attempt to avoid ambiguous faults and increase the value of FDR and FIR (see, (35)-(36)).However, considering the fact that an important aspect in the linear circuit fault diagnosis problem is constituted by the concepts of testability, the voltage phasor-based ambiguous groups determination will be in a discussion in this section. One can find that it is quite effective when it is compared with the “Jacobian rank approach” [15] based on transfer function coefficients. Futhermore, one uses the maximum error evaluation technology to evaluate the multifrequencies test previously listed in the second circuit example. This technology used to appear in [16].


### 4.1. Ambiguous Groups in Voltage Phasor Investigation

In this section, the ambiguous groups determination is based on voltage phasor investigation and the corresponding principle is shown as follows.According to (9), test equations are constructed and circuit parameter functions P (impedance value functions vector for potential faulty circuit network components) are related to test measurements U (voltage phasor values vector) through the testability matrix B as shown in the following equation:
(38)BP=U,
where B=[Uoox1i→-Uosx1i→,Uoox2i→-Uosx2i→,…,Uooxpi→-Uosxpi→], P=diag⁡([1/(•αx1•Zx1F-1),…,1/(•αxp•ZxpF-1)]), and U=[UoFi→-Uoox1i→,UoFi→-Uoox2i→,…,UoFi→-Uooxpi→].Due to general method of ambiguous groups determination illustrated in [15–17], testability matrix B can be used to locate ambiguous groups based on Definition 11 and to locate the canonical ambiguous groups because of Definition 12. Besides, the concept of “linearly dependent” here is in Definition 13; thus, the columns C1,C2,…,CD of the testability matrix are linearly dependent which means the corresponding components constitute an ambiguity group, for the whole complex test-frequencies domain.Definition 11.
A set ofD components constitutes an ambiguity group of order D if the corresponding D columns of the testability matrix are linearly dependent.Definition 12.
A set ofD components constitutes a canonical ambiguity group of order D if the corresponding columns of the testability matrix are linearly dependent and every subset of this group of columns is constituted by linearly independent columns.Definition 13.
The columnsC1,C2,…,CD of the testability matrix are linearly dependent, if ∑ı=1DκıCı=0, where the constant κı,1≤ı≤D includes at least one nonzero real value.With Definitions11–13, let us talk about the ambiguous groups division in the circuit examples of previous section; in the first circuit example, it should be noted that the output voltage phasors caused by R3, R4, R5, and C2 satisfy UooR3i→-UosR3i→=UooR4i→-UosR4i→=-(UooR5i→-UosR5i→)=-(UooC2i→-UosC2i→). Therefore, the corresponding columns are linearly dependent in matrix B. Therefore, in this circuit, one can make the conclusion that R3,R4,R5, and C2 constitute an ambiguous group, only through the direct observation of Uosi→ and Uooi→ as mentioned above, and such ambiguous group can be divided into two canonical ambiguous groups, [R4,R5] and [R3,C2].Then the similar process is used to the components set[R1, R2, R6, C1], here the [R1, R2, R6, C1] constitutes an ambiguous group due to the corresponding linearly dependent columns in B: 0×(1.13-ı1.06×106)+1×(1.11+ı0.12)-12×(-0.99+ı0.10)+(-10.77/8.95)×(-8.95+ı0.00)=0, where the voltage-phasors are from Table 3. Then [R1,R2,R6,C1] can be found as a canonical ambiguous group according to Matlab computation and Definition 12. Eventually, the final canonical ambiguous group division in the circuit of Figure 3 is [R4, R5][R3, C2] and [R1, R2, R6, C1].The similar process of determination of ambiguous group is given in the second example; then it is obvious that componentsR4 and R5 constitute a canonical ambiguous group, when UooR4i→-UosR4i→=-1×(UooR5i→-UosR5i→) and UooR4i→-UosR4i→≠0(UooR5i→-UosR5i→≠0). Besides, the columns corresponding to R2,R3,C2 in testability matrix B is B′=[B′(1),B′(2),B′(3)]=[1.199-ı0.2470,1.098-ı2.070,-1.324+ı1.248], and nonzero real constants in Definition 13 cannot be found under the circuit test with the test-frequency being Q1F=30kHz, which means these columns in B′ are not linearly dependent. After that, [R2,R3,C2] does not constitute an ambiguous group for the whole complex frequencies domain. At last, the final canonical ambiguous group division in the Sallen-key circuit is [R4,R5], [R1], [C1], [R2], [R3], and [C2] for the whole complex frequencies domain.The results of ambiguous groups division above can be verified through “Jacobian rank approach.” Therefore, in the first circuit example, the transfer function of the Thomas filter is shown in (39), while Jacobian matrix B is given by (40)–(42):
(39)ħTo(s)=-R5/R1R3R4C1C2s2+(1/R2C1)s+R5/R3R4R6C1C2(40)B=C×[B1,B2](41)B1=[R3R4R5C2R5R1R3R5R1R4-1R1R5R1C2-R5R3R6-R5R4R61R6-R5R6C20000](42)B2=[R1R2R6C1R5R1200R5R1C100-R5R62-R5R6C10R3R4C2R220R3R4C2C1R2],
where C=1/R3R4C1C2.In (41), the corresponding columns for components R3,R4,R5, and C2 in B are linearly dependent. Furthermore, in (42), the corresponding columns forR1,R2,R6, and C1 are linearly dependent too. Therefore, the effectiveness of ambiguous group division through proposed method is finally verified in a point of view of “Jacobian rank approach” when it is applied in the following special circuit example: [R3,R4,R5,C2] and [R1,R2,R6,C1]. Morevoer, a canonical ambiguous groups division in the corresponding circuit is [R4,R5], [R3,C2], and [R1,R2,R6,C1].The similar verification can be proceeded in the second circuit example; then the corresponding testability matrixB is established in (43). Here, considering the testability value does not depend on component values [18], one assigns all “1” values to the circuit parameters in matrix B. Thus, one can also find the same result about canonical ambiguous groups division in the proposed method application and the transfer function-based testability matrix (Jacobian matrix in [15]) investigation in (43). As a matter of fact, considering the relationship that the output voltage phasor is actually Uoi→=UIi→×ħTo(s)|s=ıω, where ω is the complex frequencies, it is not hard to understand such consistency. Consider
(43)B=[R1R2R3R4R5C1C2-200-11-12-1-1-200-2-2-11-21-1-1-1].


### 4.2. Multifrequencies Circuit Test Determination and Evaluation

In the last paragraph, one could conclude that the ambiguous group based on voltage phasor is the same as the result of testability analysis based on “Jacobian rank approach,” where the testability matrix is a Jacobian matrix determined by coefficients in transfer function. Such conclusion is not surprise, because the output voltage phasor can be expressed asUoi→=UIi→×ħTo(s)|s=ıω, where ω is the complex frequencies.Such conclusion implies that the sensitive test-frequency selected according to voltage phasor is also the sensitive one for fault diagnosis, which is searched out through the transfer function-based method. This solution has been verified geometrically through the frequency sweeping results shown in Figure7, when the parametric faults in Table 7 need diagnosis.In fact, the result of Table6 and Figure 6 further tells us that one should select appropriate test-signal to distinguish the given ambiguous faults between R4 and R5, which is also what canonical ambiguous groups result would like to tell us. Moreover, one can find out the same results in the ambiguous faults determination through Figure 6 or the ambiguous groups analysis in the last subsection.At last, the observation of Figure7 and Tables 7 and 8 points out that the test-frequencies QF1= 30 kHz, QF2= 20 kHz, and QF3= 60 kHz are sensitive to all the parametric faults that need diagnosis; even the tolerance is in consideration. And the values of FDR and FIR in the circuit example are 100%≥90%.In spite of the values ofFDR and FIR, the maximum error evaluation technology involved in [16] can also be used to evaluate the effectiveness of multifrequencies test in fault diagnosis, where the maximum error for the potential fault components R1 is represented by max⁡{ΔR1%}=max⁡{∥(R1F′-R1F)/R1F∥}. And R1F is a real fault value, while R1F′ is a calculated value based on (9).In order to simplify the illustration, let us firstly assume that the fault component in investigation owns a given discrete fault parameter while other fault-free components suffered from tolerance. For instance,R1 is single fault in a tolerance-influencing circuit, whose faulty value is R1= 1.5 kΩ. In addition, the tolerance of other fault-free components is ±2% and the test-frequency is 20 kHz. Then one of representative output voltage phasor measurement at test-node To in such tolerance circumstance is given as UoFi→=1.300-ı0.4438, while the corresponding estimated hard-fault output responses based on the measurements in Theorems 4 and 5 are UosR1i→=1.943+ı1.031 and UooR1i→=0.000+ı0.000. Therefore, the calculated fault value of R1 is 1.4995 kΩ according to (9), which means the error for R1 is ΔR1%=∥(1.500-1.4995)/1.500∥=0.033%. Such error tells us the test-frequency 20 kHz is a sensitive one to the fault R1= 1.5 kΩ, because it is accurate enough to accomplish the parametric diagnosis of R1= 1.5 kΩ when the corresponding input stimulus with test-frequency 20 kHz is used. Besides, it also means the estimated hard-fault responses are accurate through the methods in Theorems 4 and 5.The similar test and diagnosis process can be extended to the multifrequency measurements (QF={QF1= 20 kHz, QF2= 30 kHz, and QF3= 60 kHz}) for the fault diagnosis of R1; therefore, max⁡{ΔR1%}= 0.067%. As a matter of fact, if the calculations based on the proposed fault modeling are done for all other potential faults, the maximum errors for all these fault values are much less than the component tolerance influence (2%).


## 4.1. Ambiguous Groups in Voltage Phasor Investigation

In this section, the ambiguous groups determination is based on voltage phasor investigation and the corresponding principle is shown as follows.According to (9), test equations are constructed and circuit parameter functions P (impedance value functions vector for potential faulty circuit network components) are related to test measurements U (voltage phasor values vector) through the testability matrix B as shown in the following equation:
(38)BP=U,
where B=[Uoox1i→-Uosx1i→,Uoox2i→-Uosx2i→,…,Uooxpi→-Uosxpi→], P=diag⁡([1/(•αx1•Zx1F-1),…,1/(•αxp•ZxpF-1)]), and U=[UoFi→-Uoox1i→,UoFi→-Uoox2i→,…,UoFi→-Uooxpi→].Due to general method of ambiguous groups determination illustrated in [15–17], testability matrix B can be used to locate ambiguous groups based on Definition 11 and to locate the canonical ambiguous groups because of Definition 12. Besides, the concept of “linearly dependent” here is in Definition 13; thus, the columns C1,C2,…,CD of the testability matrix are linearly dependent which means the corresponding components constitute an ambiguity group, for the whole complex test-frequencies domain.Definition 11.
A set ofD components constitutes an ambiguity group of order D if the corresponding D columns of the testability matrix are linearly dependent.Definition 12.
A set ofD components constitutes a canonical ambiguity group of order D if the corresponding columns of the testability matrix are linearly dependent and every subset of this group of columns is constituted by linearly independent columns.Definition 13.
The columnsC1,C2,…,CD of the testability matrix are linearly dependent, if ∑ı=1DκıCı=0, where the constant κı,1≤ı≤D includes at least one nonzero real value.With Definitions11–13, let us talk about the ambiguous groups division in the circuit examples of previous section; in the first circuit example, it should be noted that the output voltage phasors caused by R3, R4, R5, and C2 satisfy UooR3i→-UosR3i→=UooR4i→-UosR4i→=-(UooR5i→-UosR5i→)=-(UooC2i→-UosC2i→). Therefore, the corresponding columns are linearly dependent in matrix B. Therefore, in this circuit, one can make the conclusion that R3,R4,R5, and C2 constitute an ambiguous group, only through the direct observation of Uosi→ and Uooi→ as mentioned above, and such ambiguous group can be divided into two canonical ambiguous groups, [R4,R5] and [R3,C2].Then the similar process is used to the components set[R1, R2, R6, C1], here the [R1, R2, R6, C1] constitutes an ambiguous group due to the corresponding linearly dependent columns in B: 0×(1.13-ı1.06×106)+1×(1.11+ı0.12)-12×(-0.99+ı0.10)+(-10.77/8.95)×(-8.95+ı0.00)=0, where the voltage-phasors are from Table 3. Then [R1,R2,R6,C1] can be found as a canonical ambiguous group according to Matlab computation and Definition 12. Eventually, the final canonical ambiguous group division in the circuit of Figure 3 is [R4, R5][R3, C2] and [R1, R2, R6, C1].The similar process of determination of ambiguous group is given in the second example; then it is obvious that componentsR4 and R5 constitute a canonical ambiguous group, when UooR4i→-UosR4i→=-1×(UooR5i→-UosR5i→) and UooR4i→-UosR4i→≠0(UooR5i→-UosR5i→≠0). Besides, the columns corresponding to R2,R3,C2 in testability matrix B is B′=[B′(1),B′(2),B′(3)]=[1.199-ı0.2470,1.098-ı2.070,-1.324+ı1.248], and nonzero real constants in Definition 13 cannot be found under the circuit test with the test-frequency being Q1F=30kHz, which means these columns in B′ are not linearly dependent. After that, [R2,R3,C2] does not constitute an ambiguous group for the whole complex frequencies domain. At last, the final canonical ambiguous group division in the Sallen-key circuit is [R4,R5], [R1], [C1], [R2], [R3], and [C2] for the whole complex frequencies domain.The results of ambiguous groups division above can be verified through “Jacobian rank approach.” Therefore, in the first circuit example, the transfer function of the Thomas filter is shown in (39), while Jacobian matrix B is given by (40)–(42):
(39)ħTo(s)=-R5/R1R3R4C1C2s2+(1/R2C1)s+R5/R3R4R6C1C2(40)B=C×[B1,B2](41)B1=[R3R4R5C2R5R1R3R5R1R4-1R1R5R1C2-R5R3R6-R5R4R61R6-R5R6C20000](42)B2=[R1R2R6C1R5R1200R5R1C100-R5R62-R5R6C10R3R4C2R220R3R4C2C1R2],
where C=1/R3R4C1C2.In (41), the corresponding columns for components R3,R4,R5, and C2 in B are linearly dependent. Furthermore, in (42), the corresponding columns forR1,R2,R6, and C1 are linearly dependent too. Therefore, the effectiveness of ambiguous group division through proposed method is finally verified in a point of view of “Jacobian rank approach” when it is applied in the following special circuit example: [R3,R4,R5,C2] and [R1,R2,R6,C1]. Morevoer, a canonical ambiguous groups division in the corresponding circuit is [R4,R5], [R3,C2], and [R1,R2,R6,C1].The similar verification can be proceeded in the second circuit example; then the corresponding testability matrixB is established in (43). Here, considering the testability value does not depend on component values [18], one assigns all “1” values to the circuit parameters in matrix B. Thus, one can also find the same result about canonical ambiguous groups division in the proposed method application and the transfer function-based testability matrix (Jacobian matrix in [15]) investigation in (43). As a matter of fact, considering the relationship that the output voltage phasor is actually Uoi→=UIi→×ħTo(s)|s=ıω, where ω is the complex frequencies, it is not hard to understand such consistency. Consider
(43)B=[R1R2R3R4R5C1C2-200-11-12-1-1-200-2-2-11-21-1-1-1].


## 4.2. Multifrequencies Circuit Test Determination and Evaluation

In the last paragraph, one could conclude that the ambiguous group based on voltage phasor is the same as the result of testability analysis based on “Jacobian rank approach,” where the testability matrix is a Jacobian matrix determined by coefficients in transfer function. Such conclusion is not surprise, because the output voltage phasor can be expressed asUoi→=UIi→×ħTo(s)|s=ıω, where ω is the complex frequencies.Such conclusion implies that the sensitive test-frequency selected according to voltage phasor is also the sensitive one for fault diagnosis, which is searched out through the transfer function-based method. This solution has been verified geometrically through the frequency sweeping results shown in Figure7, when the parametric faults in Table 7 need diagnosis.In fact, the result of Table6 and Figure 6 further tells us that one should select appropriate test-signal to distinguish the given ambiguous faults between R4 and R5, which is also what canonical ambiguous groups result would like to tell us. Moreover, one can find out the same results in the ambiguous faults determination through Figure 6 or the ambiguous groups analysis in the last subsection.At last, the observation of Figure7 and Tables 7 and 8 points out that the test-frequencies QF1= 30 kHz, QF2= 20 kHz, and QF3= 60 kHz are sensitive to all the parametric faults that need diagnosis; even the tolerance is in consideration. And the values of FDR and FIR in the circuit example are 100%≥90%.In spite of the values ofFDR and FIR, the maximum error evaluation technology involved in [16] can also be used to evaluate the effectiveness of multifrequencies test in fault diagnosis, where the maximum error for the potential fault components R1 is represented by max⁡{ΔR1%}=max⁡{∥(R1F′-R1F)/R1F∥}. And R1F is a real fault value, while R1F′ is a calculated value based on (9).In order to simplify the illustration, let us firstly assume that the fault component in investigation owns a given discrete fault parameter while other fault-free components suffered from tolerance. For instance,R1 is single fault in a tolerance-influencing circuit, whose faulty value is R1= 1.5 kΩ. In addition, the tolerance of other fault-free components is ±2% and the test-frequency is 20 kHz. Then one of representative output voltage phasor measurement at test-node To in such tolerance circumstance is given as UoFi→=1.300-ı0.4438, while the corresponding estimated hard-fault output responses based on the measurements in Theorems 4 and 5 are UosR1i→=1.943+ı1.031 and UooR1i→=0.000+ı0.000. Therefore, the calculated fault value of R1 is 1.4995 kΩ according to (9), which means the error for R1 is ΔR1%=∥(1.500-1.4995)/1.500∥=0.033%. Such error tells us the test-frequency 20 kHz is a sensitive one to the fault R1= 1.5 kΩ, because it is accurate enough to accomplish the parametric diagnosis of R1= 1.5 kΩ when the corresponding input stimulus with test-frequency 20 kHz is used. Besides, it also means the estimated hard-fault responses are accurate through the methods in Theorems 4 and 5.The similar test and diagnosis process can be extended to the multifrequency measurements (QF={QF1= 20 kHz, QF2= 30 kHz, and QF3= 60 kHz}) for the fault diagnosis of R1; therefore, max⁡{ΔR1%}= 0.067%. As a matter of fact, if the calculations based on the proposed fault modeling are done for all other potential faults, the maximum errors for all these fault values are much less than the component tolerance influence (2%).


## 5. Conclusion

This paper gives a voltage phasor-based viewpoint in order to solve the fault diagnosis problem in linear-analogue circuit: in the process of fault diagnosis, the accurate voltage phasor feature discussion let us know the accurate faulty response. Furthermore, the tolerance problem is also demonstrated, in which the proposed statistics-based point of view is used to estimate the voltage phasor varying range. After that, the test-frequencies are determined in fault diagnosis applications in order to obtain the required FDR/FIR values. In fact, all of these aspects have been analyzed and tested in the experiment of representative linear-circuit benchmarks.In sum up, the method presented in this paper, is an acute view to handle the fault diagnosis problem of the component parametric alteration in linear-analogue circuit. Furthermore, it has been discussed in the problem of ambiguous groups determination, too. And it can be extended to the following cases (a)–(c). All of these aspects should be done in further research.(a)
Although the hard-fault is not the main focus in this paper, the proposed method could be generalized to hard-fault case through hard-fault influence estimation based on voltage phasor measurement in Theorems4 and 5.(b)
In fact, the discussion of accurate analysis can be used in the nonlinear circuit cases, as long as the linear-wise segments modeling can be established. In this case, the equivalent linear circuit network is used to replace the nonlinear circuit component. This means the parametric fault influence caused by nonlinear components can be transformed to the parametric alterations of corresponding linear circuit network.(c)
The ambiguous groups determination could benefit us in the test-nodes selection problem.



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*Source: 879402-2014-08-28.xml*