Determination of poles and zeros#
Knowledge of poles and zeros is indispensable for the analysis and design of the stability of amplifiers. A network of which the solutions of \(s\) of its characteristic equation are located in the left half of the complex plane is stable. Instability cannot always be observed at the circuits’s output. In a case of non-observable instability, zeros with the same frequency as the right half plane poles cover up instability. Hence, observation of a stable time-domain response of a circuit does not guarantee the circuit’s stability.
In order to get a clear picture of both observable and non-observable instability, only the solutions of the characteristic equations should be evaluated. These solutions are the poles of the network.
We will discuss three ways for the determination of the poles:
Direct solution of the characteristic equation; this will be discussed in section Solving the characteristic equation.
Determination of the eigenvalues of the time-constant matrix; this will be discussed in section Eigenvalues of the time-constant matrix.
Direct estimation from a network; this will be discussed in section Symbolic estimation of poles and zeros.
Solving the characteristic equation#
Consider a network of which the MNA equations are given as
in which \(\mathbf{I}\) is the vector with nodal currents and branch voltages, \(\mathbf{M}\) the MNA matrix, and \(\mathbf{V}\) the vector with nodal voltages and branch currents. The poles of the network can be found by solving the equation
The zeros are found by solving
in which \(\mathbf{M}^{\prime}\)is obtained from \(\mathbf{M}\) and \(\mathbf{I}\), after application of Cramer’s rule.
In practice, calculation of the determinant of a matrix using Gaussian elimination may generate extra poles due to rounding effects. For accurate determination of the poles and the zeros, powers of the Laplace variable \(s^{n}\) \(\left( n>0\right) \) should be kept symbolically in the matrix elements and the determinant should be evaluated through expansion of minors. In order to keep \(\mathbf{M}\) or \(\mathbf{M}^{\prime}\) as small as possible, transfer functions can be implemented as discussed in section Numerator and denominator substitution. Symbolic expansion of minors, however, is a time-consuming process and can be used for small circuits only. This method has been implemented in the current version of SLICAP.
Another way is to calculate the poles and zeros from a state equation representation of the network. Formulation of the state equations for a network with passive elements and controlled sources, however, is complex, and the procedure for it differs considerably from setting up the MNA matrix. The reader is referred to Chua [53] for more information on setting up the state equations of a network.
Alternatively, poles can be found as the eigenvalues of the so-called time-constant matrix, which can be derived from the MNA matrix. This will be discussed in section Eigenvalues of the time-constant matrix.
Eigenvalues of the time-constant matrix#
Consider the network equations written as a set of first order differential equations
The matrix \(\mathbf{M}\) then has the form:
The poles are obtained by solving the equation:
This method for determination of the poles, solves the so-called generalized eigenvalue problem as stated in ((231)). This equation can be written in the form
in which \(\mathbf{T}=\mathbf{G}^{-1}\mathbf{C}\). Notice that \(\det\left( \mathbf{I}+s\mathbf{T}\right) =0\) can be written in the form of the generalized eigenvalue problem: \(\det\left( \mathbf{A-}\lambda\mathbf{I} \right) =0\), with \(\lambda=-\frac{1}{s}\) Hence, if \(\tau_{i}\) is an eigenvalue of \(\mathbf{T}\), the complex frequency of the corresponding pole \(p_{i}\) is \(s=-\frac{1}{\tau_{i}}\).
However, in most cases the number of eigenvalues of \(\mathbf{T}\) exceeds the number of independent state variables.
Thus, evaluated in this way, the \(n\) poles found from the eigenvalues of \(\mathbf{T}\) are the \(k\) system poles plus \(n-k\) poles at infinity, or after numeric solution, at very high frequencies.
For systems with only finite nonzero eigenvalues
, the modified nodal equations can be transformed into the state equations, using the transformation technique described by Haley [54]. The time constant matrix can then be written as
in which \(\mathcal{C}\) is a diagonal matrix, having \(\mathcal{C}_{ii}=C_{i}\) \(\left( i=1\cdots p\right) \) for a circuit with \(p\) capacitors, and \(\mathcal{C}_{p+j,p+j}=-L_{j}\) \(\left( j=1\cdots q\right) \) for a circuit having \(q\) inductors:
The resistance matrix \(\mathcal{R}\) is obtained as
where \(\mathcal{I}\) is the incidence matrix. This matrix holds the positions of the elements of \(\mathcal{C}\) in \(\mathbf{C}\). Its number of rows \(i\) equals that of \(\mathbf{G}\) and its number of columns \(j\) equals the sum of the number of capacitors and the number of inductors. The elements of \(\mathcal{I}\) can be obtained as follows:
\(\mathcal{I}_{ij}=1\)
if the positive terminal of a capacitor at \(\mathcal{C}_{jj}\) is connected to node \(i\).
or if \(\mathbf{V}_{i}\) is the current through an inductor at \(\mathcal{C}_{jj}\)
\(\mathcal{I}_{ij}=-1\) if the negative terminal of a capacitor at \(\mathcal{C}_{jj}\) is connected to node \(i\).
Otherwise, \(\mathcal{I}_{ij}=0\)
Example example-IncidenceMatrix shows the application of the incidence matrix.
The time-constant matrix can only be defined if, \(\mathbf{G}^{-1}\) exists. This means that the circuit needs to have a unique DC solution.
Structure and meaning of the resistance matrix \(\mathcal{R}\)#
If we have a network with \(p\) capacitors and \(q\) inductors, the network can be drawn as a \(p+q\) port network in which a capacitor \(C_{i}:i=1\cdots p\) is connected to its port \(i,\) and an inductor \(L_{j}:j=1\cdots q\) is connected to its port \(p+j\), as shown in Fig. 571A.
The resistance matrix can be found from the \(\left( p+j\right) ^{2}\)\ DC transfers that exist in this \(p+q\) port. To find these transfers, we replace all capacitors \(C_{i}\) with independent current sources \(I_{i}\), and all inductors \(L_{j}\) with independent voltage sources \(V_{p+j}\), as shown in Fig. 571B. Then, the voltage \(V_{i}\) across the independent current source \(I_{i}\) at port \(i,\) and the current \(I_{p+j}\) through\ the independent voltage source \(V_{p+j}\) at port \(p+j\), can be found from the following matrix equation:
in which \(\mathcal{R}\)\ is the resistance matrix.
The resistance matrix \(\mathcal{R}\) can be decomposed into four sub-matrices:
Expression ((235)) shows this decomposition in:
A \(p\times p\) transresistance matrix {\(r\) }
A \(q\times q\) transconductance matrix \(g\)
A \(p\times q\) voltage transfer matrix \(\mu\)
A \(p\times q\) current transfer matrix \(\alpha\)
The resistance matrix \(\mathcal{R}\) thus has the form:
Fig. 571 Interpretation of the resistance matrix (A) The complete network is drawn as a resistive multi-port network with capacitors and inductors connected to the ports. (B) The capacitors are replaced with independent current sources, and the inductors with independent voltage sources. The resistance matrix describes the relation between the independent sources and the dependent port variables.#
The physical meaning of the elements of \(\mathcal{R}\) is as follows:
A diagonal element \(r_{ii},\) with \(i=1\cdots p\), represents the port resistance at port \(i.\)
If an independent current source \(I_{i}\) is replaced with the capacitor \(C_{i}\), and all other independent sources have been set to zero, \(r_{ii}\) is the resistance that will dissipate the energy stored in \(C_{i}.\)
A diagonal element \(g_{jj},\) with \(j=1\cdots q\), represents the port conductance of port \(p+\ j.\)
If an independent voltage source \(V_{p+j}\) is replaced with the inductor \(L_{j}\), and all other independent sources have been set to zero, \(g_{jj}\) is the conductance that will dissipate the energy stored in \(L_{j}\).
An off-diagonal element \(r_{i,k},\) with \(i=1\cdots p,k=1\cdots p\) and \(i\neq k,\) represents the transresistance from port \(k\) to port \(i.\)
A nonzero coefficient \(r_{i,k}\) implies charge exchange between \(C_{i}\) and \(C_{k}\) in the original network.
An off-diagonal element \(g_{j,\ell},\) with \(j=1\cdots q,\ell=1\cdots q\) and \(j\neq\ell,\) represents the transconductance from port \(p+\ell\) to port \(p+j.\)
A nonzero coefficient \(g_{j,\ell}\) implies flux exchange between inductor \(L_{j}\) and \(L_{\ell}\) in the original network.
An off-diagonal element \(\mu_{i,j},\) with \(i=1\cdots p\) and \(j=1\cdots q,\) represents the voltage transfer from port \(p+j\) to port \(i.\)
A nonzero coefficient \(\mu_{i,j}\), indicates the exchange of energy storage between \(L_{j}\) and \(C_{i}\) in the original network.
An off-diagonal element \(\alpha_{i,j},\) with \(i=1\cdots q\) and \(j=1\cdots p,\) represents the current transfer from port \(i\) to port \(p+j.\)
A nonzero coefficient \(\alpha_{i,j}\), indicates the exchange of energy storage between \(C_{i}\) and \(L_{j}\) in the original network.
Example
\label{example-IncidenceMatrix}
We will evaluate the poles of the circuit from Fig. 572, with the aid of the time-constant matrix. The MNA equations for this circuit are:
in which
and
We find rank \(\mathbf{C}\) is \(3\), which is equal to the number of reactive elements. Hence, there are no zero eigenvalues, and three finite eigenvalues. All capacitor voltages and inductor currents can be taken as state variables.
We will define the diagonal matrix \(\mathcal{C}\) as:
Please notice the minus sign for \(L\); it corresponds with the definition of the positive directions of the port variables as shown in Fig. 571B.
The incidence matrix \(\mathcal{I}\) can be obtained from the netlist as illustrated in Fig. 573. For the given circuit we obtain
Note that \(\mathbf{C}\) can be obtained from \(\mathcal{I}\) and \(\mathcal{C}\) as:
With the aid of ((234)), we find the resistance matrix \(\mathcal{R}\) as
which yields
The time-constant matrix of the circuit from Fig. 572 is according to its definition from (233), obtained as
which yields:
with \(\tau_{i}\) being an eigenvalue of \(\mathbf{T}\). The poles \(p_{i}\) are found from these eigenvalue:
Let us now substitute the numeric element values for the elements from Fig. 572 and evaluate the pole frequencies:
We then have
and the eigenvalues of \(\mathbf{T}\) are obtained as
Hence, the poles of the circuit are
From this example, we see that the poles can easily be found from the eigenvalues of the time-constant matrix \(\mathbf{T}\). A similar procedure can be followed for zeros. The time-constant matrix is then derived from \(\mathcal{R}^{\prime}\) and \(\mathcal{C}^{\prime}\) which are obtained from \(\mathcal{R}\) and \(\mathcal{C}\) after application of Cramer’s rule.
In the next example, we will demonstrate the determination of the resistance matrix \(\mathcal{R}\)\ by network inspection.
Example
Fig. 574 Setting up the R-matrix by network inspection.#
Fig. 574 shows the circuit from Fig. 572\ redrawn as a multi-port network. The left side of the figure shows the original circuit with the capacitors and the inductor connected to a resistive multi-port. Capacitor \(C_{1}\) is connected to port \(1\), capacitor \(C_{2}\) to port \(2\) and inductor \(L_{1}\) to port \(3\).
The right side of the figure shows the same network with the capacitors replaced with current sources and the inductor with a voltage source. The resistor matrix \(\mathcal{R}\) holds the relations between the dependent and independent port variables:
The coefficients \(R_{ij}\) are found from network inspection:
Hence, we obtain:
This result corresponds to ((236)).
As stated earlier, loops of capacitors and/or capacitors with voltage sources, and cut sets of inductors and/or inductors with current sources, result in eigenvalues with zero value
Due to numerical rounding effects, these poles will appear at finite frequencies rather than at infinity. However, in most cases, these frequencies will be outside the frequency range of interest, and can easily be separated from the desired ones, and ignored. According to Haley [54], the method is robust if the number of capacitor loops and inductor cut sets is relatively low with respect to the number of capacitors and inductors.
Symbolic estimation of poles and zeros#
During the design of the frequency response of electronic circuits, designers often need to manipulate poles and zeros into their desired positions. To do so, the designer needs to know which network elements determine the positions of specific poles and zeros. One method for this is to perform a numeric sensitivity analysis. If the value of a specific pole or zero depends on a component value, then the value of that pole or zero can be changed with that component. This requires analyses of the influence of the values of all the components that constitute the elements of the time-constant matrix. This approach may be used for computer analysis, but it is not suited for quick hand estimations.
In this section, we will demonstrate the way in which poles and zeros of a network can be estimated. Limitations of the presented technique will also be shown.
Number of poles#
The number of poles of a network equals the sum of the number of independent capacitor voltages and the number of independent inductor currents.
The number of independent capacitor voltages equals the number of capacitors minus the number of loops of capacitors or loops of capacitors and voltage sources.
The number of independent inductor currents equals the number of inductors minus the number of cut sets of inductors or cut sets of inductors and current sources.
In networks with controlled sources, there is no straightforward way to determine the number of poles:
The presence of feedback across controlled sources may effectively add or remove loops of controlled voltage sources and capacitors, or cut sets of controlled current sources and inductors. If there are no feedback loops involving controlled sources, then these sources can be treated as independent sources.
In networks with ideally coupled inductors
:
The number of independent inductor currents is reduced by the number of unity coupling factors;
The number of independent capacitor voltages is reduced by one for each capacitor driven from voltage-driven, ideally coupled inductors.
Some situations sketched above will be elucidated in the next example.
Example
Fig. 575A shows a circuit with two coupled inductors.
One of the coupled inductors is driven by an ideal voltage source, while the other is connected to a capacitor \(C\).
If \(k=1\), the number of independent inductor currents is \(1\), and there exists a loop of a controlled voltage source and a capacitor. This can be seen from the equivalent circuit shown in Fig. 575B. Having \(k=1\), the inductance \(\left( 1-k^{2}\right) L_{p}\) in series with the voltage source \(V_{s}\) becomes zero. As a consequence,. the current feedback \(\alpha\) cannot affect the driving impedance for \(C\), which makes \(C\) driven from a voltage source.
This can also be seen by evaluating the determinant \(\Delta\) of the MNA matrix. The MNA matrix equations for this circuit is
The determinant \(\Delta\) can be found as:
Hence, the order of the circuit is \(3\) for \(k<1\) and \(1\) for \(k=1\).
This example also shows the effect of a controlled source on the number of independent capacitor voltages. Although Fig. 575B shows a loop of a controlled voltage source and the capacitor \(C\), this loop does not reduce the number of independent capacitor voltages if \(k<1\). This is due to the current feedback with the current-controlled current source. This feedback creates a nonzero drive impedance for \(C\). Hence, due to this feedback, the voltage across \(C\) is an independent capacitor voltage.
Number of zero-frequency poles#
The number of poles at zero frequency equals the number of independent cut sets of capacitors and independent cut sets of capacitors and current sources, plus the number of independent loops of inductors and independent loops of inductors and voltage sources.
The number of independent loops of certain types of elements is equal to the minimum number of branches with these elements, which needs to be removed to break all the loops.
The number of independent cut sets of certain types of elements is equal to the minimum number of branches that have to be added to connect all the sub-networks, that were disconnected from each other, after all branches that were part of a cut set were removed.
Example
The circuit depicted in Fig. 576 has two capacitors and two inductors. All capacitor voltages and inductor currents are independent., so the circuit has four poles. There exists one cut set of capacitors and current sources and one loop of inductors and voltage sources. Hence, two of the four poles have zero frequency.
Note that feedback loops that include controlled sources may also affect the number of zero-frequency poles.
Estimation of pole positions for circuits without feedback#
Now that we know the total number of poles and the number of zero-frequency poles, we will discuss a method for finding symbolic expressions for the finite nonzero poles of a network. In order to provide design information, these expressions must be easy to interpret, in other words, they must be relatively simple. Simple expressions are obtained if there is no charge exchange between capacitors, flux exchange between inductors and charge and flux exchange between capacitors and inductors. For circuits with controlled sources, an extra requirement is the absence of negative feedback. In these cases, the time-constant matrix comprises only diagonal elements that are the eigenvalues of the matrix.
Example
We will estimate the poles of the circuit shown in Fig. 577.
Fig. 577 Example of a circuit with no charge exchange between capacitors.#
The unilateral transfer of the controlled sources prevents charge exchange between the capacitors. As a consequence, the time-constant matrix \(\mathcal{T}\) of this circuit comprises nonzero diagonal elements only:
This time-constant matrix is the product of the resistance matrix \(\mathcal{R}\) and the capacitance matrix \(\mathcal{C}\). The resistance matrix can easily be found from network inspection, as discussed in section Eigenvalues of the time-constant matrix.
The poles of the circuit are the negative reciprocal values of the diagonal elements of \(\mathcal{T}\):
If the exchange of charge, flux or charge and flux is limited between pairs of components, the situation becomes more difficult, but it many cases, useful symbolic approximations for the poles can still be found. This will be illustrated with the aid of the circuit from Fig. 578.
The circuit from Fig. 578A has two capacitors whose voltages are independent. Hence, the circuit has two poles. Since there exists one cut set of capacitors, one of the poles has zero frequency.
Fig. 578 Circuit with two poles. Since there exists a cut set of capacitors, it has one zero-frequency pole.#
The second pole can be found as follows:
Replace the largest capacitor with a short and find the second pole from the time constant \(R_{1}C_{2}\) as shown in Fig. 578B1. In this case, this is sufficiently accurate if \(C_{1}\gg C_{2}\).
Replace the two capacitors by their series connection and find the second pole from the time constant \(R_{1}\frac{C_{1}C_{2}}{C_{1}+C_{2}}\) as shown in Fig. 578B2.
Since the circuit has no DC solution, the time-constant matrix is not defined and cannot be of any help for finding the poles.
First order approximation for finite nonzero poles#
For symbolic analysis of the poles of a network that has a DC solution, we use a procedure based on the evaluation and interpretation of the time-constant matrix, as illustrated in Fig. 571. Consider hereto the network of Fig. 579A.
Fig. 579 Circuit with two poles and interaction between capacitor voltages.#
The procedure applied to this network is as follows:
Find the largest time constant in the circuit, by evaluating the DC port resistances for the capacitors and the DC port conductances for the inductors. The dominant pole is determined by this time constant.
For the given circuit, the DC port resistance for \(C_{1}\) is \(R_{1}+R_{2}\). The DC port resistance for \(C_{2}\) is \(R_{2}\). The largest time constant of the circuit is \(C_{1}\left( R_{1}+R_{2}\right) \); the dominant pole \(p_{1}\) is thus found from the equivalent circuit shown Fig. 579A:
\[p_{1}=-\frac{1}{C_{1}\left( R_{1}+R_{2}\right) }\text{ [rad/s].}\]Now, short the port with the capacitor that caused the largest time constant.
This is justified because, at frequencies above the frequency of the dominant pole, the capacitor impedance is lower than the DC port impedance. Thus, above the frequency of the dominant pole, the capacitor that determines the dominant pole acts as a short for its driving resistance.
Similarly, if an inductor was connected to this port, leave the port open.
Now find the largest time constant of the modified circuit.
With \(C_{1}\) shorted, we obtain the circuit according to Fig. 579C, and thus obtain the second pole \(p_{2}\) as
\[p_{2}=-\frac{R_{1}+R_{2}}{C_{2}R_{1}R_{2}}.\]
The above presented method only gives exact results if there is no exchange of charge and/or flux between capacitors and/or inductors.
Second order approximation for finite nonzero poles#
For second order systems, a more accurate solution is obtained by using all elements of the circuit’s time-constant matrix. According to Haley and Hurst \cite[-1cm]{Haley1989}, the sum of the eigenvalues of the time-constant matrix equals its first order trace \(T_{1},\) and the sum of the mutual products of the eigenvalues equals its second order trace \(T_{2}\).
For a second order system this yields
We will demonstrate this for the circuit from Fig. 579A. For this circuit, we have
Hence, according to its definition, the time-constant matrix is found as
From the first order trace, we find the sum of the eigenvalues:
From the second order trace, we find the product of the eigenvalues:
If the two poles are well separated: \(p_{1}\ll p_{2}\), the sum of the eigenvalues is dominated by the largest. We then obtain the dominant pole \(p_{1}\) as
With \(R_{1}\ll R_{2}\), this can be simplified to
The non-dominant pole is then found from the product \(p_{1}p_{2}\) and the dominant pole \(p_{1}\):
An overview of the numeric and symbolic results is given in Table 30. The most accurate numeric values are those found from the eigenvalues of the time-constant matrix.
The first order approximation for this circuit has an inaccuracy of about \(10\%\). However, it provides symbolic results that can be interpreted clearly:
The dominant pole is predominantly caused by
: \(C_{1}\Vert(R_{1}+R_{2}).\)
The non-dominant pole is predominantly caused by: \(C_{2}\Vert R_{1}\Vert R_{2}.\)
Such simple expressions are useful during design.
The second order approach still gives relatively simple expressions, but, compared to the first order approximation, the poles cannot be calculated from parallel connections of a an equivalent resistor and capacitor. The accuracy of the second order approximation, however, is much better.
Exact symbolic expressions for the eigenvalues of the time-constant matrix can be found for first, second and third order networks. However, the complexity of those expressions often decreases their usefulness for deriving design conclusions. For this reason, exact symbolic expressions for the poles are not listed in Table 30.
Eigenvalues (T) |
1-st order approx. |
numeric |
2-nd order approx |
numeric |
|
\(p_{1}\) |
\(-873\times10^{3}\) |
\(-\frac{1}{C_{1}\left( R_{1}+R_{2}\right) }\) |
\(-954\times10^{3}\) |
\(-\frac{1}{C_{1}\left( R_{1}+R_{2}\right) +C_{2} R_{2}}\) |
\(-870\times10^{3}\) |
\(p_{2}\) |
\(-230\times10^{6}\) |
\(-\frac{R_{1}+R_{2}}{C_{2}R_{1}R_{2}}\) |
\(-210\times10^{6}\) |
\(-\frac{C_{1}\left( R_{1}+R_{2}\right) +C_{2}R_{2} }{C_{1}C_{2}R_{1}R_{2}}\) |
\(-230\times10^{6}\) |
LCR resonators#
In passive circuits and in circuits with controlled sources that are not part of a feedback loop, only interaction between capacitors and inductors may result in complex pole pairs. If so, one speaks of resonance. Fig. 580 shows four passive second order \(LCR\) networks in which resonance may occur. The general representation of the characteristic equation \(\Delta\) of these networks is
in which \(\omega_{0}\) is the resonance frequency and \(Q\) the quality factor. This factor is a measure for the losses that occur during exchange of energy between the capacitor and the inductor:
Fig. 580 Second-order LCR networks.#
Expression ((239)) can alternatively be written as
The poles \(p_{1}\) and \(p_{2}\) of these second order \(LCR\) resonators can thus be found from their product and their sum:
Table 31 gives the product and the sum of the poles, as well as the resonance frequency and the quality factor for the networks from Figure.Fig. 580.
\(p_{1}p_{2}\) |
\(\frac{1}{LC}\) |
\(\frac{1}{LC}\) |
\(\frac{1}{LC}\frac{R_{p}+R_{s}}{R_{p}}\) |
\(\frac{1}{LC}\frac{R_{p}}{R_{p}+R_{s}}\) |
\(p_{1}+p_{2}\) |
\(-\frac{R_{s}}{L}\) |
\(-\frac{1}{R_{p}C}\) |
\(-\frac{1}{R_{p} C}-\frac{R_{s}}{L}\) |
\(-\frac{1}{C\left( R_{p}+R_{s}\right) }-\frac{R_{p}R_{s}}{L\left( R_{p}+R_{s}\right) }\) |
\(\omega_{0}^{2}\) |
\(\frac{1}{LC}\) |
\(\frac{1}{LC}\) |
\(\approx\frac{1}{LC};~R_{p}\gg R_{s}\) |
\(\approx\frac{1}{LC};~R_{p}\gg R_{s}\) |
\(Q\) |
\(\frac{1}{R_{s}}\sqrt{\frac{L}{C}}=Q_{s}\) |
\(R_{p}\sqrt{\frac{C}{L} }=Q_{p}\) |
\(\approx\frac{1}{\frac{1}{Q_{s}}+\frac{1}{Q_{p}}};~R_{p}\gg R_{s}\) |
\(\approx\frac{1}{\frac{1}{Q_{s}}+\frac{1}{Q_{p}}};~R_{p}\gg R_{s}\) |
Estimation of pole positions for circuits with feedback#
Feedback in circuits that comprise controlled sources can have a significant influence on the position of the poles. In section Bandwidth design considerations we will discuss a graphical method for determination of the poles in a feedback circuit.
Estimation of zeros#
We have seen that determination of the poles of a network, does not require the definition of a source-to-load transfer. As a matter of fact, the poles of the network are found from the solution of the set of homogeneous differential equations that describe the network. For determination of the poles, all independent sources have been set to zero: all current sources become open circuits and voltage sources short circuits.
This differs for the determination of the zeros: this requires the definition of a source and a detector. Zeros appear at the complex frequencies for which the transfer from the source to the detector equals zero. The physical causes for zero transfer can be:
At some complex frequency, there exists an open circuit in series with the signal path.
Example
Fig. 581 At some complex frequency, an open circuit in series with the signal path causes a zero.#
Let us consider the transfer from \(V_{s}\) to \(V_{\ell}\) for the circuit from Fig. 581A. If, at some complex frequency, the current through \(R_{1}\) is cancelled by an opposite current through \(C_{1}\), no signal transfer is possible through the parallel connection of \(R_{1}\) and \(C_{1}\). This occurs if both branches have opposite conductance: \(sC_{1}=-\frac{1}{R_{1}}\), hence at a complex frequency \(s=-\frac{1}{R_{1}C_{1}}\). This is the frequency of a zero.
For the circuit from Fig. 581B, this situation occurs if \(\frac{1}{sL_{1}}=-\frac{1}{R_{1}}\), or if \(s=-\frac{R_{1}}{L_{1}}\).
For the circuit from Fig. 581C, no transfer is possible for \(s=0\). A capacitor in series with the signal path causes a zero-frequency zero.
The circuit from Fig. 581D has a parallel resonance circuit in series with the signal path. Here, signal transfer becomes zero if \(\frac{1}{R_{1}}+\frac{1}{sL_{1}}+sC_{1}=0\). This results in two zeros, \(z_{1}\) and \(z_{2}\). They can be found from their product, \(z_{1}z_{2}=\frac{1}{L_{1}C_{1}}\) and their sum, \(z_{1}+z_{2}=-R_{1}C_{1}\). The two zeros are complex conjugated if the quality factor of the parallel resonator exceeds \(0.5\).
At some complex frequency, there exists a short in parallel with the signal path. \vfill
Example
Fig. 582 At some complex frequency, a short circuit in parallel with the signal path causes a zero.#
Let us now consider the transfer from \(V_{s}\) to\ \(V_{\ell}\) for the circuit from Fig. 582A. If, at some complex frequency, the series connection of \(R_{2}\) and \(C_{1}\) becomes a short; \(V_{\ell}\) will then be zero. This occurs if \(\frac{1}{sC_{1}}=-R_{2}\). Hence, a zero is thus found at \(s=-\frac{1}{R_{2}C_{1}}\)
Similarly, the circuit from Fig. 582B has a zero at \(s=-\frac{R_{2}}{L_{1}}\).
The circuit from Fig. 582C has a zero-frequency zero,\ caused by the inductor in parallel with the signal path.
The circuit from Fig. 582D has a series resonance circuit in parallel with the signal path. Here, signal transfer becomes zero if \(R_{2}+\frac{1}{sC_{1}}+sL_{1}=0\). This results in two zeros, \(z_{1}\) and \(z_{2}\). They can be found from their product, \(z_{1}z_{2}=\frac{1}{L_{1} C_{1}}\) and their sum, \(z_{1}+z_{2}=-\frac{L_{1}}{R_{2}}\). The two zeros are complex conjugated if the quality factor of the series resonator exceeds \(0.5\).
At some complex frequency, multi-path source-load transfers cancel each other out.
Example
Fig. 583 Zeros due to multiple transmission paths.#
Fig. 583 shows a dual-path input-output transfer. Both paths have their frequency-dependent transfer modeled by a ratio of two polynomials of the Laplace variable \(s\). The two transfers cancel each other out if \(\frac{N_{1}(s)}{D_{1}(s)}=-\frac{N_{2}(s)}{D_{2}(s)}\). The zeros are the solutions for \(s\) of this equation.
If a pole of a network cannot be observed in a specific transfer, then a zero cancels out that pole.
Example
Consider the circuit from Fig. 584. Let us estimate the poles and zeros of the transfer from \(I_{1}\) to \(V_{1}\). The poles of the circuit can easily be found from network inspection. The circuit has one independent inductor current and one independent capacitor voltage. Hence, it has two poles. There exists no loop of inductors (or of inductors and voltage sources) and no cut sets of capacitors (or capacitors and current sources). So, we have no poles in the origin.
With the aid of the time-constant method we find two time-constants \(\tau _{1}=R_{1}C_{1}\) and \(\tau_{2}=\frac{L}{R_{2}},\) and since there is no exchange of energy between the inductor and the capacitor, we have two real poles: \(p_{1}=-\frac{1}{\tau_{1}}\) and \(p_{2}=-\frac{1}{\tau2}\).
At zero frequency, \(L_{1}\) shorts the output so we have a zero \(z_{1}=0\).
Let us now consider the transfer for \(I_{1}\) to \(V_{1}\). The current \(I_{1}\) flows through the parallel connection of \(L\) and \(R_{2}\) and circuit analysis yields
(240)#\[\frac{V_{1}}{I_{1}}=\frac{sL}{1+s\frac{L}{R_{2}}}. \label{eq-pzCancelTransfer}\]From this expression, we clearly see one pole and the zero in the origin. The pole caused by \(C\) is not observable in this transfer. So, the transfer from \(I_{1}\) to \(V_{1}\) has two poles and two zeros. One of the zeros cancels out one of the poles and makes it non-observable in the transfer function ((240)). A proper expression for the transfer would be
\[\frac{V_{1}}{I_{1}}=\frac{sL(1+sR_{1}C)}{\left( 1+s\frac{L}{R_{2}}\right) (1+sR_{1}C)}.\]