Application of The Laplace Transform to the Network Stability

So far we have considered three applications of Laplace's transform: circuit analysis in general, obtaining transfer functions, and solving linear integrodifferential equations. The Laplace transform also finds application in other areas in circuit analysis, signal processing, and control systems. Here we will consider two more important applications: network stability and network synthesis.

Network Stability

A circuit is stable if its impulse response $ h(t) $ is bounded (i.e., $ h(t) $ converges to a finite value) as $ t \rightarrow \infty $; it is unstable if $ h(t) $ grows without bound as $ t \rightarrow \infty $. In mathematical terms, a circuit is stable when
$$\lim _{t \rightarrow \infty}|h(t)|=\text { finite } \tag{1}$$
Since the transfer function $ H(s) $ is the Laplace transform of the impulse response $ h(t), H(s) $ must meet certain requirements in order for Eq. (1) to hold. Recall that $ H(s) $ may be written as
$$H(s)=\frac{N(s)}{D(s)} \tag{2}$$
where the roots of $ N(s)=0 $ are called the zeros of $ H(s) $ because they make $ H(s)=0 $, while the roots of $ D(s)=0 $ are called the poles of $ H(s) $ since they cause $ H(s) \rightarrow \infty $.
Fig. 1: The complex s plane: (a) poles and zeros plotted, (b) left-half plane.
The zeros and poles of $ H(s) $ are often located in the $ s $ plane as shown in Fig. 1(a). Recall from Eqs. (2) that $ H(s) $ may also be written in terms of its poles as
$$H(s)=\frac{N(s)}{D(s)}=\frac{N(s)}{\left(s+p_{1}\right)\left(s+p_{2}\right) \cdots\left(s+p_{n}\right)} \tag{3}$$
$ H(s) $ must meet two requirements for the circuit to be stable. First, the degree of $ N(s) $ must be less than the degree of $ D(s) $; otherwise, long division would produce
$$H(s)=k_{n} s^{n}+k_{n-1} s^{n-1}+\cdots+k_{1} s+k_{0}+\frac{R(s)}{D(s)} \tag{4}$$
where the degree of $ R(s) $, the remainder of the long division, is less than the degree of $ D(s) $. The inverse of $ H(s) $ in Eq. (3) does not meet the condition in Eq. (1). Second, all the poles of $ H(s) $ in Eq. (2) (i.e., all the roots of $ D(s)=0 $ ) must have negative real parts; in other words, all the poles must lie in the left half of the $ s $ plane, as shown typically in Fig. 1(b). The reason for this will be apparent if we take the inverse Laplace transform of $ H(s) $ in Eq. (2). Hence,
$$h(t)=\left(k_{1} e^{-p_{1} t}+k_{2} e^{-p_{2} t}+\cdots+k_{n} e^{-p_{n} t}\right) \tag{5}$$
We see from this equation that each pole $ p_{i} $ must be positive (i.e., pole $ s= $ $ -p_{i} $ in the left-half plane) in order for $ e^{-p_{i} t} $ to decrease with increasing $ t $. Thus,
A circuit is stable when all the poles of its transfer function H(s) lie in the left half of the s plane.
An unstable circuit never reaches steady state because the transient response does not decay to zero. Consequently, steady-state analysis is only applicable to stable circuits.
A circuit made up exclusively of passive elements $ (R, L $, and $ C) $ and independent sources cannot be unstable, because that would imply that some branch currents or voltages would grow indefinitely with sources set to zero.
Passive elements cannot generate such indefinite growth. Passive circuits either are stable or have poles with zero real parts. To show that this is the case, consider the series $ R L C $ circuit in Fig. 2.
Fig. 2: A typical RLC circuit.
The transfer function is given by
$$H(s)=\frac{V_{o}}{V_{s}}=\frac{1 / s C}{R+s L+1 / s C} \tag{6}$$
$$H(s)=\frac{1 / L C}{s^{2}+s R / L+1 / L C} \tag{7}$$
Notice that $ D(s)=s^{2}+s R / L+1 / L C=0 $ is the same as the characteristic equation obtained for the series $ R L C $ circuit. The circuit has poles at
$$p_{1,2}=-\alpha \pm \sqrt{\alpha^{2}-\omega_{0}^{2}} \tag{8}$$
$$\alpha=\frac{R}{2 L}, \quad \omega_{0}=\frac{1}{L C} \tag{9}$$
For $ R, L, C>0 $, the two poles always lie in the left half of the $ s $ plane, implying that the circuit is always stable. However, when $ R=0, \alpha=0 $ and the circuit becomes unstable. Although ideally this is possible, it does not really happen, because $ R $ is never zero.
On the other hand, active circuits or passive circuits with controlled sources can supply energy, and they can be unstable. In fact, an oscillator is a typical example of a circuit designed to be unstable. An oscillator is designed such that its transfer function is of the form
$$H(s)=\frac{N(s)}{s^{2}+\omega_{0}^{2}}=\frac{N(s)}{\left(s+j \omega_{0}\right)\left(s-j \omega_{0}\right)} \tag{10}$$
so that its output is sinusoidal.
Example 1: Determine the values of $ k $ for which the circuit in Fig. $ 3 $ is stable.
Fig. 3: Example 1.
Solution: Applying mesh analysis to the first-order circuit in Fig. $ 3 $ gives
$$V_{i}=\left(R+\frac{1}{s C}\right) I_{1}-\frac{I_{2}}{s C} \tag{1.1}$$
$$0=-k I_{1}+\left(R+\frac{1}{s C}\right) I_{2}-\frac{I_{1}}{s C} $$
$$0=-\left(k+\frac{1}{s C}\right) I_{1}+\left(R+\frac{1}{s C}\right) I_{2} \tag{1.2}$$
We can write Eqs. (1.1) and (1.2) in matrix form as
$$\left[\begin{array}{c}V_{i} \\0\end{array}\right]=\left[\begin{array}{cc}\left(R+\frac{1}{s C}\right) & -\frac{1}{s C} \\-\left(k+\frac{1}{s C}\right) & \left(R+\frac{1}{s C}\right)\end{array}\right]\left[\begin{array}{c}I_{1} \\I_{2}\end{array}\right]$$
The determinant is
$$\Delta=\left(R+\frac{1}{s C}\right)^{2}-\frac{k}{s C}-\frac{1}{s^{2} C^{2}}=\frac{s R^{2} C+2 R-k}{s C} \tag{1.3}$$
The characteristic equation ($\Delta = 0$) gives the single pole as
$$p = {k − 2R \over R^2C}$$
which is negative when $k < 2R$. Thus, we conclude the circuit is stable when $k < 2R$ and unstable for $k > 2R$.

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