For Practice Prob. 16.9. 16.5 State Variables

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For Practice Prob. 16.9. 16.5 State Variables

Thus far in this book we have considered techniques for analyzing systems with only one input and only one output. Man y engineering systems have many inputs and many outputs, as shown in Fig. 16.21. The state variable method is a v ery important tool in analyzing systems and understanding such highly complex systems. Thus, the state variable model is more gen eral than the single-input, single-output model, such as a transfer function. Although the topic cannot be adequately co vered in one chapter, let alone one section of a chapter, we will cover it briefly at this point.

In the state variable model, we specify a collection of variables that describe the internal behavior of the system. These variables are known as the state variables of the system. They are the variables that determine the future behavior of a system when the present state of the system and the input signals are kno wn. In other w ords, the y are those v ariables which, if known, allow all other system parameters to be determined by using only algebraic equations.

A state variable is a physical property that characterizes the state of a system, regardless of how the system got to that state.

Common examples of state variables are the pressure, volume, and temperature. In an electric circuit, the state v ariables are the inductor current and capacitor voltage since they collectively describe the energy state of the system.

The standard way to represent the state equations is to arrange them as a set of first-order differential equations:

$$\dot{x} = Ax + Bz \tag{16.20}$$

where

$$\dot{\mathbf{x}}(t) = \begin{bmatrix} x_1(t) \\ x_2(t) \\ \vdots \\ x_n(t) \end{bmatrix} = \text{state vector representing } n \text{ state vectors}$$

and the dot represents the first derivative with respect to time, i.e.,

$$\dot{\mathbf{x}}(t) = \begin{bmatrix} \dot{x}_1(t) \\ \dot{x}_2(t) \\ \vdots \\ \dot{x}_n(t) \end{bmatrix}$$

and

$$\mathbf{z}(t) = \begin{bmatrix} z_1(t) \\ z_2(t) \\ \vdots \\ z_m(t) \end{bmatrix} = \text{input vector representing } m \text{ inputs}$$

A and B are respectively n × n and n × m matrices. In addition to the state equation in Eq. (16.20), we need the output equation. The complete state model or state space is

$$\dot{x} = Ax + Bz$$

(16.21a)
y = Cx + Dz (16.21b)

where

$$\mathbf{y}(t) = \begin{bmatrix} y_1(t) \\ y_2(t) \\ \vdots \\ y_p(t) \end{bmatrix} = \text{the output vector representing } p \text{ outputs}$$

and C and D are, respectively, p × n and p × m matrices. For the special case of single-input single-output, n = m = p = 1.

Assuming zero initial conditions, the transfer function of the system is found by taking the Laplace transform of Eq. (16.21a); we obtain

$$s\mathbf{X}(s) = \mathbf{A}\mathbf{X}(s) + \mathbf{B}\mathbf{Z}(s) \qquad \rightarrow \qquad (s\mathbf{I} - \mathbf{A})\mathbf{X}(s) = \mathbf{B}\mathbf{Z}(s)$$

or

$$\mathbf{X}(s) = (s\mathbf{I} - \mathbf{A})^{-1} \mathbf{B} \mathbf{Z}(s)$$

(16.22)

where I is the identity matrix. Taking the Laplace transform of Eq. (16.21b) yields

$$\mathbf{Y}(s) = \mathbf{C}\mathbf{X}(s) + \mathbf{D}\mathbf{Z}(s) \tag{16.23}$$

Substituting Eq. (16.22) into Eq. (16.23) and di viding by Z(s) gives the transfer function as

$$H(s) = \frac{Y(s)}{Z(s)} = C(sI - A)^{-1}B + D$$

(16.24)

where

A = system matrix B = input coupling matrix C = output matrix D = feedforward matrix

In most cases, D = 0, so the degree of the numerator of H(s) in Eq. (16.24) is less than that of the denominator. Thus,

$$H(s) = C(sI - A)^{-1}B$$

(16.25)

Because of the matrix computation in volved, MATLAB can be used to find the transfer function.

To apply state variable analysis to a circuit, we follow the following three steps.

Steps to Apply the State Variable Method to Circuit Analysis:

• 1. Select the inductor current i and capacitor voltage v as the state variables, making sure they are consistent with the passive sign convention.
• 2. Apply KCL and KVL to the circuit and obtain circuit variables (voltages and currents) in terms of the state v ariables. This should lead to a set of first-order differential equations necessary and sufficient to determine all state variables.
• 3. Obtain the output equation and put the final result in state-space representation.

Steps 1 and 3 are usually straightforward; the major task is in step 2. We will illustrate this with examples.

Solution:

We select the inductor current i and capacitor voltage v as the state variables.

$$v_L = L \frac{di}{dt} \tag{16.10.1}$$ $$i_C = C \frac{dv}{dt} \tag{16.10.2}$$

Applying KCL at node 1 gives

$$i = i_x + i_C$$

$\rightarrow$ $C\frac{dv}{dt} = i - \frac{v}{R}$

or

$$\dot{v} = -\frac{v}{RC} + \frac{i}{C}$$

(16.10.3)

since the same voltage v is across both R and C. Applying KVL around the outer loop yields

$$v_s = v_L + v \rightarrow L\frac{di}{dt} = -v + v_s$$ $$\dot{i} = -\frac{v}{L} + \frac{v_s}{L}$$

(16.10.4)

Equations (16.10.3) and (16.10.4) constitute the state equations. If we regard ix as the output,

$$i_x = \frac{v}{R} \tag{16.10.5}$$

Putting Eqs. (16.10.3), (16.10.4), and (16.10.5) in the standard form leads to

$$\begin{bmatrix} \dot{v} \\ \dot{i} \end{bmatrix} = \begin{bmatrix} \frac{-1}{RC} & \frac{1}{C} \\ \frac{-1}{L} & 0 \end{bmatrix} \begin{bmatrix} v \\ i \end{bmatrix} + \begin{bmatrix} 0 \\ \frac{1}{L} \end{bmatrix} v_s$$

(16.10.6a)

$$i_x = \begin{bmatrix} \frac{1}{R} & 0 \end{bmatrix} \begin{bmatrix} v \\ i \end{bmatrix}$$

(16.10.6b)

If R = 1, C = _1 4 , and L = _1 ² , we obtain from Eq. (16.10.6) matrices

$$\mathbf{A} = \begin{bmatrix} \frac{-1}{RC} & \frac{1}{C} \\ \frac{-1}{L} & 0 \end{bmatrix} = \begin{bmatrix} -4 & 4 \\ -2 & 0 \end{bmatrix}, \qquad \mathbf{B} = \begin{bmatrix} 0 \\ \frac{1}{L} \end{bmatrix} = \begin{bmatrix} 0 \\ 2 \end{bmatrix},$$ $$\mathbf{C} = \begin{bmatrix} \frac{1}{R} & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \end{bmatrix}$$ $$s\mathbf{I} - \mathbf{A} = \begin{bmatrix} s & 0 \\ 0 & s \end{bmatrix} - \begin{bmatrix} -4 & 4 \\ -2 & 0 \end{bmatrix} = \begin{bmatrix} s+4 & -4 \\ 2 & s \end{bmatrix}$$

Taking the inverse of this gives

inverse of this gives
\n

$$(s\mathbf{I} - \mathbf{A})^{-1} = \frac{\text{adjoint of } \mathbf{A}}{\text{determinant of } \mathbf{A}} = \frac{\begin{bmatrix} s & 4\\ -2 & s+4 \end{bmatrix}}{s^2 + 4s + 8}$$

Thus, the transfer function is given by

Thus, the transfer function is given by

\n

$$\mathbf{H}(s) = \mathbf{C}(s\mathbf{I} - \mathbf{A})^{-1} \mathbf{B} = \frac{\begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} s & 4 \\ -2 & s+4 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \end{bmatrix}}{s^2 + 4s + 8} = \frac{8}{s^2 + 4s + 8}$$

\n

$$= \frac{8}{s^2 + 4s + 8}$$

which is the same thing we would get by directly Laplace transforming the circuit and obtaining H(s) = Ix(s)∕Vs(s). The real advantage of the state variable approach comes with multiple inputs and multiple outputs. In this case, we have one input vs and one output ix. In the next example, we will have two inputs and two outputs.

Example 16.11 Consider the circuit in Fig. 16.24, which may be regarded as a two- input, two-output system. Determine the state v ariable model and find the transfer function of the system.

Figure 16.24 For Example 16.11.

Solution:

In this case, we have two inputs vs and vi and two outputs vo and io. Again, we select the inductor current i and capacitor voltage v as the state variables. Applying KVL around the left-hand loop gives

$$-v_s + i_1 + \frac{1}{6}i = 0 \quad \to \quad i = 6v_s - 6i_1 \quad (16.11.1)$$

We need to eliminate i1. Applying KVL around the loop containing vs,1-Ω resistor, 2-Ω resistor, and _1 ³ -F capacitor yields

$$v_s = i_1 + v_o + v \tag{16.11.2}$$

But at node 1, KCL gives

$$i_1 = i + \frac{v_o}{2} \rightarrow v_o = 2(i_1 - i)$$

(16.11.3)

For Practice Prob. 16.10.

Substituting this in Eq. (16.11.2),

$$v_s = 3i_1 + v - 2i \rightarrow i_1 = \frac{2i - v + v_s}{3}$$

(16.11.4)

Substituting this in Eq. (16.11.1) gives

$$\dot{i} = 2v - 4i + 4v_s \tag{16.11.5}$$

which is one state equation. To obtain the second one, we apply KCL at node 2.

$$\frac{v_o}{2} = \frac{1}{3} \dot{v} + i_o \rightarrow \dot{v} = \frac{3}{2} v_o - 3i_o \tag{16.11.6}$$

We need to eliminate vo and io. From the right-hand loop, it is evident that

$$i_o = \frac{v - v_i}{3} \tag{16.11.7}$$

Substituting Eq. (16.11.4) into Eq. (16.11.3) gives

i

$$v_o = 2\left(\frac{2i - v + v_s}{3} - i\right) = -\frac{2}{3}(v + i - v_s) \tag{16.11.8}$$

Substituting Eqs. (16.11.7) and (16.11.8) into Eq. (16.11.6) yields the second state equation as

$$\dot{v} = -2v - i + v_s + v_i \tag{16.11.9}$$

The two output equations are already obtained in Eqs. (16.11.7) and (16.11.8). Putting Eqs. (16.11.5) and (16.11.7) to (16.11.9) together in the standard form leads to the state model for the circuit, namely,

$$\begin{bmatrix} \dot{v} \\ \dot{i} \end{bmatrix} = \begin{bmatrix} -2 & -1 \\ 2 & -4 \end{bmatrix} \begin{bmatrix} v \\ i \end{bmatrix} + \begin{bmatrix} 1 & 1 \\ 4 & 0 \end{bmatrix} \begin{bmatrix} v_s \\ v_i \end{bmatrix}$$

(16.11.10a)

$$\begin{bmatrix} v_o \\ i_o \end{bmatrix} = \begin{bmatrix} -\frac{2}{3} & -\frac{2}{3} \\ \frac{1}{3} & 0 \end{bmatrix} \begin{bmatrix} v \\ i \end{bmatrix} + \begin{bmatrix} \frac{2}{3} & 0 \\ 0 & -\frac{1}{3} \end{bmatrix} \begin{bmatrix} v_s \\ v_i \end{bmatrix}$$

(16.11.10b)

For the electric circuit in Fig. 16.25, determine the state model. Take vo Practice Problem 16.11 and io as the output variables.

Answer:

Figure 16.25 For Practice Prob. 16.11.

Example 16.12 Assume we have a system where the output is y(t) and the input is z(t). Let the following differential equation describe the relationship between the input and the output.

$$\frac{d^2y(t)}{dt^2} + 3\frac{dy(t)}{dt} + 2y(t) = 5z(t)$$

(16.12.1)

Obtain the state model and the transfer function of the system.

x .

Solution:

First, we select the state variables. Let x1 = y(t), therefore

$$x_1 = \dot{y}(t) \tag{16.12.2}$$

Now let

$$x_2 = \dot{x}_1 = \dot{y}(t) \tag{16.12.3}$$

Note that at this time we are looking at a second-order system that would normally have two first-order terms in the solution.

Now we have x . ² = y ..(t), where we can find the value ^x . 2 from Eq. (16.12.1), i.e.,

$$\dot{x}_2 = \ddot{y}(t) = -2y(t) - 3\dot{y}(t) + 5z(t) = -2x_1 - 3x_2 + 5z(t)$$

(16.12.4)

From Eqs. (16.12.2) to (16.12.4), we can now write the following matrix equations:

$$\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} 0 \\ 5 \end{bmatrix} z(t)$$

(16.12.5)

$$\mathbf{y}(t) = \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \tag{16.12.6}$$

We now obtain the transfer function.

$$s\mathbf{I} - \mathbf{A} = s \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} = \begin{bmatrix} s & -1 \\ 2 & s + 3 \end{bmatrix}$$

The inverse is

$$(s\mathbf{I} - \mathbf{A})^{-1} = \frac{\begin{bmatrix} s+3 & 1\\ -2 & s \end{bmatrix}}{s(s+3)+2}$$

The transfer function is

ne transfer function is
\n

$$\mathbf{H}(s) = \mathbf{C}(s\mathbf{I} - \mathbf{A})^{-1} \mathbf{B} = \frac{(1 \quad 0) \begin{bmatrix} s+3 & 1 \\ -2 & s \end{bmatrix} \begin{pmatrix} 0 \\ 5 \end{pmatrix}}{s(s+3)+2} = \frac{(1 \quad 0) \begin{pmatrix} 5 \\ 5s \end{pmatrix}}{(s+1)(s+2)}$$

To check this, we directly apply the Laplace transfer to each term in Eq. (16.12.1). Given that initial conditions are zero, we get

$$[s2 + 3s + 2]Y(s) = 5Z(s) \rightarrow H(s) = \frac{Y(s)}{Z(s)} = \frac{5}{s2 + 3s + 2}$$

which is in agreement with what we got previously.

$$\frac{d^3y}{dt^3} + 18\frac{d^2y}{dt^2} + 20\frac{dy}{dt} + 5y = z(t)$$

Answer:

$$\mathbf{A} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -5 & -20 & -18 \end{bmatrix}, \quad \mathbf{B} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, \quad \mathbf{C} = \begin{bmatrix} 1 & 0 & 0 \end{bmatrix}.$$

16.6 Applications

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 tw o more important applications: netw ork stability and network synthesis.

16.6.1 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 → ∞; it is unstable if h(t) grows without bound as t → ∞. In mathematical terms, a circuit is stable when

$$\lim_{t \to \infty} |h(t)| = \text{finite} \tag{16.26}$$

Because the transfer function H(s) is the Laplace transform of the impulse response h(t), H(s) must meet certain requirements for Eq. (16.26) to hold. Recall that H(s) may be written as

$$H(s) = \frac{N(s)}{D(s)}$$

(16.27)

where the roots of N(s) = 0 are called the zeros of H(s) because the y make H(s) = 0, while the roots of D(s) = 0 are called the poles of H(s) since they cause H(s) → ∞. The zeros and poles of H(s) are often located in the s plane as sho wn in Fig. 16.26(a). Recall from Eqs. (15.47) and (15.48) that H(s) may also be written in terms of its poles as _____________________ (^s + p1)(s + p2) ⋯ (s + pn)

$$H(s) = \frac{N(s)}{D(s)} = \frac{N(s)}{(s+p_1)(s+p_2)\cdots(s+p_n)}$$

(16.28)

H(s) must meet tw o 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

$$H(s) = k_n s^n + k_{n-1} s^{n-1} + \dots + k_1 s + k_0 + \frac{R(s)}{D(s)}$$

(16.29)

where the de gree of R(s), the remainder of the long di vision, is less than the degree of D(s). The inverse of H(s) in Eq. (16.29) does not meet the condition in Eq. (16.26). Second, all the poles of H(s) in

Figure 16.26 The complex s plane: (a) poles and zeros plotted, (b) left-half plane.

Eq. (16.27) (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. 16.26(b). The reason for this will be apparent if we take the inverse Laplace transform of H(s) in Eq. (16.27). Because Eq. (16.27) is similar to Eq. (15.48), its partial fraction expansion is similar to the one in Eq. (15.49) so that the inverse of H(s) is similar to that in Eq. (15.53). Hence,

$$h(t) = (k_1 e^{-p_1 t} + k_2 e^{-p_2 t} + \dots + k_n e^{-p_n t}) u(t)$$

\n(16.30)

We see from this equation that each pole pi must be positive (i.e., pole s = −pi in the left-half plane) for e⁻pi 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 w ould imply that some branch currents or v oltages 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 RLC circuit in Fig. 16.27. The transfer function is given by

$$H(s) = \frac{V_o}{V_s} = \frac{1/sC}{R + sL + 1/sC}$$ $$H(s) = \frac{1/L}{s^2 + sR/L + 1/LC}$$

(16.31)

Notice that D(s) = s 2 + sR∕L + 1∕LC = 0 is the same as the characteristic equation obtained for the series RLC circuit in Eq. (8.8). The circuit has poles at _______

$$p_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2}$$

(16.32)

where

or

$$\alpha = \frac{R}{2L}, \qquad \omega_0 = \frac{1}{LC}$$

For R, L, C > 0, the tw o poles always lie in the left half of the s plane, implying that the circuit is al ways stable. However, when R = 0, α = 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

imple of a circuit designed to be unstable. An oscula-
ch that its transfer function is of the form

$$H(s) = \frac{N(s)}{s^2 + \omega_0^2} = \frac{N(s)}{(s + j\omega_0)(s - j\omega_0)}$$

(16.33)

so that its output is sinusoidal.

Figure 16.27 A typical RLC circuit.

Determine the values of k for which the circuit in Fig. 16.28 is stable. Example 16.13

Solution:

Applying mesh analysis to the first-order circuit in Fig. 16.28 gives

$$V_i = \left(R + \frac{1}{sC}\right)I_1 - \frac{I_2}{sC}$$

(16.13.1)

and

$$0 = -kI_1 + \left(R + \frac{1}{sC}\right)I_2 - \frac{I_1}{sC}$$

or

$$0 = -\left(k + \frac{1}{sC}\right)I_1 + \left(R + \frac{1}{sC}\right)I_2\tag{16.13.2}$$

We can write Eqs. (16.13.1) and (16.13.2) in matrix form as

$$\begin{bmatrix} V_i \\ 0 \end{bmatrix} = \begin{bmatrix} \left(R + \frac{1}{sC}\right) & -\frac{1}{sC} \\ -\left(k + \frac{1}{sC}\right) & \left(R + \frac{1}{sC}\right) \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \end{bmatrix}$$

The determinant is

inant is
\n

$$\Delta = \left(R + \frac{1}{sC}\right)^2 - \frac{k}{sC} - \frac{1}{s^2C^2} = \frac{sR^2C + 2R - k}{sC}$$

\n(16.13.3)

The characteristic equation (∆ = 0) gives the single pole as

$$p = \frac{k - 2R}{R^2 C}$$

which is negative when k < 2R. Thus, we conclude the circuit is stable when k < 2R and unstable for k > 2R.

Answer: β > −1∕R.

For Practice Prob. 16.13.

An active filter has the transfer function Example 16.14

$$H(s) = \frac{k}{s^2 + s(4 - k) + 1}$$

For what values of k is the filter stable?

1

I¹ I² kI¹

R R

‒ +

Solution:

As a second-order circuit, H(s) may be written as

$$H(s) = \frac{N(s)}{s^2 + bs + c}$$

where b = 4 − k, c = 1, and N(s) = k. This has poles at p² + bp + c = 0; that is,

$$p_{1,2} = \frac{-b \pm \sqrt{b^2 - 4c}}{2}$$

For the circuit to be stable, the poles must be located in the left half of the s plane. This implies that b > 0.

Applying this to the given H(s) means that for the circuit to be stable, 4 − k > 0 or k < 4.

Practice Problem 16.14 A second-order active circuit has the transfer function

circuit has the transfer function

$$H(s) = \frac{1}{s^2 + s(25 + \alpha) + 25}$$

Find the range of the values of α for which the circuit is stable. What is the value of α that will cause oscillation?

Answer: α > −25, α = −25.