154 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS

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154 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS

In this case the root λ repeats twice. Observe that the characteristic modes in this case are eλ*^t* and teλ*^t* . Continuing this pattern, we can show that for the differential equation

$$(D - \lambda)^r y_0(t) = 0$$

the characteristic modes are eλ*^t* , teλ*^t* , t ²eλ*^t* , …, t ^r−1eλ*^t* , and that the solution is

$$y_0(t) = (c_1 + c_2t + \dots + c_rt^{r-1})e^{\lambda t}$$

Consequently, for a system with the characteristic polynomial

$$Q(\lambda) = (\lambda - \lambda_1)^r (\lambda - \lambda_{r+1}) \cdots (\lambda - \lambda_N)$$

the characteristic modes are e^λ1*^t* , te^λ1*^t* , …, t ^r−1e^λ1*^t* , e^λr+1*^t* , …, e^λNt and the solution is

$$y_0(t) = (c_1 + c_2t + \dots + c_rt^{r-1})e^{\lambda_1 t} + c_{r+1}e^{\lambda_{r+1} t} + \dots + c_N e^{\lambda_N t}$$

COMPLEX ROOTS

The procedure for handling complex roots is the same as that for real roots. For complex roots, the usual procedure leads to complex characteristic modes and the complex form of solution. However, it is possible to avoid the complex form altogether by selecting a real-form of solution, as described next.

For a real system, complex roots must occur in pairs of conjugates if the coefficients of the characteristic polynomial Q(λ) are to be real. Therefore, if α + jβ is a characteristic root, α − jβ must also be a characteristic root. The zero-input response corresponding to this pair of complex conjugate roots is

$$y_0(t) = c_1 e^{(\alpha + j\beta)t} + c_2 e^{(\alpha - j\beta)t}$$

\n(2.7)

For a real system, the response y0(t) must also be real. This is possible only if c¹ and c² are conjugates. Let

$$c_1 = \frac{c}{2}e^{j\theta} \qquad \text{and} \qquad c_2 = \frac{c}{2}e^{-j\theta}$$

This yields

$$y_0(t) = \frac{c}{2}e^{i\theta}e^{(\alpha+j\beta)t} + \frac{c}{2}e^{-j\theta}e^{(\alpha-j\beta)t}$$

=

$$\frac{c}{2}e^{\alpha t}[e^{i(\beta t + \theta)} + e^{-j(\beta t + \theta)}]$$

=

$$ce^{\alpha t}\cos(\beta t + \theta)$$

(2.8)

Therefore, the zero-input response corresponding to complex conjugate roots α ± jβ can be expressed in a complex form [Eq. (2.7)] or a real form [Eq. (2.8)].

EXAMPLE 2.1 Finding the Zero-Input Response

Find y0(t), the zero-input response of the response for an LTIC system described by

• (a) the simple-root system (D² + 3D + 2)y(t) = Dx(t) with initial conditions y0(0) = 0 and y˙0(0) = −5.
• (b) the repeated-root system (D² + 6D + 9)y(t) = (3D + 5)x(t) with initial conditions y0(0) = 3 and y˙0(0) = −7.
• (c) the complex-root system (D² + 4D + 40)y(t) = (D + 2)x(t) with initial conditions y0(0) = 2 and y˙0(0) = 16.78.

(a) Note that y0(t), being the zero-input response (x(t) = 0), is the solution of (D² +3D+ 2)y0(t) = 0. The characteristic polynomial of the system is λ² + 3λ + 2. The characteristic equation of the system is therefore λ² + 3λ + 2 = (λ + 1)(λ + 2) = 0. The characteristic roots of the system are λ¹ = −1 and λ² = −2, and the characteristic modes of the system are e−^t and e−2*^t* . Consequently, the zero-input response is

$$y_0(t) = c_1 e^{-t} + c_2 e^{-2t}$$

Differentiating this expression, we obtain

$$\dot{y}_0(t) = -c_1 e^{-t} - 2c_2 e^{-2t}$$

To determine the constants c¹ and c2, we set t = 0 in the equations for y0(t) and y˙0(t) and substitute the initial conditions y0(0) = 0 and y˙0(0) = −5, yielding

$$0 = c_1 + c_2$$ $$-5 = -c_1 - 2c_2$$

Solving these two simultaneous equations in two unknowns for c¹ and c² yields

$$c_1 = -5 \qquad \text{and} \qquad c_2 = 5$$

Therefore,

$$y_0(t) = -5e^{-t} + 5e^{-2t}$$

(2.9)

This is the zero-input response of y(t). Because y0(t) is present at t = 0−, we are justified in assuming that it exists for t ≥ 0.†

(b) The characteristic polynomial is λ² + 6λ + 9 = (λ + 3)2, and its characteristic roots are λ¹ = −3, λ² = −3 (repeated roots). Consequently, the characteristic modes of the system are e−3*^t* and te−3*^t* . The zero-input response, being a linear combination of the characteristic modes, is given by

$$y_0(t) = (c_1 + c_2 t)e^{-3t}$$

^† *^y*0(t) may be present even before ^{t = 0}−. However, we can be sure of its presence only from ^{t = 0−} onward.

We can find the arbitrary constants c¹ and c² from the initial conditions y0(0) = 3 and y˙0(0) = −7 following the procedure in part (a). The reader can show that c¹ = 3 and c² = 2. Hence,

$$y_0(t) = (3+2t)e^{-3t} \qquad t \ge 0$$

(c) The characteristic polynomial is λ² + 4λ + 40 = (λ + 2 − j6)(λ + 2 + j6). The characteristic roots are −2 ± j6.† The solution can be written either in the complex form [Eq. (2.7)] or in the real form [Eq. (2.8)]. The complex form is y0(t) = c1eλ1*^t* + c2eλ2*^t* , where λ¹ = −2+j6 and λ² = −2−j6. Since α = −2 and β =6, the real-form solution is [see Eq. (2.8)]

$$y_0(t) = ce^{-2t}\cos(6t+\theta)$$

Differentiating this expression, we obtain

$$\dot{y}_0(t) = -2ce^{-2t}\cos{(6t + \theta)} - 6ce^{-2t}\sin{(6t + \theta)}$$

To determine the constants c and θ, we set t = 0 in the equations for y0(t) and y˙0(t) and substitute the initial conditions y0(0) = 2 and y˙0(0) = 16.78, yielding

$$2 = c \cos \theta$$

16.78 = $-2c \cos \theta - 6c \sin \theta$

Solution of these two simultaneous equations in two unknowns ccos θ and csinθ yields

ccos θ = 2 and csinθ = −3.463

Squaring and then adding these two equations yield

$$c^{2} = (2)^{2} + (-3.464)^{2} = 16 \Longrightarrow c = 4$$

Next, dividing csinθ = −3.463 by ccos θ = 2 yields

$$\tan \theta = \frac{-3.463}{2}$$

and

$$\theta = \tan^{-1} \left( \frac{-3.463}{2} \right) = -\frac{\pi}{3}$$

Therefore,

$$y_0(t) = 4e^{-2t}\cos\left(6t - \frac{\pi}{3}\right)$$

For the plot of y0(t), refer again to Fig. B.11c.

$$\lambda^2 + 4\lambda + 40 = (\lambda^2 + 4\lambda + 4) + 36 = (\lambda + 2)^2 + (6)^2 = (\lambda + 2 - j6)(\lambda + 2 + j6)$$

^† The complex conjugate roots of a second-order polynomial can be determined by using the formula in Sec. B.8-10 or by expressing the polynomial as a sum of two squares. The latter can be accomplished by completing the square with the first two terms, as follows:

EXAMPLE 2.2 Using MATLAB to Find Polynomial Roots

Find the roots λ¹ and λ² of the polynomial λ² + 4λ + k for three values of k: (a) k = 3, (b) k = 4, and (c) k = 40.

(a) >> r = roots([1 4 3]).’ r = -3 -1 For k = 3, the polynomial roots are therefore λ¹ = −3 and λ² = −1. (b) >> r = roots([1 4 4]).’ r = -2 -2 For k = 4, the polynomial roots are therefore λ¹ = λ² = −2. (c) >> r = roots([1 4 40]).’ r = -2.00+6.00i -2.00-6.00i For k = 40, the polynomial roots are therefore λ¹ = −2+j6 and λ² = −2−j6.

EXAMPLE 2.3 Using MATLAB to Find the Zero-Input Response

Consider an LTIC system specified by the differential equation

$$(D^2 + 4D + k)y(t) = (3D + 5)x(t)$$

Using initial conditions y0(0) = 3 and y˙0(0) = −7, apply MATLAB’s dsolve command to determine the zero-input response when: (a) k = 3, (b) k = 4, and (c) k = 40.