3.7 THE UNIT IMPULSE [RESPONSE](#page-9-0) h[n]

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3.6 SYSTEM RESPONSE TO INTERNAL CONDITIONS: THE ZERO-INPUT RESPONSE

The zero-input response y0[n] is the solution of Eq. (3.20) with x[n] = 0; that is,

$$Q[E]y_0[n] = 0$$

or

$$(EN + a1EN-1 + \dots + aN-1E + aN)y0[n] = 0$$

\n(3.21)

Although we can solve this equation systematically, even a cursory examination points to the solution. This equation states that a linear combination of y0[n] and advanced y0[n] is zero, not for some values of n, but for all n. Such a situation is possible if and only if y0[n] and advanced y0[n] have the same form. Only an exponential function γ ⁿ has this property, as the following equation indicates:

$$E^k\{\gamma^n\} = \gamma^{n+k} = \gamma^k\gamma^n$$

This expression shows that γ ⁿ advanced by k units is a constant (γ ^k) times γ ⁿ. Therefore, the solution of Eq. (3.21) must be of the form†

$$y_0[n] = c\gamma^n \tag{3.22}$$

To determine c and γ , we substitute this solution in Eq. (3.21). Since Eky0[n] = y0[n+k] = cγ ⁿ+^k, this produces

$$c(\gamma^{N} + a_1 \gamma^{N-1} + \cdots + a_{N-1} \gamma + a_N) \gamma^{n} = 0$$

For a nontrivial solution of this equation,

$$\gamma^{N} + a_{1}\gamma^{n-1} + \dots + a_{N-1}\gamma + a_{N} = 0$$

\n(3.23)

or

Q[γ ] = 0

Our solution cγ ⁿ [Eq. (3.22)] is correct, provided γ satisfies Eq. (3.23). Now, Q[γ ] is an Nth-order polynomial and can be expressed in the factored form (assuming all distinct roots):

$$(\gamma - \gamma_1)(\gamma - \gamma_2) \cdots (\gamma - \gamma_N) = 0$$

Clearly, γ has N solutions γ1, γ2, … , γ*^N* and, therefore, Eq. (3.21) also has N solutions c1γ ⁿ 1 , c2γ ^{n 2} , … , cnγ ^{n N}. In such a case, we have shown that the general solution is a linear combination

^† A signal of the form nmγ ⁿ also satisfies this requirement under certain conditions (repeated roots), discussed later.

3.6 System Response to Internal Conditions: The Zero-Input Response 271

of the N solutions (see the footnote on page 153). Thus,

$$y_0[n] = c_1 \gamma_1^n + c_2 \gamma_2^n + \cdots + c_n \gamma_N^n$$

where γ1, γ2, … , γ*ⁿ* are the roots of Eq. (3.23) and c1, c2, … , cn are arbitrary constants determined from N auxiliary conditions, generally given in the form of initial conditions. The polynomial Q[γ ] is called the characteristic polynomial of the system, and Q[γ ] = 0 [Eq. (3.23)] is the characteristic equation of the system. Moreover, γ1, γ2, … , γN, the roots of the characteristic equation, are called characteristic roots or characteristic values (also eigenvalues) of the system. The exponentials γ ^{n i} (i = 1, 2,…,N) are the characteristic modes or natural modes of the system. A characteristic mode corresponds to each characteristic root of the system, and the zero-input response is a linear combination of the characteristic modes of the system.

EXAMPLE 3.13 Zero-Input Response of a Second-Order System with Real Roots

The LTID system described by the difference equation

$$y[n+2] - 0.6y[n+1] - 0.16y[n] = 5x[n+2]$$

has input x[n] = 4−nu[n] and initial conditions y[−1] = 0 and y[−2] = 25/4. Determine the zero-input response y0[n]. The zero-state response of this system is considered later, in Ex. 3.21.

The system equation in operator notation is

$$(E^2 - 0.6E - 0.16)y[n] = 5E^2x[n]$$

The characteristic polynomial is

$$\gamma^2 - 0.6\gamma - 0.16 = (\gamma + 0.2)(\gamma - 0.8)$$

The characteristic equation is

$$(\gamma + 0.2)(\gamma - 0.8) = 0$$

The characteristic roots are γ¹ = −0.2 and γ² = 0.8. The zero-input response is

$$y_0[n] = c_1(-0.2)^n + c_2(0.8)^n \tag{3.24}$$

To determine arbitrary constants c¹ and c2, we set n = −1 and −2 in Eq. (3.24), then substitute y0[−1] = 0 and y0[−2] = 25/4 to obtain†

\begin{array}{c}\n0 = -5c_1 + \frac{5}{4}c_2 \\ \frac{25}{4} = 25c_1 + \frac{25}{16}c_2\n\end{array}\n\right\} \quad \Longrightarrow \quad c_1 = \frac{1}{5}

\n

$$c_2 = \frac{4}{5}$$

^† The initial conditions ^y[−1] and ^y[−2] are the conditions given on the total response. But because the input does not start until n = 0, the zero-state response is zero for n < 0. Hence, at n = −1 and −2 the total response consists of the zero-input component only so that y[−1] = y0[−1] and y[−2] = y0[−2].

272 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS

Therefore,

$$y_0[n] = \frac{1}{5}(-0.2)^n + \frac{4}{5}(0.8)^n$$

$n \ge 0$

The reader can verify this solution by computing the first few terms using the iterative method (see Exs. 3.11 and 3.12).

DR ILL 3.11 Zero-Input Response of First-Order Systems

Find and sketch the zero-input response for the systems described by the following equations:

• (a) y[n+1] −0.8y[n] = 3x[n+1]
• (b) y[n+1] +0.8y[n] = 3x[n+1]

In each case the initial condition is y[−1] = 10. Verify the solutions by computing the first three terms using the iterative method.

ANSWERS

• (a) 8(0.8)ⁿ
• (b) −8(−0.8)ⁿ

DR ILL 3.12 Zero-Input Response of a Second-Order System with Real Roots

Find the zero-input response of a system described by the equation

$$y[n] + 0.3y[n-1] - 0.1y[n-2] = x[n] + 2x[n-1]$$

The initial conditions are y0[−1] = 1 and y0[−2] = 33. Verify the solution by computing the first three terms iteratively.

ANSWER

y0[n] = (0.2)ⁿ +2(−0.5)ⁿ

Section 3.5-1 introduced the method of recursion to solve difference equations. As the next example illustrates, the zero-input response can likewise be found through recursion. Since it does not provide a closed-form solution, recursion is generally not the preferred method of solving difference equations.

REPEATED ROOTS

So far we have assumed the system to have N distinct characteristic roots γ1, γ2, … , γ*^N* with corresponding characteristic modes γ ^{n 1} , γ ^{n 2} , … , γ ^{n N}. If two or more roots coincide (repeated roots), the form of characteristic modes is modified. Direct substitution shows that if a root γ repeats r times (root of multiplicity r), the corresponding characteristic modes for this root are γ ⁿ, nγ ⁿ, n²γ ⁿ, … , nr−¹γ ⁿ. Thus, if the characteristic equation of a system is

$$Q[\gamma]=(\gamma-\gamma_1)^r(\gamma-\gamma_{r+1})(\gamma-\gamma_{r+2})\cdots(\gamma-\gamma_N)$$

then the zero-input response of the system is

$$y_0[n] = (c_1 + c_2n + c_3n^2 + \dots + c_rn^{r-1})\gamma_1^n + c_{r+1}\gamma_{r+1}^n + c_{r+2}\gamma_{r+2}^n + \dots + c_n\gamma_N^n$$

EXAMPLE 3.15 Zero-Input Response of a Second-Order System with Repeated Roots

Consider a second-order difference equation with repeated roots:

$$(E2 + 6E + 9)y[n] = (2E2 + 6E)x[n]$$

Determine the zero-input response y0[n] if the initial conditions are y0[−1]=−1/3 and y0[−2]=−2/9.

The characteristic polynomial is γ ² +6γ +9 = (γ +3)2, and we have a repeated characteristic root at γ = −3. The characteristic modes are (−3)ⁿ and n(−3)ⁿ. Hence, the zero-input response is

y0[n] = (c¹ +c2n)(−3) n

Although we can determine the constants c¹ and c² from the initial conditions following a procedure similar to Ex. 3.13, we instead use MATLAB to perform the needed calculations.

c = inv([(-3)^(-1) -1*(-3)^(-1);(-3)^(-2) -2*(-3)^(-2)])*[-1/3;-2/9] c=4 3

Thus, the zero-input response is

$$y_0[n] = (4+3n)(-3)^n$$

$n \ge 0$

COMPLEX ROOTS

As in the case of continuous-time systems, the complex roots of a discrete-time system will occur in pairs of conjugates if the system equation coefficients are real. Complex roots can be treated exactly as we would treat real roots. However, just as in the case of continuous-time systems, we can also use the real form of solution as an alternative.

First we express the complex conjugate roots γ and γ ^∗ in polar form. If |γ | is the magnitude and β is the angle of γ , then

$$\gamma = |\gamma|e^{i\beta}$$

and $\gamma^* = |\gamma|e^{-j\beta}$

The zero-input response is given by

$$y_0[n] = c_1 \gamma^n + c_2(\gamma^*)^n = c_1 |\gamma|^n e^{i\beta n} + c_2 |\gamma|^n e^{-j\beta n}$$

For a real system, c¹ and c² must be conjugates so that y0[n] is a real function of n. Let

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

Then

$$y_0[n] = \frac{c}{2} |\gamma|^n \left[ e^{j(\beta n + \theta)} + e^{-j(\beta n + \theta)} \right] = c |\gamma|^n \cos(\beta n + \theta)$$

\n(3.25)

where c and θ are arbitrary constants determined from the auxiliary conditions. This is the solution in real form, which avoids dealing with complex numbers.

EXAMPLE 3.16 Zero-Input Response of a Second-Order System with Complex Roots

Consider a second-order difference equation with complex-conjugate roots:

$$(E2 - 1.56E + 0.81)y[n] = (E + 3)x[n]$$

Determine the zero-input response y0[n] if the initial conditions are y0[−1] = 2 and y0[−2] = 1.

The characteristic polynomial is (γ ² − 1.56γ + 0.81) = (γ − 0.78 − j0.45)(γ − 0.78 + j0.45). The characteristic roots are 0.78 ± j0.45; that is, 0.9e±j(π/6) . We could immediately write the solution as

$$y_0[n] = c(0.9)^n e^{j\pi n/6} + c^*(0.9)^n e^{-j\pi n/6}$$

Setting n = −1 and −2 and using the initial conditions y0[−1] = 2 and y0[−2] = 1, we find c = 1.1550−j0.2025 = 1.1726e−j0.1735 and c^∗ = 1.1550+j0.2025 = 1.1726ej0.1735.

gamma = roots([1 -1.56 0.81]); >> c = inv([gamma(1)^(-1) gamma(2)^(-1);gamma(1)^(-2) gamma(2)^(-2)])*[2;1] c = 1.1550 - 0.2025i 1.1550 + 0.2025i

Alternately, we could also find the unknown coefficient by using the real form of the solution, as given in Eq. (3.25). In the present case, the roots are 0.9e±j(π/6) . Hence, |γ | = 0.9 and β = π/6, and the zero-input response, according to Eq. (3.25), is given by

$$y_0[n] = c(0.9)^n \cos\left(\frac{\pi}{6}n + \theta\right)$$

To determine the constants c and θ, we set n = −1 and −2 in this equation and substitute the initial conditions y0[−1] = 2 and y0[−2] = 1 to obtain

$$2 = \frac{c}{0.9} \cos\left(-\frac{\pi}{6} + \theta\right) = \frac{c}{0.9} \left[\frac{\sqrt{3}}{2} \cos\theta + \frac{1}{2} \sin\theta\right]$$ $$1 = \frac{c}{(0.9)^2} \cos\left(-\frac{\pi}{3} + \theta\right) = \frac{c}{0.81} \left[\frac{1}{2} \cos\theta + \frac{\sqrt{3}}{2} \sin\theta\right]$$

or

$$\frac{\sqrt{3}}{1.8}c\cos\theta + \frac{1}{1.8}c\sin\theta = 2$$ $$\frac{1}{1.62}c\cos\theta + \frac{\sqrt{3}}{1.62}c\sin\theta = 1$$

These are two simultaneous equations in two unknowns ccos θ and csin θ. Solution of these equations yields

$$c \cos \theta = 2.308$$ $$c \sin \theta = -0.397$$

Dividing csin θ by ccos θ yields

$$\tan \theta = \frac{-0.397}{2.308} = \frac{-0.172}{1}$$ $$\theta = \tan^{-1}(-0.172) = -0.17 \text{ rad}$$

Substituting θ = −0.17 radian in ccos θ = 2.308 yields c = 2.34 and

$$y_0[n] = 2.34(0.9)^n \cos\left(\frac{\pi}{6}n - 0.17\right)$$

$n \ge 0$

Observe that here we have used radian units for both β and θ. We also could have used the degree unit, although this practice is not recommended. The important consideration is to be consistent and to use the same units for both β and θ.

DR ILL 3.13 Zero-Input Response of a Second-Order System with Complex Roots

Find the zero-input response of a system described by the equation

$$y[n] + 4y[n-2] = 2x[n]$$

The initial conditions are y0[−1]=−1/(2 ^√2) and *^y*0[−2] = ¹/(⁴ √ 2). Verify the solution by computing the first three terms iteratively.

ANSWER

^y0[n] = (2)ⁿ cos ^{π 2 n− 3π} 4

3.7 THE UNIT IMPULSE RESPONSE ^h[n]

Consider an nth-order system specified by the equation

$$(EN + a1EN-1 + \dots + aN-1E + aN)y[n] = (b0EN + b1EN-1 + \dots + bN-1E + bN)x[n]$$

or

$$Q[E]y[n] = P[E]x[n]$$

The unit impulse response h[n] is the solution of this equation for the input δ[n] with all the initial conditions zero; that is,

$$Q[E]h[n] = P[E]\delta[n] \tag{3.26}$$

subject to initial conditions

$$h[-1] = h[-2] = \cdots = h[-N] = 0$$

Equation (3.26) can be solved to determine h[n] iteratively or in a closed form. The following example demonstrates the iterative solution.