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496 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM

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496 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM

From Table 5.1, pair 7, we obtain

x[n]=[3(2)n−1+5(3)n−1]u[n−1]x[n] = [3(2)^{n-1} + 5(3)^{n-1}]u[n-1]

\n(5.8)

If we expand rational X[z] into partial fractions directly, we shall always obtain an answer that is multiplied by u[n − 1] because of the nature of pair 7 in Table 5.1. This form is rather awkward as well as inconvenient. We prefer the form that contains u[n] rather than u[n−1]. A glance at Table 5.1 shows that the z-transform of every signal that is multiplied by u[n] has a factor z in the numerator. This observation suggests that we expand X[z] into modified partial fractions, where each term has a factor z in the numerator. This goal can be accomplished by expanding X[z]/z into partial fractions and then multiplying both sides by z. We shall demonstrate this procedure by reworking part (a). For this case,

X[z]z=8z−19z(z−2)(z−3)=(−19/6)z+(3/2)z−2+(5/3)z−3\frac{X[z]}{z} = \frac{8z - 19}{z(z - 2)(z - 3)} = \frac{(-19/6)}{z} + \frac{(3/2)}{z - 2} + \frac{(5/3)}{z - 3}

Multiplying both sides by z yields

X[z]=−196+32(zz−2)+53(zz−3)X[z] = -\frac{19}{6} + \frac{3}{2} \left( \frac{z}{z-2} \right) + \frac{5}{3} \left( \frac{z}{z-3} \right)

From pairs 1 and 6 in Table 5.1, it follows that

x[n]=−196δ[n]+[32(2)n+53(3)n]u[n](5.9)x[n] = -\frac{19}{6}\delta[n] + \left[\frac{3}{2}(2)^n + \frac{5}{3}(3)^n\right]u[n] \tag{5.9}

The reader can verify that this answer is equivalent to that in Eq. (5.8) by computing x[n] in both cases for n = 0, 1, 2, 3,…, and comparing the results. The form in Eq. (5.9) is more convenient than that in Eq. (5.8). For this reason, we shall always expand X[z]/z rather than X[z] into partial fractions and then multiply both sides by z to obtain modified partial fractions of X[z], which have a factor z in the numerator.

(b)

X[z]=z(2z2−11z+12)(z−1)(z−2)3X[z] = \frac{z(2z^2 - 11z + 12)}{(z - 1)(z - 2)^3}

and

X[z]z=2z2−11z+12(z−1)(z−2)3=kz−1+a0(z−2)3+a1(z−2)2+a2(z−2)\frac{X[z]}{z} = \frac{2z^2 - 11z + 12}{(z - 1)(z - 2)^3} = \frac{k}{z - 1} + \frac{a_0}{(z - 2)^3} + \frac{a_1}{(z - 2)^2} + \frac{a_2}{(z - 2)}

where

k=2z2−11z+12(z−1)(z−2)3∣z=1=−3k = \frac{2z^2 - 11z + 12}{(z - 1)(z - 2)^3}\Big|_{z=1} = -3 a0=2z2−11z+12(z−1)(z−2)3∣z=2=−2a_0 = \frac{2z^2 - 11z + 12}{(z - 1)(z - 2)^3}\Big|_{z=2} = -2

Therefore,

X[z]z=2z2−11z+12(z−1)(z−2)3=−3z−1−2(z−2)3+a1(z−2)2+a2(z−2)\frac{X[z]}{z} = \frac{2z^2 - 11z + 12}{(z - 1)(z - 2)^3} = \frac{-3}{z - 1} - \frac{2}{(z - 2)^3} + \frac{a_1}{(z - 2)^2} + \frac{a_2}{(z - 2)}

(5.10)

We can determine a1 and a2 by clearing fractions. Or we may use a shortcut. For example, to determine a2, we multiply both sides of Eq. (5.10) by z and let z → ∞. This yields

0=−3−0+0+a2  ⟹  a2=30 = -3 - 0 + 0 + a_2 \implies a_2 = 3

This result leaves only one unknown, a1, which is readily determined by letting z take any convenient value, say, z = 0, on both sides of Eq. (5.10). This produces

128=3+14+a14−32\frac{12}{8} = 3 + \frac{1}{4} + \frac{a_1}{4} - \frac{3}{2}

which yields a1 = −1. Therefore,

X[z]z=−3z−1−2(z−2)3−1(z−2)2+3z−2\frac{X[z]}{z} = \frac{-3}{z-1} - \frac{2}{(z-2)^3} - \frac{1}{(z-2)^2} + \frac{3}{z-2}

and

X[z]=−3zz−1−2z(z−2)3−z(z−2)2+3zz−2X[z] = -3\frac{z}{z-1} - 2\frac{z}{(z-2)^3} - \frac{z}{(z-2)^2} + 3\frac{z}{z-2}

Now the use of Table 5.1, pairs 6 and 10, yields

x[n]=[−3−2n(n−1)8(2)n−n2(2)n+3(2)n]u[n]x[n] = \left[ -3 - 2\frac{n(n-1)}{8} (2)^n - \frac{n}{2} (2)^n + 3(2)^n \right] u[n]

=

−[3+14(n2+n−12)2n]u[n]- \left[ 3 + \frac{1}{4} (n^2 + n - 12) 2^n \right] u[n]

(c) Complex Poles.

X[z]=2z(3z+17)(z−1)(z2−6z+25)=2z(3z+17)(z−1)(z−3−j4)(z−3+j4)X[z] = \frac{2z(3z+17)}{(z-1)(z^2-6z+25)} = \frac{2z(3z+17)}{(z-1)(z-3-j4)(z-3+j4)}

The poles of X[z] are 1, 3 + j4, and 3 − j4. Whenever there are complex-conjugate poles, the problem can be worked out in two ways. In the first method we expand X[z] into (modified) first-order partial fractions. In the second method, rather than obtain one factor corresponding to each complex-conjugate pole, we obtain quadratic factors corresponding to each pair of complex-conjugate poles. This procedure is explained next.

MENT-ORDER FACTORS

\n

X[z]z=2(3z+17)(z−1)(z2−6z+25)=2(3z+17)(z−1)(z−3−j4)(z−3+j4)\frac{X[z]}{z} = \frac{2(3z+17)}{(z-1)(z^2-6z+25)} = \frac{2(3z+17)}{(z-1)(z-3-j4)(z-3+j4)}

We find the partial fraction of X[z]/z using the Heaviside “cover-up” method:

X[z]z=2z−1+1.6e−j2.246z−3−j4+1.6ej2.246z−3+j4\frac{X[z]}{z} = \frac{2}{z-1} + \frac{1.6e^{-j2.246}}{z-3-j4} + \frac{1.6e^{j2.246}}{z-3+j4}

and

X[z]=2zz−1+(1.6e−j2.246)zz−3−j4+(1.6ej2.246)zz−3+j4X[z] = 2\frac{z}{z-1} + (1.6e^{-j2.246})\frac{z}{z-3-j4} + (1.6e^{j2.246})\frac{z}{z-3+j4}

The inverse transform of the first term on the right-hand side is 2u[n]. The inverse transform of the remaining two terms (complex conjugate poles) can be obtained from pair 12b (Table 5.1) by identifying r/2 = 1.6, θ = −2.246 rad, γ = 3 + j4 = 5ej0.927, so that |γ | = 5, β = 0.927. Therefore,

x[n]=[2+3.2(5)ncos⁡(0.927n−2.246)]u[n]x[n] = [2 + 3.2(5)n \cos(0.927n - 2.246)]u[n]

METHOD OF QUADRATIC FACTORS

X[z]z=2(3z+17)(z−1)(z2−6z+25)=2z−1+Az+Bz2−6z+25\frac{X[z]}{z} = \frac{2(3z+17)}{(z-1)(z^2-6z+25)} = \frac{2}{z-1} + \frac{Az+B}{z^2-6z+25}

Multiplying both sides by z and letting z → ∞, we find

0=2+A⟹A=−20 = 2 + A \Longrightarrow A = -2

and

2(3z+17)(z−1)(z2−6z+25)=2z−1+−2z+Bz2−6z+25\frac{2(3z+17)}{(z-1)(z^2-6z+25)} = \frac{2}{z-1} + \frac{-2z+B}{z^2-6z+25}

To find B, we let z take any convenient value, say, z = 0. This step yields

−3425=−2+B25⟹B=16\frac{-34}{25} = -2 + \frac{B}{25} \Longrightarrow B = 16

Therefore,

X[z]z=2z−1+−2z+16z2−6z+25\frac{X[z]}{z} = \frac{2}{z-1} + \frac{-2z+16}{z^2-6z+25}

and

X[z]=2zz−1+z(−2z+16)z2−6z+25X[z] = \frac{2z}{z-1} + \frac{z(-2z+16)}{z^2 - 6z + 25}

We now use pair 12c, where we identify A = −2, B = 16, |γ | = 5, and a = −3. Therefore,

r=100+256−19225−9=3.2r = \sqrt{\frac{100 + 256 - 192}{25 - 9}} = 3.2

, β=cos⁡−1(35)=0.927\beta = \cos^{-1}\left(\frac{3}{5}\right) = 0.927 rad

and

θ=tan⁡−1(−10−8)=−2.246 rad\theta = \tan^{-1}\left(\frac{-10}{-8}\right) = -2.246 \,\text{rad}

so that

x[n]=[2+3.2(5)ncos⁡(0.927n−2.246)]u[n]x[n] = [2 + 3.2(5)n \cos (0.927n - 2.246)]u[n]

DR ILL 5.2 Inverse z**-Transform by Partial Fraction Expansion**

Find the inverse z-transform of the following functions:

(a)

z(2z−1)(z−1)(z+0.5)\frac{z(2z-1)}{(z-1)(z+0.5)}

\n(b)

1(z−1)(z+0.5)\frac{1}{(z-1)(z+0.5)}

\n(c)

9(z+2)(z−0.5)2\frac{9}{(z+2)(z-0.5)^2}

\n(d)

5z(z−1)z2−1.6z+0.8\frac{5z(z-1)}{z^2-1.6z+0.8}

\n[Hint: 0.8=2/5\sqrt{0.8} = 2/\sqrt{5} .]
\nANSWERS
\n(a) [23+43(−0.5)n]u[n]\left[\frac{2}{3} + \frac{4}{3}(-0.5)^n\right]u[n]
\n(b) −2δ[n]+[23+43(−0.5)n]u[n]-2\delta[n] + \left[\frac{2}{3} + \frac{4}{3}(-0.5)^n\right]u[n]

  • (c) 18δ[n]−[0.72(−2)n +17.28(0.5)n −14.4n(0.5)n]u[n]
  • (d) 5 √5 2 √ 2 5 n cos(0.464n+0.464)u[n]

5.1-2 Inverse z**-Transform by Power Series Expansion**

By definition,

X[z]=∑n=0∞x[n]z−nX[z] = \sum_{n=0}^{\infty} x[n]z^{-n}

= x[0]+x[1]z+x[2]z2+x[3]z3+⋯x[0] + \frac{x[1]}{z} + \frac{x[2]}{z^2} + \frac{x[3]}{z^3} + \cdots
= x[0]z0+x[1]z−1+x[2]z−2+x[3]z−3+⋯x[0]z^0 + x[1]z^{-1} + x[2]z^{-2} + x[3]z^{-3} + \cdots

This result is a power series in z−1. Therefore, if we can expand X[z] into the power series in z−1, the coefficients of this power series can be identified as x[0], x[1], x[2], x[3], … A rational X[z] can be expanded into a power series of z−1 by dividing its numerator by the denominator. Consider, for example,

X[z]=z2(7z−2)(z−0.2)(z−0.5)(z−1)=7z3−2z2z3−1.7z2+0.8z−0.1X[z] = \frac{z^2(7z-2)}{(z-0.2)(z-0.5)(z-1)} = \frac{7z^3 - 2z^2}{z^3 - 1.7z^2 + 0.8z - 0.1}

500 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM

To obtain a series expansion in powers of z−1, we divide the numerator by the denominator as follows:

z3−1.7z2+0.8z−0.1)7z3−2z2‾z^{3}-1.7z^{2}+0.8z-0.1\overline{)7z^{3}-2z^{2}} 7z3−11.9z2+5.60z−0.7‾\underline{7z^{3}-11.9z^{2}+5.60z-0.7} 7z3−11.9z2+5.60z−0.7‾\underline{7z^{3}-11.9z^{2}+5.60z-0.7} 9.9z2−5.60z+0.7‾\underline{9.9z^{2}-5.60z+0.7} 9.9z2−16.83z+7.92−0.99z−1‾\underline{9.9z^{2}-16.83z+7.92-0.99z^{-1}} 11.23z−7.22+0.99z−1‾\underline{11.23z-7.22+0.99z^{-1}} 11.23z−19.09+8.98z−1‾\underline{11.23z-19.09+8.98z^{-1}} 11.87−7.99z−1‾\underline{11.87-7.99z^{-1}}

Thus,

X[z]=z2(7z−2)(z−0.2)(z−0.5)(z−1)=7+9.9z−1+11.23z−2+11.87z−3+⋯X[z] = \frac{z^2(7z-2)}{(z-0.2)(z-0.5)(z-1)} = 7 + 9.9z^{-1} + 11.23z^{-2} + 11.87z^{-3} + \cdots

Therefore,

x[0]=7,x[1]=9.9,x[2]=11.23,x[3]=11.87,…x[0] = 7, x[1] = 9.9, x[2] = 11.23, x[3] = 11.87, \dots

Although this procedure yields x[n] directly, it does not provide a closed-form solution. For this reason, it is not very useful unless we want to know only the first few terms of the sequence x[n].

DR ILL 5.3 Inverse z**-Transform by Long Division**

Using long division to find the power series in z−1, show that the inverse z-transform of z/(z−0.5) is (0.5)nu[n] or (2)−nu[n].

RELATIONSHIP BETWEEN h[n] AND H[z]

For an LTID system, if h[n] is its unit impulse response, then from Eq. (3.39), where we defined H[z], the system transfer function, we write

H[z]=∑n=−∞∞h[n]z−nH[z] = \sum_{n=-\infty}^{\infty} h[n]z^{-n}

(5.11)

For causal systems, the limits on the sum are from n = 0 to ∞. This equation shows that the transfer function H[z] is the z-transform of the impulse response h[n] of an LTID system; that is,

h[n]⟺H[z]h[n] \Longleftrightarrow H[z]

This important result relates the time-domain specification h[n] of a system to H[z], the frequency-domain specification of a system. The result is parallel to that for LTIC systems.

DR ILL 5.4 Impulse Response by Inverse z**-Transform**

Redo Drill 3.14 by taking the inverse z-transform of H[z], as given by Eq. (3.41).