Question 3.21: Convolution by Tables Using Table 3.1, find the (zero-state)......
Convolution by Tables
Using Table 3.1, find the (zero-state) response y[n] of an LTID system described by the equation
y[n+2]-0.6 y[n+1]-0.16 y[n]=5 x[n+2]
if the input x[n]=4^{-n} u[n].
| TABLE 3.1 Select Convolution Sums | |||
| No. | x_1[n] | x_2[n] | x_1[n] * x_2[n]=x_2[n] * x_1[n] |
| 1 | δ[n − k] | x[n] | x[n − k] |
| 2 | \gamma^n u[n] | u[n] | \left[\frac{1-\gamma^{n+1}}{1-\gamma}\right] u[n] |
| 3 | u[n] | u[n] | (n + 1)u[n] |
| 4 | \gamma_1^n u[n] | \gamma_2^n u[n] | \left[\frac{\gamma_1^{n+1}-\gamma_2^{n+1}}{\gamma_1-\gamma_2}\right] u[n] \quad \gamma_1 \neq \gamma_2 |
| 5 | u[n] | nu[n] | \frac{n(n+1)}{2} u[n] |
| 6 | \gamma^n u[n] | nu[n] | \left[\frac{\gamma\left(\gamma^n-1\right)+n(1-\gamma)}{(1-\gamma)^2}\right] u[n] |
| 7 | nu[n] | nu[n] | \frac{1}{6} n(n-1)(n+1) u[n] |
| 8 | \gamma^n u[n] | \gamma^n u[n] | (n+1) \gamma^n u[n] |
| 9 | n\gamma_1^n u[n] | \gamma_2^n u[n] | \frac{\gamma_1 \gamma_2}{\left(\gamma_1-\gamma_2\right)^2}\left[\gamma_2^n-\gamma_1^n+\frac{\gamma_1-\gamma_2}{\gamma_2} n \gamma_1^n\right] u[n] \quad \gamma_1 \neq \gamma_2 |
| 10 | \left|\gamma_1\right|^n \cos (\beta n+\theta) u[n] | \left|\gamma_2\right|^n u[n] | \frac{1}{R}\left[\left|\gamma_1\right|^{n+1} \cos [\beta(n+1)+\theta-\phi]-\left|\gamma_2\right|^{n+1} \cos (\theta-\phi)\right] u[n] |
| R=\left[\left|\gamma_1\right|^2+\left|\gamma_2\right|^2-2\left|\gamma_1\right|\left|\gamma_2\right| \cos \beta\right]^{1 / 2} | |||
| \phi=\tan ^{-1}\left[\frac{\left(\left|\gamma_1\right| \sin \beta\right)}{\left(\left|\gamma_1\right| \cos \beta-\left|\gamma_2\right|\right)}\right] | |||
| 11 | \gamma_1^n u[-(n+1)] | \gamma_2^n u[n] | \frac{\gamma_2}{\gamma_1-\gamma_2} \gamma_2^n u[n]+\frac{\gamma_1}{\gamma_1-\gamma_2} \gamma_1^n u[-(n+1)] \quad\left|\gamma_1\right|>\left|\gamma_2\right| |
The input can be expressed as x[n]=4^{-n} u[n]=(1 / 4)^n u[n]=(0.25)^n u[n] . The unit impulse response of this system, obtained in Ex. 3.18, is
h[n]=\left[(-0.2)^n+4(0.8)^n\right] u[n]
Therefore,
y[n]=x[n] * h[n]
=(0.25)^n u[n] *\left[(-0.2)^n u[n]+4(0.8)^n u[n]\right]
=(0.25)^n u[n] *(-0.2)^n u[n]+(0.25)^n u[n] * 4(0.8)^n u[n]
We use pair 4 (Table 3.1) to find the foregoing convolution sums.
y[n]=\left[\frac{(0.25)^{n+1}-(-0.2)^{n+1}}{0.25-(-0.2)}+4 \frac{(0.25)^{n+1}-(0.8)^{n+1}}{0.25-0.8}\right] u[n]
=\left(2.22\left[(0.25)^{n+1}-(-0.2)^{n+1}\right]-7.27\left[(0.25)^{n+1}-(0.8)^{n+1}\right]\right) u[n]
=\left[-5.05(0.25)^{n+1}-2.22(-0.2)^{n+1}+7.27(0.8)^{n+1}\right] u[n]
Recognizing that
\gamma^{n+1}=\gamma(\gamma)^n
we can express y[n] as
y[n]=\left[-1.26(0.25)^n+0.444(-0.2)^n+5.81(0.8)^n\right] u[n]
=\left[-1.26(4)^{-n}+0.444(-0.2)^n+5.81(0.8)^n\right] u[n]