Question 1

Read Lathi section 5.3. Do exercise E5.9, page 371.

Solution.

Shift right by 2,

$$y[k]-\frac{5}{6}y[k-1]+\frac{1}{6}y[k-2]$$ = 5f[k-1]-f[k-2] (1)

Using unilateral z-transform,

$$\Longrightarrow$$ y[k] Y(z) (2)

$$\Longrightarrow \frac{1}{z}Y(z)+y[-1]$$ y[k-1] (3)

$$\Longrightarrow \frac{1}{z^2}Y(z)+\frac{1}{z}y[-1]+y[-2]$$ y[k-2] (4)

$$\Longrightarrow \frac{1}{z}F(z)+f[-1]$$ f[k-1] (5)

$$\Longrightarrow \frac{1}{z^2}F(z)+\frac{1}{z}f[-1]+f[-2]$$ f[k-2] (6)

z-transform of Eq.(1) is

$$Y(z)-\frac{5}{6}[\frac{1}{z}Y(z)+2]+\frac{1}{6}[\frac{1}{z^2}Y(z)+\frac{2}{z}] 5[\frac{1}{z}\frac{z}{z-1}] - [\frac{1}{z^2} \frac{z}{z-1}]$$ = (7)

$$(1-\frac{5}{6}\frac{1}{z}+\frac{1}{6}\frac{1}{z^2})Y(z) -\underbrace{(\frac{10}{6}-\frac{2}{6z})}_{initial-condition-terms} \underbrace{\frac{5}{z-1}-\frac{1}{z(z-1)}}_{input-terms}$$ = (8)

$$(z^2-\frac{5}{6}z+\frac{1}{6})Y(z) \underbrace{z(\frac{5}{3}z-\frac{1}{3})}_{initial-condition-terms}+\underbrace{\frac{z(5z-1)}{z-1}}_{input-terms}$$ = (9)

Now Y(z) is

$$\underbrace{\frac{2z(5z-1)}{(2z-1)(3z-1)}}_{initial-condition-terms} + \underbrace{\frac{6z(5z-1)}{(2z-1)(3z-1)(z-1)}}_{input-terms}$$ Y(z) = (10)

Break Y(z) into partial fractions

$$\frac{Y(z)}{z} \frac{2(5z-1)}{(2z-1)(3z-1)} + \frac{6(5z-1)}{(2z-1)(3z-1)(z-1)}$$ = (11)

$$\frac{A}{2z-1} + \frac{B}{3z-1} + \frac{C}{2z-1}+\frac{D}{z-1}+\frac{E}{3z-1}$$ = (12)

$$\begin{eqnarray*}A &=& 6\\ B &=& -4\\ C &=& -36\\ D &=& 12\\ E &=& 18 \end{eqnarray*}$$

$$\frac{Y(z)}{z} \frac{6}{2z-1} - \frac{4}{3z-1} - \frac{36}{2z-1}+\frac{12}{z-1}+\frac{18}{3z-1}$$ = (13)

$$\frac{3z}{z-\frac{1}{2}} - \frac{\frac{4}{3}z}{z-\frac{1}{3}} - \frac{18z}{z-\frac{1}{2}}+\frac{12z}{z-1}+\frac{6z}{z-\frac{1}{3}}$$ Y(z) = (14)

Taking inverse z-transform

$$\{ \underbrace{[3(\frac{1}{2})^k-\frac{4}{3}(\frac{1}{3})^k]}_{in... ...erms}+\underbrace{[12-18(\frac{1}{2})^k+6(\frac{1}{3})^k]}_{input-terms} \}u[k]$$ y[k] = (15)

$$[12-15(\frac{1}{2})^k+\frac{14}{3}(\frac{1}{3})^k]u[k]$$ = (16)