# Homework 6, ECE438, Fall 2011, Prof. Boutin

## Question 1

a)

$y[n]= \frac{x[n]+x[n-2]}{2}$

Applying Z-transform on both sides and grouping terms, we can obtain the transfer function

\begin{align} Y[z]&= \frac{X[z]+X[z]z^{-2}}{2} \\ \frac{Y[z]}{X[z]}&= \frac{1+z^{-2}}{2} \\ H[z] &= \frac{1+z^{-2}}{2} \\ \end{align}

Frequency Response $H(\omega)$

\begin{align} H[e^{j\omega }] &= \frac{1+e^{-2j\omega }}{2} \\ &= e^{-j\omega } \left( \frac{e^{j\omega}+e^{-j\omega}}{2} \right) \\ &= e^{-j\omega } cos \left( \omega \right) \\ \end{align}

Zero-pole plot

Frequency response

b)

$y[n]= \frac{x[n]-x[n-1]}{2}$

Applying Z-transform on both sides and grouping terms, we can obtain the transfer function

\begin{align} Y[z]&= \frac{X[z]-X[z]z^{-1}}{2} \\ \frac{Y_2[z]}{X[z]}&= \frac{1-z^{-1}}{2} \\ H[z] &= \frac{1-z^{-1}}{2} \\ \end{align}

Frequency Response $H(\omega)$

\begin{align} H[e^{j\omega }] &= \frac{1-e^{-j\omega }}{2} \\ &= e^{-j\frac{\omega }{2}} \left( \frac{e^{j\frac{\omega }{2}}-e^{-j\frac{\omega }{2}}}{2} \right) \\ &= je^{-j\frac{\omega }{2}} \left( \frac{e^{j\frac{\omega }{2}}-e^{-j\frac{\omega }{2}}}{2j} \right) \\ &= je^{-j\frac{\omega }{2}} sin \left( \frac{\omega }{2} \right) \\ \end{align}

Zero-pole plot

Frequency response

## Question 2

a. Substituting values directly would yield the following -

\begin{align} n &= ...\text{ -3, -2, -1, 0, 1, 2, 3, 4, 5} ...\\ y\left[n\right] &= ...\text{ 0, -2, -4, -1, 2, -1, -4, -2, 0} ...\\ \end{align}

b. h[n] = ?
Substitute x[n] = δ[n] to obtain y[n] = h[n],
h[n] = δ[n] + 2δ[n − 1] + δ[n − 2]

Now x[n] = -2δ[n + 2] + δ[n] - 2δ[n − 2]

\begin{align} y[n] &= x[n] * h[n] \\ &= (-2\delta[n+2] + \delta[n] - 2\delta[n-2]) * (\delta[n] + 2\delta[n-1] + \delta[n-2]) \\ &= -2\delta[n+2] - 4\delta[n+1] -2\delta[n] + \delta[n] + 2\delta[n-1] + \delta[n-2] - 2\delta[n-2] - 4\delta[n-3] - 2\delta[n-4] \\ &= -2\delta[n+2] - 4\delta[n+1] - \delta[n] + 2\delta[n-1] - \delta[n-2] - 4\delta[n-3] - 2\delta[n-4] \\ \end{align}

c. x[n] = ejωn
(i)
\begin{align} y[n] &= e^{j\omega n} + 2 e^{j\omega (n-1)} + e^{j\omega (n-2)} \\ &= e^{j\omega n}(1 + 2 e^{-j\omega } + e^{-2j\omega }) \\ H(e^{j\omega}) &= 1 + 2 e^{-j\omega } + e^{-2j\omega } \\ \end{align}

(ii)
h[n] = δ[n] + 2δ[n − 1] + δ[n − 2]
H(ejω) = 1 + 2 ejω + e − 2jω

(i) and (ii) are the same.

d. x[n] = -2δ[n + 2] + δ[n] - 2δ[n − 2]
X(ejω) = -2 e2jω + 1 - 2e − 2jω

\begin{align} Y(e^{j\omega}) &= X(e^{j\omega})H(e^{j\omega}) \\ &= (-2e^{2j\omega } + 1 - 2e^{-2j\omega } ).(1 + 2e^{-j\omega } + e^{-2j\omega }) \\ &= -2e^{2j\omega } - 4e^{j\omega } - 2 + 1 + 2e^{-j\omega } + 2e^{-2j\omega } - 2e^{-2j\omega } - 4e^{-3j\omega } - 2e^{-4j\omega } \\ &= -2e^{2j\omega } - 4e^{j\omega } - 1 - 2e^{-j\omega } - e^{-2j\omega } - 4e^{-3j\omega } - 2e^{-4j\omega } \\ \\ \text{Using Inverse DTFT,} \\ y[n] &= -2\delta[n+2] - 4\delta[n+1] - \delta[n] + 2\delta[n-1] - \delta[n-2] - 4\delta[n-3] - 2\delta[n-4] \\ \end{align}

All 3 approaches lead to the same y[n].

## Question 3

a. Sketch the locations of the poles and zeros.

\begin{align} H(z) &= \frac{1-\frac{1}{2}z^{-2}}{1-\frac{1}{\sqrt{2}} z^{-1} +\frac{1}{4} z^{-2}} \\ H(z) &= \frac{(z+\frac{1}{\sqrt{2}})(z-\frac{1}{\sqrt{2}})} { (z-(\frac{1}{2\sqrt{2}} + j\frac{1}{2\sqrt{2}}))(z-(\frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}})) } \\ \end{align}

Zeros:
$z_1 = \frac{1}{\sqrt{2}}, z_2 = -\frac{1}{\sqrt{2}}$
Poles:
$p_1 = \frac{1}{2\sqrt{2}} + j\frac{1}{2\sqrt{2}}, p_2 = \frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}}$

b. Determine the magnitude and phase of the frequency response H(ω), for

$\omega = 0$
$\left| H(e^{j\omega}) \right| = \left| H(e^{j0}) \right| = \left| H(z=1) \right|$
$= \left| \frac{(1+\frac{1}{\sqrt{2}})(1-\frac{1}{\sqrt{2}})} { (1-(\frac{1}{2\sqrt{2}} + j\frac{1}{2\sqrt{2}}))(1-(\frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}})) } \right| = 0.921$
$\angle H(e^{j0}) = \angle c + \angle d - \angle a - \angle b = 0$

$\omega =\frac{\pi}{4}$
$\left| H(e^{j\omega}) \right| = \left| H(e^{j\frac{\pi}{4}}) \right| = \left| H(z=\frac{1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}) \right|$
$= \left| \frac{(\frac{1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})} { (\frac{1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}-(\frac{1}{2\sqrt{2}} + j\frac{1}{2\sqrt{2}}))(\frac{1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}-(\frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}})) } \right| = 2$
$\angle H(e^{j\frac{\pi}{4}}) = \angle c + \angle d - \angle a - \angle b = \frac{\pi}{2} + arctan^{-1} \left( \frac{\frac{1}{\sqrt{2}}}{\sqrt{2}} \right) - \frac{\pi}{4} - arctan^{-1} \left( \frac{\frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}}}{1-\sqrt{2}-\frac{1}{2\sqrt{2}}} \right) = 0$

$\omega =\frac{\pi}{2}$
$\left| H(e^{j\omega}) \right| = \left| H(e^{j\frac{\pi}{2}}) \right| = \left| H(z=j) \right|$

$= \left| \frac{(j+\frac{1}{\sqrt{2}})(j-\frac{1}{\sqrt{2}})} { (j-(\frac{1}{2\sqrt{2}} + j\frac{1}{2\sqrt{2}}))(j-(\frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}})) } \right| = 1.455$

$\angle H(e^{j\frac{\pi}{2}}) = (\angle c + \angle d) - \angle a - \angle b = (\pi) - (arctan^{-1}\left( \frac{1-\frac{1}{2\sqrt{2}}}{\frac{-1}{\sqrt{2}}} \right) + \pi) - (arctan^{-1}\left( \frac{1+\frac{1}{2\sqrt{2}}}{1-\frac{1}{\sqrt{2}}} \right) + \pi) = -0.7563$

$\omega =\frac{3\pi}{4}$
$\left| H(e^{j\omega}) \right| = \left| H(e^{j\frac{3\pi}{4}}) \right| = \left| H(z=\frac{-1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}) \right|$

$= \left| \frac{(\frac{-1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})(\frac{-1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})} { (\frac{-1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}-(\frac{1}{2\sqrt{2}} + j\frac{1}{2\sqrt{2}}))(\frac{-1}{\sqrt{2}} + j\frac{1}{\sqrt{2}}-(\frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}})) } \right| = \frac{2}{3}$

$\angle H(e^{j\frac{3\pi}{4}}) = \angle c + \angle d - \angle a - \angle b = (arctan^{-1}\left( \frac{\frac{1}{\sqrt{2}}}{\frac{-2}{\sqrt{2}}} \right) + \pi) + \frac{\pi}{2} - (arctan^{-1}\left( \frac{\frac{1}{\sqrt{2}} - \frac{1}{2\sqrt{2}}}{\frac{1}{\sqrt{2}} - \frac{1}{2\sqrt{2}}} \right) + \pi) + \frac{3\pi}{4} = -0.9273$

$\omega =\pi$
$\left| H(e^{j\omega}) \right| = \left| H(e^{j\pi}) \right| = \left| H(z=-1) \right|$
$= \left| \frac{(-1+\frac{1}{\sqrt{2}})(-1-\frac{1}{\sqrt{2}})} { (-1-(\frac{1}{2\sqrt{2}} + j\frac{1}{2\sqrt{2}}))(-1-(\frac{1}{2\sqrt{2}} - j\frac{1}{2\sqrt{2}})) } \right| = 0.255$

$\angle H(e^{j\pi}) = (\angle c + \angle d) - \angle a - \angle b = 2\pi - 0 - 0 = 2\pi$

c. Is the system stable? Explain why or why not?
The system is causal and the ROC extends outwards from the outermost pole since |$p_1$| = |$p_2$| < 1 and this ROC contains the unit circle. Therefore the system is stable.

d. Find the difference equation for y[n] in terms of x[n], corresponding to this transfer function H(z).

$H(z) = \frac{Y(z)}{X(z)} = \frac{1-\frac{1}{2}z^{-2}}{1-\frac{1}{\sqrt{2}} z^{-1} +\frac{1}{4} z^{-2}}$
$Y(z)(1-\frac{1}{\sqrt{2}} z^{-1} +\frac{1}{4} z^{-2}) = X(z)(1-\frac{1}{2}z^{-2})$

Taking inverse,
$y[n]-\frac{1}{\sqrt{2}}y[n-1] + \frac{1}{4}y[n-2] = x[n] - \frac{1}{2}x[n-2]$
$y[n] = x[n] - \frac{1}{2}x[n-2] +\frac{1}{\sqrt{2}}y[n-1] - \frac{1}{4}y[n-2]$

## Question 4

$y[n]=\frac{1}{8} \left( x[n]+x[n-1]+x[n-2]+x[n-3]+x[n-4]+x[n-5]+x[n-6]+x[n-7]\right)$

a.
$h[n]=\frac{1}{8} \left( \delta[n]+\delta[n-1]+\delta[n-2]+\delta[n-3]+\delta[n-4]+\delta[n-5]+\delta[n-6]+\delta[n-7] \right)$
This is a finite duration response.

b. $H[z]=\frac{1}{8} \left( 1+z^{-1}+z^{-2}+z^{-3}+z^{-4}+z^{-5}+z^{-6}+z^{-7} \right)$
$H[z]=\frac{1}{8} \left( \frac{1-z^{-8}}{1-z^{-1}} \right)$

c.

$H[z]=\frac{1}{8} \left( \frac{z^{8}-1}{z^{7}(z-1)} \right)$

Poles:
$p_1 = 1$
$p_2 = p_3 = ... = p_8 = 0$

Zeros:
$z^{8} - 1 = 0$
$z^{8} = e^{j2\pi }$

Generalizing,
$z_k = e^{j2\pi k/8 }$ for k = 0,1,2,...,7

## Question 5

a.
$y[n]= \frac{1}{8} \left( x[n]-x[n-8]+y[n-1] \right)$

Using z-transform,
$Y(z) = \frac{1}{8} \left( X(z)-X(z)z^{-8} + Y(z)z^{-1} \right)$
$Y(z) (1 - \frac{1}{8}z^{-1}) = X(z) \frac{1}{8}(1 - z^{-8})$
$H(z) = \frac{Y(z)}{X(z)} = \frac{1}{8} \left( \frac{ 1 - z^{-8} } {1 - \frac{1}{8}z^{-1}} \right)$
b.

c. H(z) can be re-written as
$H[z]=\frac{1}{8} \left( \frac{1}{1-\frac{1}{8}z^{-1}} - \frac{z^{-8}}{1-\frac{1}{8}z^{-1}} \right)$
Assuming the ROC: $|z|>\frac{1}{8}$Taking inverse Z-transform of H(z) -
$h[n]=\frac{1}{8} \left( (\frac{1}{8})^n u[n] - (\frac{1}{8})^{n-8} u[n-8] \right)$

This is a infinite duration response.

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