Revision as of 10:36, 29 October 2011

Homework 6, ECE438, Fall 2011, Prof. Boutin

Question 1

a)

$ 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[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) \\ &= e^{-j\frac{\omega }{2}} cos \left( \frac{\omega }{2} \right) \\ \end{align} $

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} $

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] = e^jωn
(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(e^jω) = 1 + 2 e ^{− jω} + e ^{− 2jω}

(i) and (ii) are the same.

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

File:HW6Q4fig2.jpg

Question 5

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