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

Question 4

File:HW6Q4fig2.jpg

Question 5

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