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Sample Midterm Examination 2

ECE 438

Fall 2016

Instructor: Prof. Mireille Boutin

All questions below are derived from the homework of Prof. Mireille Boutin, all rights reserved by Prof. Mireille Boutin.

(15 pts) 1. List at least three properties of an LTI system.

Solution:

where $ {\mathcal H}(\omega) $ is DTFT of unit impulse response h[n]

(2)$ y[n] = x[n] * h[n] $

(3)$ {\mathcal y}(\omega) $=$ {\mathcal X}(\omega) $$ {\mathcal H}(\omega) $

(4)$ Y(z) = X(z)H(z) $

(30 pts) 2. For each ROAC, determine which of these system properties apply. (Just list the letters of the properties that apply.) Below we describe the ROAC of the transfer function of an LTI system.

a) the system is causal;

b) the system is BIBO stable;

c) the system has a well defined and finite frequency response function;

d) the system is an FIR filter;

e) The system is an IIR filter;

f) the unit impulse response of the system is right-sided;

g) the unit impulse response of the system is left-sided;

1.1 ROAC= all finite complex numbers, but not infinity.

1.2 ROAC= all complex numbers, including infinity.

1.3 ROAC= all complex numbers z with |z|>0.5, including infinity.

1.4 ROAC= all finite complex numbers z with |z|>3, but not infinity.

1.5 ROAC= all complex numbers z with |z|<0.5.

1.6 ROAC= all complex numbers z with 2<|z|<3.

Solution:

1.1 ROAC= all finite complex numbers, but not infinity. (b,c,d,f,g)

1.2 ROAC= all complex numbers, including infinity. (a,b,c,d,f,g)

1.3 ROAC= all complex numbers z with |z|>0.5, including infinity. (a,b,c,e,f)

1.4 ROAC= all finite complex numbers z with |z|>3, but not infinity. (e,f)

1.5 ROAC= all complex numbers z with |z|<0.5. (e,g)

1.6 ROAC= all complex numbers z with 2<|z|<3. (e)

Note:

1. The system is causal: |z| > r, including z = infinity.

2. The system is BIBO stable: including unit circle.

3. The system has a well defined and finite frequency response function: same as the system is BIBO stable (including unit circle).

4. The system is an FIR filter: H(z) has no poles in finite complex plane and h[n] is finite duration;

5. The system is an IIR filter: H(z) has poles in finite complex plane and h[n] is infinite duration;

6. The unit impulse response of the system is right-sided: |z| > r (out of unit circle);

7. The unit impulse response of the system is left-sided: |z| < r (in unit circle);

(20 pts) 3. Compute the z-transform of the signal

                  $ x[n]= 6^{n}u[n] + 8^{n}u[-n+1]  \  $

Solution:

$ X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n} = \sum_{n=-\infty}^{\infty} (6^{n}u[n]+ 8^{n}u[-n+1]) z^{-n} = \sum_{n=-\infty}^{\infty} 6^{n}u[n] z^{-n} + \sum_{m=-\infty}^{\infty} 8^{m}u[-m+1] z^{-m} $

$ X(z) = \sum_{n=-\infty}^{\infty} (\frac{6}{z})^n u[n] + \sum_{m=-\infty}^{\infty} (\frac{8}{z})^{m}u[-m+1] $

Let k = -m+1, then

$ X(z) = \sum_{n=-\infty}^{\infty} (\frac{6}{z})^n u[n] + \sum_{k=-\infty}^{\infty} (\frac{z}{8})^{k-1}u[k] = \sum_{n=0}^{\infty} (\frac{6}{z})^n + \frac{8}{z} \sum_{k=0}^{\infty} (\frac{z}{8})^{k} $

$ X(z) = \frac{1}{1-\frac{6}{z}} + \frac{8}{z} \frac{1}{1-\frac{z}{8}} = \frac{z}{z-6} + \frac{8}{z} \frac{8}{8-z} , if \quad 6 < |z| < 8 $

$ X(z) = \left\{ \begin{array}{l l} \frac{z}{z-6} + \frac{8}{z} \frac{8}{8-z} &, if \quad 6 < |z| < 8 \\ \text{diverges} &, \quad \text{otherwise} \end{array} \right. $

(20 pts) 4. Compute the inverse z-transform of

                  $ X(z)=\frac{1}{(1+ z)(6-z)}, \text{ ROC }  |z|>6 $

Solution:

$ X(z)=\frac{1}{(1+ z)(6-z)} =\frac{1}{4} (\frac{1}{1+z} + \frac{1}{6-z}) =\frac{1}{4} (\frac{1}{z} \frac{1}{1+\frac{1}{z}} - \frac{1}{z} \frac{1}{1-\frac{6}{z}}) =\frac{1}{4z} (\frac{1}{1-(-\frac{1}{z})} - \frac{1}{1-\frac{6}{z}}) $??????????????????????????????????????

So for $ |z| > 6 $, we have

$ X(z)=\frac{1}{4z} [\sum_{k=0}^{\infty} (-\frac{1}{z})^k - \sum_{k=0}^{\infty} (\frac{6}{z})^k] $

$ X(z)=\frac{1}{4} \sum_{k=-\infty}^{\infty} (-1)^k z^{-k-1} u[k] - \frac{1}{4} \sum_{k=-\infty}^{\infty} 6^k z^{-k-1} u[k] $

Let n = k+1, then

$ X(z)=\frac{1}{4} \sum_{k=-\infty}^{\infty} (-1)^{n-1} u[n-1] z^{-n} + \frac{1}{4} \sum_{k=-\infty}^{\infty} 6^{n-1} u[n-1] z^{-n} $

$ x[n]=[-\frac{1}{4} (-1)^n - \frac{1}{12} 6^n] u[n-1] $

(15 pts) 5.