Pastpaper Solution 201819 Semester 2

Q1(a)

3-point DFT can be obtained by taking 3 samples @ $\omega=2n\pi/3$ for n=0,1,2

$Y[k] = \{3,2e^{1.5j}, 2e^{-1.5j}\}$

Note :

Y[0] phase is 0.

Y[2] = Y[-1] because conjugate.

Q1(b)(i)

Q1(b)(ii)

The average value of x[n] is 15 as the dc coefficient of the spectrum is 15.

Q1(b)(iii)

Sequence x[n] is not a constant sequence as it has non-zero ac coefficients.

Q2 (a)

You can also use Z-Transform approach for the question.

Q2 (b)

The system is not a FIR filter because it is not indepndent to previous output.

Q2 ©

System is not a memoryless system because it depends on previous input x[n-2].

Q3

Note, the h[n] is the system.

Q4 (a)(b)©

Q4 (d)

First find $x[n] = x(nT_s)$ . Make it in $\pm \pi$ . <- This will determine the change of reconstructed frequency

Then find the frequencies of the reconstructed frequency components.

Finally plug the frequencies into signal to get reconstructed signal.

$x[n] = x(nT_s) = x(\frac{n}{500})$

$x[n] = 4 + 5cos(\frac{150}{500}2\pi n - \frac{\pi}{2}) + 2cos(\frac{400}{500}2\pi n) = 4 + 5cos(\frac{3}{10}2\pi n) + 2cos(\frac{4}{5}2\pi n - \frac{\pi}{2})$

$x[n] = 4+5cos(\frac{3}{10}2\pi n) + 2cos(2n\pi - (\frac{8\pi n}{5} - \frac{\pi}{2})) = 4+5cos(\frac{3}{10}2\pi n) + 2cos(\frac{10\pi n}{5} - \frac{8\pi n}{5} + \frac{\pi}{2})$

$x[n] = 4+5cos(\frac{3}{10}2\pi n) + 2cos(\frac{2\pi n}{5} + \frac{\pi}{2}) = 4+5cos(\frac{3}{10}2\pi n) + 2cos(\frac{1}{5}2\pi n + \frac{\pi}{2})$

Note that, when reconstructing a sinusoidal signal $g(t)$ from its sampled version $g[n]$ , unless extra information is given, we always assume that

$g[n] = cos(2n\pi (\frac{f_o}{f_s})+\phi)$

Where $f_o$ is the lowest frequency of all possible frequencies that can producing $g[n]$ .

Therefore-

For $5cos(\frac{3}{10}2\pi n)$ :

$\frac{f_o}{500} = \frac{3}{10}$

$f_o = 150$ Hz

For $2cos(\frac{4}{5}2\pi n - \frac{\pi}{2})$

$\frac{f_o}{500} = \frac{1}{5}$

$f_o = 100$ Hz

Therefore the reconstructed signal = $4 + 5cos(300\pi t) + 2cos(200\pi t + \frac{\pi}{2})$

Q5 (a)

Q5 (b)

Q6 (a)(b)

Note, Sometimes the $H(\omega)$ is not simplified yet. We might need to simplify it.

Q7 (a)

$H = H_2 - H_1 - H_3H_2$

Note: $H_3H_2$ means the frequency response is between 300 and 500.

Q7 (b)(i)(ii)

From the construction diagram, this is clearly a FIR filter.

Q8 (a)

The frequencies of the rejected components are $cos(200\pi t - \frac{\pi}{2})$ and $cos(800\pi t)$ .

After sampling, they are in the form of $cos(\frac{200}{1000}\pi n - \frac{\pi}{2}) = cos(\frac{1}{5}\pi n - \frac{\pi}{2})$ and $cos(\frac{800}{1000}\pi n) = cos(\frac{4}{5}\pi n)$ respectively.

To reject $cos(\frac{1}{5}\pi n - \frac{\pi}{2})$ , we need a FIR having zeros @ $e^{-j\frac{\pi}{5}}$ and $e^{j\frac{\pi}{5}}$

$H_1(z) = (1-e^{j\frac{\pi}{5}}z^{-1})(1-e^{-j\frac{\pi}{5}}z^{-1})$

To reject $cos(\frac{4}{5}\pi n)$ , we need a FIR having zeros @ $e^{-j\frac{4\pi}{5}}$ and $e^{j\frac{4\pi}{5}}$

$H_2(z) = (1-e^{j\frac{4\pi}{5}}z^{-1})(1-e^{-j\frac{4\pi}{5}}z^{-1})$

We need 2 filters $H_1(z)$ and $H_2(z)$ . Then we cascade them.

$H(z) = H_1(z)H_2(z) = (1-e^{j\frac{\pi}{5}}z^{-1})(1-e^{-j\frac{\pi}{5}}z^{-1})(1-e^{j\frac{4\pi}{5}}z^{-1})(1-e^{-j\frac{4\pi}{5}}z^{-1})$

$H(z) = (1-2cos(\pi/5)z^{-1} + z^{-2})(1- 2cos(4\pi/5)z^{-1} + z^{-2})$

Q8 (b)

Q8 ©

$H(\omega) = H(z)|_{z=e^{j\omega}}$

$= (1-2cos(\pi/5)e^{-j\omega} + e^{-j2\omega})(1- 2cos(4\pi/5)e^{-j\omega} + e^{-j2\omega})$

=>

$H(0) = (1-2cos(\pi/5)+1)(1-2cos(4\pi/5)+1) = 1.38$

$H(\pi) = (1-2cos(\pi/5)e^{-j\pi} + e^{-j2\pi})(1- 2cos(4\pi/5)e^{-j\pi} + e^{-j2\pi})$

$H(\pi) = (1-2cos(\pi/5)(-1) + 1)(1- 2cos(4\pi/5)(-1) + 1) = 1.38$

It is a bandpass filter.