Homework 2 Solution, ECE438, Fall 2015, Prof. Boutin

1) Pick a note frequency f0 = 392Hz

x(t) = 'cos'(2πf0t) = 'cos'(2π ⋅ 392t)
$ a.\ Assign\ sampling\ period\ T_1=\frac{1}{1000} $
$ 2f_0<\frac{1}{T_1}, \ No\ aliasing\ occurs. $

$ \begin{align} x_1(n) &=x(nT_1)=cos(2\pi \cdot 392nT_1)=cos(2\pi \cdot\frac{392}{1000}n) \\ &=\frac{1}{2}\left( e^{-j2\pi \cdot \frac{392}{1000}n} + e^{j2\pi \cdot\frac{392}{1000}n} \right) \\ \end{align} $

$ 0<2\pi \cdot\frac{392}{1000}<\pi $
$ -\pi<-2\pi \cdot\frac{392}{1000}<0 $

$ \begin{align} \mathcal{X}_1(\omega) &=2\pi \cdot\frac{1}{2} \left[\delta (\omega -2\pi \cdot\frac{392}{1000}) + \delta (\omega + 2\pi \cdot\frac{392}{1000})\right] \\ &=\pi \left[\delta (\omega -2\pi \cdot\frac{392}{1000}) + \delta (\omega + 2\pi \cdot\frac{392}{1000})\right] \\ \end{align} $

Xw1 singleperiod.jpg

$ for\ all\ \omega $
$ \mathcal{X}_1(\omega)=\pi\cdot rep_{2\pi} \left[\delta (\omega -2\pi \cdot\frac{392}{1000}) + \delta (\omega + 2\pi \cdot\frac{392}{1000})\right] $

Xw1 multiperiod.jpg

In this situation, no aliasing occurs. In the interval of [ − π,π], which represents one period, the frequcy spectrum remains the same as Fig a-1.
$ b.\ Assign\ sampling\ period\ T_2=\frac{1}{500} $
$ 2f_0>\frac{1}{T_2}, \ Aliasing\ occurs. $

$ \begin{align} x_2(n) &=x(nT_2)=cos(2\pi \cdot 392nT_2)=cos(2\pi \cdot\frac{392}{500}n) \\ &=\frac{1}{2}\left( e^{-j2\pi \cdot\frac{392}{500}n} + e^{j2\pi \cdot\frac{392}{500}n} \right) \\ \end{align} $

$ \pi<2\pi \cdot\frac{392}{500}<2\pi $
$ -2\pi<-2\pi \cdot\frac{392}{500}<\pi $
$ \mathcal{X}_2(\omega)=\pi \left[\delta (\omega -2\pi \cdot\frac{392}{500}) + \delta (\omega + 2\pi \cdot\frac{392}{500})\right] $
$ X_2(f)=\frac{1}{2}\left[\delta (f -\frac{392}{500}) + \delta (f + \frac{392}{500})\right] $

Xw2 singleperiod.jpg

$ for\ all\ \omega $
$ \mathcal{X}_2(\omega)=\pi\cdot rep_{2\pi} \left[\delta (\omega -2\pi \cdot\frac{392}{500}) + \delta (\omega + 2\pi \cdot\frac{392}{500})\right] $
$ X_2(f)=\frac{1}{2}rep_2\left[\delta (f -\frac{392}{500}) + \delta (f + \frac{392}{500})\right] $

Xw2 multiperiod.jpg

In this situation, aliasing DO occurs. In the interval of [ − π,π], which represents one period, the frequency spectrum is different from Fig b-1.

Xf2 multiperiod.jpg

2) Write MATLAB code to play the two DT signals from part a) for 2 seconds. Briefly comment on how each signal "sounds like".

f0 = 392;

n1 = 1:20000;
T1 = 1/1000;  % above Nyquist freq
x1 = cos(2*pi*f0*n1*T1);

n2 = 1:20000;
T2 = 1/500;  % below Nyquist freq
x2 = cos(2*pi*f0*n2*T2);

sound(x1)
pause(3)  % pause for 3 seconds
sound(x2)


x1 makes higher notes/sounds than x2.


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