Homework 3 Solution, ECE438, Fall 2015, Prof. Boutin

Question 1

Let

$ x(t)=7 \text{sinc } \left ( \frac{t-5}{2} \right). $

a) Obtain the Fourier transform X(f) of the signal and sketch the graph of |X(f)|.

Solution

To solve this, we will use the transform of a sinc, as well as the scaling and shifting properties:

$ \begin{align} \text{sinc}(t) &\leftrightarrow \text{rect}(f) \\ x(at) &\leftrightarrow \frac{1}{|a|} X\left (\frac{f}{a} \right ) \\ x(t-t_0) &\leftrightarrow e^{-j2\pi f t_0}X(f) \end{align} $

In our case, t_0=5 and a=2, so

$ \begin{align} X(f) &= CTFT \left \{7 \text{sinc}\left ( \frac{x-5}{2} \right ) \right \} \\ &= 7\cdot 2 e^{-j2\pi f 5} \text{rect}(2f) \\ &= 14 e^{-j10\pi f} \text{rect}(2f) \\ \end{align} $

b) What is the Nyquist rate $ f_0 $ for this signal?

Solution

The Nyquist rate is 1/2 Hz (twice the maximum non-zero frequency).

c) Let

$ T = \frac{1}{4 f_0}. $

Write a mathematical expression for the Fourier transform $ X_s(f) $ of

$ x_s(t)= \text{ comb}_T \left( x(t) \right). $

Sketch the graph of $ |X_s(f)| $.

Solution

$ \begin{align} X_s(f) &= \text{CTFT} \left \{ \text{comb}_T \left \{ x(t) \right \} \right \} \\ &= \frac{1}{T} \text{rep}_{\frac{1}{T}} \left \{ X(f) \right \} \\ &= 4f_0 \text{rep}_{4f_0} \left \{ X(f) \right \} \\ &= \frac{4}{2} \text{rep}_{\frac{4}{2}} \left \{ 14 e^{j10\pi f}\text{rect}( 2f )\right \} \\ &= 28\sum_{k=-\infty}^{\infty} e^{j10\pi (f-2k)} \text{rect}\left ( 2(f - 2k)\right ) \\ &= 28\sum_{k=-\infty}^{\infty} e^{j10\pi f} \text{rect}\left ( 2(f - 2k)\right ) \end{align} $

d) Let

$ T = \frac{2}{f_0}. $

Write a mathematical expression for the Fourier transform $ {\mathcal X}_d(\omega) $ of $ x_d[n]= x(nT) $ and sketch the graph of $ |{\mathcal X}_d(\omega)| $.

Solution

The relationship between the DTFT of $ x_d[n] $ and the CTFT of $ x(t) $ is

$ \mathcal{X}_d(\omega) = \text{rep}_{2\pi} \left ( X\left ( \frac{\omega f_s}{2\pi} \right ) \right ) $

Plugging in $ f_s = \frac{1}{T} = \frac{f_0}{2} = \frac{1}{4} $ gives

$ \begin{align} \mathcal{X}_d(\omega) &= \text{rep}_{2\pi} \left ( X\left ( \frac{\omega}{8\pi} \right ) \right ) \\ &= \text{rep}_{2\pi} \left (14e^{j5\omega/4} \text{rect} \left ( \frac{\omega}{4\pi} \right ) \right ) \end{align} $

Notice that the rect is $ 4\pi $ wide, and repeats every $ 2\pi $. This means that $ |\mathcal{X}(\omega)|=28\; \forall\; \omega $.