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Homework 3 Solution, ECE438, Fall 2014

Question 1

Let x(t) be a continuous-time signal and let y[n]=x(nT) be a sampling of that signal with period T>0. We would like to interpolate the samples (i.e., "connect the dots") in order to try to recover x(t).

a) Derive a formula for a band-limited interpolation of the samples (i.e., an expression for a continuous signal z(t) in terms of the samples y[n]). (Do not simply write down the formula; show how to derive it.)

Solution

To begin, we can write express the ideally sampled signal, x_s(t), as

$ \begin{align} x_s(t)&=\text{comb}_T \left \{ x(t) \right \}\\ &= x(t)\sum_{n=-\infty}^{\infty} \delta(t-kT) \end{align} $

We can then get the spectrum as

$ \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 \}\\ &=\frac{1}{T} \sum_{k=-\infty}^{\infty}X(f-k\frac{1}{T}) \end{align} $

The interpolation filter should filter out the copies of the original spectrum while preserving the one at baseband. Conceptually, the simplest filter is

$ H_r(f)=\begin{cases} T \mbox{, where } |f| \leq \frac{1}{2T} \\ 0 \mbox{, else}\end{cases} $

$ h_r(t) = \text{sinc}\left ( \frac{t}{T} \right ) $

Performing the convolution in the time domain

$ \begin{align} z(t) &= \text{comb}_{T_s} \{ x(t) \} \ast \text{sinc} \left ( \frac{t}{T} \right ) \\ &=\left ( \sum_{n=-\infty}^{\infty} x(nT) \delta (t-nT) \right )\ast \text{sinc} \left ( \frac{t}{T} \right ) \\ &=\sum_{n=-\infty}^{\infty} x(nT) \text{sinc} \left (\frac{t-nT }{T}\right) \\ \end{align} $

b) Show that your interpolation is equal to the original signal at all sample points.

Solution

Looking at the mth sample point

$ \begin{align} z(mT) &=\sum_{n=-\infty}^{\infty} x(nT) \text{sinc} \left (\frac{mT-nT }{T}\right) \\ &=\sum_{n=-\infty}^{\infty} x(nT) \text{sinc} \left (m-n \right) \mbox{, where } \text{sinc}(m-n) = \begin{cases} 1 \mbox{, where } m=n \\ 0 \mbox{, else}\end{cases} \\ &= x(mT) \end{align} $

c) Under what circumstances is your interpolation equal to the original signal x(t)? Explain.

Solution

The original signal, x(t), must be bandlimited, so that X(f) = 0 for all f>1/2T. Otherwise, there might be aliasing.

Question 2

Again, we consider a continuous-time signal x(t) and a sampling y[n]=x(nT) of that signal.

a) Write a formula for a piece-wise constant interpolation of the samples.

Solution

The basic reconstruction segment we will use for the piece-wise constant interpolation is

$ s_R(t) = \text{rect} \left ( \frac{t-\frac{T}{2}}{T} \right ) $

This segment will be weighted and shifted for each sample point. For example, the nth sample point would produce

$ \begin{align} s_{R,n} (t) &= x(nT) \text{rect}\left( \frac{t-\frac{T}{2} - nT}{T} \right)\\ &= x(nT) \left ( \text{rect} \left ( \frac{t-\frac{T}{2}}{T} \right ) \ast \delta(t-nT) \right ) \\ &= \text{rect} \left ( \frac{t-\frac{T}{2}}{T} \right ) \ast \left (x(nT) \delta(t-nT)\right) \end{align} $

Extending this for all n, we get

$ \begin{align} x_R(t) &= \sum_{n=-\infty}^{\infty} \text{rect} \left ( \frac{t-\frac{T}{2}}{T} \right ) \ast \left (x(nT) \delta(t-nT)\right) \\ &= \text{rect} \left ( \frac{t-\frac{T}{2}}{T} \right ) \ast \sum_{n=-\infty}^{\infty} \left (x(nT) \delta(t-nT)\right) \\ &= \text{rect} \left ( \frac{t-\frac{T}{2}}{T} \right ) \ast \text{comb}_T \left \{ x(t) \right \} \end{align} $

The last line helps with part b.

b) Derive the relationship between the Fourier transform of the interpolation you wrote in 2a) and the Fourier transform of x(t). (Do not simply write down the formula; show how to derive it.)

Solution

$ \begin{align} X_R(f) &= \text{CTFT} \left \{ \text{rect} \left ( \frac{t-\frac{T}{2}}{T} \right ) \right \} CTFT \left \{ \text{comb}_T \left \{ x(t) \right \} \right \} \\ &= e^{-j2\pi f \frac{T}{2}} T \text{sinc}(Tf) \frac{1}{T} rep_{\frac{1}{t}}\left \{ X(f) \right \} \\ & = e^{-j \pi f T} \text{sinc}(Tf) rep_{\frac{1}{t}}\left \{ X(f) \right \} \end{align} $

c) Is the interpolation you wrote in 2a) band-limited? Explain.

Solution

No, it is not band-limited. For one, the rep function replicates the spectrum every 1/T, and the sinc is not bandlimited.

Question 3

Let

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

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(af) \\ 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}{3 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 \} \\ &= 3f_0 \text{rep}_{3f_0} \left \{ X(f) \right \} \\ &= \frac{3}{2} \text{rep}_{\frac{3}{2}} \left \{ 14 \text{rect}( 2f )\right \} \\ &= 21\sum_{k=-\infty}^{\infty} \text{rect}\left ( 2f - \frac{3}{2}k\right ) \end{align} $

d) Let

$ T = \frac{1}{5 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_s(t) $ is

$ \mathcal{X}_d(\omega) = X_s\left (\frac{\omega F_s}{2\pi} \right ) $

Therefore

$ \begin{align} \mathcal{X}_d(\omega) &= 35\sum_{k=-\infty}^{\infty} \text{rect}\left ( \frac{5\omega}{2\pi} - \frac{5}{2}k\right ) \\ &= 35 \sum_{k=-\infty}^{\infty} \text{rect}\left ( \frac{5}{2} \frac{1}{2\pi} \left ( 2\omega- 2 \pi k \right )\right ) \\ &= 35 \text{rep}_{2\pi} \left \{ \text{rect} \left (\frac{5}{2\pi} \omega \right ) \right \} \end{align} $

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