Revision as of 19:16, 11 October 2015

Homework 6 Solution, ECE438, Fall 2015, Prof. Boutin

Questions 1

Compute the DFT of the following signals x[n] (if possible). How does your answer relate to the Fourier series coefficients of x[n]?

a) $ x[n] = \left\{ \begin{array}{ll} 1, & n \text{ multiple of } N\\ 0, & \text{ else}. \end{array} \right. $

The period of the input is N, so we will calculate the N-point DFT:

$ \begin{align} X_n[k]&=\sum_{n=0}^{N-1} x[n] e^{-j2\pi kn /N} \\ &= 1e^{-j2\pi k 0 /N} + 0e^{-j2\pi k1 /N} + \ldots + 0e^{-j2\pi k(N-1) /N} \\ &= 1 \text{ for all } k \end{align} $

b) $ x[n]= e^{j \frac{2}{5} \pi n} $

Notice that the period is 5, so we will calculate the 5-point DFT. Beginning with the inverse-DFT:

$ \begin{align} x[n]&=\frac{1}{5} \sum_{k=0}^{4} X_5[k] e^{j2\pi kn/5} \\ &= \frac{1}{5} \left ( X_5[0]e^{j2\pi k0/5} + X_5[1]e^{j2\pi k1/5} + X_5[2]e^{j2\pi k2/5} + X_5[3]e^{j2\pi k3/5} + X_5[4]e^{j2\pi k4/5} \right ) \\ &= e^{j2\pi n/5} \end{align} $

From this we can see that

$ X_5[1]=5 \mbox{, and } X_5[0]=X_5[2]=X_5[3]=X_5[4]=0 $

or

$ X_5[k]=\begin{cases} 5\mbox{, }k=1\\ 0\mbox{, } k=0, 2, 3, 4 \end{cases} \mbox{ , periodic with } = 5 $

c) $ x[n]= e^{-j \frac{2}{5} \pi n} $

Notice that the period is 5, so we will calculate the 5-point DFT. Beginning with the inverse-DFT:

$ x[n]= e^{-j \frac{2}{5} \pi n} = e^{-j \frac{2}{5} \pi n} e^{2\pi n} = e^{j2\pi n \frac{4}{5}} $

$ \begin{align} x[n]&=\frac{1}{5} \sum_{k=0}^{4} X_5[k] e^{j2\pi kn 4/5} \\ &= \frac{1}{5} \left ( X_5[0]e^{-j2\pi k0/5} + X_5[1]e^{-j2\pi k1/5} + X_5[2]e^{-j2\pi k2/5} + X_5[3]e^{-j2\pi k3/5} + X_5[4]e^{-j2\pi k4/5} \right ) \\ &= e^{-j2\pi n/5} \end{align} $

From this we can see that

$ X_5[1]=5 \mbox{, and } X_5[0]=X_5[2]=X_5[3]=X_5[4]=0 $

or

$ X_5[k]=\begin{cases} 5\mbox{, }k=1\\ 0\mbox{, } k=0, 2, 3, 4 \end{cases} \mbox{ , periodic with } = 5 $

d) $ x[n]= e^{j \frac{2}{\sqrt{3}} \pi n} $

The period of the input is $ \sqrt{3} $. We cannot take a $ \sqrt{3} $-point DFT (only integer values).

e) $ x[n]= \cos\left( \frac{2}{1000} \pi n\right) ; $

First, use Euler's formula

$ x[n]=\frac{1}{2} e^{j2\pi n/1000} + \frac{1}{2}e^{-j2\pi n/1000} $

As in d), change the exponents so they are both positive:

$ \begin{align} x[n]&=\frac{1}{2} e^{j2\pi n/1000} + \frac{1}{2}e^{-j2\pi n/1000}e^{2\pi n} \\ &=\frac{1}{2} e^{j2\pi n/1000} + \frac{1}{2}e^{j2\pi 999n/1000} \end{align} $

Then looking at the 1000-point inverse DFT:

$ \begin{align} x[n] &= \frac{1}{1000} \sum_{k=0}^{999} X_{1000}[k]e^{j2\pi kn/1000} \\ &= \frac{1}{2} e^{j2\pi n/1000} + \frac{1}{2}e^{j2\pi 999n/1000} \end{align} $

Again, by matching terms we can see that

$ X_{1000}[k] = \begin{cases} 500&\mbox{, if } k=1 \mbox{ or }k=999 \\ 0 &\mbox{, if } k=0 \mbox { or } k=2,\ldots,998 \end{cases} \mbox{, periodic with period } 1000 $

f) $ x[n]= e^{j \frac{\pi}{3} n } \cos ( \frac{\pi}{6} n ) $

Using Euler's formula

$ \begin{align} x[n] &= e^{j\frac{\pi}{3}n} \left ( \frac{1}{2} e^{j\frac{\pi}{6}} + \frac{1}{2}e^{-j\frac{\pi}{6}}\right ) \\ &= \frac{1}{2}e^{\frac{\pi}{2}n} + \frac{1}{2}e^{j\frac{\pi}{6}n} \\ &= \frac{1}{2}e^{\frac{2\pi}{12}3n} + \frac{1}{2}e^{j\frac{2\pi}{12}n} \mbox{, this will make comparing with the IDFT easier} \end{align} $

The period for the signal is 12. Looking at the 12-point IDFT:

$ \begin{align} x[n]&=\frac{1}{12}\sum_{k=0}^{11} X_{12}[k] e^{j2\pi kn /12} \\ &= \frac{1}{2}e^{j2\pi 3n/12} + \frac{1}{2}e^{j2\pi n/12} \end{align} $

We can see that

$ X_{12}[k]=\begin{cases} 6 &\mbox{, if } k=1 \mbox{ or } k=3 \\ 0 &\mbox{, if} k=0 \mbox{, } k=2 \mbox{, or} k=4,\ldots,11 \end{cases}\mbox{, periodic with period } 12 $

g) $ x[n]= (-j)^n . $

First, rewrite the signal as

$ \begin{align} x[n] &= (-j)^n \\ &= e^{j3\pi n/2} \\ &= e^{j2\pi 3n /4} \mbox{, again, this makes comparison with the IDFT easier} \end{align} $

The period of the signal is 4. You can see this by observing that the sequence is {-j, -1, j, 1, -j, ...}. Or you can find it using the general form $ e^{j\omega_0n} $. Then you can solve for the period M by solving $ \omega_0M=2\pi m $ for some integer m. This signal, for example, would be

$ \frac{3\pi}{2} M = 2\pi n \Rightarrow M=\frac{4}{3}m \mbox{, where M and m are both integers} $

When m=3, M is the integer 4.

So, looking at the 4-point IDFT

$ \begin{align} x[n]&= \sum_{k=0}^{3} X_4[k] e^{j2\pi kn/4} \\ &= e^{j2\pi 3n/4} \end{align} $

From this, we can see that

$ X_4[k]=\begin{cases} 4 &\mbox{, if }k=3 \\ 0 &\mbox{, if } k=0,1,2 \end{cases} \mbox{, periodic with period } 4 $

h) $ x[n] =(\frac{1}{\sqrt{2}}+j \frac{1}{\sqrt{2}})^n $

First, rewrite the signal

$ \begin{align} x[n] &=(\frac{1}{\sqrt{2}}+j \frac{1}{\sqrt{2}})^n \\ &=(e^{j\pi/4})^n \\ &=e^{j2\pi n/8} \end{align} $

Now, looking at the 8-point IDFT

$ \begin{align} x[n] &= \sum_{k=0}^{7} X_8[k] e^{j2\pi kn /8} \\ &=e^{j2\pi n/8} \end{align} $

We can see that

$ X_8[k] = \begin{cases} 8 &\mbox{, if } k=1 \\ 0 &\mbox{, if} k=0 \mbox{ or } k=2,\ldots,7 \end{cases}\mbox{, periodic with period } 8 $

Question 2

Compute the inverse DFT of $ X[k]= e^{j \pi k }+e^{-j \frac{\pi}{2} k} $.

Rewrite so that the exponents are negative:

$ \begin{align} x[n] &= e^{j\pi k}e^{-j2\pi k} + e^{-j\pi k/2} \\ &= e^{-j\pi k} + e^{-j\pi k/2} \end{align} $

The period of the signal is 4. We can again rewrite the signal so that the period is in the denominator of the exponent (this makes the next steps easier):

$ x[n] = e^{-j2\pi 2 k/4} + e^{-j2\pi k/4} $

Looking at the 4-point DFT

$ \begin{align} X_4[k] &= \sum_{k=0}^{3} x[n] e^{-j2\pi kn /4} \\ &= e^{-j2\pi 2 k/4} + e^{-j2\pi k/4} \end{align} $

As in question 1, we can see that

$ \begin{align} x[n]&=\begin{cases} 1 &\mbox{, if }n=1 \mbox{ or } n=2 \\ 0 &\mbox{, if } n=0 \mbox{ or } n=3 \end{cases} \mbox{, periodic with period } 4 \\ &=\delta(n-1) + \delta(n-2) \mbox{, periodic with 4} \end{align} $

Question 3

Prove the time shifting property of the DFT.

Method 1

Given that

$ X_n[k] = \text{DFT} \left \{ x[n] \right \} $

Then the shifted form can be written as

$ x[n-n_0] = x[n]\ast \delta[n-n_0] $

Then it follows that

$ \begin{align} \text{DFT}\left \{ x[n-n_0] \right \} &= \text{DFT} \left \{ x[n] \right \} \text{DFT} \left \{\delta[n-n_0] \right \} \\ &= X_n[k] e^{-j2\pi kn_0/N} \end{align} $

Method 2

Or, you can use a change of variables. Let $ X_N[k] = \text{DFT} \left \{x[n] \right \} $, then

$ \begin{align} \text{DFT}\left \{ x[n-n_0] \right \} &= \sum_{k=0}^{N-1} x[n-n_0] e^{-j2\pi kn /N} \\ &= \sum_{n'=-n_0}^{N-n_0-1} x[n']e^{-j2\pi k(n' + n_0)/N} \mbox{ using the variable substitution } n'=n-n_0 \\ &=e^{-j2\pi k n_0/N} \sum_{n'=-n_0}^{N-n_0-1}x[n'] e^{-j2\pi k n'/N} \\ &=e^{-j2\pi k n_0/N} \sum_{n'=-0}^{N-1}x[n'] e^{-j2\pi k n'/N} \mbox{ (details below)}\\ &=e^{-j2\pi k n_0/N} X_N[k] \end{align} $

The change in the limits follows because $ x[n'] e^{j2\pi k n'/N} $ has a period of N. If we're summing over one full period, it doesn't matter where we start the summation.

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