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

This is the long way. Do not do this if you can help it!!! 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}

This is the short way: write your signal as a sum of complex exponentials, and then compare with IDFT formula to extract the DFT coefficients.

\begin{align} x[n] &=s_N[n] \text{ (Remember, that function we defined when looking at downsampling in the frequency domain?) } \\ &=\frac{1}{N} \sum_{k=0}^{N-1} e^{jk \frac{2\pi}{N} n} \text{ (Writing s_N[n] as its Fourier series.) \\ \end{align}

By comparison with the IDFT expression for x[n], namely

$x[n]=\frac{1}{N}\sum_{k=0}^{N-1} Xkn] e^{j \frac{2\pi}{N} kn }$

we find that $X[k]=1$, for k=1,…,N-1. Using the periodicity of X[k] (period N), we conclude that X[k]=1, for all k.

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_5e^{j2\pi n0/5} + X_5e^{j2\pi n1/5} + X_5e^{j2\pi n2/5} + X_5e^{j2\pi n3/5} + X_5e^{j2\pi n4/5} \right ) \\ &= e^{j2\pi n/5} \end{align}

From this we can see that

$X_5=5 \mbox{, and } X_5=X_5=X_5=X_5=0$

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/5} \\ &= \frac{1}{5} \left ( X_5e^{j2\pi n0/5} + X_5e^{j2\pi n1/5} + X_5e^{j2\pi n2/5 } + X_5e^{j2\pi n3/5} + X_5e^{j2\pi n4/5} \right ) \\ &= e^{j2\pi n \frac{4}{5}} \end{align}

From this we can see that

$X_5=5 \mbox{, and } X_5=X_5=X_5=X_5=0$

or

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

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

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

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

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

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

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

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

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 $X[k]= e^{j \pi k }+e^{-j \frac{\pi}{2} k}$.

Solution

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.

## Question 4

Under which circumstances can one recover the DTFT of a finite duration signal from the DFT of its periodic repetition? Justify your answer mathematically.

Solution

Suppose the length of finite duration signal $x[n]$ is $M$. The number of points of its DFT is $N$.

Using IDFT we have

$x'[n]=\frac{1}{N}\sum_{k=0}^{N-1}X(k)e^{\frac{j2\pi nk}{N}}$

Noticing that $x'[n]=x'[n+N]$, so the $x'[n]$ obtained by IDFT is periodic with $N$.

We can reconstruct DTFT by substituting $x[n]$ by $x'[n]$ as long as $x[n]=x'[n]$ for $n=0,1,...,M-1$. i.e.

\begin{align} X(e^{j\omega}) &= \sum_{n=0}^{M-1}x[n]e^{-j\omega n} \\ &= \sum_{n=0}^{M-1}x'[n]e^{-j\omega n} \\ &= \sum_{n=0}^{M-1}\frac{1}{N}\sum_{k=0}^{N-1}X(k)e^{\frac{j2\pi nk}{N}}e^{-j\omega n} \end{align}

Since $x'[n]$ is periodic with $N$, we must guarantee that $N\ge M$ in order to fully reconstruct DTFT using DFT.

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