Revision as of 14:58, 7 October 2014

Homework 5 Solution, ECE438, Fall 2014

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_1[n] = \left\{ \begin{array}{ll} 1, & n \text{ multiple of } N\\ 0, & \text{ else}. \end{array} \right. $

Solution

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_1[n]= e^{j \frac{2}{3} \pi n} $

Solution

Notice that the period is 3, so we will calculate the 3-point DFT. Beginning with the inverse-DFT:

$ \begin{align} x[n]&=\frac{1}{3} \sum_{k=0}^{2} X_3[k] e^{j2\pi kn/3} \\ &= \frac{1}{3} \left ( X_3[0]e^{j2\pi k0/3} + X_3[1]e^{j2\pi k1/3} + X_3[2]e^{j2\pi k2/3} \right ) \\ &= e^{j2\pi n/3} \end{align} $

From this we can see that

$ X_3[1]=3 \mbox{, and } X_3[0]=X_3[2]=0 $

or

$ X_3[k]=\begin{cases} 3\mbox{, }k=1\\ 0\mbox{, else} \end{cases} \mbox{ , periodic with period} = 3 $

c) $ x_5[n]= e^{-j \frac{2}{1000} \pi n} $

Solution

The period of this signal is 1000. To make life easier, we will multiple by a factor (noting that the factor is always 1, so it doesn't change the signal):

$ \begin{align} x_5[n]&=e^{-j \frac{2}{1000} \pi n}e^{j2\pi n} \\ &= e^{j2\pi \frac{1000-1}{1000}} \\ &=e^{j2\pi \frac{998}{1000}} \end{align} $

The positive exponent is easier to deal with.

Now we can use the inverse transform as before, using a 1000-point IDFT:

$ \begin{align} x_5[n] &= \frac{1}{1000} \sum_{k=0}^{k=999} X_{1000}[k]e^{j2\pi k n/1000} \\ &= e^{j2\pi \frac{999}{1000}} \end{align} $

By matching terms, we can see that

$ X_{1000}[k]=\begin{cases} 1000&\mbox{, if }k=999 \\ 0 &\mbox{, else} \end{cases} $

d) $ x_2[n]= e^{j \frac{2}{\sqrt{3}} \pi n} $

Solution

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

e) $ x_6[n]= \cos\left( \frac{2}{1000} \pi n\right) ; $

Solution

First, use Euler's formula

$ x_6[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_6[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_6[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{, else} \end{cases} $

f) $ x_2[n]= e^{j \frac{\pi}{3} n } \cos ( \frac{\pi}{6} n ) $

Solution

Using Euler's fomula

$ \begin{align} x_2[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_2[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{, else} \end{cases} $

g) $ x_8[n]= (-j)^n . $

First, rewrite the signal as

$ \begin{align} x_8[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{, else} \end{cases} $

h) $ x_3[n] =(\frac{1}{\sqrt{2}}+j \frac{1}{\sqrt{2}})^n $

Solution

First, rewrite the signal

$ \begin{align} x_3 &=(\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{, else} \end{cases} $

Note: All of these DFTs are VERY simple to compute. If your computation looks like a monster, look for a simpler approach!

Question 2

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

Note: Again, this is a VERY simple problem. Have pity for your grader, and try to use a simple approach!

Question 3

Prove the time shifting property of the DFT.

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