Revision as of 14:01, 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} \end{align} $

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

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

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.

Discussion

You may discuss the homework below.

• write comment/question here
  • answer will go here

Back to ECE438, Fall 2014, Prof. Boutin