Line 58: Line 58:
  
 
c) <math>x_5[n]= e^{-j \frac{2}{1000} \pi n}</math>
 
c) <math>x_5[n]= e^{-j \frac{2}{1000} \pi n}</math>
 +
 +
'''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):
 
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):
Line 64: Line 66:
 
\begin{align}
 
\begin{align}
 
x_5[n]&=e^{-j \frac{2}{1000} \pi n}e^{j2\pi n} \\
 
x_5[n]&=e^{-j \frac{2}{1000} \pi n}e^{j2\pi n} \\
&= e^{j2\pi \frac{1000-2}{1000}} \\
+
&= e^{j2\pi \frac{1000-1}{1000}} \\
 
&=e^{j2\pi \frac{998}{1000}}
 
&=e^{j2\pi \frac{998}{1000}}
 
\end{align}</math>
 
\end{align}</math>
  
 
The positive exponent is easier to deal with.
 
The positive exponent is easier to deal with.
 +
 +
Now we can use the inverse transform as before, using a 1000-point IDFT:
 +
 +
<math>
 +
\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}
 +
</math>
 +
 +
By matching terms, we can see that
 +
 +
<math>X_{1000}[k]=\begin{cases} 1000&\mbox{, if }k=999 \\ 0 &\mbox{, else} \end{cases}</math>
 +
 
d) <math>x_2[n]= e^{j \frac{2}{\sqrt{3}} \pi n}</math>
 
d) <math>x_2[n]= e^{j \frac{2}{\sqrt{3}} \pi n}</math>
  

Revision as of 10:18, 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) ; $

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

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
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Back to ECE438, Fall 2014, Prof. Boutin

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang