Line 20: Line 20:
 
<math>
 
<math>
 
= 1 + (-j) \cdot (-j)^k + (-1) \cdot (1)^k + (j) \cdot (j)^k  
 
= 1 + (-j) \cdot (-j)^k + (-1) \cdot (1)^k + (j) \cdot (j)^k  
</math>
 
<math>
 
= (-j)^{k+1} + (j)^{k+1}
 
</math>
 
  
 +
= (-j)^{k+1} + (j)^{k+1} = 0, -2, 0, 2</math>
 +
 +
, when k = 0, 1 ,2 ,3. And it is periodic with K = 4.
 +
 +
Ouch... This is not right. since    <math> x[n] = (-j)^n = e^{((-j\pi/2) \cdot n )}</math>
 +
 +
it's fft should be only an impulse. And Matlab told me:
 +
 +
x = [1 -j -1 j];
 +
 +
fft(x)
 +
 +
ans =
 +
 +
    0    0    0    4
 +
 +
I'll fix it tomorrow. Or someone can point out my error?
  
 
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Revision as of 15:38, 29 September 2011


Practice Problem

Compute the discrete Fourier transform of the discrete-time signal

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

How does your answer related to the Fourier series coefficients of x[n]?

Share your answers below

You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!


Answer 1

$ X_k = \sum_{n=0}^{N-1} x_n \cdot e^{-j 2 \pi \frac{k}{N} n} = \sum_{n=0}^{3} (-j)^n \cdot e^{-j 2 \pi \frac{k}{4} n} = 1 + (-j \cdot e^{-j \frac{\pi k}{2}} ) + (-1 \cdot e^{-j \frac{2\pi k}{2}} ) + (j \cdot e^{-j \frac{3\pi k}{2}} ) $

$ = 1 + (-j) \cdot (-j)^k + (-1) \cdot (1)^k + (j) \cdot (j)^k = (-j)^{k+1} + (j)^{k+1} = 0, -2, 0, 2 $

, when k = 0, 1 ,2 ,3. And it is periodic with K = 4.

Ouch... This is not right. since $ x[n] = (-j)^n = e^{((-j\pi/2) \cdot n )} $

it's fft should be only an impulse. And Matlab told me:

x = [1 -j -1 j];

fft(x)

ans =

    0     0     0     4

I'll fix it tomorrow. Or someone can point out my error?


Answer 2

Write it here


Back to ECE438 Fall 2011 Prof. Boutin

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva