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= 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 | ||
| − | = (-j)^{k+1} + (j)^{k+1} = 0, | + | = (-j)^{k+1} + (j)^{k+1} = 0, 0, 0, 4</math> |
, when k = 0, 1 ,2 ,3. And it is periodic with K = 4. | , when k = 0, 1 ,2 ,3. And it is periodic with K = 4. | ||
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==Answer 2== | ==Answer 2== | ||
Revision as of 16:49, 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]?
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, 0, 0, 4 $
, when k = 0, 1 ,2 ,3. And it is periodic with K = 4.
Answer 2
Write it here
