-1
$\begingroup$

Suppose $x[n]$ ($n$ integers) is periodic with period 8 and its Fourier coefficients are $$ a_k = \cos(k\pi /4) + \sin(3k\pi /4). $$

Prove $x[n] = 4\delta[n-1] + 4\delta[n-7] + 4j\delta[n-3] - 4j\delta[n-5] $ where $\delta[n]$ takes zero everywhere except $n=0$ at which it takes $1$.

$\endgroup$
  • $\begingroup$ Your $x[n]$ isn't periodic. Do you mean the period-8 extension of it? $\endgroup$ – Austin A. Oct 20 '15 at 12:35
  • $\begingroup$ Yes. $x[n]$ is the extension one. $\endgroup$ – jachilles Oct 20 '15 at 12:38
0
$\begingroup$

Since the Discrete Fourier Transform (DFT) is a bijection, you can prove this by taking the forward transform of $x[n]$ rather than the inverse transform of $a_k$. I might recommend this approach as it is easy to transform the $\delta$ functions.

I take the sum from $n = 0$ to $n = 7$; the function $x[n]$ is automatically extended periodically with period 8. $$\begin{align} \mathcal{F}(x[n]) &:= \frac{1}{8} \sum_{n = 0}^8 x[n] e^{-2 \pi i k n / 8} \\ &= \frac{1}{8} \left( 4 e^{-2\pi ik/8} + 4i e^{-6\pi ik/8} - 4ie^{-10\pi ik/8} + 4e^{-14\pi ik/8} \right) \\ &= \frac{1}{8} \left( (4 e^{-\pi i k/4} + 4 e^{-7\pi i k/4}) + (4i e^{-3\pi i k/4} - 4i e^{-5\pi i k/4}) \right)\\ &= \frac{1}{8} \left( 4(e^{-\pi i k /4} + e^{\pi i k/4}) + 4i(e^{-3\pi i k/4} - e^{3\pi i k/4})\right)\\ &= \frac{1}{8} \left( 8\cos(\pi k /4) + 8\sin(3\pi k/4) \right)\\ &= \cos(\pi k /4) + \sin(3\pi k/4). \end{align}$$

Incidentally, there are more than one convention for the DFT. You should specify which one you are using when you ask your question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.