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Applied Dsp by dimitris

My attempt :

a) Summation of all values?

b)c)d) Failed

e) Parserval's theorem

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  • $\begingroup$ Hint: Observe that $x[-n] = -x[n]$ for all $n$. This should help in getting at the answers. $\endgroup$ – Dilip Sarwate Sep 1 '15 at 15:50
  • $\begingroup$ @DilipSarwate Yeah i noticed that. I managed to deduce that the frequency response is img and odd. a)zero c) zero b) even phase but can't figure out exact value. Any other hints for part b and d? $\endgroup$ – usfmohy Sep 1 '15 at 15:58
  • $\begingroup$ If a complex number $z$ (such as $X(e^{j\omega}$) is known to be wholly imaginary, can you deduce something about $\angle(z)$? $\endgroup$ – Dilip Sarwate Sep 1 '15 at 16:02
  • $\begingroup$ @DilipSarwate atan(X/0)=pi/2? $\endgroup$ – usfmohy Sep 1 '15 at 16:13
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You just need to know the formulas for the DTFT and its inverse:

$$X(e^{j\omega})=\sum_{n=-\infty}^{\infty}x[n]e^{-jn\omega}\tag{1}$$

and

$$x[n]=\frac{1}{2\pi}\int_{-\pi}^{\pi}X(e^{j\omega})e^{jn\omega}d\omega\tag{2}$$

From $(1)$ you see that your answer for $(a)$ is correct. Also the answer for $(d)$ follows immediately, if you realize that $e^{-jn\pi}=(-1)^n$.

For $(b)$ it's important to see that the coefficients are asymmetrical. Looking at $(1)$ you see that you can pair the terms for indices $\pm 1$, $\pm 2$, etc., which allows you to rewrite the sum $(1)$ as a sum of weighted sines time $j$ (imaginary unit), because $e^{jx}-e^{-jx}=2j\sin(x)$. So you'll have a purely imaginary expression. Computing the phase should then be easy.

$(c)$ is easily solved using formula $(2)$ (which value of $n$ do you need to plug into $(2)$?). And finally, for $(e)$ you're right that you need Parseval's theorem.

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  • $\begingroup$ Plug in 0 as the value of n to make it equal to 1?. Also part (d) means adding all numbers but this time odd values have a negative sign so (1-2+3-4)*2= -4? $\endgroup$ – usfmohy Sep 1 '15 at 16:23
  • $\begingroup$ @YoussefMohy: Why "make it equal to 1"? $x[0]=0$, isn't it? You're right about the odd indices getting a negative sign, but your numerical result is wrong. $\endgroup$ – Matt L. Sep 1 '15 at 16:31
  • $\begingroup$ i meant to remove the exponential (Make it equal to 1) So overall value = 0. Yeah i was mistaken (1+2+3+4-4-3-2-1) = 0 $\endgroup$ – usfmohy Sep 1 '15 at 16:38

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