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Hey all I have the following question and I am having trouble getting to the correct answer. Here is what I have: (sorry I do not understand how to format the question)

Consider a linear time-invariant system such that

$$H(e^{j\omega}) = \frac{1}{(1-\frac{1}{2}e^{j\omega})^2}$$

If the input x ̃[n] is periodic with period N0 = 8, then determine the output Fourier series coefficient y4 if x4 = 9.

I have this so far:

$$y_4 = x_4 H(e^{jk\frac{2\pi}{N_0}})$$

$$= 9 H(e^{j\pi})$$

Goes back into the given function:

$$= \frac{9}{(1-\frac{1}{2}e^{-j\omega})^2}$$

I am stuck here. How would I get a solid answer from this. I know that in the end the answer is 4.

Thanks.

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I wasn't really able to follow your reasoning, but I can tell you how you can simplify your second equation:

$$9 H(e^{j\pi})$$

Note that $$e^{j\pi} = -1$$ (Euler's formula).

Inserting that into the equation above gives:

$$ \frac{9}{(1-\frac{1}{2}(-1))^2} = \frac{9}{1.5^2}= 4$$ I hope this helps ;)

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  • \$\begingroup\$ I've cancelled my downvote - It's probably best to remove all the comments now. If you make a trivial edit like I did it removes me from the "edited by" section! \$\endgroup\$ – Andy aka Nov 1 '13 at 9:46
  • \$\begingroup\$ So is that always equal to -1? I have never heard of that but it does make sense. \$\endgroup\$ – user081608 Nov 1 '13 at 15:30
  • \$\begingroup\$ e is a constant, j is a constant, pi is a constant, so yes, $$e^{-j\pi}$$ is also a constant which is always equal to -1. Euler's fomula, which I linked to above, is really the most basic equation for everything related to complex numbers. \$\endgroup\$ – Geier Nov 1 '13 at 18:56

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