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A discrete-time linear and time-invariant (LTI) system is given by the following input-output relationship: y[n] = 2x[n] − x[n − 1], where x[n] is the input and y[n] is the output.

(a) What is the frequency response?

(b) What is the response of the system to input signal 1 + e^(j*(pi/2)*n)

For the problem above, I figured out that the frequency response is H(jw)=2-e^(-jw). So for the new response in part b, I know that x[n]= e^(j*0*n) + e^(j*(pi/2)*n). How should I proceed from here?

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This question belongs better to dsp.stackexchange.

The input signal in part (b) has 2 terms, and is the sum of 2 exponentials, one with frequency $$\omega_1 = 0$$ and the other with frequency $$\omega_2 = \pi/2$$

Since the system is linear, then the output is the sum of the output to each individual input. That is: $$y[n] = y_1[n] + y_2[n]$$

Also, since complex exponentials are eigenvectors of LTI systems, then the output will be the same exponential multiplied by the frequency response (i.e. the eigenvalue).

So,

$$ y_1[n] = 1 \times H(e^{j 0}) $$

and

$$ y_2[n] = e^{j n \pi/2} \times H(e^{j \pi/2}) $$

Where $$H(e^{j0}) = 1$$ and $$H(e^{j\pi/2})= 2+j$$.

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  • \$\begingroup\$ Thanks for the response, I didn't even know DSP existed, I'll check that out! But quick question, isn't H(e^(j*0))=2? Because H(jw)=2-e^(-jw), where w=0? Also, what if the number of terms don't match up between the frequency response and the input? Thank you.. \$\endgroup\$ – Shankar Kumar Mar 24 '13 at 21:31
  • \$\begingroup\$ please ignore the question about where omega is 0. I figured it out. With regards to my second question, is this procedure only useful when the input & response terms have the same number of terms? \$\endgroup\$ – Shankar Kumar Mar 24 '13 at 21:43
  • \$\begingroup\$ This procedure is independent on the number of terms. Note that the frequency response H is completely generic (can correspond to an IIR filter). \$\endgroup\$ – Juancho Mar 24 '13 at 22:30

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