Practice Question on Causal LTI systems defined by a linear, constant coefficient difference equation

Consider the LTI system defined by the difference equation

$ y[n]-\frac{1}{2}y[n-1]=x[n]\ $

a) What is the frequency response of this system?

b) What is the unit impulse response of this system?

c) What is the system's response to the input $ x[n] = \left( \frac{1}{5}\right)^n u[n] \ $?

c) What is the system's response to the input $ x[n] =\cos \left(\frac{\pi}{2} n \right) \ $?


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Answer 1

a)

$ \mathfrak F (y[n]-\frac{1}{2}y[n-1]) = \mathfrak F (x[n]) $

$ \mathcal Y (\omega) - \frac{1}{2}\mathfrak F (y[n-1]) = \mathcal X (\omega) $

$ \mathcal Y (\omega) - \frac{1}{2}e^{-j\omega} \mathcal Y (\omega) = \mathcal X (\omega) $

$ \mathcal Y (\omega) = \frac{1}{1-\frac{1}{2}e^{-j\omega}}\mathcal X (\omega) $

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

b)

$ h[n]=\mathfrak F ^{-1} (\mathcal H (\omega))= \mathfrak F ^{-1} \Big( \frac{1}{1-\frac{1}{2}e^{-j\omega}} \Big) $

$ use \mathfrak F (a^n u[n]) = \frac{1}{1-ae^{-j\omega}} $

$ h[n] = \Big(\frac{1}{2}\Big)^n u[n] $

c)

use table formula from last part

$ \mathcal X (\omega)= \frac{1}{1-\frac{1}{5}e^{-j\omega}} $

$ \mathcal Y (\omega)= \Big(\frac{1}{1-\frac{1}{2}e^{-j\omega}}\Big)\Big(\frac{1}{1-\frac{1}{5}e^{-j\omega}}\Big) = \Big(\frac{\frac{5}{3}}{1-\frac{1}{2}e^{-j\omega}}\Big) + \Big(\frac{\frac{-2}{3}}{1-\frac{1}{5}e^{-j\omega}}\Big) $

$ y[n] = \frac{5}{3}\Big(\frac{1}{2}\Big)^n u[n] - \frac{2}{3}\Big(\frac{1}{5}\Big)^n u[n] $

d)

$ x[n] = \frac{1}{2}e^{j\frac{\pi}{2}n}+\frac{1}{2}e^{-j\frac{\pi}{2}n} $

guess for $ e^{j\omega_0n} \, $  : $ \mathcal X (\omega) = 2\pi \delta (\omega-\omega_0)\, $

check $ \frac{1}{2\pi}\int_0^{2\pi} 2\pi \delta (\omega-\omega_0)e^{j\omega n}d\omega = e^{j\omega_0n} $

but the answer needs to be periodic. Apply to x[n],

$ \mathcal X (\omega) = \frac{1}{2} \sum_{m=-\infty}^\infty 2\pi \delta (\omega-\frac{\pi}{2}+2\pi m)+\frac{1}{2} \sum_{m=-\infty}^\infty 2\pi \delta (\omega+\frac{\pi}{2}+2\pi m) $

for $ 0\le\omega\le 2\pi $

$ \mathcal Y (\omega) = \pi \delta (\omega-\frac{\pi}{2}) \Big(\frac{1}{1-\frac{1}{2}e^{-j\omega}}\Big) + \pi \delta (\omega-\frac{3\pi}{2}) \Big(\frac{1}{1-\frac{1}{2}e^{-j\omega}}\Big) $

$ y[n] = \frac{\pi}{2\pi}\int_0^{2\pi}\delta (\omega-\frac{\pi}{2}) \Big(\frac{1}{1-\frac{1}{2}e^{-j\omega}}\Big)e^{j\omega n}d\omega +\frac{\pi}{2\pi}\int_0^{2\pi}\delta (\omega-\frac{3\pi}{2}) \Big(\frac{1}{1-\frac{1}{2}e^{-j\omega}}\Big)e^{j\omega n}d\omega $

$ y[n] = \frac{1}{2} \Big(\frac{e^{j\frac{\pi}{2} n}}{1-\frac{1}{2}e^{-j\frac{\pi}{2}}}\Big) + \frac{1}{2} \Big(\frac{e^{j\frac{3\pi}{2} n}}{1-\frac{1}{2}e^{-j\frac{3\pi}{2}}}\Big) $

--Cmcmican 21:17, 8 March 2011 (UTC)

Instructor's note: Actually, there is a much easier way to answer part d). Remember what happens when a complex exponential is the input of an LTI system? -pm

I think that it is multiplied by a complex constant. But I don't know how to compute that constant from the fourier transform. Is there something I'm missing? --Cmcmican 21:59, 9 March 2011 (UTC)

TA's comment: The response of an LTI system to a complex exponential is the same complex exponential multiplied by the transfer function evaluated at the frequency of the complex exponential. The following is a block diagram of this property:
$ e^{j\omega_0 n} \to \Bigg[ \mathcal{H}(\omega) \Bigg] \to \mathcal{H}(\omega_0) e^{j\omega_0 n} $


So, the answer would be this:

$ y[n] = \mathcal H (\omega_0)x[n] = \Big(\frac{1}{1-\frac{1}{2}e^{-j\frac{\pi}{2}}}\Big)\Big(\frac{1}{2}e^{j\frac{\pi}{2}n}+\frac{1}{2}e^{-j\frac{\pi}{2}n}\Big) $

is that right? --Cmcmican 15:22, 10 March 2011 (UTC)

yes, that's right. Notice that it is the same answer as you got using the previous method. -pm
TA's comment: I think you are missing something here. I suggest you apply the property on each complex exponential alone and then add the responses up.

Answer 2

Write it here.

Answer 3

Write it here.


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