Consider the LTI system with frequency response $$H(e^{j\omega}) = \frac{1-e^{-j2\omega}}{1+\frac{1}{2}e^{-j4 \omega}}, -\pi < \omega < \pi$$ Determine the output $y[n]$ for all $n$ if the input $x[n]$ for all $n$ is $$x[n] = \sin \left(\frac{\pi n}{4}\right)$$

My attempt: $x[n]$ is an eigenfunction of the LTI system, so the output have the form $$y[n]=|H(e^{j \omega})|\sin\left(\frac{\pi n}{4} + arg(H(e^{j \omega})\right)$$ But I dont know how to determine the phase and modulus of the frequency response with this form. For example, I think that $$| 1 - e^{-j 2 \omega} | = \sqrt{(1-\cos(2 \omega))^2 + \sin^2(2 \omega)}$$ Analogous, doing for the denominator, I could not simplificate the result.

The answer:

$$y[n]=2 \sqrt{2} \sin \left( \frac{\pi(n+1)}{4} \right) $$


1 Answer 1


You're definitely on the right track. The way you're trying to solve the problem is the best and simplest. You just need to realize that you need to evaluate the magnitude and phase of the frequency response just for one frequency, namely the frequency of the sinusoidal input signal:


with $\omega_0=\pi/4$ and $H(e^{j\omega})=\left|H(e^{j\omega})\right|e^{j\phi(\omega)}$.

Evaluating the given frequency response at $\omega_0=\pi/4$ gives

$$\frac{1-e^{-j\pi/2}}{1+\frac12 e^{-j\pi}}=\frac{1+j}{1-\frac12}=2(1+j)=2\sqrt{2}e^{j\pi/4}$$

from which the correct result follows.

As a final note, you incorrectly called the sinusoidal input signal an eigenfunction of the LTI system. This is not true, it is the complex exponential $e^{jn\omega_0}$ that is an eigenfunction, because it appears at the output unchanged other than scaled by a complex constant (the eigenvalue, which happens to be the value of the complex frequency response evaluated at the given frequency). This is not true for a sinusoid, which is not simply scaled by the LTI system.


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