0
$\begingroup$

We have a linear time-invariant system described by the input-output relation

$$y[n] = x[n] + 2x[n - 1] + x[n - 2]$$

Below is my approach to analyze this system.

The impulse response of this system $h[n]$ can be found if we input $δ[n]$ in $x[n]$'s position.

So $$h[n] = δ[n] + 2δ[n-1] + δ[n-2]$$

From above, I can get an expression for $H(e^{j\omega})$ using the Fourier transform. $$H(e^{j\omega}) = (1 + 2e^{-j\omega} + e^{-j2\omega}) = (1+e^{-j\omega})^2$$

My question is how to simplify this expression using trigonometric identities.

It would be pleasure if I can get some help.

$\endgroup$
  • $\begingroup$ This looks already pretty simple. How would it need to look like so it would be "simple enough" for your purpose. $\endgroup$ – Hilmar Oct 15 '18 at 16:12
3
$\begingroup$

HINT:

Write $H(e^{j\omega})=e^{-j\omega}G(e^{j\omega})$ and use $(e^{j\omega}+e^{-j\omega})=2\cos(\omega)$. $G(e^{j\omega})$ should turn out to be real-valued.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.