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.

  • $\begingroup$ 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


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.


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