Prove that $1+2\cos(\omega)$ is the same as $\frac{\sin(\frac{3\omega}{2})}{\sin(\frac{\omega}{2})}$

Knowing that $$p[n]=\sum_{k=-1}^1 {\delta[n-k]}=\delta[n-1]+\delta[n]+\delta[n+1]$$

Doing the FT yields $$\begin{align} P(e^{j\omega}) &=\sum_{k=-\infty}^\infty p[n]e^{-j\omega n} \\ &=e^{-j\omega}+1+e^{j\omega} \\&=1+2\cos(\omega) \end{align}$$

But, how does this fit into the general formula of $\frac{\sin(\omega(M+\frac{1}{2}))}{sin(\frac{\omega}{2})}$ where, in this case my $M=1$?

Thank you.


I have come across my ways.

$$\begin{align} P(e^{j\omega})&=1+2\cos(\omega) \\ &=\sum_{k=-\infty}^\infty \sum_{k=-1}^1 \delta[n-k]e^{-j\omega n} \end{align}$$

Knowing that non-zero only when $n=k$,

$$P(e^{j\omega})=\sum_{k=-1}^1 e^{-j\omega n}$$

Let $M=n+1$,

$$\require{cancel} \begin{align} P(e^{j\omega}) &=\sum_{k=0}^2 e^{-j\omega (M-1)} \\ &=e^{-j\omega (-1)} \sum_{k=0}^2 e^{-j\omega M} \\ &= e^{j\omega} \left(\frac {1-e^{-j\omega 3}} {1-e^{-j\omega}}\right ) \\&=\cancel{e^{j\omega}} \left( \frac {e^{j\omega 3/2}} {e^{-j \omega 3/2}} \right) \left( \frac {e^{j\omega 3/2} - e^{-j\omega 3/2}} {\cancel{e^{j\omega}} e^{j\omega 1/2} - e^{-j\omega 1/2} } \right) \\&= \frac{\sin(\frac{3\omega}{2})}{\sin(\frac{\omega}{2})} \end{align} $$

  • $\begingroup$ Welcome! It is great to see you could answer your question and share the solution with us. $\endgroup$ – Royi Jan 4 '20 at 10:31

The real-valued amplitude function of a rectangular window of length $N$ is


So what you get is exactly what you would expect according to $(1)$, because in your case you have $N=3$. The constant $M$ in your formula is related to $N$ by $N=2M+1$ or $M=(N-1)/2$.

If you used that rectangular sequence as an impulse of a causal filter, then $M$ would be the group delay caused by the filter.

And, of course, for proving the given equation you don't need any DSP knowledge, just multiply both sides with $\sin(\omega/2)$ and use the trigonometric identity $2\sin(x)\cos(y)=\sin(x+y)+\sin(x-y)$ with $x=\omega/2$ and $y=\omega$.


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