Signal Windowing Homework: x[n]w[n], M=1 & 10

[Cursed]

Homework Statement

$p[n]=x[n]w[n]$, where $w[n]$ is the rectangular window.



$x[n]=\sum_{k=-∞}^{∞} δ[n-k]$

$w[n]= 1,$ for $-M≤n≤M;$

$= 0$ otherwise​

1. What is $X(e^{jw})$?
2. What is $P(e^{jw})$ when...

M=1?
M=10?​

$j$ is the imaginary number. (it's the same as $i$.)

Homework Equations

N/A

The Attempt at a Solution

$X(e^{jw})=2\pi\sum_{k=-∞}^{∞}δ[n-2\pi k]$

I found $X(e^{jw}$ fairly easily, but I can't find $P(e^{jw})$ for either case of M. The answers from the book are below. Can someone tell me how I solve for this? I don't see why $X(e^{jw})$ is so different from $P(e^{jw})$

$P(e^{jw})=e^{jw}+1+e^{-jw}=1+2cos(\omega)$ when M=1
$P(e^{jw})=\frac{sin(21 \omega/2}{\omega/2}$ when M=1

 

[KataKoniK]

First, let's clarify the problem statement. The forum post is asking for the value of P(e^{jw}), not X(e^{jw}). So the first step is to rewrite the equation for P(e^{jw}). We know that p[n] = x[n]w[n], so we can substitute in the given equations for x[n] and w[n]:

p[n] = \sum_{k=-∞}^{∞} δ[n-k] * 1, for -M≤n≤M;
= 0 otherwise

Next, we can rewrite the equation using the definition of the rectangular window:

p[n] = \sum_{k=-M}^{M} δ[n-k]

Now, we can plug this equation into the definition of P(e^{jw}):

P(e^{jw}) = \sum_{n=-∞}^{∞} p[n]e^{-jwn}

= \sum_{n=-∞}^{∞} \sum_{k=-M}^{M} δ[n-k]e^{-jwn}

= \sum_{k=-M}^{M} \sum_{n=-∞}^{∞} δ[n-k]e^{-jwn}

= \sum_{k=-M}^{M} e^{-jwk}

= e^{-jw(-M)} + e^{-jw(-M+1)} + ... + e^{-jw(M-1)} + e^{-jwM}

= e^{jwM} + e^{jw(M-1)} + ... + e^{jw(-1)} + e^{jw(0)} + e^{jw(1)} + ... + e^{jw(M-1)} + e^{jwM}

= \sum_{k=-M}^{M} e^{jwk}

= \sum_{k=-M}^{M} cos(kw) + jsin(kw)

Now, we can plug in the values for M=1 and M=10 to get the answers from the book:

P(e^{jw}) = cos(w) + jsin(w) + 1 + cos(2w) + jsin(2w) + ... + cos(10w) + jsin(10w)

= 1 + 2cos(w) + 2cos(2w) + ... + 2cos