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enter image description here$$y[n]=(-1)^n w[n]+w[n]$$

$n$ ranges from $-\infty$ to $\infty$

where $w[n]$ is output of frequency response block with cutoff frequency = $\pi/2$, and input is unit impulse function.

I know output of frequency response block is $w[n]$ is $h[n]$ because $\delta[n]$ conv with $h[n]$ = $h[n]$ since delta exists only at zero

May be we can solve this in frequency domain?

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  • $\begingroup$ a) that is not a low-pass filter. b) no, the output of this is most definitely not $w[n]$. c) are your maybe confusing frequency domain, impulse response, white noise realization and white noise autocorrelation? $\endgroup$ – Marcus Müller May 22 at 11:23
  • $\begingroup$ ah, using $*$ here was misleading: in the context of filters, that usually denotes convolution. You need multiplication. Anyway, still not a low-pass filter. Try making a single term out of everything you multiply with $w[n]$ and then look for possible values. $\endgroup$ – Marcus Müller May 22 at 17:43
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The half-band lowpass filter, has the following impulse response : $$ h[n] = \frac{\sin(\frac{\pi}{2} n) }{ \pi n} $$

And for $x[n] = \delta[n]$ , $w[n] = h[n]$.

Then the output $y[n]$ will be :

$$y[n] = w[n] \big( 1 + (-1)^n \big) $$

which can be simplified by expanding the parenthesis as:

$$ \begin{align} y[n] &= \frac{\sin(\frac{\pi}{2} n) }{ \pi n} \cdot \big( e^{j 2 \pi n} + e^{j \pi n} \big) \\ \\ y[n] &= \frac{\sin(\frac{\pi}{2} n) }{ \pi n} \cdot e^{j \frac{3 \pi}{2} n} \cdot \big( e^{j \frac{\pi}{2} n} + e^{-j \frac{\pi}{2} n} \big)\\ \\ y[n] &= \frac{ \sin(\frac{\pi}{2} n) }{ \pi n} \cdot 2 \cdot \cos( \frac{\pi}{2} n) \cdot e^{j \frac{3 \pi}{2} n} \\ \\ y[n] &= \frac{\sin(\pi n) }{ \pi n} \cdot e^{j \frac{3 \pi}{2} n} \end{align} $$ Which can be shown to be :

$$y[n] = \begin{cases}{ 1 ~~~~, n = 0 \\ 0 ~~~~ , n \neq 0 } \end{cases} $$

Therefore $y[n] = \delta[n]$. The system is a zero-phase all pass filter. And finally, the required quantity is $$ M = \sum_{n=-\infty}^{\infty} y[n] = \sum_{n=-\infty}^{\infty} \delta[n] = 1 $$

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