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?

  • $\begingroup$ 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
  • $\begingroup$ This clearly is the homework problem from Vetterli's DSP course on Coursera. You should show your effort first. Hint is clearly stating that you should solve it in frequency domain using pictures of spectrum of $x[n]$ and ideal LPF $h[n]$. Multiplying with $-1^n$ simply is shifting the spectrum at $\pi$ and $-\pi$. You need to figure out how. $\endgroup$ – DSP Rookie Jun 22 at 7:29

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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