I can't seem to get the right answer for this. Could someone show me the procedure?

enter image description here

As shown in the figure a system is constructed by cascading four FIR filters having the following frequency responses,

If the input signal is $x[n] = 0.9n(u[n] - u[n - 20])$, compute the value of $y[10]$.

[Edit] working thus far

H[z] = (1 - e-jω)x(1 + e-jω)x(1 + e-2jω)x(1 + e-4jω)

 = (1 - e<sup>-2jω</sup>)x(1 + e<sup>-2jω</sup>)x(1 + e<sup>-4jω</sup>)
 = (1 - e<sup>-4jω</sup>)x(1 + e<sup>-4jω</sup>)
 = (1 - e<sup>-8jω</sup>)

y[n] = x[n] - x[n-20]

From this point i'm not sure how to proceed.

  • 1
    $\begingroup$ You'll improve your chances of getting a good answer if you show what you have tried, and where you got stuck; IOW, ask a more specific question instead of "tell me how to solve this". $\endgroup$ – MBaz May 1 '16 at 18:15
  • $\begingroup$ @MBaz Thanks i didn't consider that when i posted the question. I have since updated my original post. $\endgroup$ – sky knight May 1 '16 at 19:07
  • $\begingroup$ @MBaz is quite right, but you are also quite lucky !! so I will provide you a solution. $\endgroup$ – Fat32 May 1 '16 at 19:11

In this problem there are a cascade of 4 LTI systems whose multiplication of frequency responses (or convolution of impulse responses equivalently) provides the overall system Frequency Response (or impulse response equivalently) and the output $y[n]$ at n = 10 is being asked for an input of $x[n] = 0.9 \times n \times (u[n] - u[n-20])$

Now I would like to proceed by finding the overall impulse response $h[n]$ from the serial convolution of those four individual impulse responses as it seems simpler to find the output in time domain then in the Frequency domain: $$ h[n] = (\delta[n] - \delta[n-1]) \star (\delta[n]+\delta[n-1]) \star (\delta[n] + \delta[n-2]) \star (\delta[n] + \delta[n-4])$$

Where $\star$ denotes convolution operation and those impulses are directly derived from the Frequency Responses $H_i(e^{j\omega})$ of the individual systems. I hope you can see the simple corresponce with each one of them.

Then convolving from left to right, we can show in 3 steps that: $$h[n] = (\delta[n] - \delta[n-2]) \star (\delta[n] + \delta[n-2]) \star (\delta[n] + \delta[n-4])$$

$$h[n] = (\delta[n] - \delta[n-4]) \star (\delta[n] + \delta[n-4])$$

$$h[n] = (\delta[n] - \delta[n-8])$$

I hope also that you have no difficulty in obtainig shifted impulses as a result of convolution with a shifted impulse.

Therefore now the question becomes; given LTI impulse response $h[n] = \delta[n] - \delta[n-8]$ , compute $y[n]$ for the given input $x[n]$. It's now easy to show that: $$y[n] = h[n] \star x[n] = (\delta[n] - \delta[n-8])*x[n]$$ $$y[n] = x[n] -x[n-8]$$

therefore with $x[n] = 9 \times n \times(u[n]-u[n-20]$ , $y[10]$ becomes: $$y[10] = x[10] - x[2]$$ $$y[10] = 0.9 \times 10 \times (u[10]-u[-10]) - 0.9 \times 2 \times (u[2]-u[-18])$$ $$y[10] = 9 - 1.8 = 7.2$$

Where before the last line I have used the famous property of unit step $u[n]$.

  • $\begingroup$ Thanks so much for this explanation. There are some clear gaps in my knowledge which you pointed out from this and i can focus on those thanks to you. $\endgroup$ – sky knight May 1 '16 at 20:22
  • $\begingroup$ @skyknight which step(s) exactly do you think you have a problem with? $\endgroup$ – Fat32 May 1 '16 at 21:30
  • $\begingroup$ i had issues with convolution and wasn't sure on how the LTI impulse response was used to compute y[n] when i was working on it by myself. Though i understand that now. $\endgroup$ – sky knight May 2 '16 at 5:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.