# Response of cascading FIR filters

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

 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.

• 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". – MBaz May 1 '16 at 18:15
• @MBaz Thanks i didn't consider that when i posted the question. I have since updated my original post. – sky knight May 1 '16 at 19:07
• @MBaz is quite right, but you are also quite lucky !! so I will provide you a solution. – 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 $x[n] = 9 \times n \times(u[n]-u[n-20]$ , $y$ becomes: $$y = x - x$$ $$y = 0.9 \times 10 \times (u-u[-10]) - 0.9 \times 2 \times (u-u[-18])$$ $$y = 9 - 1.8 = 7.2$$

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

• 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. – sky knight May 1 '16 at 20:22
• @skyknight which step(s) exactly do you think you have a problem with? – Fat32 May 1 '16 at 21:30
• 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. – sky knight May 2 '16 at 5:59