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I'm trying to solve the following question, but I've been told I'm doing it incorrectly. Could anyone here give me a hint as to where I'm tripping up? Below follows my current solution:

enter image description here

I do the following to solve it:

a) $ y[n] = 3 ( x[n] + \frac{1}{3}x[n-1] + 2x[n]) + 6x[n-1] = 3x[n] + x[n-1] +6x[n] + 6x[n-1] = 9x[n] + 7x[n-1]$

b) I find the system function: $H(z) = \frac{Y(z)}{X(z)} = \frac{9X(z)+7z^{-1}X(z)}{X(z)} = 9 + 7 z^{-1}$

Then I apply the inverse z-transform and find the impulse response: $h[n] = 9\delta[n] + 7\delta[n-1]$

Thanks!

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There is clearly recursion in there and your solution does not reflect it since its purely FIR.

I recommend doing this as a coupled difference equation: Define a new variable $v[n]$ at the point where Peter drew the circle and than derive the transfer function from (1) X->V and then (2) X and V -> Y. Then you can eliminate V in eq (2) by using equation (1).

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  • $\begingroup$ Thank you, I see what I did wrong now. And thanks for getting me started on a solution : ) $\endgroup$ – enpi Oct 22 '15 at 22:02
  • $\begingroup$ I haven't seen a signal graph drawn this way before. Are the circles both adders and branchpoints? If that's the case, does $u[n] = x[n] + \frac{1}{3}x[n-1]$ and $y[n] = 2x[n] + 3u[n] + 6x[n-1]$ or is there a $u[n-1]$ or $y[n-1]$ I'm missing? $\endgroup$ – panthyon Oct 22 '15 at 22:16
  • $\begingroup$ Whats the final answer for this? $\endgroup$ – user18837 Dec 22 '15 at 0:32
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Have a look at the cycle around the indicated node. Your equations don't appear to take that into account.

enter image description here

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  • $\begingroup$ @PeterK. can we use mason's gain formula and write the transfer function of the system instead of this summing point and intermediate nodes issue? $\endgroup$ – Karan Talasila Oct 23 '15 at 6:12

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