# Discrete time system signal flow graph problem

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:

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!

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).
• 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? – panthyon Oct 22 '15 at 22:16