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Calculate the difference equation and then draw the simulation diagram of the below transfer function. $$ H(z) = \frac{Y(z)}{X(z)} = \frac{0.4142 + 0.4142z^{-1}}{1.4142 - 0.5858z^{-1}} $$

I performed the normal procedure to find the difference equation, by cross multiplying and using the delay property of the $\mathcal Z$-transforms, I finally ended up with:

$$y[n] = 0.2929 x[n] + 0.2929 x[n-1] + 0.4137 y[n-1]$$

How do I draw the simulation diagram?

I want to know how to draw the block diagram of the difference equation.

I will try to draw the diagram but I am not sure if it is correct.

I don't require the answer but guidance on how to derive the diagram and final answer.

This is what I have done,

I have attached my solution.

is this correct?

I have modified my answer again. enter image description here

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  • $\begingroup$ What is a "simulation diagram"? $\endgroup$ – MBaz Sep 3 '16 at 17:38
  • $\begingroup$ Could you explain where exactly you're stuck when drawing a diagram? It's just a visual representation of the difference equation. You need 3 multipliers, and you need a delayed version of the input signal, and a delayed version of the output signal. $\endgroup$ – Matt L. Sep 3 '16 at 17:41
  • $\begingroup$ I need to know how to draw the block diagram of the difference equation. $\endgroup$ – joe Sep 3 '16 at 20:52
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Your equation is: $$y[n] = 0.2929 x[n] + 0.2929 x[n−1] + 0.4137 y[n−1]$$ which has input $x[n]$ and output $y[n]$.

You also need $x[n-1]$, so there is a delay block required for that and $y[n-1]$ so a delay block is required to remember the last output.

You also need constant multipliers for the coefficients 0.2929 and 0.4137.

Read up on signal flow graphs.

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  • $\begingroup$ Is my difference equation correct? $\endgroup$ – joe Sep 4 '16 at 10:17
  • $\begingroup$ Close, but not quite. For example: you appear to be multiplying $x$ by 0.2929 twice --- once on input, once after a delay. Multiplying it once by 0.2929 on input should be enough to keep it there. $\endgroup$ – Peter K.♦ Sep 4 '16 at 16:27
  • $\begingroup$ Hi Peter, I have modified my answer. Is it correct now? thanks. $\endgroup$ – joe Sep 4 '16 at 16:48
  • $\begingroup$ @joe Looks good. The only thing I'd do differently is to put the second delay element back underneath so the output goes straight out. $\endgroup$ – Peter K.♦ Sep 4 '16 at 22:31

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