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

  • $\begingroup$ What is a "simulation diagram"? $\endgroup$
    – MBaz
    Sep 3, 2016 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, 2016 at 17:41
  • $\begingroup$ I need to know how to draw the block diagram of the difference equation. $\endgroup$
    – joe
    Sep 3, 2016 at 20:52

1 Answer 1


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.

  • $\begingroup$ Is my difference equation correct? $\endgroup$
    – joe
    Sep 4, 2016 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, 2016 at 16:27
  • $\begingroup$ Hi Peter, I have modified my answer. Is it correct now? thanks. $\endgroup$
    – joe
    Sep 4, 2016 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, 2016 at 22:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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