1
$\begingroup$

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

Improve this question
$\endgroup$
3
  • $\begingroup$ 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
1
$\begingroup$

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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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