1
$\begingroup$

enter image description here

Can you help me with this question. Can you explain where i should start to solve the problem.

$\endgroup$
  • 1
    $\begingroup$ Hi! You can start by rotating your posted image counterclockwise :-)) $\endgroup$ – Fat32 Nov 26 '18 at 16:33
  • 1
    $\begingroup$ hahahahahahahahhaha $\endgroup$ – ahmad-hrm Nov 26 '18 at 16:37
0
$\begingroup$

The block diagram you have posed is composed of serial cascade of three IIR sections each having its own input output relationship. The first two are second order stages and the last one is first order.

If you know about Z-transforms, you could easily find the algebraic relation between those signal variables in terms of their Z-transforms.

I would denote the intermediate summation nodes as $w[n]$ and $v[n]$ together with the input $x[n]$ and output $y[n]$ the Z-domain relations become:

  • $W(z) (1 - a_1 z^{-1} - a_2 z^{-2}) = X(z)$

  • $V(z) (1 - b_1 z^{-1} - b_2 z^{-2}) = W(z)$

  • $Y(z) (1 - c_1 z^{-1} ) = V(z)$

where I have denoted the system coefficients as $a,b,c$ respectively. (Do not mix this $a,b$ with the standard LCCDE expression coefficients $a,b$.)

Then simple algebra tells us the relation between $X(z)$ and $Y(z)$ as:

$$Y(z) = 2 \left( \frac{1}{1 - c_1 z^{-1}} ~ \frac{1}{1 - b_1 z^{-1} - b_2 z^{-2}} ~\frac{1}{1 - a_1 z^{-1} - a_2 z^{-2}} \right) X(z)$$

By first expanding the product polynomials and then mapping back from Z-domain to time domain you can express $y[n]$ as a function of $x[n]$ (and feedback)

$\endgroup$

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.