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Pardon my paint skills, I did my best enter image description here

My attempt is short and seems to fail, I have no idea why:

$$\begin{align} \alpha &= in - a_{1}\beta - a_{2}\gamma \\ \beta &= \alpha z^{-1} \\ \gamma &= \beta z^{-1} = \alpha z^{-2} \end{align}$$

inputting 2nd and 3rd equation into the first one I get:

$$\begin{align} \alpha &= in - a_{1}\alpha z^{-1} - a_{2}\alpha z^{-2} \\ in &= \alpha + a_{1}\alpha z^{-1} + a_{2}\alpha z^{-2} \end{align}$$

I can write the output as:

$$\begin{align} out &= b_{2}\gamma + b_{1}\beta + b_{0}\alpha \\ out &= b_{2}\alpha z^{-2} + b_{1}\alpha z^{-1} + b_{0}\alpha \end{align}$$

I have output and input in terms of alpha, but I can't figure what to do from here.

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You are going in the right direction! Lets take these two equations: $$(1) \quad in = \alpha+a_1\alpha z^{-1}+a_2\alpha z^{-2}$$ $$(2) \quad out = b_0\alpha+b_1\alpha z^{-1}+b_2\alpha z^{-2}$$ now rewrite (1) such that it becomes a function of $\alpha$: $$in = \alpha\left(1+a_1z^{-1}+a_2z^{-2}\right)$$ $$\alpha = \frac{in}{1+a_1z^{-1}+a_2z^{-2}}$$ Substitute $\alpha$ in equation (2): $$out = in\frac{b_0+b_1 z^{-1}+b_2 z^{-2}}{1+a_1z^{-1}+a_2z^{-2}}$$ And derive proper discrete transfer function from it: $$H(z) = \frac{out}{in} = \frac{b_0+b_1 z^{-1}+b_2 z^{-2}}{1+a_1z^{-1}+a_2z^{-2}}$$

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  • $\begingroup$ Thank you kind sir. I'm ashamed I didn't figure it out myself. But you made a little mistake I believe, in the second (2) equation the last term should have b2 instead of a2, and that mistakes propagates to the results as well $\endgroup$ Sep 17 '20 at 11:14
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    $\begingroup$ Oops, small copy-past error haha. fixed it! $\endgroup$
    – Petrus1904
    Sep 17 '20 at 11:15
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Just wanted to add that you could use this to draw diagrams next time. Much easier to use than paint. You can even use Latex.

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  • $\begingroup$ This really should be posted as a comment and not an answer. $\endgroup$
    – Eric S
    Oct 13 '20 at 15:19
  • $\begingroup$ Don't have the reputation to do so, just need 4 more to get to 50 lol $\endgroup$
    – mhdadk
    Oct 13 '20 at 19:08
  • $\begingroup$ Just one good answer will get you there! $\endgroup$
    – Eric S
    Oct 13 '20 at 22:13
  • $\begingroup$ If you used this website to draw a new diagramm for thr op and then answer the OPs question i’d definitely +1 $\endgroup$
    – morbo
    Oct 14 '20 at 8:59

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