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I am trying to get transfer function $H_{xy}$ from this graph. I managed to get the other one but I am struggling with this one. Any help or guidance?

Thank you

enter image description here

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  • $\begingroup$ homework! (set $e[n]=0$ and do it as the textbook says.) $\endgroup$ Feb 20 '15 at 18:39
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    $\begingroup$ hi Robert, this isnt the homework but part of online course . deadline passed so i am no getting the extra poitns at all this is to me only because I want to understand $\endgroup$ Feb 20 '15 at 19:01
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If you mark with $v[n]$ the signal between summing nodes for input $x[n]$ and output $y[n]$, then:

$v[n] = x[n] -8v[n-2] -4v[n-3]$

Re-arranging sides:

$v[n]+8v[n-2]+4v[n-3]=x[n] $

Using the Z-transform:

$V(z)[1+8z^{-2}+4z^{-3}]=X(z)$

Now let's calculate the transfer function between $v[n]$ and $x[n]$:

$H_{x\to v}(z)=\dfrac{V(z)}{X(z)}=\dfrac{1}{1+8z^{-2}+4z^{-3}}$

Now moving to the relation between output $y[n]$ and $v[n]$:

$y[n]=v[n]+2v[n-1]$

Again calculating the Z-transform:

$Y(z)=V(z)[1+1+2z^{-1}]$

Now converting the transfer function $H_{vy}(z)$:

$H_{v\to y}(z)=1+2z^{-1}$

Final step is to merge these two together. We know that in time domain it is a convolution: $h_{x\to y}[n]=h_{x\to v}[n]\star h_{v\to y}[n]$, thus in Z-transform domain it will be the multiplication, yielding the final result:

$H_{x\to y}(z)=H_{x\to v}(z)H_{v\to y}(z)=\dfrac{1+2z^{-1}}{1+8z^{-2}+4z^{-3}}$

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  • $\begingroup$ i wouldn't have done this for the OP, jo. it's pretty obviously homework. (or what if it's a "take-home exam"? we used to have those when i was an EE student.) $\endgroup$ Feb 20 '15 at 18:41

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