Tell me more ×
Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It's 100% free, no registration required.

I have the next system, and I want to find the transfer function from d to y.

enter image description here

So I've got the next equations

v = Ce = C(r-y)

e = r - y

u = d + v

y = Pu = Pd + PCr - PCy

Now I know that:

\$e/r = \frac{1}{1+PC}\$

So eventaully if I am not mistaken I arrive at:

$$y/d = \frac{P}{1+PC}+\frac{PC}{1+PC} \frac{r}{d}$$

How do I eliminate the dependence on \$r/d\$? i.e, I want y/d to be a function of P and C.

Thanks in advance.

Edit: actually, I arrive at:

$$ y = \frac{P}{1+PC} d + \frac{PC}{1+PC} r$$

So the transfer function from d to y should be: \$P/(1+PC)\$, correct?

share|improve this question
2  
If you want to find the transfer function of y/d you have to ignore r, ie, set it to 0. It has no place in the y/d transfer function. – AngryEE Dec 17 '12 at 19:07

1 Answer

$$ y = \frac{P}{1+PC} d + \frac{PC}{1+PC} r $$ r is desired input and d is disturbance so we have to reduce effect of d if gain of C picked big so we have: $$ if \space C \uparrow \hspace{8 mm} \frac{P}{1+PC} \downarrow \hspace{8mm} and \hspace{8mm} \frac{PC}{1+PC} \simeq 1 $$ so $$ y \rightarrow r $$ warning : be careful about stability.if C picked very big stability of system would be at risk

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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