[circuit theory] Transfer function of feedback network

Status
Not open for further replies.
M

melkord

Full Member level 3
Full Member level 3
Joined
May 18, 2018
Messages
151
Helped
0
Reputation
0
Reaction score
1
Trophy points
18
Activity points
1,772
In the system with negative feedback, what is described by G(s) actually? is it G(s)=A(s)/Y(s) ?

If I want to derive Y(s)/X(s), is it a correct way to derive G(s)=A(s)/Y(s) and then I can plug it in to the equation (8.2)?

With simple resistor as feedback, I usually derive the transfer function directly.
But right now, I have a quite complicated feedback network with some frequency-dependent components.
That's why I want to derive G(s) first.


1639570544585.png

1639570196508.png
 

Attachments

  • 1639570082378.png
    1639570082378.png
    138.3 KB · Views: 174

Sort by date Sort by votes

Attachments

  • AN154.pdf
    843.2 KB · Views: 201
Last edited:
Upvote 0 Downvote
danadakk said:
Just derive the G(s) element expression, and plug that
into 8.2, just as you do for H(s).
Click to expand...

sorry it is a bit unclear for me.
I think I should derive G(s)=A(s)/Y(s) when the feedback network is detached from the feedforward. Is this correct?


I will check your links. Thanks
 

Upvote 0 Downvote
Yes, you treat G(s) as a stand alone network with an input
and output. But dont use Y(s) as input because that would make
it a f( G(s), X(s), Y(s) ). So make input Z(s), output A(s), and then
solve for is T(s) = A(s) / Z(s) and plug that into the general T(s)
for the overall problem.

Regards, Dana.
 
Upvote 0 Downvote
I wonder what you mean with "derive G(s)"? The feedback factor is intentionally chosen in the design process.
 

Upvote 0 Downvote
You are stating in post #1:
I have a quite complicated feedback network with some frequency-dependent components.
That's why I want to derive G(s) first.
Click to expand...
In this case, G(s) will be described by a rational function with numerator and denominator coefficients (zeros and poles). Consider that the "detached" feedback network must be loaded similarly to the final circuit.
 

Upvote 0 Downvote

Upvote 0 Downvote
Status
Not open for further replies.

Similar threads