Engineering Circuit Analysis Help: Find Frequency Response Equation

AI Thread Summary
To find the frequency response equation for the given circuit, the nodal analysis method is recommended, requiring four equations. The first equation establishes that the voltage at node v1 equals the input voltage Vi. Subsequent equations are derived from Kirchhoff's Current Law (KCL) applied to nodes v3 and v4, with specific relationships between the node voltages. It is crucial to ensure that all node voltages are represented with single coefficients to simplify the equations. Redrawing the circuit may help clarify the relationships between components and facilitate the analysis.
jaderberg
Messages
28
Reaction score
0
I have to find the frequency response equation for this circuit in the attatched photo, but i don't know how to go about analysing it as I cannot see how to do voltage loop and node analysis does not work as the two nodes are not related so nothing can be eliminated from the generated equations... any suggestions?

thanks
http://img361.imageshack.us/img361/3217/dsc00275ls9.jpg
http://g.imageshack.us/img361/dsc00275ls9.jpg/1/
 
Last edited by a moderator:
Physics news on Phys.org
You can use the nodal method. You will need 4 equations. Designate the node between the two capacitors as v3 and the one between the two resistors as v4.

Assume that a voltage source, Vi, is connected to node v1. Then the transfer function will be the ratio v2/Vi, in terms of the Laplace variable s.

The first equation will be a constraint equation, simply saying that v1 is equal to Vi:

1*v1 + 0*v2 + 0*v3 +0*v4 = Vi

The third equation is derived from an application of KCL to node 3:

(-s*C)*v1 + (-s*C)*v2 + (s*C+s*C+2/R)*v3 + 0*v4 = 0

You should be able to fill in the 2nd and 4th equations.

Then solve the linear system that results and the result you get for the voltage at node v2 will be Vi times a fraction in powers of s. That fraction is the transfer function.

If you still have problems, show your work and you'll get more help.
 
yeah that's what i started doing, then the next equation at node v4 will be:

v1 + v2 - v4*(2RCs + 2) = 0

which does not include any v3 terms so i cannot relate the two equations to eliminate v3.

i cannot see any other nodes to analyse at this point so this is where i am stuck!
 
If you redraw the circuit, you will see that the first capacitor C, the R/2 resistor, the capacitor 2C and the first resistor R form a loop.
In the same way, the second capacitor C, the R/2 resistor, the capacitor 2C and the second resistor R form a loop too.
The voltages v1 and v2 are in the diagonals of the loops.
 
jaderberg said:
yeah that's what i started doing, then the next equation at node v4 will be:

v1 + v2 - v4*(2RCs + 2) = 0

which does not include any v3 terms so i cannot relate the two equations to eliminate v3.

i cannot see any other nodes to analyse at this point so this is where i am stuck!

You can get an equation for each of the 4 nodes.

The equation at node 4 should be:

(-1/R)*v1 + (-1/R)*v2 + 0*v3 + (2/R+2*s*c)*v4 = 0

This equation has been simplified somewhat. You add all the currents in each component connected to the node and equate the sum to zero. Here's the very lowest level equation:

1/R*(v4-v1) + 1/R*(v4-v2) + 2*s*C*(v4) = 0

If you rearrange so that each node voltage, V1, v2, v3 and v4 has a single coefficient, you should get what I gave above.

Use the same method to get the equation for node 2
 

Similar threads

Replies
4
Views
993
Replies
7
Views
2K
Replies
25
Views
3K
Replies
2
Views
1K
Replies
6
Views
4K
Replies
3
Views
3K
Replies
3
Views
3K
Replies
5
Views
2K
Replies
2
Views
1K
Back
Top