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Transfer function for RC circuit

evol_w10lv

Feb 19, 2013
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Hey!
I have got this circuit:
dh05vz73lkwnpryuc8.png


So.. I have to find transfer function to graph logarathmic amplitude-frequency response and logarathmic phase-frequency response.

As we know H(jw) = output phasor/input phasor
Input phasor is total impedance (Z). The impedance of a capacitor is 1/(s*C) = 1/(j*w*C)
By my calculation, Z should be:
y28jv8jsetp0ga7bdeu6.png


Is it correct? And how can I find output phasor? I have to use voltage divider somehow to find it?
 

Laplace

Apr 4, 2010
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Suppose the capacitors were shorted out of the circuit. Then your formula would be Z=(1/R2 + 1/R1). What does that give you?
 

evol_w10lv

Feb 19, 2013
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Then R1 || R2 are loads and transfer function looks something like that:
5ofsvir5ekh3tdykf5ki.png

And:
v8bfd70pro7s4vnsd04n.png


Or am I wrong?
 

Laplace

Apr 4, 2010
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Just by looking at the circuit you should be able to determine that the magnitude plot is wrong. There should be two corner frequencies, separated by three decades of frequency on a log scale. The first corner @ -3dB will have a frequency of 1/(R2C2). From there the plot will slope down to the left at -20dB/decade until it reaches the second corner frequency, 1/(R1C1). From there the plot will slope down to the left at -40dB/decade. You should not need to do any calculations at all in order to sketch the Bode magnitude plot of this circuit, but the transfer function must provide a refined version of the basic sketch.
 

evol_w10lv

Feb 19, 2013
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But then it suggests that transfer function H(jw) is wrong.
tc0iz1vlp9jhslvwnry2.png


I draw Bode graph using Matlab:
lplycaveo5ucqdncne3k.png


There I can see only one corner frequency. What am I doing wrong?
 

Laplace

Apr 4, 2010
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Can you show how the transfer function was derived? Did you start with the node equations? What were they?
 

evol_w10lv

Feb 19, 2013
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It seems to me that I were doing completely wrong before.
This time I calculated H(s) using quadripole method and it is:
6mhtdcuzpjjpc2in2asc.png


It gives me:
c54l1249hb71uqesw4mj.png


Here I can see two corner frequencies as you said it should be. Is it looks correct for you as well?
 

Laplace

Apr 4, 2010
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Looks pretty good to me. I've never used the quadripole method but it seems to work for you.
 

evol_w10lv

Feb 19, 2013
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I still have some non understanding here. I have to find coefficient k and time constants T1 and T2. Actually, as you said before T1 =R1*C1 and T2=R2*C2, but is it possible in this circuit or it's just theoretically time constants? I'm not sure, but as there is no amplifier in that circuit, capacitors interacts between each other, that's why solution of time constants not so "clear" as we would like to have. Or is it absolutely wrong?

And now about calculations.
This is how I found transfer function before:
cbt0z6vzk9mdv183h7bh.png


But I guess, circuit suggests, there is series of two ''inertial differential sections''. Then:
ochz4sb7b6alcpzbr9km.png


When I compare these transfer functions:
72j1yad83y3kaf6d2t3z.png


I'm quite sure, that my first transfer transfer should be correct. Maybe second one is more theoretical? What's your opinion about it?
 
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