# RC Bandpass transfer function

Discussion in 'Homework Help' started by fireface, Apr 21, 2014.

1. ### fireface Thread Starter New Member

Apr 21, 2014
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0
I'm having trouble getting the right transfer function for this using the OC and SC method for the time constants and getting the poles.

Am I doing this right ...
Open circuit
1/s1 = (R1+R2)*C1
1/s2 = C2*R2
1/s3 = (R2+R3+R4)*C3
1/s4 = C4*R4

Short circuit
1/s5 = R1*C1
1/s6 = (R1||R2 +R3)*C2
1/s7 = R3*C3
1/s8 = R3||R4 *C4

Where do I go after this?

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2. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,516
515
Is this in terms of jw or s?

3. ### fireface Thread Starter New Member

Apr 21, 2014
3
0
's'... i think. sorry, been a long time

s = jw?

Did I calculate the poles?

When I run it through multisim with the given values I get a pole at 500 rads/s. In my calcs I'm getting two SC poles at 1000 rads/sec and one OC at 333 rads/s. Is the combining effect giving me a corner frequency at 500 or am I just making things up?

4. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,516
515
Replace the capacitors with their impedance [Zc=1/(sC)=1/(jwC)]. Then use node voltage or mesh current method to get Vout in terms of Vin, so you will have Vout=(something)*Vin. Then carry Vin to the left side of the equal sign so you will have:
Vout/Vin=(something)=Transfer Function=H(s)=H(jw)

Done!

5. ### WBahn Moderator

Mar 31, 2012
18,089
4,917
Are you trying to do an analysis based on a dominant-pole assumption?

6. ### fireface Thread Starter New Member

Apr 21, 2014
3
0
^Yes

Indeed approximating the poles and cutoff frequency. So the dominant pole is the smallest for high pass and the opposite for low pass?