Electronics > Beginners
Add a zero to this RC transfer function
bonzer:
Hello everyone! Please help me to find out how to add a zero at higher frequency to this transfer function (the first one). One solution I found is to insert Rx in series with that capacitor but someone already did it I would like to discover new ways. If what you think of adds another pole too, it's not a problem unless it's at an even higher frequency like 2 decades after.
Benta:
Try adding a cap in series with the first resistor. But the argument that "someone else already did it" leaves me somewhat baffled.
MrAl:
--- Quote from: bonzer on January 05, 2019, 04:49:04 pm ---Hello everyone! Please help me to find out how to add a zero at higher frequency to this transfer function (the first one). One solution I found is to insert Rx in series with that capacitor but someone already did it I would like to discover new ways. If what you think of adds another pole too, it's not a problem unless it's at an even higher frequency like 2 decades after.
--- End quote ---
Hello,
I dont think you get another zero by placing a cap in series with the first resistor in the first diagram.
However, i think you do get another zero if you place a cap in parallel to either the first resistor or the third resistor.
I think the idea is to get it to behave more 'high pass like'.
Also note that either of those caps would 'bridge' either a source and cap or two caps so probably no extra pole.
We could verify with some formulas.
In the numerator, s*a+1 is one zero, (s*a+1)*(s*b+1) is a second zero, so you should get two factors like that in the numerator.
The transfer function of the original circuit is:
Vout/Vin=(s*C2*R3+1)/(s^2*C1*C2*R1*R3+s*C2*R3+s^2*C1*C2*R1*R2+s*C2*R2+s*C2*R1+s*C1*R1+1)
and note only one factor in the numerator.
We could go over this in detail if you like and change components and check.
And yes that extra resistor in series with the bottom (second) resistor creates another zero too.
Wimberleytech:
--- Quote ---I dont think you get another zero by placing a cap in series with the first resistor in the first diagram.
--- End quote ---
Yes, it puts the zero at the origin.
MrAl:
--- Quote from: Wimberleytech on January 06, 2019, 02:20:08 am ---
--- Quote ---I dont think you get another zero by placing a cap in series with the first resistor in the first diagram.
--- End quote ---
Yes, it puts the zero at the origin.
--- End quote ---
Hi,
Thanks for the reply. You are right, except it doesnt look like it is at the origin. You can show your transfer function if you disagree.
Here is the transfer function i get with both the 'lower' R8 added and the series 'C8' added:
((s*C8*R1+1)*(s*C2*R3+1)*(s*C1*R8+1))/(s^3*C1*C2*C8*R1*R3*R8+
s^2*C1*C2*R3*R8+s^3*C1*C2*C8*R1*R2*R8+s^2*C1*C2*R2*R8+s^2*C1*C8*R1*R8+s^2*C1*
C2*R1*R8+s*C1*R8+s^2*C2*C8*R1*R3+s^2*C1*C2*R1*R3+s*C2*R3+s^2*C2*C8*R1*R2
+s^2*C1*C2*R1*R2+s*C2*R2+s*C8*R1+s*C2*R1+s*C1*R1+1)
Here we can see three zeros because that lower R8 is still in the circuit. If we short that out, we get two zeros but it doesnt look like one of those is at the origin. Take a look if you can.
Here's what i get with all R's the same and all C's the same:
(s*C*R+1)^2/(4*s^2*C^2*R^2+5*s*C*R+1)
I see a double zero there but not at the origin so would be interesting to see your transfer function.
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