# Simple low pass filter calculation...

Discussion in 'Math' started by dannybeckett, Oct 31, 2011.

1. ### dannybeckett Thread Starter Active Member

Dec 9, 2009
163
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Just a quick question...

Using Fc=1/(2*pi*R*C) to find out the cut-off frequency of a low pass filter of say a filter with values of 10K and 500pF, you get an answer of 31,831Hz. This gives you the -3dB point I believe. When I try and prove this by working out the reactance of the capacitor at that frequency, then using the potential divider rule to find the voltage loss, my sums don't add up:

Xc=1/(2*pi*f*c)=10,000R

A potential divider with a resistance of 10K and 10K gives you half the voltage you put in to it, not the Vin*0.707 I was expecting. Any idea on where I am going wrong here?

Thanks a lot!

Feb 17, 2009
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3. ### MrChips Moderator

Oct 2, 2009
13,353
3,745
Your numbers are correct. You cannot just add the R and Xc as in a voltage divider. The voltages across R and Xc are 90 degrees out of phase. See:

Since R and the magnitude of Xc are both 10K, the resultant vector will be at 45 degrees. Vr will be 0.707*Vin.

4. ### dannybeckett Thread Starter Active Member

Dec 9, 2009
163
0
Should have realised this. Thanks guys

5. ### dannybeckett Thread Starter Active Member

Dec 9, 2009
163
0
OK so I've worked everything out and it works fine. I'm a little stuck on one more part. Well, kind of...

A resistance was added to the output of this circuit to ground. When working out the new characteristics of the low pass filter, what I did was work out the capacitor impedance in parallel with the new resistor added (because it physically is in parallel with the capacitor). Carried out the sums and everything worked out (tested the circuit in multisim to verify my numbers). I have been told that doing it this way is wrong, and what I really needed to do was put the output resistor in parallel with the input resistor, because of nodal impedances.

So when working out the total circuit impedance using vectors (pythagoras), you have the impedance = sqrt[(R1||R2)^2+Xc^2]. By this definition, it doesn't matter if you swap around R1 and R2, since they are in parallel with eachother. R1||R2 = R2||R1.

However, when you swap around the 50k and 100k resistor, the output of the system drops significantly. This calculation does not account for that drop, can anyone shed some light onto this subject? When I do it my way (parallel Xc with whatever the output resistance is) the value I get is very close to what multisim tells me, but not quiite right.

6. ### justtrying Active Member

Mar 9, 2011
329
571
from what i know, your first approach seems to be right (assume you are going Zt = R2llZc+R1 and then apply vector math). I guess you could put the load resistor in parallel with R1 but that changes the circuit completely and of course once you do that, swapping R1 and R2 will have no effect on your total impedance... does not make much sense to me, different circuit implemenation.

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784

The calculation can be done either way.

1. Calculate the Thevenin equivalent for the Source, Zs & Zin

Vth=(100k/150k)*Vs=0.667Vs
Zth=Zs||Zin=50k||100k=33.33k

So
Vout=Vth*-jXc/(Zth-jXc)=0.667Vs*-j10k/(33.33k-j10k)

or

Vout=(0.055-j0.1836)*Vs

Using rectangular to polar conversion the output magnitude is therefore found to be ~0.192*Vs. The phase shift is -73.3°

2. Find the parallel equivalent of -jXc and Zin

Zp=-jXc*Zin/(-jXc+Zin)=-j10k*100k/(-j10k+100k)=(0.99-j9.9)k

Vout=Vs*Zp/(Zs+Zp)=Vs*(0.99-j9.9)k/((0.99-j9.9)k+50k)

or

Vout=(0.055-j0.1835)*Vs

Again, using rectangular to polar conversion, the output magnitude is then found to be ~0.192*Vs