calculating RC corner frequency the hard way

Thread Starter

adav84

Joined Apr 28, 2012
17
I try to calculate the corner frequency of an RC lowpass in but the result I'm getting is wrong. Here's what I did:
R=17k, C=15n
The lowpass is basically a voltage divider, hence Vout=Vin*Z/Xc
rearrange: Vout/Vin=Z/Xc
The corner frequency is where the voltage drops by sqrt(2) (and power by half):
1/sqrt(2)=Z/Xc
reactance Xc is 1/wC
impedance Z:
Z=17k-jXc=17k-j(1/wC)
substitute:
1/sqrt(2)=(17k-j(1/wC)) / (-j(1/wC))
solve for w:
w=-j(sqrt(2)-2)*(1000/51)=-j1148.60
f=w/(2*pi)=182.806i
to polar: 182.806 Hz<90°

The actual corner frequency is 624.1 Hz

what am I doing wrong? how to do it properly?
 

WBahn

Joined Mar 31, 2012
24,854
Look at your very first line:

Vout=Vin*Z/Xc

If Z is the series impedance of the R and C (which is seems to be based on what you have further down, but it would be nice if you would define your terms before you use them in an equation), then does this expression for Vout make sense? What if you have just a resistive voltage divider?

Also, you have an impedance divided by a reactance. Wouldn't it make more sense if you had an impedance divided by an impedance?

You are just plugging stuff into formulas without taking the time to ask if the formulas you are coming up with make sense.

A HUGE clue that you have done something wrong is that you got a negative imaginary value for w. Does that make sense? If not, STOP! There is not point going any further until you resolve the problem(s) up to this point.

How did you then divide a negative imaginary number by 2pi and somehow, through some magical intervention, end up with a positive imaginary number? And don't switch back and forth between i and j as the imaginary unit -- pick one and stick with it.

What does it mean for a corner frequency to be imaginary (or for a frequency to have an angle of 90°)?
 

MrAl

Joined Jun 17, 2014
6,649
Hi,

Yeah i'd have to ask where did that formula come from too.

The way i used to explain it was to start with a voltage divider we all know and love, the resistive voltage divider made up out of just two resistors R1 and R2 in series, with one end of R2 grounded and one end of R1 going to the positive voltage source:
Vcc o---R1---+---R2---GND

and we want to calculate the voltage at the junction of R1 and R2.

Most people learn the formula as:
Vout=Vcc*R2/(R1+R2)

So it is the resistance connected to ground divided by the sum of both resistors multiplied by the voltage.

Now since impedances work the same way as resistances except they are complex, we can replace R2 with the impedance of C1 when we have a resistor R1 in series with C1 instead of two resistors. Since the impedance of C1 is just 1/(j*w*C1) once we replace R2 with zC1 we get;
Vout=Vcc*zC1/(R1+zC1)

and since zC1=1/(j*w*C1) we have before simplification:
Vout=Vcc*1/(j*w*C1)/(R1+1/(j*w*C1))

and after simplifying that we get the formula for an RC low pass filter. We could then go on to find the amplitude of that and then finally go on to find the -3db point.

So the voltage divider formula comes in handy again :)

To generalize, the final algebraic result with a given topology is the same as that of a network of resistors with the same topology but with the appropriate resistors changed into the appropriate complex impedances followed by an algebraic simplification.

To the OP, if you care to, try again and see what you get. It ends up being a very simple formula for the -3db point.
 
Last edited:

Thread Starter

adav84

Joined Apr 28, 2012
17
\(\frac{V_{out}}{V_{in}}=\frac{Z_C}{R+Z_C}=\frac{\frac{1}{j{\omega}C}}{R+\frac{1}{j{\omega}C}}\)
\(\omega=\frac{j(V_{out}-V_{in})}{V_{out}RC}
=\frac{jV_{out}}{V_{out}RC}-\frac{jV_{in}}{V_{out}RC}
=(1-\frac{V_{in}}{V_{out}})\frac{j}{RC}\)

ok, and now?


EDIT nope, that's wrong, please ignore...


Wouldn't it make more sense if you had an impedance divided by an impedance?
I wrote Z/X, but I (tacitly, sorry) converted X to impedance later:
1/sqrt(2)=(17k-j(1/wC)) / (-j(1/wC))
-j/(1/a)=1/(j*a). Or did you mean something else?

What does it mean for a corner frequency to be imaginary (or for a frequency to have an angle of 90°)?
why, what does it mean?
 
Last edited:

MrAl

Joined Jun 17, 2014
6,649
\(\frac{V_{out}}{V_{in}}=\frac{Z_C}{R+Z_C}=\frac{\frac{1}{j{\omega}C}}{R+\frac{1}{j{\omega}C}}\)
\(\omega=\frac{j(V_{out}-V_{in})}{V_{out}RC}
=\frac{jV_{out}}{V_{out}RC}-\frac{jV_{in}}{V_{out}RC}
=(1-\frac{V_{in}}{V_{out}})\frac{j}{RC}\)

ok, and now?


EDIT nope, that's wrong, please ignore...




I wrote Z/X, but I (tacitly, sorry) converted X to impedance later:

-j/(1/a)=1/(j*a). Or did you mean something else?



why, what does it mean?

Hi,

You were on the right track there...

jwC=j*w*C and jwRC=j*w*R*C here

Vout/Vin=(1/jwC)/(R+1/jwC)

and simplifying we get:
Vout/Vin=1/(1+jwRC)

Now to simplify further, we multiply top and bottom of that by the complex conjugate of the denominator which is:
cc=1-jwRC

so we would then have:
Vout/Vin=(1-jwRC)/((1+jwRC)*(1-jwRC))=(1-jwRC)/(1-j^2*w^2*C^2*R^2)

and for this last expression:
(1-jwRC)/(1-j^2*w^2*C^2*R^2)

we note that j^2=-1 so we can simplify that more:
(1-jwRC)/(1-j^2*w^2*C^2*R^2)=(1-jwRC)/(1+w^2*C^2*R^2)

and note that the denominator no longer has any imaginary part, and we can then easily separate the real and imaginary parts of that expression to get:
RP=realpart=1/(1+w^2*C^2*R^2)
IP=imagpart=-jwRC/(1+w^2*C^2*R^2)

and now we can calculate the amplitude and phase shift with:
Amplitude=sqrt(RP^2+IP^2)
Phase=atan2(IP,RP)

and that allows us to calculate Vout amplitude by simply multiplying by the peak of Vin and also the phase shift,
for any frequency we want to test where w=2*pi*f where f is frequency.

So what we would do next would be to set the frequency to some value like 100Hz, then calculate w=2*pi*100, then use that w in the expression for RP and IP, then calculate the amplitude and phase shift and that would tell us that information for that particular frequency of 100Hz. To do 200Hz we'd do the same thing except use 200Hz as f.
 
Last edited:
Top