Half power frequency of passive filters

[jb885]

Please can anyone find the half power freq. of both of these circuits step by step so I can finally understand how to do it?
for the first one Vout is across R2.

 

[Winterstone]

Both circuits are a first order passive lowpass with a general transfer function

H(s)=Ao/(1+sT) T=RC time constant=1/wo (wo=cut-off frequency equivalent to the desired half-power frequency).

You can derive the transfer functions for each circuit by yourself simply by applying the complex voltage divider concept.
Rewrite these functions to get the form of the general function H(s) as shown above. Then compare the time constants.
Take the inverse and you have the wanted result.

 

[MrAl]

Hi,

Winterstone:
I see you like these kinds of circuits too huh?

jb885:
Just so you know, the half power frequency is the frequency where the output is 3db down from the norm. This means the gain is 1/sqrt(2) down from the normal output level.

 

[Winterstone]

Hi,
Winterstone:
I see you like these kinds of circuits too huh?
.

Yes, I like filters (analog please!) and passive RC circuits are basic.

 

[jb885]

Im' not sure how to do this with the complex voltage divider..........
For the second can't I just use fc=1/2piRC where R is R1|R2??? but on the other hand R1 is parallel to R2 in series with C1 right?

 

[jb885]

Im' not sure how to do this with the complex voltage divider..........
For the second can't I just use fc=1/2piRC where R is R1|R2??? but on the other hand R1 is parallel to R2 in series with C1 right?

 

[Winterstone]

Im' not sure how to do this with the complex voltage divider..........
For the second can't I just use fc=1/2piRC where R is R1|R2??? but on the other hand R1 is parallel to R2 in series with C1 right?

Voltage divider: Vout/Vin=(R2||Xc)/(R1 +R2||Xc) with R2=10k and Xc=1/jwC

 

[jb885]

I think I' am starting to understand but what do I do with j since I don't have complex numbers? Secondly what does s represent and thirdly how was the transfer function derived(Is it standard?)?

 

[Winterstone]

I think I' am starting to understand but what do I do with j since I don't have complex numbers? Secondly what does s represent and thirdly how was the transfer function derived(Is it standard?)?

OK, here are my explanations:
Starting with the equation as given in post#7 (for the second circuit):

Vout/Vin=(R2||Xc)/(R1 +R2||Xc)=N(s)/D(s) with N(s)=1/(1/R2+sC) and D(s)=R1+1/(1/R2+sC).

It is useful to rewrite this equation (by multiplying N and D by suitable terms) to get the so called "normal form" with s-polynoms in both numerator and denominator - the latter containing "1" as an absolute value.
This leads to

H(s)=Vout/Vin=[R2/(R1+R2)]/(1+s*RpC) with Rp=R1||R2

Now, if you compare this expression with the general first order transfer function (as given in post#2)

H(s)=Ao/(1+sT)

the actual time constant is T=RpC=1/wo ; Thus: wo=1/RpC (wo=cut-off frequency equivalent to the desired half-power frequency).

*Additional explanation: In this case, "s" is nothing else than an abbreviation for jw. That means: s=jw.
For s=jwo=j*(1/T) the denominator is D(s=jwo)=(1+j) and the magnitude is sqrt(1+1)=sqrt(2)=1.414.
That means: For s=jwo the numerator N(s) is multiplied by 1/1.414=0.707 which is eqivalent to a 3dB loss.
This justifies the definition of the cut-off frequency.

 

[jb885]

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[jb885]

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[jb885]

One last thing...............I can find fc using the transfer function(finally) but I'm wondering if its possible to use the formula Xc=1/2πfc and argue that at the break point for the first filter R2=R1 + (1/2πfC) and solve for f. Similarly for the second
argue that at the break point Xc= R1||R2 and solve for f. Is it possible?

 

[Winterstone]

One last thing...............I can find fc using the transfer function(finally) but I'm wondering if its possible to use the formula Xc=1/2πfc and argue that at the break point for the first filter R2=R1 + (1/2πfC) and solve for f. Similarly for the second
argue that at the break point Xc= R1||R2 and solve for f. Is it possible?

No, you can't do that. Because you forgot that the capacitive impedance is imaginary (you suppressed the "j"). For example, assume R1=R2 and the result would be 1/2Pi*fc=0.

But another important point: Following your question I again had a look on your first circuit as well as my first reply - and I must confess that it contains a (nearly catastrophic) error:
Of course, the first circuit is no lowpass but a highpass with the transfer function

H(s)=sR2C1/[1+sC1(R1+R2)]

and the cut-off frequency (inverse of the "s" factor in the denominator): wo=1/[C1(R1+R2)] .

 

[MrAl]

One last thing...............I can find fc using the transfer function(finally) but I'm wondering if its possible to use the formula Xc=1/2πfc and argue that at the break point for the first filter R2=R1 + (1/2πfC) and solve for f. Similarly for the second
argue that at the break point Xc= R1||R2 and solve for f. Is it possible?

Hi,

Maybe what you meant here was that for the second circuit you want to simplify by finding an equivalent R for use in the RC time constant. I think you can do that for this circuit because the resistances act together as if they were in parallel so we get R=R1||R2
and then we can reason that the RC time constant is:
tau=R*C
The only difference though is that we also have to calculate the gain, which is the voltage divider formed by R1 and R2:
K=R2/(R1+R2)

So the entire transfer function then would come out to a more familiar form:
K/(s*R*C+1)

where R is given above and we had to include the gain K above too.

If that's what you meant this works.

 

[Winterstone]

Similarly for the second
argue that at the break point Xc= R1||R2 and solve for f. Is it possible?

Hi jb885, I strongly recommend not trying to follow this approach. I know what you mean: You follow the rule: Imaginary part=real part at f=fo.
Sometimes this may work (for very simple circuits like the one under discussion) but in most cases it will not.
And therefore my recommendation: Always find the pole frequencies using the transfer function. Remember: That was the way the cut-off frequency was defined!

 

[jb885]

Hi,
In the general function A0/1+sC what does A0 represent? For both circuits actually we don't care what it is right? we only care finding the values near s(?) Also for total impedance I noticed that we don't use the formula for finding the impedance of a resistor with a capacitor in parallel (Z=RXC/sqr(R^2 + xc^2))) or in series formula
(Just a bit confused with all these complex numbers && (things!))

 

[Winterstone]

Jb885,
there is no reason to be confused.
Applying the basic rules for circuit analysis (Voltage divider, Kirchhoff's current and voltage laws) you can find the transfer function H(s) for all linear passive and active circuits
And - the transfer function tells you everything.
*Set s=jw=0 and you get the dc value (in your above example: Ao).
*Let s=jw approach infinite and you see the behaviour for very large frequencies.
*Try to write H(s) in a suitable form - and you can identify by visual inspection poles, zeros, time constants.
*For second order (or higher order) circuits this requires to calculate poles (and perhaps zeros); this can be involved or time-consuming. Thus, for this purpose there are simulation programs to support these analyses.

 

[MrAl]

Hi,

I have to agree fully with Winterstone in that we shouldnt take shortcuts. The other method i presented was more for curiosity really and to show how these simpler circuits can be viewed when time is important and you need a quick idea what will happen.


Winterstone:
Very nice explanations from you i see

 

[jb885]

Hi,
With what terms did you multiply the voltage divider (circuit2) to simplify it?????? I can't get it how you simplified it to get it to be in the desired format although I can understand all the other steps as well as the method. The other circuit is much easier to simplify and I get it why the transfer function is H(s)=sR2C1/[1+sC1(R1+R2)]. When i try to simplify the second I get stuck.

circuit 2 Vout/Vin=(R2||Xc)/(R1 +R2||Xc)=N(s)/D(s) with N(s)=1/(1/R2+sC) and D(s)=R1+1/(1/R2+sC).
Fine with this but with what terms do you multiply to get it to be in the format we want?

 

[Winterstone]

In order to find suitable multiplication terms it is important to know what you want - that means: Which form for H(s) is desired?
Answer: H(s) should assume the so called "normal form" which means:

N(s)=a + b*s +d*s^2 (note: not all factors must exist)
and D(s)=1 + d*s + e*s^2 (important: c=1 , e=0 for first order)

That means for the present case:
step 1: multiply with [(1/R2)+sC] >>> now N(s)=1 (but c=(R1+R2)/R2 in the denominator)
step 2: multiply with R2/(R1+R2) >>> now: N(s)=a=R2/(R1+R2) and D(s)=1 + RpC with Rp=R1||R2.

In short: The transfer function consists of two polynominals in "s" and the constant term in the denominator must be "1".
This is the standard form for first and second order functions.
(a) In case of first order you can directly identify the cut-off frequency wc (wc=1/d with d=time constant)
(b) In case of second order the solutions of D(s)=0 give the two (complex ?) poles.