Graphing Transfer Function

[MrAl]

Hi Eno,


Well, there is no bandwidth for a low pass filter, but the cutoff freqeuncy increases as the RC time constant decreases.

Increasing R while lower C to maintain the same cutoff frequency causes the impedance to rise. Increasing C while lowering R causes the impedance to fall.

I guess you mean increasing input voltage. Increasing the input voltage at a given frequency causes the output to increase by the same proportion.

You're right about the RC and the impedance.

 

[EN0]

Hey MrAl,

Cool, thanks for those notes.

I'll work on the equations for a high-pass filter, perhaps you can check my math?

Thanks for your help so far,

Austin

 

[EN0]

The impedance will vary with frequency though; it isn't just equivalent to R, correct?

 

[EN0]

First-Order Passive RC High-Pass Filter

Hey MrAl,

Do the equations so far look good?

[LATEX]V_O = \frac{Z_R}{Z_R + Z_C}[/LATEX]

[LATEX]V_O = \frac{Z_R}{Z_R + \frac{1}{j \omega C}}[/LATEX]

Conjugate:

[LATEX]= \frac{V_I_N * Z_R}{Z_R + \frac{1}{j \omega C}} * \frac{Z_R - \frac{1}{j \omega C}}{Z_R - \frac{1}{j \omega C}}[/LATEX]

Simplify:

[LATEX]= \frac{V_I_N * Z_R^2 - \frac{V_I_N * Z_R}{j \omega C}}{Z_R^2 + \frac{1}{j \omega C}}[/LATEX]

[LATEX]= \frac{V_I_N * Z_R^2 - \frac{V_I_N * Z_R}{j \omega C}}{Z_R^2 + \frac{1}{\omega^2 C^2}}[/LATEX]

Separate:

[LATEX]Real = \frac{V_I_N * Z_R^2}{Z_R^2 + \frac{1}{\omega^2 C^2}}[/LATEX]

[LATEX]Imaginary = -\frac{\frac{V_I_N * Z_R}{\omega C}}{Z_R^2 + \frac{1}{\omega^2 C^2}}[/LATEX]

[LATEX]= -\frac{V_I_N * Z_R}{Z_R^2 \omega C + \frac{1}{\omega C}}[/LATEX]

Hopefully that's all correct!

Thanks,

Austin

 

[MrAl]

Hi again,


Yes the impedance varies with frequency, but for any frequency F1 we can specify an impedance Z1 at that frequency with one set of R and C, and another impedance Z2 at that same frequency with another different set of R and C. For example, with R1 and C1 we get Z=10 at some frequency F1, and at that same frequency F1 we get Z=100 with R2 and C2.

You got close to the right answer for the high pass filter, but there is one little problem:
The conjugate of R+1/(j*w*C) is not equal to R-1/(j*w*C).
Remember that to come up with the conjugate we have to take the opposite sign of the imaginary part, and the imaginary part is not 1/(j*w*C) because j is in the denominator and the proper form to start with is:
RealPart+j*ImagPart

not:
RealPart+1/(j*ImagPart)

which is much different.

One way around this when we have j in the denominator of the imag part is to multiply top and bottom of the imag part by j such as with:
1/(j*w*C)

we multiply top and bottom by j and we get:
(j)/(j^2*w*C)

which is the same as:
j/(-1*w*C)

which is equal to:
-j/(w*C)

and now we have j in the numerator as required, and now the sign is negative. So that means for R+1/(j*w*C) we would first have to fix this into the required form of Real+j*Imag and multiplying the imag top and bottom by j we would end up with this:
R+(-j/(w*C))

which of course is the same as:
R-j/(w*C)

Now we can inspect and see that the real part is R and the imag part is -1/(w*C).

You can try it that way if you want, but sometimes it's simpler to simplify the entire equation by multiplying top and bottom by j*w*C and get the answer that way.

Try it again having this new information and im sure you'll get it right as your procedure otherwise looks good.

 

[EN0]

Alright, I'll try that again:

[LATEX]V_O = \frac{Z_R}{Z_R + \frac{1}{j \omega C}}[/LATEX]

[LATEX]V_O = \frac{Z_R}{Z_R + \frac{1}{j \omega C}} * \frac{j \omega C}{j \omega C}[/LATEX]

[LATEX]V_O = \frac{Z_R j \omega C}{Z_R j \omega C + 1}[/LATEX]

[LATEX]Real = 1[/LATEX] - This doesn't seem right.

[LATEX]Imaginary = \frac{Z_R \omega C}{Z_R \omega C + 1}[/LATEX]

Does that all look correct?

Thanks,

Austin

Edit: Forgot to get rid of the complex number on the denominator!

I guess it looks like this:

[LATEX]V_O = \frac{j - Z_R \omega C}{-Z_R \omega C}[/LATEX]

 

[MrAl]

Hi again,


Well, take your:



and now find the conjugate of the denominator and multiply top and bottom, then take it from there. I think you'll have it after this.

 

[colors]

hi,

a little question,
if i have the circuit , i can get the response graph,
is there a way to get the circuit, if i have the desired response graph, the ckt might not be a filter,
actually i was looking for a way to change a transfer function to an equivalent ckt.

 

[MrAl]

hi,

a little question,
if i have the circuit , i can get the response graph,
is there a way to get the circuit, if i have the desired response graph, the ckt might not be a filter,
actually i was looking for a way to change a transfer function to an equivalent ckt.

Hello there,


Well, if it's not a filter then what would it be. If it's something else non linear then it could be very difficult to tell the exact circuit that made it, and in fact there may be many circuits that could do the same thing.

If it's a filter of some type then we can use Bode analysis to get the approximate poles and zeros and work backwards from there.

 

[colors]

thanks,
and moreover, the circuit that i develop that way will have only L,C,R , i suppose.

 

[EN0]

thanks,
and moreover, the circuit that i develop that way will have only L,C,R , i suppose.

My grandfather has this type of thing at his work. Where you simulate all the effects of parasitic capacitance, inductance, and resistance that non-perfect components exhibit. Eventually, from the beginning of the circuit to the end, it will produce a transfer function.

It doesn't come cheap, though.

 

[EN0]

Hi again,


Well, take your:



and now find the conjugate of the denominator and multiply top and bottom, then take it from there. I think you'll have it after this.

Hi,

Ok, hopefully this looks better:

[LATEX]V_O = \frac{Z_R j \omega C + Z_R^2 \omega^2 C^2}{1 + Z_R^2 \omega^2 C^2}[/LATEX]

[LATEX]Real = \frac{Z_R^2 \omega^2 C^2}{1 + Z_R^2 \omega^2 C^2}[/LATEX]

[LATEX]Imaginary = \frac{Z_R \omega C}{1 + Z_R^2 \omega^2 C^2[/LATEX]

Thanks

 

[EN0]

Forgot Vin!

I think this is it:

[LATEX]V_O = V_I_N \frac{Z_R}{Z_R + \frac{1}{j \omega C}}[/LATEX]

Conjugate:

[LATEX]V_O = V_I_N \frac{Z_R}{Z_R + \frac{1}{j \omega C}} * \frac{j \omega C}{j \omega C}[/LATEX]

Simplify:

[LATEX]V_O = \frac{V_I_N Z_R j \omega C}{Z_R j \omega C + 1}[/LATEX]

Conjugate:

[LATEX]V_O = \frac{V_I_N Z_R j \omega C}{Z_R j \omega C + 1} * \frac{1 - Z_R j \omega C}{1 - Z_R j \omega C}[/LATEX]

Simplify:

[LATEX]V_O = \frac{V_I_N Z_R j \omega C + V_I_N Z_R^2 \omega^2 C^2}{1 + Z_R^2 \omega^2 C^2}[/LATEX]

Separate:

[LATEX]Real = \frac{V_I_N Z_R^2 \omega^2 C^2}{1 + Z_R^2 \omega^2 C^2}[/LATEX]

[LATEX]Imaginary = \frac{V_I_N Z_R \omega C}{1 + Z_R^2 \omega^2 C^2}[/LATEX]

Amplitude:

[LATEX]Amplitude = \sqrt{(\frac{V_I_N Z_R^2 \omega^2 C^2}{1 + Z_R^2 \omega^2 C^2})^2 + (\frac{V_I_N Z_R \omega C}{1 + Z_R^2 \omega^2 C^2})^2}[/LATEX]

Square terms:

[LATEX]Amplitude = \sqrt{\frac{V_I_N^2 Z_R^4 \omega^4 C^4}{(1 + Z_R^2 \omega^2 C^2)^2} + \frac{V_I_N^2 Z_R^2 \omega^2 C^2}{(1 + Z_R^2 \omega^2 C^2)^2}}[/LATEX]

Factor:

[LATEX]Amplitude = \sqrt{\frac{V_I_N Z_R^2 \omega^2 C^2 (1 + Z_R^2 \omega^2 C^2)}{(1 + Z_R^2 \omega^2 C^2)^2}}[/LATEX]

Simplify:

[LATEX]Amplitude = \sqrt{\frac{V_I_N Z_R^2 \omega^2 C^2}{1 + Z_R^2 \omega^2 C^2}}[/LATEX]

Simplify to final answer:

[LATEX]Amplitude = \frac{V_I_N Z_R \omega C}{\sqrt{1 + Z_R^2 \omega^2 C^2}[/LATEX]

I hope that's correct!

Thanks,

Austin

 

[MrAl]

Hi again,


Looks like you got the right answer. I wondered how you dealt with Vin^2, but however you did it you got it right this time

 

[EN0]

Hi MrAl,

I'm glad that I got it correct, thank you very much for your help. It looks great on the graph I produced, I think I'll eventually have to show you guys all the graphs via MathCAD.

Eventually I intend to do the mathematics for most all passive and active filters. Then, I might create a seperate coherent thread for people who would like to learn more about designing their own filters. This stuff is excellent, and I know many would be grateful to learn it.

I'm open to advice and suggestions as to what path I, and others, should take in order to make this possible. What order do you think I should learn each of the filters? Here's a brief list, for instance:

• RC Low-Pass Filter
• RC High-Pass Filter
• RC Band-Pass Filter

That is, of course, simply with just caps and resistors. I would like to get involved with inductors as well, which shouldn't change the equations too much. Perhaps I should work on second order filters now?

Also, one more note for the RC Low-Pass Filter; the equations seem to be strange, now that I review them. Shouldn't it be as follows?

[LATEX]V_O = V_I_N (\frac{\frac{1}{j \omega C}}{\frac{1}{j \omega C} + Z_R})[/LATEX]

Instead of what you had:

[LATEX]V_O = V_I_N (\frac{j \omega C}{ j \omega C + Z_R})[/LATEX]

Thanks,

Austin

 

[EN0]

Hi,

Nevermind about that previous note, I contrived the final equation for a LPF myself:

[LATEX]Amplitude = \frac{V_I_N}{\sqrt{1 + Z_R^2 \omega^2 C^2}}[/LATEX]

Austin

 

[MrAl]

Hi again Eno,


I think you are right in going from LP to HP and then to BP filters in that order. You have the LP so now on to the HP ok?

Oh yeah, we dont want to leave out the BS (Band Stop) filter either, and if you really want an interesting challenge, check out the twin tee passive notch filter.

 

[EN0]

Hi,

I have done the equations for both LP and HP, so now I think I'll move on to BP. Essentially, I think that a BP filter is comprised of both a HP and LP filter. I'll submit my equations for the BP filter so that you can check them.

Thanks,

Austin

 

[EN0]

RC Band-Pass Filter

Hey MrAl,

Before I do too much work, I'd like to know if this is correct so far?

[LATEX]V_O = (\frac{V_I_N Z_R}{\frac{1}{j \omega C} + Z_R})(\frac{\frac{1}{j \omega C}}{\frac{1}{j \omega C} + Z_R})[/LATEX]

[LATEX]V_O = \frac{\frac{V_I_N Z_R}{j \omega C}}{(\frac{1}{j \omega C} + Z_R)^2}[/LATEX]

[LATEX]V_O = \frac{\frac{V_I_N Z_R}{j \omega C} (Z_R^2 - \frac{1}{\omega^2 C^2} + \frac{2 Z_R}{j \omega C})}{Z_R^4 + \frac{3Z_R^2 - Z_R}{\omega^2 C^2} - \frac{1}{\omega^4 C^4}}[/LATEX]

[LATEX]V_O = \frac{\frac{V_I_N Z_R^2}{j \omega C} - \frac{V_I_N Z_R}{j \omega^3 C^3} + \frac{V_I_N * 2Z_R^2}{\omega^2 C^2}}{Z_R^4 + \frac{3Z_R^2 - Z_R}{\omega^2 C^2} - \frac{1}{\omega^4 C^4}}[/LATEX]

This really is good LaTex practice!

 

[MrAl]

Hi again Eno,


I think you are definitely on the right track now, but i dont think we should assume that the R and C are the same for the high pass as for the low pass, and in fact we should assume that they are definitely not the same because we could certainly want to set the frequencies differently.

As i said, i think you are definitely on the right track here and have made a lot of progress already, but there is one additional thing you should know. If you convolve two systems like you have done (and that's perfectly fine with an active element included in the filter) that assumes that there is at least one active element or at least the impedance of the second circuit is much higher than the first. That's a valid assumption for some circuits, but it's not for others. So i guess what we should do is analyze both possibilities, if that's what you would like to do. It's a little more work to do the network as a whole rather than divide it up and convolve the two, but it may be worth it.
So i guess the question i have to ask you now is, do you want to work the filter with either:
1. At least one active gain of 1 element, or
2. The impedance of the second circuit is much higher than the first

or do you want to do a general passive analysis where we could find the solution to ANY passive BP filter, without any constraints as outlined above?