# What kind of filter can I make with this circuit?

#### omar-rodriguez

Joined Jun 24, 2015
67
This is the circuit there are two question that I really don't know how to answer

- What kind of filter can you do with this circuit?
- how many cut off frequencies can you adjust with the value of the four components?

I have found the transfer function, the poles and zeros, but then I don't know what to do #### MrAl

Joined Jun 17, 2014
10,068
Hi there,

You can create a lowpass, highpass, or bandpass filter. However, if you dont choose bandpass then you will be wasting one of the caps because the value will have to be set low or just leave one cap out. The bandpass however will require both caps so that's the most likely for this circuit topology. You might also get a bandstop too however. Once you calculate the amplitude and plug in some values you can see what kind of responses you get.

If your analysis is correct (didnt check that yet myself) then what you can do next if you want the frequency response is set s=j*w and then solve for the amplitude, then graph the amplitude vs frequency. That gives you the most complete picture unless you need the phase response too and then you find the real and imaginary parts and take the inverse tangent using the two argument inverse tangent function.

If you have the poles then you can calculate some interesting frequencies too.

Have you ever done this before? If not we can go over an example.

LATER:
I get transfer function similar to yours:
((s*C1*R1+1)*(s*C2*R2+1))/(s^2*C1*C2*R1*R2+s*C2*R2+s*C2*R1+s*C1*R1+1)
just written a little differently.

For the denominator roots i get:
s1=
(-sqrt(C2^2*R2^2+(2*C2^2-2*C1*C2)*R1*R2+(C2^2+2*C1*C2+C1^2)*R1^2)-C2*R2-C2*R1-C1*R1)/(2*C1*C2*R1*R2)
s2=
(sqrt(C2^2*R2^2+(2*C2^2-2*C1*C2)*R1*R2+(C2^2+2*C1*C2+C1^2)*R1^2)-C2*R2-C2*R1-C1*R1)/(2*C1*C2*R1*R2)

In fact, with both caps it looks more like a bandstop.

Last edited:

#### omar-rodriguez

Joined Jun 24, 2015
67
I mean, how do you know that you can do an passband filter or an stopband filter with that circuit, under what condition? it is related to the amount of zeros and poles?

#### Jony130

Joined Feb 17, 2009
5,445
First assume very low frequency signal, so Xc = ∞ it is easy to see that Vin = Vout. For high frequency Xc = 0Ω therefore Vout = Vin.
and at the middle frequency Xc ≈ R we will have some attenuation (Z2/(Z1 + Z2)). So, this look like a bandstop filter.

#### MrAl

Joined Jun 17, 2014
10,068
I mean, how do you know that you can do an passband filter or an stopband filter with that circuit, under what condition? it is related to the amount of zeros and poles?

Hello again,

Well to start you can look at the circuit and try to imagine what happens as the frequency changes from 0 to infinity, knowing that capacitors act as open circuits for low frequencies and short circuits at high frequencies, and that resistors stay the same for all frequencies.

For example, at low frequency like 0 (DC) the value of each resistor is all that matters, but because C2 is open only R1 matters, so with no load the output must be 1 (same as input).
At high frequency, C1 and C2 become shorts, so C1 conducts right to the output making R2 look insignificant. This means at w=infinity we have 1 again.
Because there are no inductors and no gain stage, anything in between these two extremes should be less than 1, which means it's probably a bandstop.
At a mid frequency, if C1 is the same as or different than C2 we get an in between value, so it looks like a bandstop.

If you make C2 very very small, it will start to look more like a high pass filter. If you make C1 very very small, it will start to look more like a low pass filter. This would be a more extreme view however because then either cap would be a waste.

To be more exact, calculate the ampltiude and plug in some well chosen values.

#### omar-rodriguez

Joined Jun 24, 2015
67

#### MrAl

Joined Jun 17, 2014
10,068
Hi again,

You're welcome, and also, if you study the asymptotic Bode method you can find out from the poles and zeros, but when the calculation of the poles or zeros is kind of complicated you'll still have to plug in some values and kind of think about it a little like we did when thinking about the zero to infinity frequency and the way caps respond to frequency.