### Author Topic: Can anyone explain this? Bandpass Circuit?  (Read 3276 times)

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#### Hairston

##### Can anyone explain this? Bandpass Circuit?
« on: September 03, 2003, 12:12:51 AM »
http://home.lehighone.com/rmetzger/images/Bandpass_Schematic.jpg

I am assuming this is some kind of bandpass filter.

I see that there is both a low-pass and a hi-pass filter which should create a narrow band but I am confused on why this works.

I used the usual passive crossover equation to try to figure out crossover frequencies:

1
Freq =  ----------------
2 * Pi * R * C

Where "R" is the value of the crossover impedance resistor and "C" is the value of the capacitor.

The calculated values of the crossover frequencies of the two filters come out to be around 430Hz and 300Hz

This is what I don't understand, To me, the first filter which is a low-pass filter will only allow frequencies BELOW 300Hz through. The second filter with is a high-pass will only let frequencies above 430Hz through. This doesn't make any sense (to me anyway )

Am I just missing something here that has to do with it being coupled with a non-inverting op-amp stage? Please help!

R.G.? Mr. Strand? do you want to take a crack at this one? Many thanx!

-HB

#### Stu

##### bandpass filter
« Reply #1 on: September 03, 2003, 01:12:59 AM »
what you're looking at is an active bandpass filter

The transfer function (Vout/ Vin) is messy.

K is 2
D is 1/3

Vo =                                           sC2R2KD
Vi       s^2C1C2R1R2D + s[C1R1D + C2R2(1-K)(1-D) +C2(R1+R2)D) + 1

where:

R1 = 11k3
R2 =7k87
C1= 0.047uF
C2= 0.047uF

get the co efficients down the bottom make into a quadratic then finding where Vo/Vi = infinte will give the two frequencies at each side of your bandpass

subing in values

Vo/Vi   =    s(2.466E-4) / s^2(6.548E- + s(1.77E-4 - 2.466E-4 + 3.003E-4) + 1

Vo/Vi   =    s(2.466E-4) / s^2(6.548E- + s(2.307E-4) + 1

and graphically i get one pole at 3908rad/sec => 622Hz and the other pretty close to it again (can't see graphically)

i hope this is right (haven't done the calc's in a long time)

so i think it's basically a bandpass filter that emphasises the region around 622Hz

grab a electronics textbook if you realy want to find out more.

Stu

#### AllyP

##### Can anyone explain this? Bandpass Circuit?
« Reply #2 on: September 03, 2003, 04:02:11 AM »
I dont really get much of the above....I'll try and work it out later....

You can get *band cut* filters as well......ie the inverse of the graphical representation of a band pass

#### Hairston

##### Re: bandpass filter
« Reply #3 on: September 03, 2003, 11:35:16 AM »
Stu!

Thank you! This was exactly what I was looking for, it is much clearer now. I still need to do a little research to fully understand it, but this helps out alot.

I'm assuming this was the quadratic you were referring to (without the smilies):

Vo/Vi   =    s(2.466E-4) / s^2(6.548E- 8) + s(1.77E-4 - 2.466E-4 + 3.003E-4) + 1

Vo/Vi   =    s(2.466E-4) / s^2(6.548E- 8) + s(2.307E-4) + 1

Unfortunately, the audio engineering textbook I had back in college is long gone (got lost in a move). Do you have another decent textbook you would recommend? I can't remember ever studying this (that doesn't mean that I didn't at one time, it's been a while)

Thanks again!
-HB

#### Rob Strand

##### Can anyone explain this? Bandpass Circuit?
« Reply #4 on: September 03, 2003, 01:18:56 PM »
That circuit is the well known (Kerwin and Huelsman) voltage controlled voltage source (VCVS) bandpass filter.

The way it works is you have a cascaded low-pass and high-pass filter as you have already noted, the first resistor is then spit and wrapped around the opamp.  This kind of connection is a Q-enhancement scheme which increases the Q of the otherwise low Q bandpass filter which results from the passive cascade arrangement.  That I guess is the inspiration for the form of the circuit.  It's kind of the bandpass equivalent of the Sallen and Key low-pass and High-pass (IIRC Sallen and Key had their own band-pass which was different).

There isn't much point undertsanding it beyong that,  at the end of the day you don't even need to know where it come from-  you just analyse the circuit and get the transfer function Stu quoted (which I haven't verified).  Most filter work pretty much degenerates into munching equations and transfer functions- it's quite mechanical really but  scares a lot of people away because of math.
The mind often distorts without gain.