Transfer Function of Active Band Pass filter

Discussion in 'General Electronics Chat' started by Mechatronical, Feb 15, 2014.

  1. Mechatronical

    Thread Starter Member

    Feb 15, 2014
    30
    0
    How do I derive the transfer function of a general non-inverting active band pass filter as shown in the link below:

    http://www.electronics-tutorials.ws/filter/fil51.gif

    The transfer function of this filter is:

    H(jω)=(jωR2C1) / [(jωR2C2+1)(jωR1C1+1)]

    Shouldn't the transfer function include the gain R2/R1 and * (-1);

    H(jω)= - (R2/R1) * [(jωR2C1) / [(jωR2C2+1)(jωR1C1+1)]] ?

    The transfer function of a low pass filter is:
    H(jω)= -R2/R1 * 1/(jωR2C2)

    The transfer function of a high pass filter is:
    H(jω) - jωR2C1 / (jωR1C1+1)

    Shouldn't the transfer function of the high pass filter also include gain;

    H(jω) - (R2/R1) * [jωR2C1 / (jωR1C1+1)] ?

    So can I find the transfer function of the band pass filter by multiplying the transfer function of the low pass and high pass filter, and the gain;

    H(jω)low * H(jω)high * - (R2/R1) ?

    Finally, how do I find the impedance input and output of this bandpass filter?
     
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,387
    497
    Transfer function is: H(jw)=Vout/Vin
    So.
    Vout=[H(jw)]Vin
    Take your circuit, write out an equation where Vout is represented in terms of Vin. Factor out Vin, you will have "something" times Vin. That "something" is the transfer function.
     
  3. Mechatronical

    Thread Starter Member

    Feb 15, 2014
    30
    0
    I figured out that the transfer function is:
    H(jω)bandpass = H(jω)low * H(jω)high

    But answer this: Shouldn't you multiply it with the gain (- R2/R1)?
     
  4. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,387
    497
    Scroll to the bottom to Active Band Pass Filter: http://www.biomedstudent.com/Transfer Functions.shtml

    Basically, the fact that you have input resistor and feedback resistor means that you already have gain structure built in.

    For the sake of simplicity of design, you could set input and feedback resistors to be equal, so gain is 1, calculated the capacitors, build the band pass filter. Then add a second stage whose job is just to provide amplification, meaning gain is more than 1. This way instead of dealing with one hard to design device, you use two simple to design devices.
     
    Mechatronical likes this.
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