Hi,
I am trying to derive the transfer function for this bandpass filter by breaking it up into two parts, the lowpass and high pass.

I am I doing something wrong in the pdf? I simulated the ciruit to get the bod plot. However with the transfer function I derived, the maximum gain matches but there seems to be a shift of about 1 decade from one plot to another.

is it a good idea to use the logarithms are as derived in the pdf, to plot the bode plots by hand? thanks

 

I see a number of errors.

In the second page, for the last expression in your derivation of the low pass function, you have:

A(jw) = 1/(jwR3C2) You dropped the 1+, but put it back in the numerical calculation.

Besides that, in the schematic, the low pass involves R1 and C1, not R3 and C2.

For the high pass calculation, you have:

Ahp(jw) = 1+ R2/(R1 + 1/(jwC1))

but when I look at the schematic, it appears to me that the high pass function involves R2, R3 and C2, not R1, C1 and R2.

In the high pass calculation, fourth line, you do a numeric calculation with this parenthetical in the numerator: (15kΩ+14kΩ)

I don't see a 14k resistor anywhere in the schematic, and besides which the result of the numerator calculation is wrong.

There may be other errors, but those are some that I see with a brief examination.

 

The bode plots are correct.
The cutoff frequencies are -3dB from the peak maximum output voltage.

Why is your math so complicated for such a simple circuit?

 

@ The Electrician Corrected the errors you mentioned. Sorry for wasting your time, but some of the errors were because I was following another schematic with different component numbers.

@Audioguru
The complicated maths is because I need to draw the bode plot by hand. I am only using the simulation bode plot to verify my work but I cannot get there with my calculations.

I am using this source: http://lpsa.swarthmore.edu/Bode/Bode.html

The source explains how to draw zeros and poles individually can I use this method in my case? Or is there a better way to simplify things? I appreciate if you can verify the whole TF. thanks

 

The circuit is very simple since it has only a single RC for a lowpass filter and another single RC for a highpass filter. The frequencies are far apart so they do not affect each other.

You should learn that a single RC is -3dB (x 0.707 times) and the level drops 6dB (half the level) for each octave so you do not need to do any math, just simple arithmatic.

The peak output is 1+ (15K/4.7K)= 4.2 (a little more than +12dB) and the cutoff frequency of the lowpass filter is 3386Hz.
Then the output is -6dB (half the peak level) at 2 times the cutoff frequency and is another half at 4 times the cutoff frequency.

The highpass filter is the same except it cuts low frequencies. -6dB (half the peak level) at half the cutoff frequency but the output cannot be less than the input level.

 

thanks a lot solve. Managed to match the simulation plot.