# Bandpass Filter component calculations.

Discussion in 'Homework Help' started by Petrucciowns, Jun 25, 2009.

1. ### Petrucciowns Thread Starter Active Member

Jun 14, 2009
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Once again I'm going through all my old homework and taking the notes I should have done in the past, and I am stuck on calculating the components on a Bandpass filter. How would I go about this. I have the answers I just want to know how to find them.
Thanks

http://img31.imageshack.us/img31/676/bandpass.jpg

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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For my you answer is wrong, because R2 must by equal R1 (Use R2 for R1).
So if R1=R2 then C1=C2
And F_center=1/(2*pi*RC)

3. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,501
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The image says to use R2 for R1, but it doesn't appear that it was done.

Furthermore, the usual way bandwidth is defined is the interval between the 3 dB down frequencies, and with the values given, the bandwidth is not 2kHz to 10 kHz.

What you must do is derive the transfer function and then find C1 and C2 to give the required bandwidth. Have you given that a try?

4. ### Petrucciowns Thread Starter Active Member

Jun 14, 2009
62
0
The image says "Use" R1 for R2 not R1=R2. This means that R1 is included in the calculation of R2. I am confident that all the answers are correct, because this is a scan from my exam and I didn't miss one question on it. The bandwidth is correct or else it would not be given. Obviously both of you have no clue what you are talking about. I will figure it out on my own. Keep in mind that this is a bandpass filter consisting of one High pass, and one low pass filter. I don't know what you are thinking of.

Thanks anyways.

Last edited: Jun 27, 2009
5. ### hobbyist Distinguished Member

Aug 10, 2008
878
85
According to my electronics course material,
Says "In both these filters (high and low pass) the cutoff frequency, fco, is the point where Xc=R.
The fco of any RC network can be found by the expression:
fco = 1/ 2*pi*R*C "

So there's your answer to your question, when R = Xc you have the value needed to solve for any RC filter.

6. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,501
393
This is only applicable to an isolated low or high pass filter, or a pair of such (single RC) filters separated by a buffer stage, perhaps a unity gain opamp stage.

When one stage is loaded by the following stage, the corner frequencies are not at a frequency where a single Xc = a single R.

In the given circuit, all 4 R's and C's are involved in the expressions for the corner frequencies.

The transfer function for this circuit is (s=jω):

$Av=\frac{s R2 C2}{s^2 R1 R2 C1 C2 + s(R1 C1+R1 C2+R2 C2)+1}$

So how shall the "corner" or "cutoff" frequencies be defined? Is it where Av is at a value of .707 times the midband value (-3dB), or some other value? However it's defined, it will not be given by an expression involving only a single R and a single C.

For this network, the output at 2000 Hz is down 1.051 dB with respect to the midband gain, and at 10000 Hz, it's down 1.036 dB.

This is not true for just any RC network; the given circuit provides a counterexample.

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Haha. For your information the -3dB bandwidth for R1=1.57K; R2=4.7K, C1=10.1nF, C2=16.9nF is F1=1.2KHz and F2=16.5Hz
And for F=2Khz is -1.05dB and for 10KHz is -1.03dB.
Don't make me laugh, obviously you're right and we all wrong.
Live yourself in this ignorance.

8. ### hobbyist Distinguished Member

Aug 10, 2008
878
85

I stand corrected::

I learned all that stuff long time ago, from CIE course, the s parameters, transient analysis, but that's way beyond me now, I look back on my notes now, and think, I used to understand all this???? I wouldn't even know where to start to get back into the full swing of all this, I gyess back then I was intensely interested in that, but now for me just designing transistor circuits is my electronics hobby, that's why I learn a lot from you guys, on this board, keep up the good work..

Sorry for any wrong info. I gave...

Last edited: Jun 28, 2009