How Is a Second Order Chebyshev Passband Filter Designed?

[cathode-ray]

Homework Statement

Get the transfer function of a second order Chebyshev passband filter, with central frequency f0 = 1 [kHz], lower cutoff frequency fc=670 [Hz], 3dB ripple in the pass-band and 30dB of gain in the central frequency.

Homework Equations

Maximum allowed variation in passband transmission $A_{max}=10log(1+ε^2)$
Transfer function of Chebyshev filter $T(s)=\frac{Kw_{p}^N}{ε2^{N-1}(s-p_{1})...(s-p_{N})}$
Chebyshev filter poles $p_{k}=-w_{p}sin(\frac{(2k-1)\pi}{2N})sinh(\frac{1}{N}sinh^{-1}(\frac{1}{ε}))+jw_{p}cos(\frac{(2k-1)\pi}{2N})cosh(\frac{1}{N}sinh^{-1}(\frac{1}{ε})), k=1,2,...,N$
N is the order of the filter
K is the gain
(expressions taken from "Microelectronic Circuits", Sedra, 5th edition)

The Attempt at a Solution

Hi,

My first attempt, to this problem was to calculate ε through $A_{max}$ and get the two poles(which are conjugated) through the expression given, considering $w_{p}$ as $670\times2\pi$. I then had all that was needed to build the transfer function. The problem is that this is the transfer function for a low pass filter. I have no idea how to get the pass band filter at this point, and I am also not sure if what i did is correct.

 

[rude man]

I recommend http://www.analog.com/library/analogDialogue/archives/43-09/EDCh 8 filter.pdf

pp. 8.21 ff.

If you wan the correct answer without the details, go to www.filterfree.com. Stipulate chebyshev bandpass type 1, 2nd order, input your upper & lower cutoff frequencies and midband gain, 3dB passband ripple, and get the answer.

 

[cathode-ray]

I made a search in the web before coming here and also found that document, but it didnt helped :S .

 

[NascentOxygen]

If you wan the correct answer without the details, go to www.filterfree.com. Stipulate chebyshev bandpass type 1, 2nd order, input your upper & lower cutoff frequencies and midband gain, 3dB passband ripple, and get the answer.

Have you visited that site recently, rude man?

 

[NascentOxygen]

The problem is that this is the transfer function for a low pass filter. I have no idea how to get the pass band filter at this point, and I am also not sure if what i did is correct.

I recall from my uni days that there are transforms to convert the LP to a HP, and also to a BP. That's why design equations deal only with the LP. Searching, I found mention on wikipedia. http://en.wikipedia.org/wiki/Frequency_transformations#Bandform_transformation

I think you are just looking for a second order system, of the form A.s / (s² + bs +c)
so it won't have ripple as such, there's only the one peak in the response, it's bandwidth being measured between the pair of -3dB points.

Beyond this, I'm of no help here, sorry.

 

[rude man]

Have you visited that site recently, rude man?

see below

 

[rude man]

I recommend http://www.analog.com/library/analogDialogue/archives/43-09/EDCh 8 filter.pdf

pp. 8.21 ff.

If you want the correct answer without the details, go to www.filterfree.com. Stipulate chebyshev bandpass type 1, 2nd order, input your upper & lower cutoff frequencies and midband gain, 3dB passband ripple, and get the answer.

SORRY, N/O. I'll try to figure out how I got their site. definitely did, only yesterday. Stay tuned, it's worth it.

OK, found the problem - the site is www.nuhertz.com. Scroll down to the "Filter Free" downlink.

 

[cathode-ray]

I finally found the answer and the mistake i was making :D. To design the filter i was considering initially a second order low pass Chebyshev filter, with the specified characteristics, and then i applied the frequency transformation to make it a band pass. However to make the transformation what i needed was a first order low pass Chebyshev filter that as a transfer function of the form:

$T(S)=\frac{1}{S-p_{1}}$​

where S is the normalized frequency with respect to the central frequency:

$S=\frac{s}{w_{p}}$​

I then calculated the pole $p_{1}$ of the filter, using the expression i took from the "Microelectronic Circuits", which gave me $p_{1}=-1$. Thus the transfer function for the low pass filter is:

$T(S)=\frac{1}{S+1}$​

To make it a band pass filter i used a frequency transformation:

$S\rightarrow\frac{S^{2}+1}{2\xi S}$​

After making the transformation i ended up with a band pass second order Chebyshev filter , with the desired characteristics, of the form:

$T(S)=\frac{s}{\frac{s^{2}}{B}+s+\frac{w_{p}^{2}}{B}}$​

B is the bandwidth

In order to have the gain of 30dB at the central frequency i multiplied the transfer function for a constant k and calculated what value it had to have at $s=jw_{p}$.Then i finally got the desired result:

$T(S)=\frac{s}{\frac{s^{2}}{Bk}+\frac{s}{k}+\frac{w_{p}^{2}}{Bk}}$​

Thanks for all the replies!

Note: Thanks for the link for that software rude man :) it will be useful