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cathode-ray
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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 [itex]A_{max}=10log(1+ε^2)[/itex]
Transfer function of Chebyshev filter [itex] T(s)=\frac{Kw_{p}^N}{ε2^{N-1}(s-p_{1})...(s-p_{N})}[/itex]
Chebyshev filter poles [itex] 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[/itex]
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 [itex]A_{max}[/itex] and get the two poles(which are conjugated) through the expression given, considering [itex]w_{p}[/itex] as [itex]670\times2\pi[/itex]. 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.