multistage Butterworth filter

[Martin]

I want to design a capacitively coupled (a) two-pole (b) three pole
bandpass butterworth filter with center frequency of 4.5 Mhz and 3 db
bandwidth of 150 Khz. The source and load resistance are 2200 ohms.

Can someone tell me that for the (a)two pole and the (a) three pole
cases, how would the center frquency and the bandwidth change (please
explain with numbers).

Thanks

Martin

 

[John Larkin]

I want to design a capacitively coupled (a) two-pole (b) three pole
bandpass butterworth filter with center frequency of 4.5 Mhz and 3 db
bandwidth of 150 Khz. The source and load resistance are 2200 ohms.

Can someone tell me that for the (a)two pole and the (a) three pole
cases, how would the center frquency and the bandwidth change (please
explain with numbers).

Thanks

Martin

If the CF is 4.5 MHz and the bandwidth is 150 KHz, what could change?

Is this homework?

John

 

[Active8]

I want to design a capacitively coupled (a) two-pole (b) three pole
bandpass butterworth filter with center frequency of 4.5 Mhz and 3 db
bandwidth of 150 Khz. The source and load resistance are 2200 ohms.

Can someone tell me that for the (a)two pole and the (a) three pole
cases, how would the center frquency and the bandwidth change (please
explain with numbers).

Thanks

Martin

IIRC it's the geometric mean of the individual center freqs.

mike

 

[Martin]

The resonant frequency changes and also the bandwidth changes.

 

[Martin]

How do you find the new center frequency?

 

[Jim Backus]

The resonant frequency changes and also the bandwidth changes.

The filter's centre frequency and the 3 dB bandwidth are the same
regardless of the number of poles. Those are the design parameters.
The resonant frequencies of each resonator and the bandwidth at
attenuations other than 3 dB do change.

--
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[John Larkin]

How do you find the new center frequency?

If the filter is designed for 4.5 MHz, the new center frequency is 4.5
MHz.

John

 

[Active8]

IIRC it's the geometric mean of the individual center freqs.

mike

uh, sorry. that's all good if you use 2 diff Cfs to get a wide
bandwidth, but this might be doable with 1 Cf.

i'll give you a hint:

w = 2.PI.F

Q=XL/R = (w.L)/R

Xnet = |XL - XC| = |wL - 1/(wL)|

the bandwidth is the difference in -3db freqs or .707V or .707I freqs.
which are the freqs where R = |X|. the center freq is where X = 0, so
you can derive:

BW = w/Q in radians/s or f/Q in Hz

but again, if you use 2 different center freqs to widen the bandwidth,
the new center freq is the geometric mean of the 2.

hint#2: the R controls the Q.

hint#3: use a table.

brs,
mike

 

[Ban]

The resonant frequency changes and also the bandwidth changes.

Martin,
there is no way doing it with active elements directly at 4.5MHz because of
the high Q required to achieve the 150k bandwidth. But you can mix down with
a VCO and a double-balanced mixer to a more suitable intermediate frequency
like 488k or even 0Hz and do the filtering there with opamps or switched
inductors.
Another possibility would be to directly digitize the signal and do the
filtering in the digital domain, but this would require a lot of skill
either.
It seems to me both solutions are far beyond your experience and skill, so
you can better buy some scanner or ham receiver with switchable bandwidth
ready made. :-(

ciao Ban

 

[maxfoo]

I want to design a capacitively coupled (a) two-pole (b) three pole
bandpass butterworth filter with center frequency of 4.5 Mhz and 3 db
bandwidth of 150 Khz. The source and load resistance are 2200 ohms.

Can someone tell me that for the (a)two pole and the (a) three pole
cases, how would the center frquency and the bandwidth change (please
explain with numbers).

Thanks

Martin

Butterworth Capacitive Coupled Resonators Table
n q1 qn k12 k22
2 1.414 1.414 0.707
3 1.0 1.0 0.707 0.707

Center freq shouldn't change, however attenuation is better with a
higher order.

AdB=10Log[1+(BWx/BW3dB)^2n]

 

[Active8]

Martin,
there is no way doing it with active elements directly at 4.5MHz because of
the high Q required to achieve the 150k bandwidth.

Q is 30.

i can't say if the they cover that freq, but MuRata makes filters for
10.7 MHz with BW way less than 150kHz. what? 4kHz and less? thats a
helluva Q. granted they're ceramic and all, but IF cans do 200kHz or
less for the FM b'cast band. that's a Q of 53.5 or so. this BW is
wiiiide

i know you may have messed with LC filters enough to be able to make
this claim, but please elaborate. i'm *all* confused. and could use some
sleep

what about an HP/LP or BP cascade?

brs,
mike

 

[Ban]

Q is 30.

i can't say if the they cover that freq, but MuRata makes filters for
10.7 MHz with BW way less than 150kHz. what? 4kHz and less? thats a
helluva Q. granted they're ceramic and all, but IF cans do 200kHz or
less for the FM b'cast band. that's a Q of 53.5 or so. this BW is
wiiiide

i know you may have messed with LC filters enough to be able to make
this claim, but please elaborate. i'm *all* confused. and could use
some sleep

what about an HP/LP or BP cascade?

brs,
mike

Mike,
the OP is talking about *adjustable* mid-frequency and bandwidth. That is
why I advised a superhet with an intermediate frequency. There the bandwidth
could be adjusted with an active filter.

At 4.5MHz with a Q of 30 this would require Q^2*fm=900*4.5MHz transit
frequency of the opamp, with a few 450Mhz opamps it would be doable with
very expensive multipliers to adjust the frequency and Q.

The digital approach seems more promising today with high-speed video-A/D
converters and some dedicated hardware DSP, but all solutions are not
trivial and scratch the "state of the art" limits.

ciao Ban

 

[Active8]

hi ban the man

Mike,
the OP is talking about *adjustable* mid-frequency and bandwidth.

that isn't clear. oh well.

mike