Confused about frequency and bandwidth in BP filter?

Question

I am trying to design a simple bandpass filter and I am confused what should be my central frequency and bandwidth should be.

I have a 1MHz square wave with 20% duty cycle, so the signal is on for 200nsec and off for 800nsec. I have a rise time and fall time of 10nsec.

The desired filter is passing this signal and eliminating all the noise. If my bandpass filter centered around 1MHz, how wide it should be so that I can maintain a decent rise time. According to the relationship between rise time and bandwidth (0.34=tr*BW) my signal bandwidth to observe 10nsec rise time is 34MHz. If my filter is 1MHz +/-5KHz, I would loose the rise time granularity (At least this is what I think since the FFT of that sharp edge will be in higher frequencies and will be cut off)

Bonus question: How can I design a BP filter that still allows for sharp rise and fall time and narrow around 1MHz?

Bonus answer :-) - [Cauer](http://en.wikipedia.org/wiki/Elliptic_filter) or [Chebyshev](http://en.wikipedia.org/wiki/Chebyshev_filter) response (or others). — Russell McMahon, Nov 08 '11 at 06:45
In what domain are you doing this project - Do you want passive components, op amps, or digital filtering? — Kevin Vermeer, Nov 08 '11 at 12:33

score 5 · Accepted Answer · answered Nov 08 '11 at 08:00

To answer your last question first, you can't have a narrow bandwidth filter around 1MHz and yet still have a fast rise time. If you think about the spectrum of a square wave, it has frequency components extending to infinity. The higher frequency components contribute to the 'squareing up' of the signal and sharpening of the edges. e.g. see http://mathworld.wolfram.com/FourierSeriesSquareWave.html Having a narrow band around 1MHz means your signal will come out looking like a 1MHz sine wave.

With that in mind you have to design a bandpass filter that does not attenuate your 1MHz fundamental frequency too much, yet includes high enough frequencies to give the desired rise time. Following your formula, 0.34 = rise time * bandwidth, you have calculated a bandwidth of 34MHz is required. The next step is to consider bandwidth = high cutoff freq. - low cutoff freq. You want the low cutoff to be less than 1MHz. Let's choose 500kHz. Thus the high cutoff would be 34.5Mhz and the centre frequency 17.25Mhz.

To get rid of the most noise, the filter should have a steep rolloff, e.g. the two filters mentioned in the comments. This means your low cutoff frequency can be very close to 1MHz without too much attenuation, and in the higher end of the spectrum rolls off very quickly after the high cutoff frequency, reducing high frequency noise.

Confused about frequency and bandwidth in BP filter?

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