Electronic – Bandwidth and frequency domain

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.

However, we are in the imaginary frequency domain, so what would the bandwidth of this filter be?

I'm not sure why you're having difficulty with the fact that the frequency domain signal is, in this case, imaginary. The symmetries of the Fourier transform are usually taught early on in signal processing courses:

• If the time domain signal is real and odd, e.g., \$\sin\omega t\$, the associated frequency domain signal is imaginary and odd.
• If the time domain signal is real and even, e.g., \$\cos\omega t\$, the associated frequency domain signal is real and even.

But, for bandwidth calculation, you're interested in the magnitude in the frequency domain (think Bode magnitude plot). Since your frequency domain signal is pure imaginary, it couldn't be any easier; simply remove the j factor.

If, however, your frequency domain signal were complex, you would need to multiply the function by its conjugate and take the square root in order to find the magnitude.