问题
I am working on doing some digital filter work using Python and Numpy/Scipy.
I'm using scipy.signal.iirdesign to generate my filter coefficents, but it requires the filter passband coefficents in a format I am not familiar with
wp, ws : float
Passband and stopband edge frequencies, normalized from 0 to 1 (1 corresponds
to pi radians / sample).
For example:
Lowpass: wp = 0.2, ws = 0.3
Highpass: wp = 0.3, ws = 0.2
(from here)
I'm not familiar with digital filters (I'm coming from a hardware design background). In an analog context, I would determine the desired slope and the 3db down point, and calculate component values from that.
In this context, how do I take a known sample rate, a desired corner frequency, and a desired rolloff, and calculate the wp, ws values from that?
(This might be more appropriate for math.stackexchange. I'm not sure)
回答1:
If your sampling rate is fs, the Nyquist rate is fs/2. This represents the highest representable frequency you can have without aliasing. It is also equivalent to the normalized value of 1 referred to by the documentation. Therefore, if you are designing a low pass filter with a corner frequency of fc, you'd enter it as fc / (fs/2).
For example, you have fs=8000 so fs/2=4000. You want a low pass filter with a corner frequency of 3100 and a stop band frequency of 3300. The resulting values would be wp=fc/(fs/2)=3100/4000. The stopband frequency would be 3300/4000.
Make sense?
回答2:
Take the function x(t) = cos(2*pi*fa*t). If we're sampling at frequency fs, the sampled function is x(n*ts) = x(n/fs) = cos(2*pi*n*fa/fs). The maximum frequency before aliasing (folding) is the Nyquist frequency fa = fs/2, which normalizes to (fs/2)/fs = 1/2. The normalized angular frequency is 2*pi*1/2 rad/sample = pi rad/sample. Thus the signal x[n] = cos[pi*n] = [1,-1,1,-1,...].
The sampled version of a given frequency such as a corner frequency 2*pi*fc rad/s would be 2*pi*fc/fs rad/sample. As a fraction of the Nyquist frequency pi, that's 2*fc/fs = fc/(fs/2).
A few formulas to live by:
exp[j*w*n] = cos[w*n] + j*sin[w*n]
x_even[n] = 0.5*x[n] + 0.5*x[-n]
cos[w*n] = 0.5*exp[j*w*n] + 0.5*exp[-j*w*n] # cos is even
x_odd[n] = 0.5*x[n] - 0.5*x[-n]
j*sin[w*n] = 0.5*exp[j*w*n] - 0.5*exp[-j*w*n] # sin is odd
The DFT of the even component (a sum of cosines) of a real-valued signal will be real and symmetric while the DFT of the odd component (a sum of sines) will be imaginary and anti-symmetric. Thus for real-valued signals such as the impulse response of a typical filter, the magnitude spectrum is symmetric while the phase spectrum is antisymmetric. Thus you only have to specify a filter for the range 0 to pi, which is normalized to [0,1].
来源：https://stackoverflow.com/questions/4697235/converting-from-samplerate-cutoff-frequency-to-pi-radians-sample-in-a-discrete-t