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Easy way to implement Root Raised Cosine (RRC) filter using Python & Numpy

SciPy / Numpy seems to support many filters, but not a cosine root filter. Is there a trick to easily create it, and not calculate the transfer function? The approximation will also be perfect.

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5 answers

The commpydelivery includes several filters. In version 0.2.0, returned variables are switched. To install, follow the instructions here .

Here is a usage example:

import numpy as np
from commpy.modulation import QAMModem
from commpy.filters import rrcosfilter
N = 1024  # output size
mod1 = QAMModem(16)  # QAM16
sB = randint(0, 2, mod1.num_bits_symbol*N*M/4)  # Random bit stream
sQ = mod1.modulate(sB)  # Modulated baud points
sPSF = rrcosfilter(N*4, 0.8, 1, 24)[1]
qW = np.convolve(sPSF, sQ) # Waveform with PSF
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commpy has not been released yet. But here is my nugget of knowledge.

beta = 0.20 # roll off factor

Tsample = 1.0 # sampling period, should at least twice the rate of the symbol

oversampling_rate = 8 # oversampling of the bit stream, this gives samples per symbol
# must be at least 2X the bit rate

Tsymbol = oversampling_rate * Tsample # pulse duration should be at least 2 * Ts
span = 50 # number of symbols to span, must be even
n = span*oversampling_rate # length of the filter = samples per symbol * symbol span

# t_step must be from -span/2 to +span/2 symbols.
# each symbol has 'sps' number of samples per second.
t_step = Tsample * np.linspace(-n/2,n/2,n+1) # n+1 to include 0 time

BW = (1 + beta) / Tsymbol
a = np.zeros_like(t_step)

for item in list(enumerate(t_step)):
    i,t = item 
    # t is n*Ts
    if (1-(2.0*beta*t/Tsymbol)**2) == 0:
        a[i] = np.pi/4 * np.sinc(t/Tsymbol)
        print 'i = %d' % i
    elif t == 0:
        a[i] = np.cos(beta * np.pi * t / Tsymbol)/ (1-(2.0*beta*t/Tsymbol)**2)
        print 't = 0 captured'
        print 'i = %d' % i 

    else:
        numerator = np.sinc( np.pi * t/Tsymbol )*np.cos( np.pi*beta*t/Tsymbol )
        denominator = (1.0 - (2.0*beta*t/Tsymbol)**2)
        a[i] =  numerator / denominator

#a = a/sum(a) # normalize total power

plot_filter = 0
if plot_filter == 1:

    w,h = signal.freqz(a)
    fig = plt.figure()
    plt.subplot(2,1,1)
    plt.title('Digital filter (raised cosine) frequency response')
    ax1 = fig.add_subplot(211)
    plt.plot(w/np.pi, 20*np.log10(abs(h)),'b')
    #plt.plot(w/np.pi, abs(h),'b')
    plt.ylabel('Amplitude (dB)', color = 'b')
    plt.xlabel(r'Normalized Frequency ($\pi$ rad/sample)')

    ax2 = ax1.twinx()
    angles = np.unwrap(np.angle(h))
    plt.plot(w/np.pi, angles, 'g')
    plt.ylabel('Angle (radians)', color = 'g')
    plt.grid()
    plt.axis('tight')
    plt.show()


    plt.subplot(2,1,2)
    plt.stem(a)
    plt.show()
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, - .

h(n) = (sinc(n/T)*cos(pi * alpha* n /T)) / (1-4*(alpha*n/T)**2)

.

output = scipy.signal.convolve(signal_in, h)
0

, CommPy, :

def rcosfilter(N, beta, Ts, Fs):
    t = (np.arange(N) - N / 2) / Fs
    return np.where(np.abs(2*t) == Ts / beta,
        np.pi / 4 * np.sinc(t/Ts),
        np.sinc(t/Ts) * np.cos(np.pi*beta*t/Ts) / (1 - (2*beta*t/Ts) ** 2))
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SciPy will support any filter. Just calculate the impulse response and use any of the appropriate scipy.signal filter / convolve functions.

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