I am simulating a signal with numpy.sin in python, and i wanna simulate different sampling frequencies, the code is as follows:

import numpy as np
import matplotlib.pyplot as plt

frequencies = [10, 20, 40, 49, 100, 101, 102, 103, 104, 301, 526, 1222] # different sampling frequencies
T = 1 #amount of time for which to simulate the signal in seconds
f = 50 #frequency of the signal

for i in frequencies:
    fig, ax1 = plt.subplots(1)
    N = T * i # number of samples needed at the sampling frequency for the elapsed time of the signal
    linear = np.linspace(0, T, N) # create the data points at which to evaluate the sin 
    y = np.sin(2 * np.pi * f * linear) # evaluate the sin for the amount of sample points
    #creat the graph
    ax1.scatter(linear, y) 
    ax1.set(ylabel = 'Amplitude', xlabel = 'Time in s')
    fig.savefig('Freq{}.png'.format(i))
    plt.close(fig)
  

The problem arises with the sampling frequency 101, here the sin function evaluates to ridiculously tiny values, something around 10^(-14) (for all the other values normal graphs come out) (Sin function Evaluated with a sampling frequency of 101 Hz) whilst if I evaluate the sine func by hand by printing out the linear array for sampling frequency 101 Hz, I get normal values. Does somebody know the problem? Maybe and approximation problem of np.sin? maybe the dtype float64 somehow breaks, because the values are kinda sus.

CodePudding user response：

When i is 101, f * linear is [0.0, 0.5, 1.0, 1.5, ...], so the expression np.sin(2 * np.pi * f * linear) is sampling the sin function at multiples of pi. If you were working with infinite precision, the values would all be 0. These sample points, however, will not be exact multiples of pi, because of normal floating point imprecision, so the values returned by that expression should be close to 0, but they won't necessarily be exactly 0.