I am trying to visualize a frequency spectrum using the BURG algroithm. The data that I am trying to visualize is the distance between heartbeats in milliseconds (e.g: [700, 650, 689, ..., 702]). Time distance is measured from R peak to R peak of next heartbeat.
Now I would like to visualize the frequency band with python's spectrum library (I'm a total noob). The minimum frequency that I am trying to display is 0.0033Hz, so all time differences in my dataset summarized are 5 Minutes long.
My approach was to first take the reciprocal of each value, then multiply by 1000, and then multiply by 60. This should get me the Bpm for each heartbeat.
This is what it looks like: [67.11409396 64.72491909 ... 64.58557589]
Afterwards I use spectrum's burg algorithm to create the PSD. The "data" list contains my BpM for each heartbeat.
AR, rho, ref = arburg(data.tolist(), 7)
PSD = arma2psd(AR, rho=rho, NFFT=1024)
PSD = PSD[len(PSD):len(PSD)//2:-1]
plot(linspace(0, 0.5, len(PSD)), 10*log10(abs(PSD)*2./(1.*pi)))
pylab.legend(['PSD estimate of x using Burg AR(7)'])
The graph that I get looks like this:
5 Minutes Spectrogram
This specific data already exists as a 3D-Spectrogram (Graph above is the equivalent to the last 5 Minutes of 3D-Spectrogram):
Long Time 3D-Spectrogram
My Graph does not seem to match the 3D-Spectrogram. My frequencies are way off.... What causes this and how can I fix it?
Also I would like the y-Axis in my Graph not in [dB] but in absolute Values. I tried with:
plot(linspace(0, 0.5, len(PSD)), abs(PSD))
but that did not really seem to work. It just drew a hyperbole.
Thank you for your help!
The spectrum package comes with a pburg class than can generate a frequencies array, this is shown below. If you want direct comparison between a spectrogram and AR PSDs, I would take the time definition used to compute the spectrogram to also compute the AR PSD per window.
Also, your spectrogram example image looks focused on very low frequencies, so you may want to increase nfft to increase frequency resolution.
import matplotlib.pyplot as plt
from scipy.signal import spectrogram
import numpy as np
from spectrum import pburg
# Parameter settings
n_seconds = 10
fs = 1000 # sampling rate, in hz
freq = 10
nfft = 4096
nperseg = fs
order = 8
# Simulate 10 hz sine wave with white noise
x = np.sin(np.arange(0, n_seconds, 1/fs) * freq * 2 * np.pi)
x += np.random.rand(len(x)) / 10
# Compute spectrogram
freqs, times, powers = spectrogram(x, fs=fs, nfft=nfft)
# Get spectrogram time definition
times = (times * fs).astype(int)
window_times = np.array((times-times[0], times+times[0])).T
# Compute Burg's spectrum per window
powers_burg = np.array([pburg(x[t[0]:t[1]], order=order,
NFFT=nfft, sampling=fs).psd for t in window_times]).T
freqs_burg = np.array(pburg(x, order=order, NFFT=nfft, sampling=fs).frequencies())
# Plot
inds = np.where(freqs < 20)
inds_burg = np.where(freqs_burg < 20)
fig, axes = plt.subplots(ncols=2, figsize=(10, 5))
axes[0].pcolormesh(times/fs, freqs[inds], powers[inds], shading='gouraud')
axes[1].pcolormesh(times/fs, freqs_burg[inds_burg], powers_burg[inds_burg], shading='gouraud')
axes[0].set_title('Spectrogram')
axes[1].set_title('Burg\'s Spectrogram')
Related
I'm trying to plot the Amplitude (dBFS) vs. Time (s) plot of an audio (.wav) file using matplotlib. I managed to do that with the following code:
def convert_to_decibel(sample):
ref = 32768 # Using a signed 16-bit PCM format wav file. So, 2^16 is the max. value.
if sample!=0:
return 20 * np.log10(abs(sample) / ref)
else:
return 20 * np.log10(0.000001)
from scipy.io.wavfile import read as readWav
from scipy.fftpack import fft
import matplotlib.pyplot as gplot1
import matplotlib.pyplot as gplot2
import numpy as np
import struct
import gc
wavfile1 = '/home/user01/audio/speech.wav'
wavsamplerate1, wavdata1 = readWav(wavfile1)
wavdlen1 = wavdata1.size
wavdtype1 = wavdata1.dtype
gplot1.rcParams['figure.figsize'] = [15, 5]
pltaxis1 = gplot1.gca()
gplot1.axhline(y=0, c="black")
gplot1.xticks(np.arange(0, 10, 0.5))
gplot1.yticks(np.arange(-200, 200, 5))
gplot1.grid(linestyle = '--')
wavdata3 = np.array([convert_to_decibel(i) for i in wavdata1], dtype=np.int16)
yvals3 = wavdata3
t3 = wavdata3.size / wavsamplerate1
xvals3 = np.linspace(0, t3, wavdata3.size)
pltaxis1.set_xlim([0, t3 + 2])
pltaxis1.set_title('Amplitude (dBFS) vs Time(s)')
pltaxis1.plot(xvals3, yvals3, '-')
which gives the following output:
I had also plotted the Power Spectral Density (PSD, in dBm) using the code below:
from scipy.signal import welch as psd # Computes PSD using Welch's method.
fpsd, wPSD = psd(wavdata1, wavsamplerate1, nperseg=1024)
gplot2.rcParams['figure.figsize'] = [15, 5]
pltpsdm = gplot2.gca()
gplot2.axhline(y=0, c="black")
pltpsdm.plot(fpsd, 20*np.log10(wPSD))
gplot2.xticks(np.arange(0, 4000, 400))
gplot2.yticks(np.arange(-150, 160, 10))
pltpsdm.set_xlim([0, 4000])
pltpsdm.set_ylim([-150, 150])
gplot2.grid(linestyle = '--')
which gives the output as:
The second output above, using the Welch's method plots a more presentable output. The dBFS plot though informative is not very presentable IMO. Is this because of:
the difference in the domains (time in case of 1st output vs frequency in the 2nd output)?
the way plot function is implemented in pyplot?
Also, is there a way I can plot my dBFS output as a peak-to-peak style of plot just like in my PSD (dBm) plot rather than a dense stem plot?
Would be much helpful and would appreciate any pointers, answers or suggestions from experts here as I'm just a beginner with matplotlib and plots in python in general.
TLNR
This has nothing to do with pyplot.
The frequency domain is different from the time domain, but that's not why you didn't get what you want.
The calculation of dbFS in your code is wrong.
You should frame your data, calculate RMSs or peaks in every frame, and then convert that value to dbFS instead of applying this transformation to every sample point.
When we talk about the amplitude, we are talking about a periodic signal. And when we read in a series of data from a sound file, we read in a series of sample points of a signal(may be or be not periodic). The value of every sample point represents a, say, voltage value, or sound pressure value sampled at a specific time.
We assume that, within a very short time interval, maybe 10ms for example, the signal is stationary. Every such interval is called a frame.
Some specific function is applied to each frame usually, to reduce the sudden change at the edge of this frame, and these functions are called window functions. If you did nothing to every frame, you added rectangle windows to them.
An example: when the sampling frequency of your sound is 44100Hz, in a 10ms-long frame, there are 44100*0.01=441 sample points. That's what the nperseg argument means in your psd function but it has nothing to do with dbFS.
Given the knowledge above, now we can talk about the amplitude.
There are two methods a get the value of amplitude in every frame:
The most straightforward one is to get the maximum(peak) values in every frame.
Another one is to calculate the RMS(Root Mean Sqaure) of every frame.
After that, the peak values or RMS values can be converted to dbFS values.
Let's start coding:
import numpy as np
import matplotlib.pyplot as plt
from scipy.io import wavfile
# Determine full scall(maximum possible amplitude) by bit depth
bit_depth = 16
full_scale = 2 ** bit_depth
# dbFS function
to_dbFS = lambda x: 20 * np.log10(x / full_scale)
# Read in the wave file
fname = "01.wav"
fs,data = wavfile.read(fname)
# Determine frame length(number of sample points in a frame) and total frame numbers by window length(how long is a frame in seconds)
window_length = 0.01
signal_length = data.shape[0]
frame_length = int(window_length * fs)
nframes = signal_length // frame_length
# Get frames by broadcast. No overlaps are used.
idx = frame_length * np.arange(nframes)[:,None] + np.arange(frame_length)
frames = data[idx].astype("int64") # Convert to in 64 to avoid integer overflow
# Get RMS and peaks
rms = ((frames**2).sum(axis=1)/frame_length)**.5
peaks = np.abs(frames).max(axis=1)
# Convert them to dbfs
dbfs_rms = to_dbFS(rms)
dbfs_peak = to_dbFS(peaks)
# Let's start to plot
# Get time arrays of every sample point and ever frame
frame_time = np.arange(nframes) * window_length
data_time = np.linspace(0,signal_length/fs,signal_length)
# Plot
f,ax = plt.subplots()
ax.plot(data_time,data,color="k",alpha=.3)
# Plot the dbfs values on a twin x Axes since the y limits are not comparable between data values and dbfs
tax = ax.twinx()
tax.plot(frame_time,dbfs_rms,label="RMS")
tax.plot(frame_time,dbfs_peak,label="Peak")
tax.legend()
f.tight_layout()
# Save serval details
f.savefig("whole.png",dpi=300)
ax.set_xlim(1,2)
f.savefig("1-2sec.png",dpi=300)
ax.set_xlim(1.295,1.325)
f.savefig("1.2-1.3sec.png",dpi=300)
The whole time span looks like(the unit of the right axis is dbFS):
And the voiced part looks like:
You can see that the dbFS values become greater while the amplitudes become greater at the vowel start point:
I am new to the fourier theory and I've seen very good tutorials on how to apply fft to a signal and plot it in order to see the frequencies it contains. Somehow, all of them create a mix of sines as their data and i am having trouble adapting it to my real problem.
I have 242 hourly observations with a daily periodicity, meaning that my period is 24. So I expect to have a peak around 24 on my fft plot.
A sample of my data.csv is here:
https://pastebin.com/1srKFpJQ
Data plotted:
My code:
data = pd.read_csv('data.csv',index_col=0)
data.index = pd.to_datetime(data.index)
data = data['max_open_files'].astype(float).values
N = data.shape[0] #number of elements
t = np.linspace(0, N * 3600, N) #converting hours to seconds
s = data
fft = np.fft.fft(s)
T = t[1] - t[0]
f = np.linspace(0, 1 / T, N)
plt.ylabel("Amplitude")
plt.xlabel("Frequency [Hz]")
plt.bar(f[:N // 2], np.abs(fft)[:N // 2] * 1 / N, width=1.5) # 1 / N is a normalization factor
plt.show()
This outputs a very weird result where it seems I am getting the same value for every frequency.
I suppose that the problems comes with the definition of N, t and T but I cannot find anything online that has helped me understand this clearly. Please help :)
EDIT1:
With the code provided by charles answer I have a spike around 0 that seems very weird. I have used rfft and rfftfreq instead to avoid having too much frequencies.
I have read that this might be because of the DC component of the series, so after substracting the mean i get:
I am having trouble interpreting this, the spikes seem to happen periodically but the values in Hz don't let me obtain my 24 value (the overall frequency). Anybody knows how to interpret this ? What am I missing ?
The problem you're seeing is because the bars are too wide, and you're only seeing one bar. You will have to change the width of the bars to 0.00001 or smaller to see them show up.
Instead of using a bar chart, make your x axis using fftfreq = np.fft.fftfreq(len(s)) and then use the plot function, plt.plot(fftfreq, fft):
import matplotlib.pyplot as plt
import pandas as pd
import numpy as np
data = pd.read_csv('data.csv',index_col=0)
data.index = pd.to_datetime(data.index)
data = data['max_open_files'].astype(float).values
N = data.shape[0] #number of elements
t = np.linspace(0, N * 3600, N) #converting hours to seconds
s = data
fft = np.fft.fft(s)
fftfreq = np.fft.fftfreq(len(s))
T = t[1] - t[0]
f = np.linspace(0, 1 / T, N)
plt.ylabel("Amplitude")
plt.xlabel("Frequency [Hz]")
plt.plot(fftfreq,fft)
plt.show()
I am trying to detect the pitch of a B3 note played with a guitar. The audio can be found here.
This is the spectrogram:
As you can see, it is visible that the fundamental pitch is about 250Hz which corresponds to the B3 note.
It also contains a good amount of harmonics and that is why I chose to use HPS from here. I am using this code for detecting the pitch:
def freq_from_hps(signal, fs):
"""Estimate frequency using harmonic product spectrum
Low frequency noise piles up and overwhelms the desired peaks
"""
N = len(signal)
signal -= mean(signal) # Remove DC offset
# Compute Fourier transform of windowed signal
windowed = signal * kaiser(N, 100)
# Get spectrum
X = log(abs(rfft(windowed)))
# Downsample sum logs of spectra instead of multiplying
hps = copy(X)
for h in arange(2, 9): # TODO: choose a smarter upper limit
dec = decimate(X, h)
hps[:len(dec)] += dec
# Find the peak and interpolate to get a more accurate peak
i_peak = argmax(hps[:len(dec)])
i_interp = parabolic(hps, i_peak)[0]
# Convert to equivalent frequency
return fs * i_interp / N # Hz
My sampling rate is 40000. However, instead of getting a result close to 250Hz (B3 note), I am getting 0.66Hz. How is this possible?
I also tried with an autocorrelation method from the same repo but I also get bad results like 10000Hz.
Thanks to an answer I understand I have to apply a filter to remove the low frequencies in the signal. How do I do that? Are there multiple methods to do that, and which one is recommended?
STATUS UPDATE:
The high-pass filter proposed by the answer is working. If I apply the function in the answer to my audio signal, it correctly displays about 245Hz. However, I would like to filter the whole signal, not only a part of it. A note could lie in the middle of the signal or a signal contain more than one note (I know a solution is onset detection, but I am curious to know why this isn't working). That is why I edited the code to return filtered_audio.
The problem is that if I do that, even though the noise has been correctly removed (see screenshot). I get 0.05 as a result.
Based on the distances between the harmonics in the spectrogram, I would estimate the pitch to be about 150-200 Hz. So, why doesn't the pitch detection algorithm detect the pitch that we can see by eye in the spectrogram? I have a few guesses:
The note only lasts for a few seconds. At the beginning, there is a beautiful harmonic stack with 10 or more harmonics! These quickly fade away and are not even visible after 5 seconds. If you are trying to estimate the pitch of the entire signal, your estimate might be contaminated by the "pitch" of the sound from 5-12 seconds. Try computing the pitch only for the first 1-2 seconds.
There is too much low frequency noise. In the spectrogram, you can see a lot of power between 0 and 64 Hz. This is not part of the harmonics, so you could try removing it with a high-pass filter.
Here is some code that does the job:
import numpy as np
from scipy.io import wavfile
from scipy import signal
import matplotlib.pyplot as plt
from frequency_estimator import freq_from_hps
# downloaded from https://github.com/endolith/waveform-analyzer/
filename = 'Vocaroo_s1KZzNZLtg3c.wav'
# downloaded from http://vocaroo.com/i/s1KZzNZLtg3c
# Parameters
time_start = 0 # seconds
time_end = 1 # seconds
filter_stop_freq = 70 # Hz
filter_pass_freq = 100 # Hz
filter_order = 1001
# Load data
fs, audio = wavfile.read(filename)
audio = audio.astype(float)
# High-pass filter
nyquist_rate = fs / 2.
desired = (0, 0, 1, 1)
bands = (0, filter_stop_freq, filter_pass_freq, nyquist_rate)
filter_coefs = signal.firls(filter_order, bands, desired, nyq=nyquist_rate)
# Examine our high pass filter
w, h = signal.freqz(filter_coefs)
f = w / 2 / np.pi * fs # convert radians/sample to cycles/second
plt.plot(f, 20 * np.log10(abs(h)), 'b')
plt.ylabel('Amplitude [dB]', color='b')
plt.xlabel('Frequency [Hz]')
plt.xlim((0, 300))
# Apply high-pass filter
filtered_audio = signal.filtfilt(filter_coefs, [1], audio)
# Only analyze the audio between time_start and time_end
time_seconds = np.arange(filtered_audio.size, dtype=float) / fs
audio_to_analyze = filtered_audio[(time_seconds >= time_start) &
(time_seconds <= time_end)]
fundamental_frequency = freq_from_hps(audio_to_analyze, fs)
print 'Fundamental frequency is {} Hz'.format(fundamental_frequency)
So after my two last questions I come to my actual problem. Maybe somebody finds the error in my theoretical procedure or I did something wrong in programming.
I am implementing a bandpass filter in Python using scipy.signal (using the firwin function). My original signal consists of two frequencies (w_1=600Hz, w_2=800Hz). There might be a lot more frequencies that's why I need a bandpass filter.
In this case I want to filter the frequency band around 600 Hz, so I took 600 +/- 20Hz as cutoff frequencies. When I implemented the filter and reproduce the signal in the time domain using lfilter the right frequency is filtered.
To get rid of the phase shift I plotted the frequency response by using scipy.signal.freqz with the return h of firwin as numerator and 1 as predefined denumerator.
As described in the documentation of freqz I plotted the phase (== angle in the doc) as well and was able to look at the frequency response plot to get the phase shift for the frequency 600 Hz of the filtered signal.
So the phase delay t_p is
t_p=-(Tetha(w))/(w)
Unfortunately when I add this phase delay to the time data of my filtered signal, it has not got the same phase as the original 600 Hz signal.
I added the code. It is weird, before eliminating some part of the code to keep the minimum, the filtered signal started at the correct amplitude - now it is even worse.
################################################################################
#
# Filtering test
#
################################################################################
#
from math import *
import numpy as np
from scipy import signal
from scipy.signal import firwin, lfilter, lti
from scipy.signal import freqz
import matplotlib.pyplot as plt
import matplotlib.colors as colors
################################################################################
# Nb of frequencies in the original signal
nfrq = 2
F = [60,80]
################################################################################
# Sampling:
nitper = 16
nper = 50.
fmin = np.min(F)
fmax = np.max(F)
T0 = 1./fmin
dt = 1./fmax/nitper
#sampling frequency
fs = 1./dt
nyq_rate= fs/2
nitpermin = nitper*fmax/fmin
Nit = int(nper*nitpermin+1)
tps = np.linspace(0.,nper*T0,Nit)
dtf = fs/Nit
################################################################################
# Build analytic signal
# s = completeSignal(F,Nit,tps)
scomplete = np.zeros((Nit))
omg1 = 2.*pi*F[0]
omg2 = 2.*pi*F[1]
scomplete=scomplete+np.sin(omg1*tps)+np.sin(omg2*tps)
#ssingle = singleSignals(nfrq,F,Nit,tps)
ssingle=np.zeros((nfrq,Nit))
ssingle[0,:]=ssingle[0,:]+np.sin(omg1*tps)
ssingle[1,:]=ssingle[0,:]+np.sin(omg2*tps)
################################################################################
## Construction of the desired bandpass filter
lowcut = (60-2) # desired cutoff frequencies
highcut = (60+2)
ntaps = 451 # the higher and closer the signal frequencies, the more taps for the filter are required
taps_hamming = firwin(ntaps,[lowcut/nyq_rate, highcut/nyq_rate], pass_zero=False)
# Use lfilter to get the filtered signal
filtered_signal = lfilter(taps_hamming, 1, scomplete)
# The phase delay of the filtered signal
delay = ((ntaps-1)/2)/fs
plt.figure(1, figsize=(12, 9))
# Plot the signals
plt.plot(tps, scomplete,label="Original signal with %s freq" % nfrq)
plt.plot(tps-delay, filtered_signal,label="Filtered signal %s freq " % F[0])
plt.plot(tps, ssingle[0,:],label="original signal %s Hz" % F[0])
plt.grid(True)
plt.legend()
plt.xlim(0,1)
plt.xlabel('Time (s)')
plt.ylabel('Amplitude')
# Plot the frequency responses of the filter.
plt.figure(2, figsize=(12, 9))
plt.clf()
# First plot the desired ideal response as a green(ish) rectangle.
rect = plt.Rectangle((lowcut, 0), highcut - lowcut, 5.0,facecolor="#60ff60", alpha=0.2,label="ideal filter")
plt.gca().add_patch(rect)
# actual filter
w, h = freqz(taps_hamming, 1, worN=1000)
plt.plot((fs * 0.5 / np.pi) * w, abs(h), label="designed rectangular window filter")
plt.xlim(0,2*F[1])
plt.ylim(0, 1)
plt.grid(True)
plt.legend()
plt.xlabel('Frequency (Hz)')
plt.ylabel('Gain')
plt.title('Frequency response of FIR filter, %d taps' % ntaps)
plt.show()'
The delay of your FIR filter is simply 0.5*(n - 1)/fs, where n is the number of filter coefficients (i.e. "taps") and fs is the sample rate. Your implementation of this delay is fine.
The problem is that your array of time values tps is not correct. Take a look
at 1.0/(tps[1] - tps[0]); you'll see that it does not equal fs.
Change this:
tps = np.linspace(0.,nper*T0,Nit)
to, for example, this:
T = Nit / fs
tps = np.linspace(0., T, Nit, endpoint=False)
and your plots of the original and filtered 60 Hz signals will line up beautifully.
For another example, see http://wiki.scipy.org/Cookbook/FIRFilter.
In the script there, the delay is calculated on line 86. Below this, the delay is used to plot the original signal aligned with the filtered signal.
Note: The cookbook example uses scipy.signal.lfilter to apply the filter. A more efficient approach is to use numpy.convolve.
Seems like you may have had this answered already, but I believe that this is what the filtfilt function is used for. Basically, it does both a forward sweep and a backward sweep through your data, thus reversing the phase shift introduced by the initial filtering. Might be worth looking into.
I use Tektronix oscilloscope to perform some signal acquisition. I get 10.000 measurement points (few signal periods) and I have to do a frequency analysis on that set of data. My signal is 8MHz sine wave. When I use either SciPy or NumPy I get the same result - frequencies are spreaded too wide. The distance between two values is 500kHz and the highest frequency is 2.5GHz (absurd). When I want to measure frequency bandwidth around 8MHz I can only get exact values of 7.5, 8.0 and 8.5 MHz. I tried to change sample spacing determined by (x[1]-x[0]) and I got nothing better.
def CalculateFFT(t_val,p_val):
x = t_val #Two parameters: [x,y] values
y = lambda x: p_val
com_signal = y(x) # Combined signal
FFT_val = abs(scipy.fft(com_signal))
freq_val = scipy.fftpack.fftfreq(len(com_signal), x[1]-x[0])
spec_val = 20*scipy.log10(FFT_val)
return freq_val, spec_val
It is worth reading in more depth how DFFTs work but you should always have the following formulae in mind. For a time series with n points and maximum time Tmax, the time resolution is given by dt = Tmax / n
A DFFT will produce n points with
Fmax = 1 / dt
dF = 1 / Tmax
You seem to suggest the maximum frequency is sufficient (so the time resolution is okay) but the frequency resolution isn't good enough: you need to collect more data, at the same time resolution.
If (1) the sampling time is too short, (2) you require higher estimation frequency accuracy, and, (3) you know that your signal is a sine wave, then you can fit the signal to a sine wave. Like in How do I fit a sine curve to my data with pylab and numpy?,
with the exception that the frequency needs to be added.
Here is an example figure with a frequency of around 8 MHz:
Below is the example code:
""" Modified from https://stackoverflow.com/a/16716964/6036470 """
from numpy import sin, linspace, pi,average;
from pylab import plot, show, title, xlabel, ylabel, subplot, scatter
from scipy import fft, arange, ifft
import scipy
import matplotlib.pyplot as plt
import numpy as np
from scipy.optimize import leastsq
ff = 8e6; # frequency of the signal
Fs = ff*128; # sampling rate
Ts = 1.0/Fs; # sampling interval
t = arange(0,((1/ff)/128)*(128)*5,Ts) # time vector
A = 2.5;
ff_0 = 8.1456e6
y = A*np.sin(2*np.pi*ff_0*t+15.38654*pi/180) + np.random.randn(len(t))/5
guess_b = 0
guess_a = y.std()*2**0.5;
guess_c = 10*pi/180
guess_d = ff*0.98*2*pi
fig = plt.figure(facecolor="white")
plt.plot(t,y,'.', label='Signal Fred. %0.4f Hz'%(ff_0/1e6))
plt.xlabel('Time')
plt.ylabel('Amplitude')
plt.grid(alpha=0.5);
optimize_func = lambda x: (x[0]*np.sin(x[2]*t+x[1]) - y);
est_a, est_c, est_d = leastsq(optimize_func, [guess_a, guess_c, guess_d])[0]
data_fit = est_a*np.sin(est_d*t+est_c) ;
plt.plot(t,data_fit,label='Fitted Est. Freq. %0.4f Hz'%(est_d/(2*pi)/1e6))
plt.legend()
plt.tight_layout();
plt.show();
fig.save("sinfit.png")