PHYCS 3730/6720 Lab Exercise
Reading and references:
Answer file Mylab15.txt.
Exercise 1. Numpy Discrete Fourier Transforms
A pure cosine signal
f(t) = cos(w*t)
is sampled 128 times at the times t = j Dt for j =
0,1,...,127. The frequency is nu = 20/(128*Dt) =
0.15625/Dt. The angular frequency is w = 2 pi times
that. So the discrete signal is
fj = cos(40*Pi*j/N)
for N = 128. Compute its discrete Fourier transform and plot
the power spectrum. Note that the power spectrum looks continuous,
but that is only because pyplot connects the 128 plot points.
Copy your Python commands (your interactive session) to the answer
file. Then also put the answers to the questions below in your answer
- Without looking at the Fourier transform, determine
the frequency (not the angular frequency!) of the original
- Where are the peaks in the power spectrum P[m]? That is,
precisely at what values of m? You can check this by
printing selected components.
- Considering that the frequency values are
related to the discrete index m through v = m/(N Dt),
what is the frequency of the peak at the lower
index m? (It should agree with the frequency of the
signal in the vector you get with np.fft.fftfreq.)
- Why are there two peaks? See the discussion in the notes about
identities for Fourier transforms of real data.
Exercise 2. Damped oscillation
Use Python to construct the power spectrum for the signal f(t) =
exp(-0.2*t/Dt)*cos(0.8*t/Dt) using the same discretization as
above. That is, sample at t/Dt = j = 0,1,...,127. The line
shape you get in power vs frequency is called a Lorentzian.
In your answer file give the Python commands you used. Then answer
- At what value of m does the positive frequency peak in the
power spectrum occur? What is the frequency in units
of 1/Dt. Compare it with the frequency of the cosine in
the original signal.
- Estimate the width of the peak at half its height and give its
value in angular frequency (with the 2*Pi
included). That width is related to the decay rate of the
signal. The slower the decay, the narrower the width. If we
write the exponential decay factor as
exp(-Gamma/2*t). Your width should be equal
to Gamma, which, in this case is 0.4/Dt. Is it?