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PHYCS 3730/6720 Lab Exercise

Reading and references:

Answer file **Mylab15.txt**.

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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
f_{j} = 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
file:
- Without looking at the Fourier transform, determine
the
*frequency* (not the angular frequency!) of the original
signal.
- 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
these questions:
- 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?