Problem 3¶

In [76]:

y = linspace(-2,2,1000)
g = y/sqrt(1-y**2)
g[abs(y) >= 1] = 0.0
plot(y,g,'k')
xlabel('y')
ylabel('g(y)')
grid()

/usr/local/lib/python2.7/dist-packages/ipython-3.0.0_b1-py2.7.egg/IPython/kernel/__main__.py:2: RuntimeWarning: invalid value encountered in sqrt
  from IPython.kernel.zmq import kernelapp as app

Problem 4¶

In [33]:

data = load('/home/bhardin2/ece458/2015.01.08.14.02.24.MST_10min.npz.npz')
print data.keys()

['timezone_hr', 'heights', 'epochtime', 'MSTdata']

In [35]:

E = data['MSTdata']
t = data['epochtime']
h = data['heights']
print shape(E)

(4800, 1287)

In [36]:

P = zeros(len(h))
for j in range(len(h)):
    Ej = E[:,j]
    P[j] = mean(abs(Ej)**2)
logP = log10(P)
figure(figsize=(2,6))
plot(logP,h,'k-')
xlabel('log10(power)')
ylabel('height [km]')

Out[36]:

<matplotlib.text.Text at 0x7f21ad87e6d0>

Secondary peak is around 150 km. Calculated below:

In [38]:

P2 = P.copy()
P2[h < 120] = 0.0
jmax = argmax(P2)
print 'Secondary peak at %.2f km' % h[jmax]

Secondary peak at 146.25 km

In [42]:

Ej = E[:,jmax]

n = 64
M = int(len(Ej)/n)

F2 = zeros(n)
for j in range(M):
    Estub = Ej[n*j:n*(j+1)]
    F = fft.fft(Estub)/n
    F2 += abs(F)**2
F2 = F2/M
F2s = fft.fftshift(F2)
x = arange(n)
xs = fftshift(x)
xs[xs>=n/2] = xs[xs>=n/2] - n
plot(xs,F2s,'k')
xlabel('FFT bin')
ylabel('Power')

Out[42]:

<matplotlib.text.Text at 0x7f21ad6cd190>

In [55]:

imax = argmax(F2s)
fftbin = xs[imax]
print 'FFT bin at maximum is %i' % fftbin
T = t[1]-t[0]
df = 1/(T*n)
ds = fftbin*df
print 'Doppler shift is %.3f Hz' % ds
v = ds*lam/2
print 'Doppler velocity is %.3f m/s' % v

FFT bin at maximum is -8
Doppler shift is -0.926 Hz
Doppler velocity is -0.116 m/s

Maximum peak is around 100 km

In [56]:

P2 = P.copy()
P2[h < 50] = 0.0
jmax = argmax(P2)
print 'Primary peak at %.2f km' % h[jmax]

Primary peak at 100.95 km

In [66]:

Ej = E[:,jmax]

n = 64
M = int(len(Ej)/n)

F2 = zeros(n)
for j in range(M):
    Estub = Ej[n*j:n*(j+1)]
    F = fft.fft(Estub)/n
    F2 += abs(F)**2
F2 = F2/M
F2s = fft.fftshift(F2)
x = arange(n)
xs = fftshift(x)
xs[xs>=n/2] = xs[xs>=n/2] - n
plot(xs,F2s,'k')
xlabel('FFT bin')
ylabel('Power')

Out[66]:

<matplotlib.text.Text at 0x7f21acfbfa90>

In [64]:

idx = logical_and(h > 90, h < 110)
figure(figsize=(2,3))
plot(P[idx],h[idx],'k-')
xlabel('Power')
ylabel('height [km]')

Out[64]:

<matplotlib.text.Text at 0x7f21ad0f5390>

This spectrum is much more wide-band than the spectrum at 150 km. This is likely because the physics of the two scattering processes are much different. The 150 km echoes are apparently caused by a coherent source with a rather constant velocity. However, the 100 km echoes are much more diffuse and have a large range of Doppler velocities, many of which are apparently aliased.

In [ ]: