Problem 3

Rectangular window, zero padding and frequency resolution.

Compute the discrete Fourier transform of the following cosine signal: $x(t) = cos(2πft)$ , for 𝑓 = 50 Hz, sampled during 1 second at every ∆= 0.001 s.
- Plot the power spectrum using the periodogram (i.e. the one-sided power spectrum based on the DFT) in dB. Use three different zero padding conditions: 1, 2 and 6 second zero padding.
The power spectrum does not look like a line spectrum, i.e. a 𝛿-function at 50 Hz. Why?
Why does the shape of the power spectrum change with different zero padding conditions? (d) Consider the following signal x(t)= cos(2πf1t)+cos(2πf2t), for 𝑓1 = 57 Hz and 𝑓2 = 60 Hz. What is the maximum frequency resolution for a Fourier transform applied to a 200 ms time window? Would you be able to discriminate between the two frequencies? Would zero padding help the discrimination? What should one do in order to discriminate the two frequencies?



In [1]:

    
using PlotlyJS









    






 
 Plotly javascript loaded.
 To load again call 
init_notebook(true)



In [5]:

    
# Define signal
f = 50 
dt = 0.0001
Fs=1/dt;   #sampling rate
fNyquist= Fs/2
t= collect(0:1*Fs-1)*dt
s = cos.(2*π*f*t);



In [20]:

    
"""
    compute_periodogram(x; nfft=length(x), fs=1)

### Arguments:
x: signal
T_pad: time to pad 

Note, this was done as an excercise. Could also use 
DSP.periodogram for a more rocust function
"""
function compute_periodogram(x; T_pad=0, fs=1, T=1)
    
    if T_pad > 0 
        T = T + 2*T_pad
        pad_size = Int(T_pad*fs)
        x = vcat(zeros(Float64, pad_size), x, zeros(Float64, pad_size))
    end
    
    N= T*Fs
    N_half = Int(round(N/2+1))



    # Periodogram (rectangular window): Discrete Fourier transform (via Fast Fourier Transform, FFT)
    x = x - mean(x) #subtract mean: zero DC shift
    xf = fft(x)

    # One-sided power spectrum (periodogram)
    Sp = dt^2 * 1/T * abs.(xf).^2 # dt^2 * 1/T * Xf.*conj(Xf);
    Sp = 2 * Sp[1:N_half] #One-sided spectrum (i.e. only positive frequencies) for even N;
    Sp_db = 10*log10.(Sp/maximum(Sp))
    
    return Sp_db
end









    



WARNING: replacing docs for 'compute_periodogram :: Tuple{Any}' in module 'Main'.






    Out[20]:





compute_periodogram



In [29]:

    
function plot_periodogram(f, Sp_db; pad=0)
    trace = PlotlyJS.scatter(x = f, y=Sp_db)
    layout = Layout(;title="Pad = $(pad)", showlegend=false,
                         xaxis=attr(title="Frequency (Hz)", zeroline=false, range=[40, 60]),
                         yaxis=attr(title="Power (dB)", zeroline=false, range=[-60, 0]))
    p = plot([trace], layout)
end









    Out[29]:





plot_periodogram (generic function with 1 method)

Question 1 (a)

*Compute the discrete Fourier transform of the following cosine signal: $x(t) = cos(2πft)$ , for 𝑓 = 50 Hz, sampled during 1 second at every ∆= 0.001 s.

R: See computation above

*Plot the power spectrum using the periodogram (i.e. the one-sided power spectrum based on the DFT) in dB. Use three different zero padding conditions: 1, 2 and 6 second zero padding.

R: See plots below



In [32]:

    
pad = 0
df = 1/(pad*2 + 1); #frequency resolution
F = collect(0:df:fNyquist)#frequency grid
p_dB = compute_periodogram(s, T_pad = pad, fs = Fs)
p0 = plot_periodogram(F, p_dB)

pad = 1
df = 1/(pad*2 + 1); #frequency resolution
F = collect(0:df:fNyquist)#frequency grid
p1_dB = compute_periodogram(s, T_pad = pad, fs = Fs)
p1 = plot_periodogram(F, p1_dB, pad=pad)

pad = 2
df = 1/(pad*2 + 1); #frequency resolution
F = collect(0:df:fNyquist)#frequency grid
p2_dB = compute_periodogram(s, T_pad = pad, fs = Fs)
p2 = plot_periodogram(F, p2_dB, pad=pad)

pad = 6
df = 1/(pad*2 + 1); #frequency resolution
F = collect(0:df:fNyquist)#frequency grid
p6_dB = compute_periodogram(s, T_pad = pad, fs = Fs)
p6 = plot_periodogram(F, p6_dB, pad=pad)

[p0 p1 p2 p6]









    Out[32]:

Question 2 (b)

The power spectrum does not look like a line spectrum, i.e. a 𝛿-function at 50 Hz. Why?

Question 3 (c)

Why does the shape of the power spectrum change with different zero padding conditions?

R: As we add more padding (up to a point), we smooth the power spectrum, because padding leads to interpolation

Question 4 (d)

Consider the following signal x(t)= cos(2πf1t)+cos(2πf2t), for 𝑓1 = 57 Hz and 𝑓2 = 60 Hz.

What is the maximum frequency resolution for a Fourier transform applied to a 200 ms time window?

R: fs = 1/0.2 = 5Hz.
Would you be able to discriminate between the two frequencies?

R: No since the f2-f1 < fs
Would zero padding help the discrimination? What should one do in order to discriminate the two frequencies?

R: No, padding does not effectively increase sampling rate. We could increase the duration of the window



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