[Math] Fourier transform of sampling function

fourier analysisfourier transformsampling

Calculate the Fourier transform of $f_{ZOH}$ (the zero-order hold reconstruction of a sampled signal).

Where $f_{ZOH} (t)= f(kT), \ \ kT \leq t < (k+1)T,$ and the sampled signal is

$$f_s = f(t) \sum^\infty_{-\infty} \delta (t – kT).$$

Attempt

First writing the $f_{ZOH} (t)$ as a convolution between impulse response and sampled signal:

$$f_{ZOH} (t) = \left( \sum^{\infty}_{k=-\infty} f(kT) \delta(t-kT) \right) * h(t) = \sum^{\infty}_{k=-\infty} f(kT)h(t-kT).$$

But the shape function for the impulse response is the rectangular pulse function:

$$f_{ZOH} (t) = \sum^{\infty}_{k=-\infty} f(kT) \Pi (t-kT).$$

Now we want to calculate the Fourier transform of $f_{ZOH}$:

$$FT \Big[ f_{ZOH} \Big] = \frac{1}{T} F(\nu ) \sum^{\infty}_{k=-\infty} \Pi \left( \nu – \frac{k}{T} \right) = \frac{1}{T} \sum^{\infty}_{k=-\infty} F\left(\frac{k}{T}\right) FT \left[ \Pi \left(\nu – \frac{k}{T}\right) \right]$$

$$= \frac{1}{T} \sum^{\infty}_{k=-\infty} F\left(\frac{k}{T}\right) \textrm{sinc} \left(\nu – \frac{k}{T}\right).$$

But I was told that this is not the correct answer. What is wrong here?

Any explanation would be greatly appreciated.

Best Answer

The Fourier transform of the sampled signal $$f_s = f(t) \sum^\infty_{-\infty} \delta (t - kT)$$ is $$\begin{align} F_s(\nu)&=F(\nu)*\mathcal F\left\{\sum^\infty_{k=-\infty} \delta (t - kT)\right\}\\ &=F(\nu)*\sum^\infty_{k=-\infty} \frac{1}{T}\delta \left(\nu - \frac{k}{T}\right)\\ &=\frac{1}{T}\sum^\infty_{k=-\infty} F\left(\nu - \frac{k}{T}\right) \end{align} $$ so the Fourier transform $F_{\mathrm{ZOH}}(\nu)$ of $f_{\mathrm{ZOH}}(t)=f_s(t)*h_{\mathrm{ZOH}}(t)$ is $$ F_{\mathrm{ZOH}}(\nu)=\mathcal F\left\{(f_s*h_{\mathrm{ZOH}})(t)\right\}=F_s(\nu)H_{\mathrm{ZOH}}(\nu)=H_{\mathrm{ZOH}}(\nu)\frac{1}{T}\sum^\infty_{k=-\infty} F\left(\nu - \frac{k}{T}\right) $$ The Fourier transform $\mathcal F\left\{f_{\mathrm{ZOH}}(t)\right\}$ can be also evaluated as $$ \begin{align} F_{\mathrm{ZOH}}(\nu)&=F(\nu)*\mathcal F\left\{\sum^\infty_{k=-\infty} h_{\mathrm{ZOH}}(t - kT)\right\}\\ &=F(\nu)*\sum^\infty_{k=-\infty} \frac{1}{T} H_{\mathrm{ZOH}}\left(\frac{k}{T}\right)\delta\left(\nu - \frac{k}{T}\right)\\ &=\sum^\infty_{k=-\infty} \frac{1}{T} H_{\mathrm{ZOH}}\left(\frac{k}{T}\right)F\left(\nu - \frac{k}{T}\right) \end{align} $$ so we have $$ F_{\mathrm{ZOH}}(\nu)=H_{\mathrm{ZOH}}(\nu)\frac{1}{T}\sum^\infty_{k=-\infty} F\left(\nu - \frac{k}{T}\right)=\sum^\infty_{k=-\infty} \frac{1}{T} H_{\mathrm{ZOH}}\left(\frac{k}{T}\right)F\left(\nu - \frac{k}{T}\right) $$ For the ZOH we use $$h_{\mathrm{ZOH}}(t)=\begin{cases}1 & 0\le t\le T\\ 0 & \text{otherwise}\end{cases}$$ with Fourier tansform $$ H_{\mathrm{ZOH}}(\nu)=\frac{\sin(\pi\nu T)}{\pi\nu}\mathrm{e}^{-\mathrm j \pi\nu T}=T\mathrm{sinc(\pi\nu T)}\mathrm{e}^{-\mathrm j \pi\nu T} $$ and then $$ F_{\mathrm{ZOH}}(\nu)=\mathrm{sinc(\pi\nu T)}\mathrm{e}^{-\mathrm j \pi\nu T}\sum^\infty_{k=-\infty} F\left(\nu - \frac{k}{T}\right) $$

"Proof"

We start by constructing a periodic signal (and its corresponding periodic sequence of samples) out of $x(t)$:

$$ x_p(t) \triangleq \sum_{n=-\infty}^\infty x(t-nT_0) \tag{6} $$$$ x_p[k] \triangleq x_p(kT_s) \tag{7} $$

Now $x_p(t)$ has a Fourier Series:

$$ x_p(t) = \sum_{n=-\infty}^\infty c[n]e^{i2\pi nf_0t} \tag{8} $$

where:

$$ c[n] \triangleq \frac{1}{T_0}\int_{T_0} x_p(t)e^{-i2\pi nf_0t}dt \tag{9} $$

And $x(t)$ has a Fourier Transform $(1)$ which looks a lot like $c[n]$. In fact it's simply:

$$ c[n] = f_0X(nf_0) \tag{10} $$

Substituting $(10)$ into $(8)$ and the result into $(6)$, we get one of the forms of the Poisson Summation Formula:

$$ \sum_{n=-\infty}^\infty x(t-nT_0) = \sum_{n=-\infty}^\infty f_0X(nf_0)e^{i2\pi nf_0t} \tag{11} $$

This allows us to write $x_p[k]$ in a new way:

$$\begin{align} x_p[k] &= \sum_{n=-\infty}^\infty f_0X(nf_0)e^{i2\pi nf_0(kT_s)} \tag{12} \\ &= \sum_{n=-\infty}^\infty f_0X(nf_0)e^{i2\pi k\frac{n}{N}} \tag{13} \end{align}$$

where $N = T_0/T_s$ is the number of samples in the finite sequence. To simplify slightly, we assume that $N$ is even. If $|X(f)|$ drops fast enough, the terms for $|n|\ge N/2$ will vanish, giving:

$$ x_p[k] \approx \sum_{n=1-N/2}^{N/2} f_0X(nf_0)e^{i2\pi k\frac{n}{N}} \tag{14} $$

Now we work on the other representation of $x_p[k]$, using $(4)$ and remembering that $x_p[k] = x[k]$ for $k=0,\dots,N-1$:

$$\begin{align} x_p[k] &= \frac{1}{N}\sum_{n=0}^{N-1} X[n]e^{i2\pi n\frac{k}{N}} \tag{15} \\ &= \frac{X[0]}{N} + \sum_{n=1}^{N/2-1} \frac{X[n]}{N}e^{i2\pi n\frac{k}{N}} + \frac{X[N/2]}{N}e^{i2\pi \frac{N}{2}\cdot\frac{k}{N}} + \underbrace{\sum_{n=N/2-1}^{N-1} \frac{X[n]}{N}e^{i2\pi n\frac{k}{N}}}_S \tag{16} \end{align}$$

$$\begin{align} S &= \sum_{n=1}^{N/2-1} \frac{X[N-n]}{N}e^{i2\pi (N-n)\frac{k}{N}} \tag{17} \\ &= \sum_{n=1}^{N/2-1} \frac{X[N-n]}{N}e^{i2\pi (-n)\frac{k}{N}}\underbrace{e^{i2\pi N\frac{k}{N}}}_{=1} \tag{18} \\ &= \sum_{n=1-N/2}^{-1} \frac{X[n]}{N}e^{i2\pi n\frac{k}{N}} \tag{19} \end{align}$$

where we use circular index for $X[n]$, i.e. $n<0 \Rightarrow X[n]=X[n+N]$. Joining all the terms together simplifies to:

$$ x_p[k] = \sum_{k=1-N/2}^{N/2} \frac{X[n]}{N}e^{i2\pi n\frac{k}{N}} \tag{20} $$

Finally, comparing $(20)$ to $(14)$, we conclude that:

$$ \frac{X[n]}{N} \approx f_0X(nf_0) = \frac{X(nf_0)}{T_0} = \frac{X(nf_0)}{N T_s} = f_s\frac{X(nf_0)}{N} \tag{21} $$

Simulations

In practice, when dealing with real signals, instead of calculating the Fourier Transform of the continuous signal, we sample the data (often the data is already in discrete form) and calculate its Fast Fourier Transform (which is exactly the same as the Discrete Fourier Transform, but computed by a faster method). Then we use this to get a decent approximation of the continuous Fourier Transform. Here is an example of such signal:

Notice how this one is not perfectly symmetric. Its Fourier Transform can be computed by applying the definition $(1)$ directly, which might take a long time:

Notice how the argument goes crazy. That's because most of the energy of the signal is shifted to $t\approx 2.3$, which adds a phase of $\approx -2\pi\cdot 2.3f$ to the spectrum relative to a more "centered" one.

We can also sample the signal and calculate the FFT for different values of $N$ and compare the result with samples of the exact Fourier Transform. The approximation gets better and better for higher sampling rates ($T_0$ is fixed here):

The simulations above were done on Python 3, with the SciPy and NumPy libraries. The code was written for personal use only, so it's a mess, but it's given below just for reference:

import matplotlib.pyplot as plt
from cmath import *
from scipy.integrate import quad
from numpy import linspace, angle, inf, nan
import matplotlib.lines as mlines
from numpy.fft import fft
from numpy import array as vec

hide_discontinuity = True
fourier_transform_plot_samples = 100 #recommended >1000, but it's very slow
signal_plot_samples = 1000

#signal(t+DT/2) = exp(t/5)*sin(2*pi*F*t)*bump(t)
F = 5.0
DT = 4.0
def bump(t):
    if abs(t)>=1:
        return 0
    return exp(-1/(1-t*t))
def centralized_signal(t):
    return (exp(t/3)*sin(2*pi*F*t)*bump(t)).real
def signal(t):
    return centralized_signal(t-DT/2)
t = linspace(0,DT,num=signal_plot_samples)
signal_samples = [signal(k) for k in t]

plt.plot(t,signal_samples)
plt.title('Signal')
plt.xlabel('Time (s)')
plt.ylabel('Amplitude')
plt.grid(True, linestyle='dotted')
fig = plt.gcf()
fig.canvas.set_window_title('Signal')
plt.show()

###############################
def integrate(func, a, b):
    def real_func(k):
        return func(k).real
    def imag_func(k):
        return func(k).imag
    real_integral = quad(real_func, a, b)
    imag_integral = quad(imag_func, a, b)
    return real_integral[0] + 1j*imag_integral[0]

def fourier(func, f, a, b):
    return integrate(lambda t: func(t)*exp(-2j*pi*f*t), a, b)

def show_fourier_transform():
    f = linspace(-2*F,2*F,num=fourier_transform_plot_samples)
    Xm = []
    Xa = []
    Xa_prev = nan
    for k in f:
        Xf = fourier(signal, k, -inf,inf)
        Xm.append(abs(Xf))
        Xa_new = angle(Xf)
        if hide_discontinuity and Xa_prev != nan and abs(Xa_prev-Xa_new) > 0.999*pi/2:
            Xa_new = nan
        Xa_prev = Xa_new
        Xa.append(Xa_new)

    ay = plt.gca()
    ay2 = ay.twinx()
    ay.plot(f,Xm, color='C0')
    ay2.plot(f,Xa, color='C1', linewidth=0.5)
    C0_line = mlines.Line2D([], [], color='C0', label='Magnitude')
    C1_line = mlines.Line2D([], [], color='C1', linewidth=0.5, label='Argument')
    plt.legend(handles=[C0_line, C1_line], loc=1)
    plt.title('Fourier Transform')
    plt.xlabel('Frequency (Hz)')
    ay.set_ylabel('Magnitude')
    ay2.set_ylabel('Argument')
    ay.grid(True, linestyle='dotted')
    fig = plt.gcf()
    fig.canvas.set_window_title('Fourier Transform')
    plt.show()

show_fourier_transform()

###############################
def show_relation(N):
    t = linspace(0,DT,num=N)

    Fs = N/DT
    Ts = 1/Fs
    f_samples = round(2*F*DT)
    f_max = (f_samples-1)/DT

    indices_range = range(0,f_samples)
    frequency_range = linspace(0,f_max,num=f_samples)
    dense_frequency_range = linspace(0,f_max,num=fourier_transform_plot_samples)

    ########## DFT ##########
    DFT = []
    for k in t:
        DFT.append(signal(k))
    DFT = fft(DFT)[indices_range]

    ########## FT ###########
    FT = []
    for k in frequency_range:
        FT.append(fourier(signal, k, -inf,inf))
    FT = vec(FT)
    dense_FT = []
    for k in dense_frequency_range:
        dense_FT.append(fourier(signal, k, -inf,inf))
    dense_FT = vec(dense_FT)

    ######## Error ##########
    dF = Ts*DFT - FT

    ######### Plot ##########
    ax = plt.gca()
    ax2 = ax.twiny()

    ax.plot(dense_frequency_range, abs(dense_FT), color='C1')    
    ax.scatter(frequency_range, abs(FT), marker='o', color='C1')
    #ax.plot(frequency_range, angle(FT), marker='o', color='C1', linewidth=0.5)
    ax2.plot(indices_range, abs(Ts*DFT), marker='d', color='C0')
    #ax2.plot(indices_range, angle(Ts*DFT), marker='d', color='C0', linewidth=0.5)
    ax2.plot(indices_range, abs(dF), color='C2')

    C0_line = mlines.Line2D([], [], color='C1', marker='o', label='Fourier Transform')
    C1_line = mlines.Line2D([], [], color='C0', marker='d', label='DFT × Tₛ')
    C2_line = mlines.Line2D([], [], color='C2', label='Error')
    plt.legend(handles=[C0_line, C1_line, C2_line], loc=1)

    ax.set_xlim((-f_max/f_samples, f_max+f_max/f_samples))
    ax2.set_xlim((-1, f_samples-1+1))
    ax.set_xlabel('Frequency (Hz)')
    #ax2.set_xlabel('Sample')
    ax.set_ylabel('Magnitude')

    plt.title('N = '+str(N)+', Tₛ = '+str(1000*Ts)+' ms', y=1.08)
    ax.yaxis.grid(True, linestyle='dotted')
    ax.xaxis.grid(True, linestyle='dotted')
    fig = plt.gcf()
    fig.canvas.set_window_title('Relation between Fourier Transform and DFT')
    plt.show()

show_relation(100)
show_relation(1000)
show_relation(100000)

Inverse Fourier transform of windowed cosine

The Fourier transform is linear, so you can compute the $rect$ function that isn't multiplied by $cos$ by itself.

For the $rect$ * $cos$ term, remember that multiplication in the frequency domain is convolution in the time domain.

So $\mathscr F^{-1} (G(\omega)*F(\omega)) = \{g \ast f\}(t)$.

Consider $cos(pi \cdot f)$ and $rect(f/2)$ as $G(\omega)$ and $F(\omega)$ respectively. Take the inverse Fourier transforms of $G$ and $F$ separately. Then, try plugging $g(t)$ and $f(t)$ into the convolution definition.

Best Answer

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[Math] Sampling, Fourier Transform, and Discrete Fourier Transform

"Proof"

Simulations

Inverse Fourier transform of windowed cosine

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