{"id":169,"date":"2023-12-17T22:47:36","date_gmt":"2023-12-17T22:47:36","guid":{"rendered":"https:\/\/neilfoxman.com\/?page_id=169"},"modified":"2024-06-04T19:18:10","modified_gmt":"2024-06-04T19:18:10","slug":"sampling","status":"publish","type":"page","link":"https:\/\/neilfoxman.com\/?page_id=169","title":{"rendered":"Sampling"},"content":{"rendered":"\n

<\/span>Continuous Time Sampled Signal<\/span><\/h2>\n\n\n\n

Consider the impulse train with impulses spaced $T$ sec apart.<\/p>\n\n\n\n

$$
\\begin{align}
p(t) &= \\sum_{n=-\\infty}^{\\infty} \\delta(t-nT) \\\\
\\end{align}
$$<\/p>\n\n\n\n

This signal then has a Sampling Frequency of $\\omega_s = \\frac{2 \\pi}{T}$, and its Fourier Transform is (TBD on reference):<\/p>\n\n\n\n

$$
\\begin{align}
P(j \\omega) &= \\frac{2 \\pi}{T} \\sum_{k=-\\infty}^{\\infty} \\delta(\\omega – \\frac{2 \\pi k}{T}) \\\\
P(j \\omega) &= \\frac{2 \\pi}{T} \\sum_{k=-\\infty}^{\\infty} \\delta(\\omega – k \\omega_s) \\\\
\\end{align}
$$<\/p>\n\n\n\n

Next consider a continuous signal, $x_c(t)$ and its sampled signal $x_p(t)$ which is formed by multiplying $x_c(t)$ by the impulse train. Recalling the sampling property ($x(t) \\delta(t-t_0) = x(t_0) \\delta(t-t_0)$) we can say that<\/p>\n\n\n\n

$$
\\begin{align}
x_p(t) &= x_c(t) p(t) \\\\
x_p(t) &= x_c(t) \\sum_{n=-\\infty}^{\\infty} \\delta(t-nT) \\\\
x_p(t) &= \\sum_{n=-\\infty}^{\\infty} x_c(t) \\delta(t-nT) \\\\
x_p(t) &= \\sum_{n=-\\infty}^{\\infty} x_c(nT) \\delta(t-nT) \\\\
\\end{align}
$$<\/p>\n\n\n\n

<\/span>Multiplication Property Analysis<\/span><\/h2>\n\n\n\n

From the multiplication property, the Fourier Transform of $x_p(t)$, $X_p(j \\omega)$ is the convolution between $X_c(j \\omega)$ and $P(j \\omega)$. Because $P(j \\omega)$ is merely an impulse train in the frequency domain, $X_p(j \\omega)$ becomes a replication of $X_c (j \\omega)$ every $\\omega_s$ in the frequency domain.<\/p>\n\n\n\n

$$
\\begin{align}
X_p(j \\omega) &= \\frac{1}{2 \\pi} \\int_{-\\infty}^{\\infty} X_c(j \\theta) P(j(\\omega – \\theta)) d \\theta \\\\

X_p(j \\omega) &= \\frac{1}{2 \\pi} \\int_{-\\infty}^{\\infty} X_c(j \\theta) \\left[ \\frac{2 \\pi}{T} \\sum_{k=-\\infty}^{\\infty} \\delta(\\omega – \\theta – k \\omega_s) \\right] d \\theta \\\\

X_p(j \\omega) &= \\frac{1}{T} \\int_{-\\infty}^{\\infty} X_c(j \\theta) \\sum_{k=-\\infty}^{\\infty} \\delta(\\omega – \\theta – k \\omega_s) d \\theta \\\\

X_p(j \\omega) &= \\frac{1}{T} \\sum_{k=-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} X_c(j \\theta) \\delta(\\omega – \\theta – k \\omega_s) d \\theta \\\\

X_p(j \\omega) &= \\frac{1}{T} \\left[ \u2026 + \\int_{-\\infty}^{\\infty} X_c(j \\theta) \\delta(\\omega – \\theta + \\omega_s) d \\theta + \\int_{-\\infty}^{\\infty} X(j \\theta) \\delta(\\omega – \\theta) d \\theta + \\int_{-\\infty}^{\\infty} X(j \\theta) \\delta(\\omega – \\theta – \\omega_s) d \\theta + \u2026\\right] \\\\

X_p(j \\omega) &= \\frac{1}{T} \\sum_{k=-\\infty}^{\\infty} X_c(j(\\omega – k \\omega_s)) \\\\

\\end{align}
$$<\/p>\n\n\n\n

The second to last line can be conceptualized as successive convolutions between $X(j \\omega)$ and impulses each offset by $k\\omega_s$and scaled by $1\/T$ .<\/p>\n\n\n\n

The summarized result here is that the sampled frequency response can be visualized easily from the frequency response of the continuous signal (copies every $\\omega_s$).<\/p>\n\n\n\n

$$
\\begin{align}
x_p(t) = \\sum_{n=-\\infty}^{\\infty} x_c(nT) \\delta(t-nT)
&\\stackrel{\\mathcal{F}}{\\leftrightarrow}
X_p(j \\omega) = \\frac{1}{T} \\sum_{k=-\\infty}^{\\infty} X_c(j(\\omega – k \\omega_s))
\\end{align}
$$<\/p>\n\n\n\n

<\/span>Time Shift Property Analysis<\/span><\/h2>\n\n\n\n

Now, consider the Continuous Fourier Transform of an impulse and modifying it in the continuous domain as shown below<\/p>\n\n\n\n

$$
\\begin{align}
\\delta(t) &\\stackrel{\\mathcal{F}}{\\leftrightarrow} 1 \\\\
\\delta(t-nT) &\\stackrel{\\mathcal{F}}{\\leftrightarrow} e^{-j \\omega nT} \\\\
x_c(nT) \\delta(t-nT) &\\stackrel{\\mathcal{F}}{\\leftrightarrow} x_c(nT)e^{-j \\omega nT} \\\\
\\sum_{n=-\\infty}^{\\infty} x_c(nT) \\delta(t-nT) &\\stackrel{\\mathcal{F}}{\\leftrightarrow} \\sum_{n=-\\infty}^{\\infty} x_c(nT)e^{-j \\omega nT} \\\\
\\end{align}
$$<\/p>\n\n\n\n

Therefore, the Fourier transform of equally spaced weighted impulses can be represented by a composition of samples multiplied by corresponding phase delays.<\/p>\n\n\n\n

$$
\\begin{align}
x_p(t) = \\sum_{n=-\\infty}^{\\infty} x_c(nT) \\delta(t-nT)
&\\stackrel{\\mathcal{F}}{\\leftrightarrow}
X_p(j \\omega) = \\sum_{n=-\\infty}^{\\infty} x_c(nT) e^{-j \\omega n T} \\\\
\\end{align}
$$<\/p>\n\n\n\n

<\/span>Nyquist Frequency<\/span><\/h2>\n\n\n\n
\"\"\n\t\t\t\n\t\t\t\t\n\t\t\t<\/svg>\n\t\t<\/button><\/figure>\n\n\n\n
\"\"\n\t\t\t\n\t\t\t\t\n\t\t\t<\/svg>\n\t\t<\/button><\/figure>\n\n\n\n

If $x(t)$ is a band-limited signal ($X(j \\omega) = 0$ when $|\\omega| > \\omega_M$) then $x(t)$ is uniquely determined by samples iff<\/p>\n\n\n\n

$$
\\omega_s > 2 \\omega_M =\\text{Nyquist Rate} \\\\
$$<\/p>\n\n\n\n

The original signal may be recovered from the sampled signal (frequency response curve placed every $\\pm k \\omega_s$) using a LPF so that only the frequency response centered at $\\omega=0$ is kept.<\/p>\n\n\n\n

\"\"\n\t\t\t\n\t\t\t\t\n\t\t\t<\/svg>\n\t\t<\/button><\/figure>\n\n\n\n
\"\"\n\t\t\t\n\t\t\t\t\n\t\t\t<\/svg>\n\t\t<\/button><\/figure>\n\n\n\n

<\/span>
$$<\/span><\/h1>\n\n\n\n

<\/span>Anti-Aliasing Filter<\/span><\/h2>\n\n\n\n

Consider an input signal of just constant noise, that is<\/p>\n\n\n\n

$$
\\begin{align}
X_c(j \\omega) = C \\\\
\\end{align}
$$<\/p>\n\n\n\n

Then due to the Multiplication Analysis of sampled signals, we have<\/p>\n\n\n\n

$$
\\begin{align}
X_p(j \\omega) &= \\frac{1}{T} \\sum_{k=-\\infty}^{\\infty} X_c(j(\\omega – k \\omega_s)) \\\\
\\end{align}
$$<\/p>\n\n\n\n

If no low-pass filtering occurs, then the result diverges. In reality, however, there is usually at least one pole associated with sampling the input. Note that in that case we have<\/p>\n\n\n\n

$$
\\begin{align}
X_c(j \\omega) &= C \\frac{1}{1 + j \\tau \\omega} \\\\
\\\\
X_p(j \\omega) &= \\frac{1}{T} \\sum_{k=-\\infty}^{\\infty} C \\frac{1}{1 + j \\tau (\\omega – k \\omega_s)} \\\\
X_p(j \\omega) &= \\frac{C}{T} \\sum_{k=-\\infty}^{\\infty} \\frac{1}{1 + j \\tau (\\omega – k \\omega_s)} \\\\
X_p(j \\omega) &= \\frac{C}{T} \\left[ \\dots + \\frac{1}{1 + j \\tau (\\omega + 2 \\omega_s)} + \\frac{1}{1 + j \\tau (\\omega + \\omega_s)} + \\frac{1}{1 + j \\tau \\omega} + \\frac{1}{1 + j \\tau (\\omega – \\omega_s)} + \\frac{1}{1 + j \\tau (\\omega – 2 \\omega_s)} + \\dots \\right] \\\\
\\end{align}
$$<\/p>\n\n\n\n

With some rearranging<\/p>\n\n\n\n

$$
\\begin{align}
X_p(j \\omega) &= \\frac{C}{T} \\left[ \\frac{1}{1 + j \\tau \\omega} + \\left(\\frac{1}{1 + j \\tau (\\omega + \\omega_s)} + \\frac{1}{1 + j \\tau (\\omega – \\omega_s)} \\right) + \\left( \\frac{1}{1 + j \\tau (\\omega + 2 \\omega_s)} + \\frac{1}{1 + j \\tau (\\omega – 2 \\omega_s)} \\right) + \\dots \\right] \\\\

X_p(j \\omega) &= \\frac{C}{T} \\left[ \\frac{1}{1 + j \\tau \\omega} + \\sum_{k=1}^{\\infty} \\left(\\frac{1}{1 + j \\tau (\\omega + k\\omega_s)} + \\frac{1}{1 + j \\tau (\\omega – k\\omega_s)} \\right) \\right] \\\\

X_p(j \\omega) &= \\frac{C}{T} \\left[ \\frac{1}{1 + j \\tau \\omega} + \\sum_{k=1}^{\\infty} \\frac{1 + j \\tau (\\omega – k\\omega_s) + 1 + j \\tau (\\omega + k\\omega_s)}{[1 + j \\tau (\\omega + k\\omega_s)] [1 + j \\tau (\\omega – k\\omega_s)]} \\right] \\\\

X_p(j \\omega) &= \\frac{C}{T} \\left[ \\frac{1}{1 + j \\tau \\omega} + \\sum_{k=1}^{\\infty} \\frac{2 + j 2 \\tau \\omega}{1 + j \\tau (\\omega – k\\omega_s) + j \\tau (\\omega + k\\omega_s) – \\tau^2(\\omega^2 – k^2 \\omega_s^2)} \\right] \\\\

X_p(j \\omega) &= \\frac{C}{T} \\left[ \\frac{1}{1 + j \\tau \\omega} + \\sum_{k=1}^{\\infty} \\frac{2 + j 2 \\tau \\omega}{1 + j 2 \\tau \\omega – \\tau^2(\\omega^2 – k^2 \\omega_s^2)} \\right] \\\\

X_p(j \\omega) &= \\frac{C}{T} \\left[ \\frac{1}{1 + j \\tau \\omega} + 2\\sum_{k=1}^{\\infty} \\frac{1 + j \\tau \\omega}{1 + j 2 \\tau \\omega – \\tau^2 \\omega^2 – \\tau^2 \\omega_s^2 k^2} \\right] \\\\

\\end{align}
$$<\/p>\n\n\n\n

As an aside, consider the Basel problem which has been proven (using advanced mathematical methods) to be<\/p>\n\n\n\n

$$
\\begin{align}
\\sum_{k=1}^{\\infty} \\frac{1}{k^2} = \\frac{\\pi^2}{6}
\\end{align}
$$<\/p>\n\n\n\n

Note that the summation above could be rewritten in the form<\/p>\n\n\n\n

$$
\\begin{align}
\\sum_{k=1}^{\\infty} \\frac{1}{a + k^2} \\\\
\\end{align}
$$<\/p>\n\n\n\n

I would like to say that the series is convergent by the comparison test<\/p>\n\n\n\n

$$
\\frac{1}{a + k^2} < \\frac{1}{k^2}
$$<\/p>\n\n\n\n

But this should be revisited with more mathematical rigor to make sure that<\/p>\n\n\n\n

    \n
  1. $a$ is indeed positive<\/li>\n\n\n\n
  2. imaginary $j$ does not impact the result.<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

    Continuous Time Sampled Signal Consider the impulse train with impulses spaced $T$ sec apart. $$\\begin{align}p(t) &= \\sum_{n=-\\infty}^{\\infty} \\delta(t-nT) \\\\\\end{align}$$ This signal then has a Sampling Frequency of $\\omega_s = \\frac{2 \\pi}{T}$, and its Fourier Transform is (TBD on reference): $$\\begin{align}P(j \\omega) &= \\frac{2 \\pi}{T} \\sum_{k=-\\infty}^{\\infty} \\delta(\\omega – \\frac{2 \\pi k}{T}) \\\\P(j \\omega) &= \\frac{2 \\pi}{T} […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":106,"menu_order":2,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-169","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages\/169","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/neilfoxman.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=169"}],"version-history":[{"count":50,"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages\/169\/revisions"}],"predecessor-version":[{"id":876,"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages\/169\/revisions\/876"}],"up":[{"embeddable":true,"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages\/106"}],"wp:attachment":[{"href":"https:\/\/neilfoxman.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=169"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}