{"id":782,"date":"2024-06-02T17:10:47","date_gmt":"2024-06-02T17:10:47","guid":{"rendered":"https:\/\/neilfoxman.com\/?page_id=782"},"modified":"2024-06-04T19:05:32","modified_gmt":"2024-06-04T19:05:32","slug":"interpolation","status":"publish","type":"page","link":"https:\/\/neilfoxman.com\/?page_id=782","title":{"rendered":"Signal Recovery"},"content":{"rendered":"\n

The process described for sampling a signal<\/a> is effectively reversed when going from discrete to continuous time, and we first focus on generating an idealized continuous time sampled signal (weighted pulses spaced by $T$) from the discrete time sampled signal (sequence in $n$). The continuous time sampled signal then can be fed through continuous time LPF filters that represent various real-world interpolation methods to generate the re-created signal.<\/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

<\/span>Discrete to Continuous Conversion<\/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

The impulse train can then be recreated for each discrete sample. That is<\/p>\n\n\n\n

$$
\\begin{align}
y_p(t) = \\sum_{n=-\\infty}^{\\infty} y_d[n] \\delta(t-nT)
&\\stackrel{\\mathcal{F}}{\\leftrightarrow}
Y_p(j \\omega) = \\sum_{n=-\\infty}^{\\infty} y_d[n] e^{-j \\omega n T} \\\\
\\end{align}
$$<\/p>\n\n\n\n

The discrete sample signal frequency response $Y_d$ will repeat every $2\\pi$ in the $\\Omega$ domain as any discrete signal would. We also know that the output continuous signal frequency response $Y_p(j \\omega)$ will repeat in the $\\omega$ domain relative to the output sampling frequency $\\omega_s = 2 \\pi \/ T$, and so $\\Omega = \\omega T$ still holds as a viable comparison between $\\Omega$ and $\\omega$ domains.<\/p>\n\n\n\n

This pulsed output signal $Y_p(j \\omega)$ contains information of the single desired result we want in the continuous domain, which we can denote as the idealized $Y_c$ which is repeated every $\\omega_s$ in the $\\omega$ domain and scaled.<\/p>\n\n\n\n

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

From this, we can see that to accurately recreate $Y_d$ in continuous time, the ideal filter applied to $Y_p$ scales all signals $0 < |\\omega| < \\omega_s \/ 2$ by $T$, and all other frequency information is removed. This idealized interpolation method is explored further below.<\/p>\n\n\n\n

Note that the following discussion reverts back to using signal $x$ instead of signal $y$ for generality.<\/p>\n\n\n\n

<\/span>Interpolation<\/span><\/h2>\n\n\n\n
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An interpolation formula describes how to fit a continuous curve between sample points $x(nT)$ using a filter $h(t) \\stackrel{\\mathcal{F}}{\\leftrightarrow} H(j \\omega)$. Note that the shape of the filter impulse response $h(t)$ is convolved with the idealized continuous sample pulse $x_p(t)$ to represent the recovered signal.<\/p>\n\n\n\n

$$
\\begin{align}
x_r(t) &= x_p(t) * h(t) \\\\
x_r(t) &= \\left[ \\sum_{n=-\\infty}^{\\infty} x(nT) \\delta(t-nT) \\right] * h(t) \\\\
x_r(t) &= \\sum_{n=-\\infty}^{\\infty} x(nT) h(t – nT) \\\\
\\\\
X_r(j \\omega) &= X_p(j \\omega) H(j \\omega) \\\\
\\end{align}
$$<\/p>\n\n\n\n

The analysis does not need to be limited to the idealized continuous time sampled signal, and the same result appears if using discrete sampled values and the discrete frequency response.<\/p>\n\n\n\n

$$
\\begin{align}
x_r(t) &= \\sum_{n=-\\infty}^{\\infty} x_d[n] * h(t – nT) \\\\
\\\\
X_r(j \\omega) &= \\sum_{n=-\\infty}^{\\infty} x_d[n] H(j \\omega) e^{-j \\omega T n} \\\\
X_r(j \\omega) &= H(j \\omega) \\sum_{n=-\\infty}^{\\infty} x_d[n] e^{-j \\omega T n} \\\\
X_r(j \\omega) &= H(j \\omega) \\sum_{n=-\\infty}^{\\infty} x_d[n] e^{-j \\Omega n} \\\\
X_r(j \\omega) &= H(j \\omega) X_d(e^{j \\Omega}) \\\\
\\end{align}
$$<\/p>\n\n\n\n

<\/span>Ideal Signal Recreation Filter \/ Ideal Low-Pass Filter<\/span><\/h2>\n\n\n\n

This filter recreates the original signal perfectly such that $x_r(t) = x(t)$ It is difficult to implement in real-life, but provides a useful foundation of analysis.<\/p>\n\n\n\n

A unity-gain low-pass filter is described by<\/p>\n\n\n\n

$$
H_u(j \\omega) = \\cases{
\\begin{align}
1 && |\\omega| < \\omega_c \\\\
0 && |\\omega| \\geq \\omega_c \\\\
\\end{align}
}
$$<\/p>\n\n\n\n

Which can be found to have a frequency response of (TBD Reference)<\/p>\n\n\n\n

$$
\\begin{align}
h_u(t) &= \\frac{\\sin(\\omega_c t)}{\\pi t} \\\\
\\end{align}
$$<\/p>\n\n\n\n

Recall that the magnitude of $X_p(j \\omega)$ is $X(j \\omega)$ scaled by $1\/T$, so in order to get the recovered signal spectrum $X_r(j \\omega)$, we would need to scale the filter by the sampling period, $T$. That is, the ideal signal recovery interpolation filter is described by<\/p>\n\n\n\n

$$
H(j \\omega) = \\cases{
\\begin{align}
T && |\\omega| < \\omega_c \\\\
0 && |\\omega| \\geq \\omega_c \\\\
\\end{align}
}
$$<\/p>\n\n\n\n

$$
\\begin{align}
H(j \\omega) = T H_u (j \\omega)
\\end{align}
$$<\/p>\n\n\n\n

$$
\\begin{align}
h(t) &= T \\frac{\\sin(\\omega_c t)}{\\pi t} \\\\
h(t) &= \\frac{\\omega_c T}{\\pi} \\frac{\\sin(\\omega_c t)}{\\omega_c t} \\\\
h(t) &= \\frac{\\omega_c T}{\\pi} sinc(\\omega_c t) \\\\
\\end{align}
$$<\/p>\n\n\n\n

So the recovered signal becomes
$$
\\begin{align}
x_r(t) &= \\sum_{n=-\\infty}^{\\infty} x(nT) \\frac{\\omega_c T}{\\pi} \\frac{\\sin(\\omega_c (t-nT))}{\\omega_c (t-nT)} \\\\
\\end{align}
$$<\/p>\n\n\n\n

In the case where we assume the signal is band-limited and no aliasing occurs, we can use $\\omega_c = \\omega_s\/2 = \\pi\/T$ to get<\/p>\n\n\n\n

$$
\\begin{align}
h(t) &= \\frac{\\sin(\\pi t \/ T)}{\\pi t \/ T} \\\\
h(t) &= sinc(\\pi t \/ T) \\\\
\\end{align}
$$<\/p>\n\n\n\n

The reconstructed signal is then described by<\/p>\n\n\n\n

$$
\\begin{align}
x_r(t) &= \\sum_{n=-\\infty}^{\\infty} x(nT) h(t – nT) \\\\
x_r(t) &= \\sum_{n=-\\infty}^{\\infty} x(nT) \\frac{\\sin[\\pi (t – nT) \/ T]}{\\pi (t – nT) \/ T} \\\\
x_r(t) &= \\sum_{n=-\\infty}^{\\infty} x[n] \\frac{\\sin[\\pi (t – nT) \/ T]}{\\pi (t – nT) \/ T} \\\\
\\end{align}
$$<\/p>\n\n\n\n

In other words, the summation of sinc functions at each time step scaled by the original signal exactly replicates the original signal.<\/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

<\/span>Zero-Order Hold<\/span><\/h2>\n\n\n\n

More common in practice than multiplying input signal by generated impulses. Output is equivalent to convolving $x_p(t)$ (same as above) with a filter, $H_0(j \\omega)$, that has impulse response, $h_0(t)$, that is unity between $t=0$ and $t=T$.<\/p>\n\n\n\n

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Note that $H_0(j \\omega)$ is very similar to the rectangular pulse, but shifted in time.<\/p>\n\n\n\n

$$
\\begin{align}
H_{sq} (j \\omega) &= \\frac{2 sin(\\omega T\/2)}{\\omega}
&& \\text{Square of width }T\\text{, centered at }t=0 \\\\
H_0 (j \\omega) &= e^{-j \\omega T\/2} \\frac{2 sin(\\omega T\/2)}{\\omega}
&& \\text{Square of width }T\\text{, delayed by }t=T\/2 \\\\
\\end{align}
$$<\/p>\n\n\n\n

If the effects of the Zero-Order hold are desired to be reversed, an ideal reconstruction filter $H_r(j \\omega)$ would be<\/p>\n\n\n\n

$$
\\begin{align}
H_0(j \\omega) H_r(j \\omega) &= 1 \\\\
H_r(j \\omega) &= \\frac{1}{H_0(j \\omega)} \\\\
H_r(j \\omega) &= \\frac{1}{e^{-j \\omega T\/2} \\frac{2 sin(\\omega T\/2)}{\\omega}} \\\\
H_r(j \\omega) &= \\frac{e^{j \\omega T\/2}}{\\frac{2 sin(\\omega T\/2)}{\\omega}} \\\\
\\end{align}
$$<\/p>\n\n\n\n


<\/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

<\/span>First-Order Hold\/Linear Interpolation<\/span><\/h2>\n\n\n\n

Consider $x_p(t)$, the series of impulses weighted by $x(t)$<\/p>\n\n\n\n

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


<\/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

An interpolating filter $h(t)$ would then generate a recreated signal, $x_r(t)$ be described as<\/p>\n\n\n\n

$$
\\begin{align}

x_r(t) &= x_p(t) * h(t) \\\\

x_r(t) &= \\left[ \\sum_{n=-\\infty}^{\\infty} x(nT) \\delta(t-nT) \\right] * h(t) \\\\

x_r(t) &= [ \\cdots + x(-2T)\\delta(t+2T) + x(-T)\\delta(t+T) + x(0)\\delta(t) + \\\\
&x(T)\\delta(t-T) + x(2T)\\delta(t-2T) + \\cdots] * h(t) \\\\

x_r(t) &= \\cdots + x(-2T)h(t+2T) + x(-T)h(t+T) + x(0)h(t) + \\\\
&x(T)h(t-T) + x(2T)h(t-2T) + \\cdots \\\\
\\end{align}
$$<\/p>\n\n\n\n

Alternatively, one can find more simply that<\/p>\n\n\n\n

$$
\\begin{align}
x_r(t) &= x_p(t) * h(t) \\\\
x_r(t) &= \\left[ \\sum_{n=-\\infty}^{\\infty} x(nT) \\delta(t-nT) \\right] * h(t) \\\\
x_r(t) &= \\sum_{n=-\\infty}^{\\infty} x(nT) \\left[ \\delta(t-nT) * h(t) \\right] \\\\
x_r(t) &= \\sum_{n=-\\infty}^{\\infty} x(nT) h(t-nT) \\\\
\\end{align}
$$<\/p>\n\n\n\n

If we confine our focus to the region between two impulses of $x_p(t)$, say at some $t_0$, we can make some conceptual simplifications to determine the interpolation function. Note that in this case, all impulses of $x_p(t)$ where $t > t_0$ are weighted by the acausal side of $h(t)$ ($t < 0$) whereas all the impulses of $x_p(t)$ where $t \\leq t_0$ are weighted by the causal half of $h(t)$ ($t \\geq 0$). Graphically, we can split $h(t)$ into two functions where<\/p>\n\n\n\n

$$ h(t) = h_{pre}(t) + h_{post}(t) $$<\/p>\n\n\n\n

<\/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

Now with knowledge of what we want as the output, we can define the interpolation function. For example if we know that we want<\/p>\n\n\n\n

$$
\\begin{align}
x_r(t_0) = x(t_0) && { t_0 | t_0=nT, n \\in \\mathbb{Z}} \\\\
\\end{align}
$$<\/p>\n\n\n\n

then<\/p>\n\n\n\n

$$
\\begin{align}
x_r(0) &= \\cdots + x(-2T)h(2T) + x(-T)h(T) + x(0)h(0) + \\\\
&x(T)h(-T) + x(2T)h(-2T) + \\cdots \\\\
\\\\
&= x(0)\\\\
\\end{align}
$$<\/p>\n\n\n\n

and<\/p>\n\n\n\n

$$
\\begin{align}
x_r(T) &= \\cdots + x(-2T)h(T+2T) + x(-T)h(T+T) + x(0)h(T) + \\\\
&x(T)h(T-T) + x(2T)h(T-2T) + \\cdots \\\\
&= \\cdots + x(-2T)h(3T) + x(-T)h(2T) + x(0)h(T) + \\\\
&x(T)h(0) + x(2T)h(-T) + \\cdots \\\\
&= x(T)\\\\
\\end{align}
$$<\/p>\n\n\n\n

then to be agnostic of the input function, $x(t)$, we must set<\/p>\n\n\n\n

$$
\\begin{align}
h(0) = 1 \\\\

h(t) = 0 && \\forall |t| \\geq T
\\end{align}
$$<\/p>\n\n\n\n

Now, let’s pick $t_0 \\in [0,T]$ and only look at this interval. In this case we can rewrite $x_r(t)$ as<\/p>\n\n\n\n

$$
\\begin{align}
x_r(t_0) &= \\cdots + x(-2T)h_{post}(t_0+2T) + x(-T)h_{post}(t_0+T) + x(0)h_{post}(t_0) + \\\\
&x(T)h_{pre}(t_0-T) + x(2T)h_{pre}(t_0-2T) + \\cdots && t_0 \\in [0,T] \\\\
\\end{align}
$$<\/p>\n\n\n\n

Applying these constraints, $x_r(t)$ has many terms that become 0 and it simplifies to<\/p>\n\n\n\n

$$
\\begin{align}
x_r(t_0) &= x(0)h_{post}(t_0) + x(T)h_{pre}(t_0-T) && t_0 \\in [0,T] \\\\
\\end{align}
$$<\/p>\n\n\n\n

Additionally, if we want $x_r(t_0)$ to resemble a line between the two samples (first-order hold\/linear interpolation), we set the equation of that line equal to $x_r(t)$ then solve for the impulse response function that makes that possible.<\/p>\n\n\n\n

$$
\\begin{align}
x(0)h_{post}(t_0) + x(T)h_{pre}(t_0-T) &= x(0) + \\frac{x(T) – x(0)}{T} t_0 && t_0 \\in [0,T] \\\\
x(0)h_{post}(t_0) + x(T)h_{pre}(t_0-T) &= x(0) + \\frac{1}{T} x(T) t_0 – \\frac{1}{T}x(0) t_0 && t_0 \\in [0,T] \\\\
x(0)h_{post}(t_0) + x(T)h_{pre}(t_0-T) &= x(0)\\left[1 – \\frac{t_0}{T} \\right] + x(T) \\frac{t_0}{T} && t_0 \\in [0,T] \\\\
\\end{align}
$$<\/p>\n\n\n\n

The impulse response function can then be isolated such that<\/p>\n\n\n\n

$$
\\begin{align}
h_{post}(t) = 1 – \\frac{t}{T} \\\\
h_{post}(t) = \\frac{T – t}{T} \\\\
\\\\
h_{pre}(t-T) = \\frac{t}{T} \\\\
h_{pre}(t) = \\frac{t+T}{T} \\\\
\\end{align}
$$<\/p>\n\n\n\n

and so the final impulse response is<\/p>\n\n\n\n

$$
h(t) = \\cases{
\\begin{align}
&\\frac{t+T}{T} &&-T < t < 0 \\\\
&\\frac{T-t}{T} &&0 \\leq t < T \\\\
&0 &&\\text{Otherwise} \\\\
\\end{align}
}
$$<\/p>\n\n\n\n


<\/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

Note that the impulse response is acausal, but $h(t) = 0 \\;\\;\\; \\forall t > T$, so one can effectively achieve linear interpolation with a latency of $T$.<\/p>\n\n\n\n

Transfer function is determined by taking derivatives<\/p>\n\n\n\n

$$
h'(t) = \\cases{
\\begin{align}
&\\frac{1}{T} &&-T < t < 0 \\\\
&\\frac{-1}{T} &&0 \\leq t < T \\\\
&0 &&\\text{Otherwise} \\\\
\\end{align}
}
$$<\/p>\n\n\n\n

$$
h”(t) = \\frac{1}{T} \\delta(t+T) – \\frac{2}{T} \\delta(t) + \\frac{1}{T} \\delta(t-T)
$$<\/p>\n\n\n\n

$$
\\begin{align}
H”(j \\omega) = \\frac{1}{T} e^{j \\omega T} – \\frac{2}{T} + \\frac{1}{T} e^{-j \\omega T} \\\\
H”(j \\omega) = \\frac{1}{T} \\left[ e^{j \\omega T} + e^{-j \\omega T} -2 \\right] \\\\
\\end{align}
$$<\/p>\n\n\n\n

$$
\\begin{align}
H(j \\omega) = \\frac{1}{(j \\omega)^2 T} \\left[2 \\cos(\\omega T) -2 \\right] &&\\omega \\neq 0 \\\\
H(j \\omega) = \\frac{2}{(j \\omega)^2 T} \\left[\\cos(\\omega T) – 1 \\right] &&\\omega \\neq 0 \\\\
\\end{align}
$$<\/p>\n\n\n\n

Half Angle Formula<\/p>\n\n\n\n

$$
\\begin{array}{ll}
sin^2(x) = \\frac{1 – cos(2x)}{2} \\\\
2 sin^2(x) = 1 – cos(2x) \\\\
cos(2x) – 1 = – 2 sin^2(x)
\\end{array}
$$<\/p>\n\n\n\n

$$
\\begin{align}
2x = \\omega T \\\\
x = \\frac{\\omega T}{2} \\\\
\\end{align}
$$<\/p>\n\n\n\n

$$
\\begin{align}
H(j \\omega) = \\frac{2}{(j \\omega)^2 T} \\left[ -2 \\sin^2 \\left( \\frac{\\omega T}{2} \\right) \\right] &&\\omega \\neq 0 \\\\
H(j \\omega) = \\frac{-4 \\sin^2 (\\omega T\/2)}{(j \\omega)^2 T} &&\\omega \\neq 0 \\\\
H(j \\omega) = \\frac{1}{T} \\frac{4 \\sin^2 (\\omega T\/2)}{ \\omega^2} &&\\omega \\neq 0 \\\\
H(j \\omega) = \\frac{1}{T} \\frac{\\sin^2 (\\omega T\/2)}{ (\\omega \/ 2)^2} &&\\omega \\neq 0 \\\\
H(j \\omega) = T \\frac{\\sin^2 (\\omega T\/2)}{ (\\omega T \/ 2)^2} &&\\omega \\neq 0 \\\\
H(j \\omega) = T sinc^2 (\\omega T\/2) &&\\omega \\neq 0 \\\\
H(j \\omega) = \\frac{2 \\pi}{\\omega_s} sinc^2 \\left(\\pi \\frac{\\omega}{\\omega_s} \\right) &&\\omega \\neq 0 \\\\
\\end{align}
$$<\/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","protected":false},"excerpt":{"rendered":"

The process described for sampling a signal is effectively reversed when going from discrete to continuous time, and we first focus on generating an idealized continuous time sampled signal (weighted pulses spaced by $T$) from the discrete time sampled signal (sequence in $n$). The continuous time sampled signal then can be fed through continuous time […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":169,"menu_order":1,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-782","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages\/782","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=782"}],"version-history":[{"count":31,"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages\/782\/revisions"}],"predecessor-version":[{"id":870,"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages\/782\/revisions\/870"}],"up":[{"embeddable":true,"href":"https:\/\/neilfoxman.com\/index.php?rest_route=\/wp\/v2\/pages\/169"}],"wp:attachment":[{"href":"https:\/\/neilfoxman.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=782"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}