interpolation — Jemoka Knowledge Base

nyquist limit is great and all, but I really don’t want to wait for all T to be able to sample all the necessary terms to solve for every a_{j},b_{j} before we can reconstruct our signal. So, even if we got our sequence of \frac{1}{2B} length of points, we need an alternative way to reconstruct the signal as we go. One way to reconstruction via interpolation is just to connect the dots; however, this is bad because it creates sharp corners. In General Suppose you have a sampling period length T_{s}:

\begin{equation} \hat{x}(t) = \sum_{m=0}^{\infty} X\left(mT_{s}\right) F\left( \frac{t-mT_{s}}{T_{s}}\right) = x(0) F \left(\frac{t}{T_{s}}\right) + x(T_{s}) f\left(\frac{t-T_{s}}{T_{s}}\right) + \dots \end{equation}

where F(t) is some interpolation function such that:

\begin{equation} \begin{cases} F(0) = 1 \\ F(k) = 0, k \in \mathbb{Z} \backslash \{0\} \end{cases} \end{equation}

Notice the above is a convolution between X and F, where y is fixed as a multiple m around mT_{s} and the convolution is centered at \frac{t}{T_{s}}. However, because we are finite valued, we just slide a window around and skip around. Consider now \hat{x} at kT_{s} \begin{align} \hat{x}(kT_{s}) &= \sum_{m=0}^{\infty} X(mT_{s}) F \left(\frac{kT_{s}- mT_{s}}{T_{s}}\right) \ &= \sum_{m=0}^{\infty} X(mT_{s}) F \left(k-m\right) \end{align} now, recall that F is 0 for all non-zero integers, so each term will only be preserved once, precisely at m = k. Meaning:

\begin{align} \hat{x}(kT_{s}) &= \sum_{m=0}^{\infty} X(mT_{s}) F \left(k-m\right) \\ &= X(kT_{s}) 1 \\ &= X(kT_{s}) \end{align}

so this is why we need F(k) = 0, k \in \mathbb{Z} \backslash \{0\} Zero-Hold Interpolation Choose F such that:

\begin{equation} F = \begin{cases} 1, \text{if}\ |x| < \frac{1}{2} \\ 0 \end{cases} \end{equation}

Infinite-Degree Polynomial Interpolation \begin{equation} F(t) = (1-t) (1+t) \left(1- \frac{t}{2}\right) \left(1+ \frac{t}{2}\right) \dots = \text{sinc}(t) = \frac{\sin(\pi t)}{\pi t} \end{equation} This is the BEST interpolation; this is because it will be stretched such that every zero crossing matches eat mT_{s}, meaning we will recover a sum of sinusoids. This gives a smooth signal; and if sampling was done correctly with the nyquist limit, interpolating with sinc interpolation will give you your original signal. Shannon’s Nyquist Theorem Let X be a Finite-Bandwidth Signal where [0, B] Hz. if:

\begin{equation} \hat{X}(t) = \sum_{m=0}^{\infty} X(mTs) \text{sinc} \left( \frac{t-mTs}{Ts}\right) \end{equation}

where:

\begin{equation} \text{sinc}(t) = \frac{\sin \left(\pi t\right)}{\pi t} \end{equation}

if Ts < \frac{1}{2B}, that is, fs > 2B, then \hat{X}(t) = X(t) (this is a STRICT inequality!) otherwise, if Ts > \frac{1}{2B}, then \hat{X}(t) \neq X(t), yet \hat{X}(mTs) = X(mTs), and \hat{X} will be bandwidth limited to [0, \frac{fs}{2}]. This second case is callled “aliasing”, or “strocoscopic effect”. Alternate way of presenting the same info:

\begin{equation} \hat{X}(t) = \sum_{m=0}^{\infty} X(mTs) \text{sinc} \left( \frac{t-mT_{s}}{T_{s}}\right) \end{equation}

Let X(t), as before, be a continuous-time, bandwidth limited, signal with Bandwidth B; let \hat{X}(t) be the reconstruction of this signal with samples taken apart by T_{s} < \frac{1}{2B}; then \hat{X}(t) = X(t). Otherwise, if T_{s} > \frac{1}{2B}, then the reconstruction \hat{X}(t) \neq X(t), but the samples at mT_{s} will still match (that is, X(m T_{s}) = \hat{X}(m T_{s})) and \hat{X}(t) will be a Baseband Signal whose spectrum is limited by [0, \frac{\frac{1}{T_{s}}}{2}] = [0, \frac{F_{s}}{2}]. This second case is callled “aliasing”, or “strocoscopic effect”.