Sampling and sinc interpolation - Shah (Dirac comb)

Definition. The shah function or Dirac comb is defined as

$$ \text{III}_T (x) = \text{comb}(t) := \sum_{k=-\infty}^{\infty} \delta(x-kT).$$

Graphically, this is an infinite train of $\delta$'s spaced $T$ apart.

 

In the case where the spacing $T$ is 1, we have

$$ \text{III}(x) = \sum_{k=\infty}^{\infty} \delta(x-k) = \sum_{k=\infty}^{\infty} \delta_k. $$

 

Periodizing

$$ \begin{align*} \sum_{k=\infty}^\infty f(x-kT) &= \sum_{k=-\infty}^\infty f(x) * \delta (x-kT) \\ &= \left( f * \text{III}_T (x) \right) (t) \end{align*} $$

Convolving with $\text{III}_T(x)$ not only shifts $f$ by multiples of $T$, but also adds up the shifted replicas.

 

Sampling

$$ f(x) \text{III}_T(x) = \sum_{k=-\infty}^\infty f(x) \delta(x- kT) = \sum_{k=-\infty}^\infty f(kT)\delta(x-kT).$$

Multiplying with $\text{III}_T(x)$ produces evenly spaced samples of $f(x)$ at the points $0, \pm T, \pm 2T, \cdots$.

 

Scaling

$$ \begin{align*} \text{III} (Tx) &= \sum_{-\infty}^{\infty} \delta (Tx-k) \\ &=\sum_{-\infty}^{\infty} \delta \left(T (x- \frac{k}{T})\right) \\ & \hspace{2cm} \left( \text{Since } \delta(ax) = \frac{1}{|a|}\delta(x) \right) \\ &= \sum_{-\infty}^{\infty} \frac{1}{T} \delta(x-\frac{k}{T}) \\ &= \frac{1}{T} \text{III}_{1/T}(x) \end{align*} $$

To give it its own display, 

$$ \text{III}(Tx) = \frac{1}{T}\text{III}_{1/T}(x).$$

Replacing $T$ by $1/T$ in the above formula gives as well

$$ \text{III}(Tx) = \frac{1}{T} \text{III}_{1/T}(x) \Rightarrow \text{III}(\frac{x}{T}) = T\text{III}_T (x) $$

$$ \therefore \text{III}_T(x) = \frac{1}{T}\text{III} ( \frac{1}{T} x ) $$

 

Fourier Series of the Shah function

It is apparent that $\text{III}_T (t)$ is periodic with period $T$, i.e.

$$ \text{III}_T(t+T) = \text{III}_T(t). $$

Hence, it can be represented via a Fourier series representation:

$$ \begin{align*} \text{III}_T(t) &= \sum_{k=-\infty}^\infty \delta (t-kT) \\ &= \sum_{-\infty}^\infty c_n e^{2\pi i n (t/T)}, \end{align*} $$

where the complex Fourier coefficients $c_n$ are given by

$$ \begin{align*} c_n &= \frac{1}{T} \int_{t_0 + T}^T e^{-2\pi n(t/T)} \text{III}_T(t)dt \\ &= \frac{1}{T} \int_{-T/2}^{T/2} e^{-2\pi n(t/T)} \left( \sum_{k=-\infty}^\infty \delta (t-kT) \right) dt \\ &= \frac{1}{T} \int_{-T/2}^{T/2} e^{-2\pi n(t/T)} \delta(t) dt \\ & =\frac{1}{T} \end{align*} \tag{1}$$

In Eq. (1), the shah function reduces to a single impulse because the only impulse within the integration range $(-T/2, T/2)$ is the impulse at $t=0.$

$$\therefore \text{III}_T(t) =\frac{1}{T} \sum_{n=-\infty}^\infty e^{2\pi i n(t/T)} $$

 

Setting $T=1$, we have

$$ \text{III}(t) = \sum_{n=-\infty}^\infty e^{2\pi int} $$

 

Fourier Transform of the Shah function

Now, $\mathcal{F}\text{III} (s)$ can be derived from the preceding Fourier series expansion.

$$ \begin{align*} \left( \mathcal{F} \text{III} \right) (s) &= \int_{-\infty}^\infty e^{-2\pi ist} \text{III} (t) dt \\ &= \int_{-\infty}^\infty e^{-2 \pi ist} \left( \sum_{n=-\infty}^\infty e^{2\pi i nt} \right) dt \\ &= \sum_{n=-\infty}^\infty \int_{-\infty}^\infty e^{-2\pi i st} \left( e^{2\pi i nt} \right) dt \\ &= \sum_{n=-\infty}^\infty \left( \mathcal{F} e^{2\pi int} \right) (s) \\ &= \sum_{n=-\infty}^\infty \delta(s-n) = \text{III}(s) \end{align*} $$

$$ \therefore \mathcal{F}\text{III} = \text{III} $$

Then, we can deduce the formula for $\mathcal{F} \text{III}_T $:

$$ \begin{align*} \mathcal{F} \text{III}_T(s) &= \frac{1}{T} \left(\mathcal{F} \text{III} \left( \frac{x}{T} \right) \right) (s) \\ &=\frac{1}{T} \cdot T \mathcal{F} \text{III} (Ts) \\ &= \mathcal{F}\text{III} (Ts) \\ &= \text{III} (Ts) \\ &= \frac{1}{T} \text{III}_{1/T}(s) \end{align*} $$

$$ \therefore \mathcal{F} \text{III}_T (s) = \frac{1}{T} \text{III}_{1/T}(s) $$

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Definition. A signal $f(t)$ is bandlimited if there is a finite number $p$ such that $\mathcal{F}f(s) = 0$ for all $|s| \geq p/2 $, i.e. $ \mathcal{F} f(s) $ has compact support. 

 

Sinc interpolation

Consider a bandlimited signal $f(t)$ with the bandwidth $B$. We can get back \mathcal{F}f by first periodizing it using $\text{III}_B$ and then applying LPF:

$$ \mathcal{F}f = \Pi_B \left( \mathcal{F}f * \text{III}_B \right), $$

where $\Pi_B$ is a rectangle function of width $B$.

 

Taking the inverse Fourier transform of $\mathcal{F}f = \Pi_B \left( \mathcal{F}f * \text{III}_B \right) $:

$$ \begin{align*} f(t) &= \mathcal{F}^{-1} \mathcal{F}f(t) = \mathcal{F}^{-1} \left( \Pi_B (\mathcal{F}f * \text{III}_B ) \right) (t) \\ &= \mathcal{F}^{-1} \Pi_B * \mathcal{F}^{-1} \left( \mathcal{F}f * \text{III}_B \right) (t) \\ &= \mathcal{F}^{-1}\Pi_B (t) * \left( \mathcal{F}^{-1} \mathcal{F} f(t) \cdot \mathcal{F}^{-1} \text{III}_B (t) \right) \\ &= B\text{sinc} Bt * \left( f(t) \cdot \mathcal{F} ^{-1} \text{III}_B (t) \right) \\ & \hspace{2cm} \left( \text{Since } \mathcal{F}^{-1} \Pi_B = \mathcal{F}^{-1}\Pi \left( \frac{s}{B} \right) = B \text{sinc} Bt \right) \\ &= B\text{sinc}Bt * \left( f(t) \cdot \mathcal{F} \left( \frac{1}{B} \text{III} (\frac{x}{B}) \right) (t) \right) \\ &= B\text{sinc} Bt * \left( f(t) \cdot \text{III} (Bt) \right) \\ &= B\text{sinc}Bt * \left( f(t) \cdot \frac{1}{B} \text{III}_{1/B} (t) \right) \\ &= \text{sinct}Bt * \left( f(t) \cdot \sum_{k=\infty}^\infty \delta(t- \frac{k}{B}) \right) \\ &= \text{sinct}Bt * \sum_{k=-\infty}^\infty f(\frac{k}{B})\delta(t-\frac{k}{B}) \\ &= \sum_{k=\infty}^\infty f(\frac{k}{B})\text{sinc}B \left( t-\frac{k}{B} \right) \\ &= \sum_{k=-\infty}^{\infty} f(\frac{k}{B}) \text{sinc} \left( Bt - k \right) \end{align*} $$

 

Nyquist-Shannon sampling theorem

If $f(t)$ is a signal with bandwidth $B$, then

$$ f(t) = \sum_{k=-\infty}^{\infty} f \left( \frac{k}{p} \right) \text{sinc} p(t - \frac{k}{p}) $$

for any $p > B.$

• The sample points are spaced $1/p$ apart.
• The sampling rate is $p$, in units of $\text{Hz}$.

 

Plug in $t_l = l/p$:

$$ \sum_{k=-\infty}^\infty f (\frac{k}{p}) \text{sinc}p \left( \frac{l}{p} - \frac{k}{p} \right) = \sum_{-\infty}^\infty f(\frac{k}{p}) \text{sinc} (l-k). $$

Now recall that the sinc function is zero at the integers, except at $\text{sinc} 0 = 1$:

$$ \begin{cases} \text{sinc} (l-k) = 1, & l=k \\ 0, & l \neq k \end{cases} $$

Hence, 

$$\sum_{k=-\infty}^{\infty} f \left( \frac{k}{p} \right) \text{sinc} (l-k) = \sum_{k=-\infty}^\infty f (\frac{k}{p} )\delta_{l,k} = f \left( \frac{l}{p} \right). $$

When evaluated at a sample point, the formula indeed returns the value of the signal at that point.