Reconstruction from samples taken at Nyquist rate

Let $g(t)$ be band-limited to $W$ Hz and sampled ideally at the Nyquist rate $f_s = 2W$ ($T_s = 1/2W$).

1. Sampled signal and its spectrum

$$g_\delta(t) = \sum_n g(nT_s)\,\delta(t - nT_s), \qquad G_\delta(f) = f_s\sum_n G(f - nf_s)$$

With $f_s = 2W$ the copies just touch without overlapping, so in $|f| < W$:

$$G_\delta(f) = 2W\,G(f) \;\Rightarrow\; G(f) = \frac{1}{2W}G_\delta(f), \quad |f| < W$$

2. Express $G(f)$ through the samples. Taking the Fourier transform of $g_\delta(t)$ directly:

$$G_\delta(f) = \sum_n g\!\left(\frac{n}{2W}\right)e^{-j\pi nf/W}$$

so

$$G(f) = \frac{1}{2W}\sum_n g\!\left(\frac{n}{2W}\right)e^{-j\pi nf/W}, \quad |f| < W$$

$G(f)$, and therefore $g(t)$, is fixed by the samples alone.

3. Inverse transform (ideal LPF reconstruction)

$$\begin{aligned} g(t) &= \int_{-W}^{W}G(f)e^{j2\pi ft}df \\ &= \sum_n g\!\left(\frac{n}{2W}\right)\frac{1}{2W}\int_{-W}^{W}e^{j2\pi f(t - n/2W)}df \\ &= \sum_n g\!\left(\frac{n}{2W}\right)\frac{\sin\big(2\pi W(t - n/2W)\big)}{2\pi W(t - n/2W)} \\ &= \sum_n g\!\left(\frac{n}{2W}\right)\text{sinc}(2\pi Wt - n\pi) \end{aligned}$$

This is the interpolation formula. Each sample is multiplied by a sinc pulse centred at its own instant; at $t = kT_s$ all other sincs are zero, so $g(kT_s)$ is reproduced exactly, and between samples the sum fills in $g(t)$. Physically this is the output of an ideal LPF (cut-off $W$, gain $T_s$) driven by the sample impulses. Hence a band-limited signal is fully recovered from samples at the Nyquist rate.