SU-ENGR76 MAY022024 — Jemoka Knowledge Base

nyquist sampling theorem Formally, the nyquist limit is states as: for X(t) a continuous-time signal with bounded frequency representation which is bounded by [0, B] Hz; if X is sampled every T seconds, then if T < \frac{1}{2B} (sampling interval is smaller than 1/2B) or equivalently \frac{1}{T} > 2B (sampling frequency is larger than 2B), then X can be reconstructed from its samples X(0), X(T), X(2T), \ldots. At every time, we can go back and fourth between X the samples and sinusoids via:

\begin{align} X(t) &= b_0 + \sum_{j=0}^{BT} a_{j} \sin \left(2\pi \frac{j}{T}t\right) + b_{j} \cos \left(2\pi \frac{j}{T}t\right) \\ &= A_{0} + \sum_{j=1}^{BT} A_{j} \sin \left(2\pi \frac{j}{T} t + \phi_{j}\right) \end{align}

We use the second representation (in particular with A_{j} = \sqrt{a_{j}^{2} + b_{j}^{2}} because its easy to actually visualize and recover. Passband Signal What if our signal, instead of being a Baseband Signal (f \in [0,B]), what if we have a Passband Signal meaning f \in [f_{\min} > 0, f_{\max}]? We actually still only need 2(f_{\max} - f_{\min}) worth of samples. Its the same nyquist limit argument due to degrees of freedom.