Spectrum of a signal and its samples

Sampling

Sampling converts a continuous-time signal $x(t)$ into samples $x(nT)$ taken every $T$ seconds ($\omega_s = 2\pi/T$). It is modelled as multiplication by an impulse train:

$$x_p(t) = x(t)\sum_{n=-\infty}^{\infty}\delta(t-nT) = \sum_n x(nT)\delta(t-nT)$$

Relation between the two spectra

The impulse train has transform $P(j\omega) = \dfrac{2\pi}{T}\sum_k \delta(\omega - k\omega_s)$. By the multiplication property,

$$\begin{aligned} X_p(j\omega) &= \frac{1}{2\pi}X(j\omega) * P(j\omega) \\ &= \frac{1}{T}\sum_{k=-\infty}^{\infty} X\big(j(\omega - k\omega_s)\big) \end{aligned}$$

So the spectrum of the sampled signal is:

• the original spectrum $X(j\omega)$ scaled by $1/T$,
• repeated periodically at every integer multiple of $\omega_s$,
• hence periodic in $\omega$ with period $\omega_s$.

(a) Original X(jw), band-limited to wM
            /\  A
           /  \
     -----/----\-----> w
        -wM    wM

(b) Sampled Xp(jw), ws > 2wM   (height A/T)
   /\        /\        /\
  /  \      /  \      /  \
-/----\----/----\----/----\--> w
  -ws         0        ws

(c) Sampled Xp(jw), ws < 2wM  -> overlap (aliasing)
    ___ ___ ___
   /   X   X   \
--/-------------\--> w

Interpretation

• If $\omega_s > 2\omega_M$, the copies do not overlap and the original spectrum can be cut out with an ideal LPF of gain $T$; $x(t)$ is recovered exactly.
• If $\omega_s < 2\omega_M$, the copies overlap and the spectrum near $\omega_s/2$ is corrupted (aliasing).
• For the discrete-time sequence $x[n] = x(nT)$, the DTFT is $X(e^{j\Omega}) = X_p(j\Omega/T)$, i.e. the same picture with frequency axis scaled so that $\omega_s$ maps to $2\pi$.