Sampling. Part II

In an earlier post we discussed how to obtain the continuous-time Laplace and Fourier transforms of a sampled signal based on the Laplace and Fourier transforms of the original signal. We shall now explore other relationships between those transforms and the Z-transform of the sampled signal.

The Z-transform

The Z-transform of the discrete-time signal $u[n]$, $n \in \mathbb{N}$, is the function of the complex variable $z$ defined by

$$U[z]=\mathcal{Z} \{ u[n] \} = \sum_{n=0}^{\infty} u[n] \, z^{-n}.$$

The Z-transform does to a discrete-time signal what the Laplace transform does to a continuous-time signal. It has many properties that are analogous to the properties of the Laplace transform. We shall discuss other properties of the Z-transform in more detail later in this and on future posts.

In the present post, our goal is to reveal a connection between the Laplace transform of the sampled signals $u_s$ and $u_t$ and the Z-transform of its associated sampled-signal $u[n]$.

The Z-transform of sampled signals

Recall the discrete-time signal

$$u[n] = u(n T_s), \quad n = 0, 1, \ldots$$

consisting of samples obtained periodically from the continuous-time signal $u(t)$, $t \geq 0$. As discussed here, consider also two related continuous-time signals: the sample-and-hold signal

u_s(t) = u(n T_s) = u[k], \quad n T_s \leq t \leq (n+1) T_s

and the modulated train of impulses

u_t(t)=\sum_{n = 0}^{\infty} u[n] \, \delta(t – n T_s) = u(t) \sum_{n = 0}^{\infty} \delta(t – n T_s).

As seen before, $u_s$ and $u_t$ are related by

$$u_s(t) = \int_{0}^{t} g_h(t – \tau) \, u_t(\tau) \, d\tau, \quad g_h(t) = 1(t) – 1(t – T_s).$$

We also already know how to calculate the transforms

U_t(s)=\frac{1}{T_s}\sum_{k=-\infty}^{\infty} U(s – j k \omega_s), \quad
U_s(s)=\frac{1-e^{-s T_s}}{s} U_t(s).

Let us now attempt to calculate the Laplace transform of $u_t$ directly using the defintion

$$U_t(s)=\int_{0}^{t} u_t(t) \, e^{- s t} \, dt.$$

Upon substitution of $u_t$ one obtains

U_t(s)&=\int_{0}^{\infty} \sum_{n = 0}^{\infty} u[n] \delta(t – n T_s) \, e^{- s t} \, dt \\
&= \sum_{n = 0}^{\infty} u[n] \int_{0}^{\infty} \delta(t – n T_s) \, e^{- s t} \, dt \\
&= \sum_{n = 0}^{\infty} u[n] \, e^{- s n T_s}

from which

$$U_t(s)= \left . U[z] \right |_{z=e^{s T_s}} = U[e^{s T_s}].$$

The ability to calculate $U_t(s)$ directly from $U[z]$ will be key in providing a formula for the inverse Z-transform. We leave that for a future post.

The above relationship suggests a strong connection between the variable $z$ and the exponential $e^{s T_s}$. Indeed, $U_t(s)$ converges in the same region as $U(s)$, that is $\operatorname{Re}\{s\} > \sigma_u$, consequently, because $U[z]$ must converge for $z=e^{s T_s}$, it follows that $U[z]$ converges in the region $|z| > \tilde{\sigma}_u = e^{\sigma_u T_s}$. Conversely, if $U[z]$ converges in $|z| > \tilde{\sigma}_u$ then $U_t(s)$ converges in $\operatorname{Re}\{s\} > \sigma_u = \log(\tilde{\sigma}_u)/T_s$.

Similar relationships hold for the sample-and-hold signal

$$U_s(s) = G_h(s) \, U_t(s) = \frac{1 – e^{-s T_s}}{s} U[e^{s T_s}],$$

and, if $\sigma_u < 0$, for the corresponding continuous- and discrete-time Fourier transforms (we will get to those eventually)

$$U_t(j \omega) = U[e^{j \omega T_s}].$$

By now, it should be apparent that the Z-transform can handle sampled signals in a much more straightforward way than the Laplace transform. We will show in future posts how this capability extends naturally to discrete-time linear systems.

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