In previous posts we have explored sampling of continuous-time signals and introduced the Z-transform as a tool to work with discrete-time signals. In this post we address the other end of the process, that is the reconstruction of a signal from its samples.

## The signal reconstruction problem

By signal reconstruction we mean the process of generating a continuous-times signal $u_r(t)$ from periodic samples $u[n]$, $n \in \mathbb{Z}$, that is the inverse operation of sampling. As with sampling, one can easily construct a physical device, for example a *zero-order holder* to produce the continuous-time signal

$$

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

$$

from given discrete samples $u[n]$. As with sampling, properly analyzing the behavior of such a device is a much more complicated task.

Note that, in general, if $u[n]$ was obtained by sampling a continuous-time signal $u(t)$, that is $u[n] = u(n T_s)$, that there is no reason to expect that $u_r(t) = u(t)$. Nor that $u_r(t) = u_s(t)$, where $u_s(t)$ is the sample-and-hold version of the signal $u(t)$ as discussed previously. Indeed, the possibility of *aliasing* will generally make it impossible to exactly reconstruct the original continuous-time signals from its samples.

Take for example the family of continuous-time sinusoidal signals

$$

\begin{aligned}

u_k(t) &= \cos(\omega t + k \omega_s t), &

\omega_s &= \frac{2 \pi}{T_s}, &

k &\in \mathbb{Z}.

\end{aligned}

$$

Because

$$

u[n] = u_0(n T_s) = u_1(n T_s) = u_2(n T_s) = \cdots

$$

it should not be possible to distinguish between any of the members of the family exclusively from the samples $u[n]$. These signals are *aliases* of each other.

We have two main reasons for studying signal reconstruction: the first is to characterize the continuous-time signal $u_r(t)$ that can be reconstructed from the samples $u[n]$, the second is to derive conditions under which it would be possible to exactly reconstruct a signal from its samples. We address the first in this post.

## Signal reconstruction and the inverse Z-transform

Recall the Z-transform of a discrete-time signal $u[n]$, $n\in\mathbb{N}$, given by

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

We have shown earlier that the existence of an exponential bound on $u[n]$ is a sufficient condition for $U[z]$ to exist and converge outside of a disk of radius, say, $\tilde{\sigma}_u$.

Our first goal is to calculate the signal $u_r(t)$ from $U[z]$. We have shown in this post that if $u_t(t)$ is the continuous-time modulated train of impulses obtained from a continuous-time signal $u(t)$ that it is possible to calculate $U_t(s)=\mathcal{L}\{u_t(t)\}$ directly from the corresponding $U[z]$ by evaluating

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

Conversely, one would expect that the Laplace inverse formula applied to $U[e^{s T_s}]$, that is

$$

u_z(t) =\mathcal{L}^{-1} \left \{ U[e^{s T_s}] \right \},

$$

should produce a continuous-time modulated train of impulses. From this train of impulses,$u_z(t)$, a time-invariant filter with impulse response $g_h(t) = 1(t) – 1(t – T_s)$, see this post for details, should provide the signal $u_r(t)$!

Moreover, the magnitude of each impulse of the modulated train of impulses, $u_z(t)$, should coincide with the samples used to produce $U[z]$. That is, solving the above signal reconstruction problem actually amounts to inverting the Z-transform!

## The inverse Z-transform

Start by noticing that

$$U[e^{(s – j k\omega_s) T_s}]= U[e^{s T_s}], \quad k \in \mathbb{Z},$$

to define

$$

\hat{U}[z] = \begin{cases}

T_s U[z], & |\angle z| < \pi \\

0, & |\angle z| \geq \pi

\end{cases}

$$

so that

$$U[e^{s T_s}]= \frac{1}{T_s} \sum_{k = -\infty}^{\infty} \hat{U}[e^{(s -j k \omega_s) T_s}].$$

Assume that $U[z]$ converges in $|z|>\tilde{\sigma}_u$ and $U[e^{s t}]$ converges in $\operatorname{Re}\{s\} > \sigma_{u} = \log(\tilde{\sigma}_u)/T_s$, and proceed by reversing the arguments discussed in this post to rewrite

$$

\begin{aligned}

U[e^{s T_s}] &= \frac{1}{T_s} \sum_{k = -\infty}^{\infty} \hat{U}[e^{(s -j k \omega_s) T_s}], \qquad \operatorname{Re}\{s\} > \sigma_{u} = \log(\tilde{\sigma}_u)/T_s \\

&= \frac{1}{2 \pi j} \lim_{\rho\rightarrow \infty} \int_{\alpha – j \rho}^{\alpha + j \rho} \frac{\hat{U}[e^{(s -z) T_s}]}{1 – e^{-z T_s}} \, ds, \qquad \alpha > 0 \\

&= \mathcal{L} \left \{ \hat{u}(t) \sum_{k=-\infty}^{\infty} \delta(t-n T_s)\right \},

\end{aligned}

$$

in which

$$\hat{u}(t)=\mathcal{L}^{-1} \{ \hat{U}(e^{s T_s}) \}.$$

Note that the magnitude of the impulses in $u_z(t) = \mathcal{L}^{-1}\{U[e^{s T_s}]\}$ should coincide with

$$u[n]=\hat{u}(n T_s)=\mathcal{Z}^{-1}\{U[z]\}.$$

All that remains is to evaluate

$$\hat{u}(t)=\mathcal{L}^{-1} \{\hat{U}(e^{s T_s}) \}=\frac{1}{2 \pi j} \lim_{\rho\rightarrow \infty} \int_{\alpha-j\rho}^{\alpha+j\rho} \hat{U}[e^{s T_s}] \, e^{s t} \, ds.$$

This can be done by using the fact that $\hat{U}[z] = 0$ for $|\angle z| \geq \pi$ to rewrite

$$

\begin{aligned}

\hat{u}(t)&=\frac{1}{2 \pi j} \lim_{\rho\rightarrow \infty} \int_{\alpha-j\rho}^{\alpha+j\rho} \hat{U}[e^{s T_s}] \, e^{s t} \, ds \\

&=\frac{T_s}{2 \pi j} \int_{\alpha-j \pi/T_s}^{\alpha+j\pi/T_s} U[e^{s T_s}] \, e^{s t} \, ds

\end{aligned}

$$

Changing variables

$$

\begin{aligned}

z &= e^{s T_s} , & s &=\frac{1}{T_s} \log(z), & ds &= \frac{dz}{T_s z},

\end{aligned}

$$

one obtains

$$

\begin{aligned}

\hat{u}(t)&=\frac{1}{2 \pi j} \oint_{C} U[z] \, z^{-1} \, z^{t/T_s} \, dz

\end{aligned}

$$

where $C$ is any circle centered at the origin with radius larger than $\tilde{\sigma}_u$. Finally

$$

\begin{aligned}

u[n]=\hat{u}(n T_s)&=\frac{1}{2 \pi j} \oint_{C} U[z] \, z^{n-1} \, dz

\end{aligned}

$$

which is the inverse Z-transform formula.

## One thought on “Sampling. Part III”