7.1: Models of Sampled-Data Systems

7.1.1 Sampled-Data Systems

The sampled-data systems operate on discrete-time, represented as integral multiples of the sampling time (\(T\)).

To model the sampled-data systems, we consider an ideal sampler that samples a physical signal \(r(t)\) every \(T\) seconds and generates a series of impulses with weights \(r(kT),\; \; k=0,1,\ldots .\)

Mathematically, the sampler output, \(r^{*} (t)\), represents multiplication of \(r(t)\) with an impulse train, given as:

\[r^{*} (t)=\sum _{k=0}^{\infty } r(kT)\delta (t-kT) \nonumber \]

By applying the Laplace transform to the sampled signal \(r^{*} (t)\), we obtain:

\[r^{*} (s)=\sum _{0}^{\infty } r(kT)e^{-skT} \nonumber \]

Since sampling time, \(T\), is constant, we can change the variable to: \(z=e^{sT}\), to represent the sampled signal as:

\[r(z)=\sum _{0}^{\infty } r(kT)z^{-k} \nonumber \]

where \(r(z)\) defines the \(z\)-transform of the sampled signal, \(r(kT)\).

7.1.2 The z-transform

The \(z\)-transform is the Laplace transform equivalent in the case of sampled-data systems. The domain of the \(z\)-transform includes real- or complex-valued number sequences.

Assuming the number sequence, \(r(kT)\), is obtained by sampling a real-valued signal \(r\left(t\right)\), the time dependence may be suppressed to represent the sequence as: \(r\left(k\right)=\left\{r\left(0\right), r\left(1\right),\dots \right\}\). The z-transformed sequence is given as:

\[r(z)=z[r(kT)]=\sum _{0}^{\infty } r(k)z^{-k} \nonumber \]

The transformed sequence, \(r(z)\), represents a rational function of a complex variable  \(z=\left|z\right|e^{j\theta }\).

Example \(\PageIndex{1}\)

The \(z\)-transform of a unit-step sequence: \(u(k)=\{ 1,\; 1,\; \ldots \}\) is given as:

\[u(z)=\sum _{0}^{\infty } z^{-k} =\frac{1}{1-z^{-1} } ;\; \; |z^{-1} |<1 \nonumber \]

The convergence of the above geometric series is conditioned on: \(|z^{-1} |<1\), or \(|z|>1\)which defines its region of convergence (ROC); the ROC is outside of the unit-circle: \(e^{j\theta },\ 0\le \theta <2\pi\) in the complex \(z\)-plane.

Example \(\PageIndex{2}\)

Let \(r(t)=e^{-at} u(t)\); then \(r(kT)=\{ 1,\; e^{-aT} ,e^{-2aT} ,\ldots \}\). Hence,

\[r(z)=\sum _{0}^{\infty } e^{-akT} z^{-k} =\frac{1}{1-e^{-aT} z^{-1} } ;\; |e^{-aT} z^{-1} |<1 \nonumber \]

The ROC is outside of the circle of radius \(e^{-aT}\) in the complex \(z\)-plane.

In digital signal processing (DSP) applications, the index '\(n\)' is commonly used to represent a real-valued sequenc, i.e., the z-transform of a sequence, \(r(n)=a^nu(n)\) is given as: \(r(z)=\frac{1}{1-az^{-1}}\). 

Example \(\PageIndex{3}\)

Let \(r(t)=e^{j\omega t} u(t);\) then \(r(kT)=\{ 1,e^{j\omega T} ,e^{2j\omega T} ,\ldots \}\). Hence,

\[r(z)=\sum _{0}^{\infty } e^{jk\omega T} z^{-k} =\frac{1}{1-e^{j\omega T} z^{-1} } ;\; \; |e^{j\omega T} z^{-1} |=|z^{-1} |<1 \nonumber \]

The ROC is outside of the unit-circle in the complex \(z\)-plane.

The \(z\)-transforms of the sampled sinusoidal signals are obtained by applying the Euler’s identity to the above \(z\)-transform of \(e^{jk\omega T}\) and separating the real and imaginary parts. Thus

\[\sin (k\omega T)\; {\mathop{\leftrightarrow }\limits^{z}} \frac{\sin (\omega T)\; z}{z^{2} -2\; \cos (\omega T)\; +1} \nonumber \]

\[\cos (k\omega T){\mathop{\leftrightarrow }\limits^{z}} \frac{z(z-\cos (\omega T)\; )}{z^{2} -2\; \rm cos(\omega T)\; +1} \nonumber \]

\[e^{-akT} \sin (k\omega T){\mathop{\leftrightarrow }\limits^{z}} \frac{e^{-aT} \sin (\omega T)z}{z^{2} -2\; \cos (\omega T)\; e^{-aT} +e^{-2aT} } \nonumber \]

\[e^{-akT} \; \cos (k\omega T){\mathop{\leftrightarrow }\limits^{z}} \frac{z(z-e^{-aT} \cos (\omega T))}{z^{2} -2\; \cos (\omega T)\; e^{-aT} +e^{-2aT} } . \nonumber \]

The \(z\)-transforms of other complex signals may be obtained by using its properties of linearity, differentiation, and translation, etc.

Inverse z-Transform 

Given the z-transform of a composite signal, the underlying sequence can be recovered by long division. Alternatively, partial fraction expansion (PFE) can be used to obtain first and second-order factors that can be inverse transformed with the help of z-transform tables.

Example \(\PageIndex{4}\)

Let \(y\left(z\right)=\frac{0.2z\left(z+0.9\right)}{\left(z-1\right)\left(z-0.6\right)\left(z-0.9\right)}\) represent the output of a feedback control system; then, the output sequence can be recovered by the following methods:

1. Expand \(\frac{y\left(z\right)}{z}\)in partial fractions to obtain: \(y\left(z\right)=\frac{4.75z}{z-1}-\frac{6z}{z-0.9}+\frac{1.25}{z-0.6}\). Hence, \(y\left(k\right)=4.75-6{\left(0.9\right)}^k+1.25{\left(0.6\right)}^k\).
2. Alternatively, use long division to obtain: \(y\left(z\right)=0.1z^{-1}+0.34z^{-2}+0.65z^{-3}+0.98z^{-4}+\dots\). Hence, \(y\left(k\right)=\left\{0,\ 0.1,\ 0.34,\ 0.65,\ 0.98,\ \dots \right\}\).