10.2: Linear Time-Invariant Models

In the case of a time-invariant linear discrete-time system, the solutions can be simplified considerably. We first examine a direct time-domain solution, then compare this with a transform-domain solution, and finally return to the time domain, but in modal coordinates.

Direct Time-Domain Solution

For a linear time-invariant system, observe that

\[\left.\begin{array}{l}
A(k)=A \\
B(k)=B
\end{array}\right\} \text { for all } \quad k \geq 0 \ \tag{10.15}\]

where \(A\) and \(B\) are now constant matrices. Thus

\[\Phi(k, l)=A(k-1) \ldots A(l)=A^{k-l}, \quad k \geq l \ \tag{10.16}\]

so that, substituting this back into (10.10), we are left with

\[\begin{aligned}
x(k)=& A^{k} x(0)+\sum_{l=0}^{k-1} A^{k-l-1} B u(l) \\
=& A^{k} x(0)+\left[A^{k-1} B\left|A^{k-2} B\right| \cdots \mid B\right]\left(\begin{array}{c}
u(0) \\
u(1) \\
\vdots \\
u(k-1)
\end{array}\right) \ (10.17)
\end{aligned}\nonumber\]

Note that the zero-state response in this case exactly corresponds to a convolution sum. Similar expressions can be worked out for the outputs, by simplifying (10.14); we leave the details to you.

Transform-Domain Solution

We know from earlier experience with dynamic linear time-invariant systems that the use of appropriate transform methods can reduce the solution of such a system to the solution of algebraic equations. This expectation does indeed hold up here. First recall the definition of the one-sided \(\mathcal{Z}\)-transform :

Definition 10.1

The one-sided \(\mathcal{Z}\)-transform, \(F (z)\), of the sequence \(f (k)\) is given by

\[F(z)=\sum_{k=0}^{\infty} z^{-k} f\tag{k}\]

for all \(z\) such that the result of the summation is well defined, denoted by the Region of Convergence (ROC).

The sequence \(f (k)\) can be a vector or matrix sequence, in which case \(F (z)\) is respectively a vector or matrix as well.

It is easy to show that the transform of a sum of two sequences is the sum of the individual transforms. Also, scaling a sequence by a constant simply scales the transform by the same constant. The following shift property of the one-sided transform is critical, and not hard to Z establish. Suppose that \(f(k) \stackrel{\mathcal{Z}}{\longrightarrow} F(z)\). Then

1. \[g(k)=\left\{\begin{array}{ll}
   f(k-1) & ; \quad k \geq 1 \\
   0 & ; \quad k=0
   \end{array} \Rightarrow G(z)=z^{-1} F(z)\right. \nonumber\]
2. \[g(k)=f(k+1) \Rightarrow G(z)=z[F(z)-f(0)] \nonumber\]

Convolution is an important operation that can be defined on two sequences \(f (k)\), \(g(k)\) as

\[f * g(k)=\sum_{m=0}^{k} g(k-m) f\tag{m}\]

whenever the dimensions of \(f\) and \(g\) are compatible so that the products are defined. The \(\mathcal{Z}\) transform of a convolutions of two sequences satisfy

\[\begin{aligned}
\mathcal{Z}(f * g) &=\sum_{k=0}^{\infty} z^{-k} f * g(k) \\
&=\sum_{k=0}^{\infty} z^{-k}\left(\sum_{m=0}^{k} f(k-m) g(m)\right) \\
&=\sum_{m=0}^{\infty} \sum_{k=m}^{\infty} z^{-k} f(k-m) g(m) \\
&=\sum_{m=0}^{\infty} \sum_{k=0}^{\infty} z^{-(k+m)} f(k) g(m) \\
&=\sum_{m=0}^{\infty} z^{-m}\left(\sum_{k=0}^{\infty} z^{-k} f(k)\right) g(m) \\
&=F(z) G(z)
\end{aligned}\nonumber\]

Now, given the state-space model (10.1), we can take transforms on both sides of the equations there. Using the transform properties just described, we get

\[\begin{aligned}
z X(z)-z x(0) &=A X(z)+B U(z) \ (10.18) \\
Y(z) &=C X(z)+D U(z) \ (10.19)
\end{aligned}\nonumber\]

This is solved to yield

\[\begin{array}{ll}
X(z)= & z(z I-A)^{-1} x(0)+(z I-A)^{-1} B U(z) \\
Y(z)= & z C(z I-A)^{-1} x(0)+\underbrace{\left[C(z I-A)^{-1} B+D\right]}_{\text {Transfer Function }} U(z)
\end{array} \ \tag{10.20}\]

To correlate the transform-domain solutions in the above expressions with the timedomain expressions in (10.10) and (10.14), it is helpful to note that

\[(z I-A)^{-1}=z^{-1} I+z^{-2} A+z^{-3} A^{2}+\cdots \ \tag{10.21}\]

as may be verified by multiplying both sides by \((zI - A)\). The region of convergence for the series on the right is all values of \(z\) outside of some sufficiently large circle in the complex plane. What this series establishes, on comparison with the definition of the \(\mathcal{Z}\)-transform, is that the inverse transform of \(z(zI - A)^{-1}\) is the matrix sequence whose value at time \(k\) is \(A^{k}\) for \(k \geq 0\) the sequence is 0 for time instants \(k < 0\). That is we can write

\[\begin{aligned}
\left(I, A, A^{2}, A^{3}, A^{4}, \ldots\right) \quad \stackrel{\mathcal{Z}} {\longleftrightarrow} z(z I-A)^{-1} \\
\left(0, I, A, A^{2}, A^{3}, \ldots\right) \quad \stackrel{\mathcal{Z}} {\longleftrightarrow}(z I-A)^{-1}
\end{aligned} \nonumber\]

Also since the inverse transform of a product such as \((z I-A)^{-1} B U(z)\) is the convolution of the sequences whose transforms are \((z I-A)^{-1} B\) and \(U(z)\) respectively, we get

\[\begin{array}{c}
\left(x(0), A x(0), A^{2} x(0), A^{3} x(0), \ldots\right) \quad \stackrel{\mathcal{Z}}{\longleftrightarrow} \quad z(z I-A)^{-1} x(0) \\
\left(0, B, A B, A^{2} B, A^{3} B, \ldots\right) *(u(0), u(1), u(2), u(3), \ldots) \stackrel{\mathcal{Z}}{\longleftrightarrow}(z I-A)^{-1} B U(z)
\end{array}\nonumber\]

Putting the above two pieces together, the parallel between the time-domain expressions and the transform-domain expressions in (10.20) should be clear.