Chapter 7: State-space form

Starting with this chapter we begin the discussion of subjects of a more theoretical nature, not in the sense that they are not useful for practical purposes, but rather that one can already do a lot without the respective knowledge. Simultaneously the mathematical level of the presented text is generally higher than in the previous chapters. Readers who are not too interested in the respective subjects may consider skipping directly to Chapter 11, where the discussion returns to the previous “practical” level.

Transfer functions fully describe the behavior of linear time-invariant systems, but, as we already have seen, once the system parameters start to vary, we find out that some important information about the system topology is lacking. The state-space form provides a mathematical way to describe a system without losing the essential information about the system’s topology.¹ Practically it’s not much different from block diagrams, just instead of a graphical representation of a system we represent it by mathematical equations. The state-space form can help to obtain new insights into the way how differential and difference systems work.

7.1 Differential state-space form

The term state-space form simply means that a differential system is written in the form of ordinary differential equations of the first order, where the differentiation is done with respect to time, and the equations have been algebraically resolved in respect to derivatives. E.g. suppose we are interested in a 2-pole allpass based on the state-variable filter (Fig. 4.1). In principle we already have the respective equations in (4.1) but for the sake of demonstration let’s reobtain them from the block diagram in Fig. 4.1.

Let $u_1=y_{\mathrm{BP}}$ denote the output of the first integrator and $u_2=y_{\mathrm{LP}}$ denote the output of the second integrator. The input of the first integrator is $x-2R{y}_{\mathrm{BP}}-y_{\mathrm{LP}}$, thus

$$u_1=\int {\omega }_c(x-2R{u}_1-u_2)dt$$

The output of the second integrator is simply $y_{\mathrm{BP}}$:

$$u_2=\int {\omega }_c{u}_{1}dt$$

According to (4.23), the allpass signal can be obtained as $y=x-4R{y}_{\mathrm{BP}}$:

$$y=x-4R{u}_1$$

Writing all three equations together:

$$\begin{array}{rl}{u}_1& =\int {\omega }_c(x-2R{u}_1-u_2)dt\\ u_2& =\int {\omega }_c{u}_{1}dt\\ y& =x-4R{u}_1\end{array}$$

we have obtained the state-space form representation of Fig. 4.1, except that we are having integral rather than differential equations. From the mathematical point of view this is no more than a matter of notation and we can equivalently rewrite the same equations as

$$\begin{array}{rl}{\dot{u}}_1& ={\omega }_c(x-2R{u}_1-u_2)\\ {\dot{u}}_2& ={\omega }_c{u}_1\\ y& =x-4R{u}_1\end{array}$$

It is common to write the state-space equations in the matrix form:

$$\frac{d}{dt}\left(\begin{array}{c}{u}_1\\ u_2\end{array}\right)=\left(\begin{array}{cc}-2R{\omega }_c& -{\omega }_c\\ {\omega }_c& 0\end{array}\right)\left(\begin{array}{c}{u}_1\\ u_2\end{array}\right)+\left(\begin{array}{c}{\omega }_c\\ 0\end{array}\right)x\text{\hspace{1em}(7.1a)}$$

$$y=\left(\begin{array}{cc}-4R& 0\end{array}\right)\left(\begin{array}{c}{u}_1\\ u_2\end{array}\right)+x\text{\hspace{1em}(7.1b)}$$

or, by introducing

$$\begin{array}{rl}A& =\left(\begin{array}{cc}-2R{\omega }_c& -{\omega }_c\\ {\omega }_c& 0\end{array}\right)\\ \mathbf{b}& ={\left(\begin{array}{cc}{\omega }_c& 0\end{array}\right)}^{\mathsf{T}}\\ {\mathbf{c}}^{\mathsf{T}}& =\left(\begin{array}{cc}-4R& 0\end{array}\right)\\ d& =1\end{array}$$

we rewrite the same in vector notation:

$$\begin{array}{rl}\dot{\mathbf{u}}& =A\mathbf{u}+\mathbf{b}x\\ y& ={\mathbf{c}}^{\mathsf{T}}\mathbf{u}+d\cdot x\end{array}$$

This is the general state-space form for a single-input single-output differential system. We can promote it further to multiple inputs and multiple outputs by promoting $x$ and $y$ to vectors and promoting $\mathbf{b}$, ${\mathbf{c}}^{\mathsf{T}}$ and $d$ to matrices:

$$\dot{\mathbf{u}}=A\mathbf{u}+B\mathbf{x}\text{\hspace{1em}(7.2a)}$$

$$\mathbf{y}=C\mathbf{u}+D\mathbf{x}\text{\hspace{1em}(7.2b)}$$

E.g. for a single-input multiple-output LP/BP/HP SVF the equation (7.1b) turns into

$$\mathbf{y}=\left(\begin{array}{cc}0& 1\\ 1& 0\\ -2R& -1\end{array}\right)\left(\begin{array}{c}{u}_1\\ u_2\end{array}\right)+\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)x$$

The term state-space form originates from the fact that the vector of differential variables $\mathbf{u}$ represents the states of the integrators, or simply the state of the system. Respectively the linear space of vectors $\mathbf{u}$ is referred to as the state space of the system.

The state-space form encodes the essential information about the system’s topology, namely, which gains precede the integrators and which follow the integrators. Specifically, $B$ is the matrix of gains occuring on the paths from the inputs to the integrators, $C$ is the matrix of gains occurring on the paths from the integrators to the outputs, $D$ is the matrix of gains bypassing the integrators and $A$ is the matrix of gains on the feedback paths, thus they simultaneously precede and follow the integrators.

Integral form

Equations (7.2) can be rewritten in the integral form, which is merely a notational switch:

$$\mathbf{u}=\int (A\mathbf{u}+B\mathbf{x})dt=\mathbf{u}(0)+{\int }_0^t(A\mathbf{u}+B\mathbf{x})d\tau \text{\hspace{1em}(7.3a)}$$

$$\mathbf{y}=C\mathbf{u}+D\mathbf{x}\text{\hspace{1em}(7.3b)}$$

The integral form also allows to convert the state-space form back to the block diagram form. Each line of (7.3a) corresponds to an integrator, the respective right-hand side describing the integrator’s input signal.

Nonlinear state-space form

The right-hand sides of the equations (7.2) actually can be arbitary nonlinear vector functions of vector arguments, in which case we could write the equations as

$$\begin{array}{rl}\dot{\mathbf{u}}& =F(\mathbf{u},\mathbf{x})\\ \dot{\mathbf{y}}& =G(\mathbf{u},\mathbf{x})\end{array}$$

The discussion of nonlinear systems has been done in Chapter 6. Most of the ideas discussed in Chapter 6 can be equally applied to the systems expressed as a state-space form, and we won’t discuss nonlinear state-space forms further.

7.2 Integratorless feedback

Before we can convert a block diagram (or an equation system, for that matter) into a state-space form we need to resolve integratorless feedback loops, if there are any. Integratorless feedback is a continuous-time version of zero-delay feedback. While zero-delay feedback loops in discrete time systems are the loops containing no unit delays, integratorless feedback loops in continuous time systems are the loops containing no integrators.

The resolution of integratorless feedback is therefore subject to the same considerations and procedures as the resolution of zero-delay feedback. We are going to demonstrate this using the TSK allpass from Fig. 5.35 as an example.

Figure 7.1: Allpass TSK filter from Fig. 5.35 with expanded 1-pole allpass structures.

Expanding the internal structures of the 1-pole allpasses in Fig. 5.35 we obtain the structure in Fig. 7.1. Denoting the 1-pole allpass states as $u_1$ and $u_2$, their output signals as $y_1$ and $y_2$ and the input of the first 1-pole allpass as $y_0$ (as shown in Fig. 7.1) we obtain the following equations:

$$\begin{array}{rl}{\dot{u}}_1& =y_0-u_1\\ y_1& =u_1-(y_0-u_1)=2u_1-y_0\\ {\dot{u}}_2& =y_1-u_2\\ y_2& =u_2-(y_1-u_2)=2u_2-y_1\\ y_0& =x-k{y}_2\\ y& =y_2+k{y}_0\end{array}$$

where this time we have assumed ${\omega }_c=1$ for simplicity.

Apparently Fig. 7.1 contains an integratorless feedback loop, starting at $y_0$, going through the highpass path of the first allpass to $y_1$, then through the highpass path of the second allpass to $y_2$ and returning via the global feedback path to $y_0$. This loop contains three inverters and a gain of $k$, thus the total gain of this integratorless feedback loop is $-k$ and it is not instantaneously unstable provided $k>-1$. Under this assumption we can resolve it algebraically. Selecting just the equations for $y_n$ we have

$$\begin{array}{rl}{y}_1& =2u_1-y_0\\ y_2& =2u_2-y_1\\ y_0& =x-k{y}_2\end{array}$$

We would like to solve for $y_0$, therefore we first eliminate $y_2$ in the third equation:

$$y_0=x-k(2u_2-y_1)$$

and then $y_1$ in the just obtained equation:

$$y_0=x-k(2u_2-(2u_1-y_0))=x-k{y}_0+2k(u_1-u_2)$$

$$(1+k)y_0=x+2k(u_1-u_2)$$

and

$$y_0=\frac{x+2k(u_1-u_2)}{1+k}$$

Notice that the denominator corresponds to the instantaneously unstable case occuring for $k<-1$.

Now that we have resolved the integratorless feedback, we need to substitute the resolution result into the remaining equations of the original equation system:

$$\begin{array}{rl}{\dot{u}}_1& =y_0-u_1=\left(\frac{2k}{1+k}-1\right)u_1-\frac{2k}{1+k}{u}_2+\frac{1}{1+k}x=\frac{(k-1)u_1-2k{u}_2+x}{1+k}\\ {\dot{u}}_2& =y_1-u_2=2u_1-y_0-u_2=\\ & =\left(2-\frac{2k}{1+k}\right)u_1-\left(1-\frac{2k}{1+k}\right)u_2-\frac{1}{1+k}x=\\ & =\frac{2u_1+(k-1)u_2-x}{1+k}\\ y& =y_2+k{y}_0=2u_2-y_1+k{y}_0=2u_2-(2u_1-y_0)+k{y}_0=\\ & =2(u_2-u_1)+(1+k)y_0=2(u_2-u_1)+x+2k(u_1-u_2)=\\ & =2(k-1)u_1-2(k-1)u_2+x\end{array}$$

Or, in the matrix form

$$\dot{\mathbf{u}}=\frac{1}{k+1}\left(\begin{array}{cc}k-1& -2k\\ 2& k-1\end{array}\right)\mathbf{u}+\frac{1}{k+1}\left(\begin{array}{c}1\\ -1\end{array}\right)x\text{\hspace{1em}(7.4a)}$$

$$y=2(k-1)\cdot \left(\begin{array}{cc}1& -1\end{array}\right)\mathbf{u}+x\text{\hspace{1em}(7.4b)}$$

7.3 Transfer matrix

If $\mathbf{x}(t)=\mathbf{X}(s)e^{st}$, all other signals in the system have the same exponential form and the system turns into

$$\begin{array}{rl}s\mathbf{U}(s)e^{st}& =A\mathbf{U}(s)e^{st}+B\mathbf{X}(s)e^{st}\\ \mathbf{Y}(s)e^{st}& =C\mathbf{U}(s)e^{st}+D\mathbf{X}(s)e^{st}\end{array}$$

or

$$\begin{array}{rl}s\mathbf{U}(s)& =A\mathbf{U}(s)+B\mathbf{X}(s)\\ \mathbf{Y}(s)& =C\mathbf{U}(s)+D\mathbf{X}(s)\end{array}$$

The first of the two equations is a linear equation system in a matrix form in respect to the unknown $\mathbf{U}(s)$ and the solution is found from

$$(s-A)\mathbf{U}(s)=B\mathbf{X}(s)$$

(where $s-A$ is a short notation for $sI-A$ where $I$ is identify matrix), and

$$\mathbf{U}(s)=(s-A{)}^{-1}B\cdot \mathbf{X}(s)\text{\hspace{1em}(7.5)}$$

and thus

$$\mathbf{Y}(s)=C\mathbf{U}(s)+D\mathbf{X}(s)=C(s-A{)}^{-1}B\cdot \mathbf{X}(s)+D\cdot \mathbf{X}(s)$$

Introducing the matrix

$$H(s)=C(s-A{)}^{-1}B+D=\frac{C\mathrm{adj}(s-A)B}{\det (s-A)}+D\text{\hspace{1em}(7.6)}$$

we have

$$\mathbf{Y}(s)=H(s)\mathbf{X}(s)$$

Thus $H(s)$ is the transfer matrix of the system, its elements being the individual transfer functions corresponding to all possible input-output pairs of the system. In case of a single-input single-output system $H(s)$ reduces to a $1\times 1$ matrix:

$$H(s)={\mathbf{c}}^{\mathsf{T}}(s-A{)}^{-1}\mathbf{b}+d=\frac{{\mathbf{c}}^{\mathsf{T}}\mathrm{adj}(s-A)\mathbf{b}}{\det (s-A)}+d\text{\hspace{1em}(7.7)}$$

being simply the familiar transfer function.

From the formula (7.6) or (7.7) we can derive why the transfer functions of system built on integrators are nonstrictly proper rational functions. Indeed, the elements of $\mathrm{adj}(s-A)$ are polynomials of $s$ of up to $(N-1)$-th order (where $N$ is the dimension of the state space, that is simply the number of integrators). On the other hand, $\det (s-A)$ is a polynomial of $s$ of $N$-th order. Therefore, the elements of $(s-A{)}^{-1}$ are rational functions of $s$ sharing the same $N$-th order denominator $\det (s-A)$ and having numerators of up to $(N-1)$-th order. Thus, if $D=0$, the elements of $H(s)$ are strictly proper rational functions.

If $D\ne 0$, (7.6) turns into

$$H(s)=\frac{C\mathrm{adj}(s-A)B}{\det (s-A)}+D=\frac{C\mathrm{adj}(s-A)B+D\det (s-A)}{\det (s-A)}$$

and thus the numerators of the elements of $H(s)$ become polynomials of order $N$, if the respective element of matrix $D$ is nonzero. Thus, the transfer function becomes nonstrictly proper only if there is a direct (in the sense that it contains no integrators) path from the input to the output.

Note that, since the denominator of the transfer function(s) is $\det (s-A)$, it follows that the roots of the $\det (s-A)$ polynomial are the system poles. At the same time the roots of $\det (s-A)=0$ are the eigenvalues of $A$. Thus, eigenvalues of $A$ are the system poles.

7.4 Transposition

Computing the transfer matrix transpose we obtain from (7.6):

$$H^{\mathsf{T}}(s)={\left(C(s-A{)}^{-1}B+D\right)}^{\mathsf{T}}=B^{\mathsf{T}}(s-A^{\mathsf{T}}{)}^{-1}{C}^{\mathsf{T}}+D^{\mathsf{T}}$$

This looks like a transfer function of another system:

$$\begin{array}{rl}{\dot{\mathbf{u}}}^{\prime }& =A^{\prime }{\mathbf{u}}^{\prime }+B^{\prime }{\mathbf{x}}^{\prime }\\ {\mathbf{y}}^{\prime }& =C^{\prime }{\mathbf{u}}^{\prime }+D^{\prime }{\mathbf{x}}^{\prime }\end{array}$$

where

$$A^{\prime }=A^{\mathsf{T}}{B}^{\prime }=C^{\mathsf{T}}{C}^{\prime }=B^{\mathsf{T}}{D}^{\prime }=D^{\mathsf{T}}$$

We will refer to this new system as transposed system. The transposition of the state-space form corresponds to the transposition of block diagrams described in Section 2.14. Particularly, we swap the input gains $B$ for the output gains $C$ and vice versa.

So, the transfer function of the transposed system is

$$H^{\prime }(s)=C^{\prime }(s-A^{\prime }{)}^{-1}{B}^{\prime }+D^{\prime }=B^{\mathsf{T}}(s-A^{\mathsf{T}}{)}^{-1}{C}^{\mathsf{T}}+D^{\mathsf{T}}=H^{\mathsf{T}}(s)$$

where

$${\mathbf{Y}}^{\prime }(s)=H^{\prime }(s){\mathbf{X}}^{\prime }(s)=H^{\mathsf{T}}(s){\mathbf{X}}^{\prime }(s)$$

or, in component form

$$Y_n^{\prime }(s)=H_{nm}^{\prime }(s)X_m^{\prime }(s)=H_{mn}(s)X_m^{\prime }(s)$$

while for the original system we have

$$Y_m(s)=H_{mn}(s)X_n(s)$$

But the input/output pair $x_m^{\prime }$, $y_n^{\prime }$, is the transposed system corresponds to the input/output pair $x_n$, $y_m$ of the original system and the transfer function for each pair is $H_{mn}$. Thus, transposition preserves the transfer function relationships between the respective input/output pairs.

7.5 Basis changes

In the process of further analysis of state-space forms it will be highly useful to be able to change the basis of the state-space. Since a basis change is equivalent to a linear transformation of the linear space, let ${\mathbf{u}}^{\prime }=T\mathbf{u}$ denote such transformation (where $T$ is some nonsingular matrix). Remember that what we are doing is changing the basis of the space, the transformation $T$ is just a way to notate the respective change of coordinates! Then $\mathbf{u}=T^{-1}{\mathbf{u}}^{\prime }$ and we can rewrite (7.2a) in terms of ${\mathbf{u}}^{\prime }$:

$$\begin{array}{rl}\frac{d}{dt}\left(T^{-1}{\mathbf{u}}^{\prime }\right)& =A{T}^{-1}{\mathbf{u}}^{\prime }+B\mathbf{x}\\ T^{-1}{\dot{\mathbf{u}}}^{\prime }& =A{T}^{-1}{\mathbf{u}}^{\prime }+B\mathbf{x}\\ {\dot{\mathbf{u}}}^{\prime }& =TA{T}^{-1}{\mathbf{u}}^{\prime }+TB\mathbf{x}\end{array}\text{\hspace{1em}(7.8)}$$

Respectively, (7.2b) in terms of ${\mathbf{u}}^{\prime }$ turns into

$$\mathbf{y}=C\mathbf{u}+D\mathbf{x}=C{T}^{-1}{\mathbf{u}}^{\prime }+D\mathbf{x}$$

Introducing

$$A^{\prime }=TA{T}^{-1}{B}^{\prime }=TB{C}^{\prime }=C{T}^{-1}\text{\hspace{1em}(7.9)}$$

we obtain

$${\dot{\mathbf{u}}}^{\prime }=A^{\prime }{\mathbf{u}}^{\prime }+B^{\prime }\mathbf{x}\text{\hspace{1em}(7.10a)}$$

$$\mathbf{y}=C^{\prime }{\mathbf{u}}^{\prime }+D\mathbf{x}\text{\hspace{1em}(7.10b)}$$

which has exactly the same form as (7.2). That is we have obtained a new state-space representation of the system in the new basis. Note that thereby we didn’t change the basis of the spaces of the input signals $\mathbf{x}$ or the output signals $\mathbf{y}$, but solely the basis of the state signals $\mathbf{u}$. Thus, the basis change is a purely internal operation and doesn’t affect the components of the vectors $\mathbf{x}$ and $\mathbf{y}$. Respectively, the transfer matrix is not affected either, which can be explicitly shown by computing the transfer matrix in the new basis:

$$\begin{array}{rl}{H}^{\prime }(s)& =C^{\prime }(s-A^{\prime }{)}^{-1}{B}^{\prime }+D=C{T}^{-1}(s-TA{T}^{-1}{)}^{-1}TB+D=\\ & =C{T}^{-1}(Ts{T}^{-1}-TA{T}^{-1}{)}^{-1}TB+D=\\ & =C{T}^{-1}(T(s-A)T^{-1}{)}^{-1}TB+D=\\ & =C{T}^{-1}T(s-A{)}^{-1}{T}^{-1}TB+D=\\ & =C(s-A{)}^{-1}B+D=H(s)\end{array}$$

7.6 Matrix exponential

Another tool which we will need is the concept of the matrix exponential. We define the matrix exponential by writing the Taylor series for an ordinary exponential:

$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$$

and replacing $x$ with a matrix:

$$e^X=1+X+\frac{X^2}{2!}+\frac{X^3}{3!}+\dots \text{\hspace{1em}(7.11)}$$

The properties of the matrix exponential are similar to the ones of the ordinary exponential, except that typically the commutativity of the involved matrices is required. Particularly, the following properties are derived from (7.11) in a straightforward manner, under the assumption $XY=YX$ and $X{X}^{\prime }=X^{\prime }X$:

$$e^{X}Y=Y{e}^X{e}^{X+Y}=e^X{e}^Y=e^Y{e}^X\frac{d}{dt}{e}^{X(t)}=e^X{X}^{\prime }=X^{\prime }{e}^X$$

The value of $e^X$ is particularly easy to compute if $X$ is diagonal:

$$X=\left(\begin{array}{cccc}{\lambda }_1& 0& \cdots & 0\\ 0& {\lambda }_2& \cdots & 0\\ 0& 0& \ddots & 0\\ 0& 0& \cdots & {\lambda }_N\end{array}\right)$$

In this case formula (7.11) turns into $N$ parallel Taylor series for $e^{{\lambda }_n}$ and we simply have

$$e^X=\left(\begin{array}{cccc}{e}^{{\lambda }_1}& 0& \cdots & 0\\ 0& e^{{\lambda }_2}& \cdots & 0\\ 0& 0& \ddots & 0\\ 0& 0& \cdots & e^{{\lambda }_N}\end{array}\right)$$

If $X$ is not diagonal, but diagonalizable by a similarity transformation $TX{T}^{-1}$, the value $e^X$ can be computed by noticing that matrix exponential commutes with similarity trasformation:

$$e^{TX{T}^{-1}}=T{e}^X{T}^{-1}\text{\hspace{1em}(7.12)}$$

which allows to express $e^X$ via $e^{TX{T}^{-1}}$. The formula (7.12) is obtainable from (7.11) in a straightforward manner as well, where we also notice that (7.12) holds for any $T$ and $X$, they don’t have to commute.

If $X$ is not diagonalizable, then Jordan normal form can be used instead. We are going to address this case slightly later.

7.7 Transient response

The differential state-space equation (7.2a) can be solved in the same fashion as we solved the differential equations for the 1-pole in Section 2.15. Indeed, the difference between (7.2a) and the Jordan 1-pole (2.21) is that the former has matrix form. Also in (7.2a) the input signal is additionally multiplied by the matrix $B$, but that doesn’t change the picture essentially.

Repeating the same steps as in in Section 2.15, we multiply both sides of (7.2a) by the matrix exponential $e^{-At}$:

$$e^{-At}\dot{\mathbf{u}}=e^{-At}A\mathbf{u}+e^{-At}B\mathbf{x}$$

or

$$e^{-At}\dot{\mathbf{u}}-e^{-At}A\mathbf{u}=e^{-At}B\mathbf{x}$$

Noticing that

$$\frac{d}{dt}\left(e^{-At}\mathbf{u}\right)=e^{-At}\dot{\mathbf{u}}-e^{-At}A\mathbf{u}$$

we rewrite the state-space differential equation further as

$$\frac{d}{dt}\left(e^{-At}\mathbf{u}\right)=e^{-At}B\mathbf{x}$$

Integrating with respect to time from 0 to $t$:

$$e^{-At}\mathbf{u}-\mathbf{u}(0)={\int }_0^t{e}^{-A\tau }B\mathbf{x}dt$$

$$\mathbf{u}=e^{At}\mathbf{u}(0)+e^{At}{\int }_0^t{e}^{-A\tau }B\mathbf{x}dt=e^{At}\mathbf{u}(0)+{\int }_0^t{e}^{A(t-\tau )}B\mathbf{x}dt\text{\hspace{1em}(7.13)}$$

The formula (7.13) is directly analogous to (2.22). Further, assuming complex exponential $\mathbf{x}(t)=\mathbf{X}(s)e^{st}$ (note that all elements of $\mathbf{x}$ share the same exponential $e^{st}$, just with different amplitudes) we continue as:

$$\begin{array}{rl}\mathbf{u}& =e^{At}\mathbf{u}(0)+e^{At}{\int }_0^t{e}^{-A\tau }B\mathbf{X}(s)e^{s\tau }dt=\\ & =e^{At}\mathbf{u}(0)+e^{At}{\int }_0^t{e}^{(s-A)\tau }dt\cdot B\mathbf{X}(s)=\end{array}$$

$$\begin{array}{rl}& =e^{At}\mathbf{u}(0)+e^{At}(s-A{)}^{-1}{e}^{(s-A)\tau }{|}_{\tau =0}^t\cdot B\mathbf{X}(s)=\\ & =e^{At}\mathbf{u}(0)+e^{At}(s-A{)}^{-1}\left(e^{(s-A)t}-1\right)B\mathbf{X}(s)=\\ & =e^{At}\left(\mathbf{u}(0)-(s-A{)}^{-1}B\mathbf{X}(s)\right)+(s-A{)}^{-1}B\mathbf{X}(s)e^{st}\end{array}$$

Comparing to the transfer matrix for $\mathbf{u}(t)$ defined by (7.5) we introduce the steady-state response

$${\mathbf{u}}_s(t)=(s-A{)}^{-1}B\cdot \mathbf{X}(s)e^{st}$$

and therefore

$$\mathbf{u}(t)=e^{At}\left(\mathbf{u}(0)-{\mathbf{u}}_s(0)\right)+{\mathbf{u}}_s(t)={\mathbf{u}}_t(t)+{\mathbf{u}}_s(t)\text{\hspace{1em}(7.14)}$$

where ${\mathbf{u}}_t(t)$ is the transient response.

Note that we have just explicitly obtained the fact (previously shown only for the system orders $N\le 2$) that, given a complex exponential input $\mathbf{X}(s)e^{st}$, the elements of the steady-state response ${\mathbf{u}}_s$ will be the same complex exponentials $e^{st}$, just with different amplitudes. An immediately following conclusion is that the steady-state signals $\mathbf{y}$, being linear combinations of $\mathbf{u}$ and $\mathbf{x}$, are also the same complex exponentials $e^{st}$. In fact, any other steady-state signal in the system, being a linear combination of $\mathbf{u}$ and $\mathbf{x}$, is the same complex exponential $e^{st}$.

In a fully analogous to the 1-pole case way we can show that (7.14) also holds for

$$\mathbf{x}(t)={\int }_{\sigma -j\infty }^{\sigma +j\infty }\mathbf{X}(s)e^{st}\frac{ds}{2\pi j}$$

in which case

$${\mathbf{u}}_s(t)={\int }_{\sigma -j\infty }^{\sigma +j\infty }(s-A{)}^{-1}B\mathbf{X}(s)e^{st}\frac{ds}{2\pi j}$$

Substituting (7.14) into (7.2b) we obtain

$$\begin{array}{rl}\mathbf{y}(t)& =C{e}^{At}\left(\mathbf{u}(0)-{\mathbf{u}}_s(0)\right)+C{\mathbf{u}}_s(t)+D\mathbf{x}(t)=\\ & =e^{At}\left(\left(C\mathbf{u}(0)+D\mathbf{x}(0)\right)-\left(C{\mathbf{u}}_s(0)+D\mathbf{x}(0)\right)\right)+C{\mathbf{u}}_s(t)+D\mathbf{x}(t)=\\ & =e^{At}\left(\mathbf{y}(0)-{\mathbf{y}}_s(0)\right)+{\mathbf{y}}_s(t)={\mathbf{y}}_t(t)+{\mathbf{y}}_s(t)\end{array}$$

where

$${\mathbf{y}}_t(t)=C{\mathbf{u}}_t(t)=e^{At}\left(\mathbf{y}(0)-{\mathbf{y}}_s(0)\right)$$

and

$$\begin{array}{rl}{\mathbf{y}}_s(t)& =C{\mathbf{u}}_s(t)+D\mathbf{x}(t)=C{\int }_{\sigma -j\infty }^{\sigma +j\infty }(s-A{)}^{-1}B\mathbf{X}(s)e^{st}\frac{ds}{2\pi j}+D\mathbf{x}(t)=\\ & ={\int }_{\sigma -j\infty }^{\sigma +j\infty }\left(C(s-A{)}^{-1}B+D\right)\mathbf{X}(s)e^{st}\frac{ds}{2\pi j}=\\ & ={\int }_{\sigma -j\infty }^{\sigma +j\infty }H(s)\mathbf{X}(s)e^{st}\frac{ds}{2\pi j}\end{array}\text{\hspace{1em}(7.15)}$$

The latter confirms the fact that ${\mathbf{y}}_s$ is the steady-state response.

7.8 Diagonal form

We have seen that the transient response part of the signals in the system consists of linear combinations of elements of the matrix $e^{At}$. The elements of $e^{At}$ can be easily found if $A$ is diagonalized by a similarity transformation. However, instead of diagonalizing the matrix $A$ taken in isolation, it will be more instructive to consider this as diagonalization of the state-space system itself.

According to (7.2a), the matrix $A$ is an operator converting vectors from the state space into vectors in the same space. This, diagonalization of $A$ can be achieved by a specific choice of the state space basis, where the basis vectors must be the eigenvectors of $A$. After the change of basis we are having exactly the same system, just expressed in different coordinates. In these coordinates the matrix $A$ becomes diagonal and its diagonal elements are eigenvalues of $A$ (which are basis-independent). Now recall that eigenvalues of $A$ are the same as the system poles. Therefore, a sufficient condition for the state-space system to be diagonalizable is that all of its poles are distinct.²

Thus, in a diagonalizing basis the elements of $A$ are simply the system poles:

$$A=\left(\begin{array}{cccc}{p}_1& 0& \cdots & 0\\ 0& p_2& \cdots & 0\\ 0& 0& \ddots & 0\\ 0& 0& \cdots & p_N\end{array}\right)$$

and the system falls apart into a set of parallel Jordan 1-poles:

$${\dot{u}}_n=p_n{u}_n+{\mathbf{b}}_n^{\mathsf{T}}\cdot \mathbf{x}\text{\hspace{1em}(7.16a)}$$

$$\mathbf{y}=C\mathbf{u}+D\mathbf{x}=\sum _n{\mathbf{c}}_n{u}_n+D\mathbf{x}\text{\hspace{1em}(7.16b)}$$

where ${\mathbf{b}}_n^{\mathsf{T}}$ are the rows of matrix $B$ (respectively ${\mathbf{b}}_n^{\mathsf{T}}\cdot \mathbf{x}$ are the input signals of the Jordan 1-poles), and ${\mathbf{c}}_n$ are the columns of matrix $C$.

Stability

We already know that the transient response $u_{tn}(t)$ of a 1-pole is an exponent $K_n{e}^{p_{n}t}$ (where $K_n$ is the exponent’s amplitude). Respectively, the transient response part of $\mathbf{y}$ in (7.16b) is a linear combination of transient responses of the Jordan 1-poles:

$${\mathbf{y}}_t=C{\mathbf{u}}_t=\sum _n{\mathbf{c}}_n{u}_{tn}=\sum _n{\mathbf{c}}_n{K}_n{e}^{p_{n}t}$$

That is, the elements of ${\mathbf{y}}_t$ are linear combinations of exponents $e^{p_{n}t}$.

Now recall that $\mathbf{y}$ is independent on the choice of basis and so must be its separation into steady-state and transient response parts. Note that this is in agreement with the fact that according to (7.15) the steady-state response depends only on the transfer matrix and thus is independent of the basis changes. This means that the fact that the elements of ${\mathbf{y}}_t$ are linear combinations of exponents $e^{p_{n}t}$ is also independent of basis choice. Respectively ${\mathbf{y}}_t\to 0$ if and only if $\mathrm{Re}{p}_n<0\forall n$. Thus we have obtained the explanation of the stability criterion for linear filters which we introduced in Section 2.9 and have partially shown for lower-order systems.³

Transfer matrix

Computing the transfer matrix for the diagonal form we notice that

$$(s-A{)}^{-1}=\left(\begin{array}{cccc}\frac{1}{s-p_1}& 0& \cdots & 0\\ 0& \frac{1}{s-p_2}& \cdots & 0\\ 0& 0& \ddots & 0\\ 0& 0& \cdots & \frac{1}{s-p_N}\end{array}\right)\text{\hspace{1em}(7.17)}$$

that is we have transfer functions of the Jordan 1-poles on the main diagonal. Respectively the main term of the transfer matrix $C(s-A{)}^{-1}B$ is just a linear combination of the Jordan 1-pole transfer functions. Apparently the common denominator of the terms of this linear combination is

$$\prod _{n=1}^N(s-p_n)$$

which is simultaneously the common denominator of the transfer matrix elements.

It can be instructive to explicitly write out the elements of the transfer matrix $H(s)$ in the diagonal case:

$$H_{nm}(s)=\sum _{k=1}^N{c}_{nk}\frac{1}{s-p_k}{b}_{km}+d_{nm}=\sum _{k=1}^N\frac{c_{nk}{b}_{km}}{s-p_k}+d_{nm}\text{\hspace{1em}(7.18)}$$

that is, we are having a partial fraction expansion of the rational function $H_{nm}(s)$ into fractions of 1st order. Thus, if a transfer matrix is given in advance, there is not much freedom in respect to the choice of the elements of $B$ and $C$. The poles $p_k$ are prescribed by the common denominator of the transfer matrix and the values of the products $c_{nk}{b}_{km}$ and of $d_{nm}$ are prescribed by the specific functions $H_{nm}(s)$ occurring in the respective elements of the transfer matrix.

In the single-input single-output case the transfer matrix has $1\times 1$ dimensions, while $C$ has $1\times N$ and $B$ has $N\times 1$ dimensions respectively. Thus there is only one equation (7.18) and we have $N$ freedom degrees in respect to the choice of $c_{nk}$ and $b_{km}$ giving the required values of the products $c_{nk}{b}_{km}$. Each such degree can be associated with a variable ${\alpha }_k$, where we replace $b_{km}$ with ${\alpha }_k{b}_{km}$ and $c_{nk}$ with $c_{nk}/{\alpha }_k$. Apparently such replacement doesn’t affect the value of the product $c_{nk}{b}_{km}$. One can also realize that ${\alpha }_k$ simply scale the levels of the signals $u_k$, which corresponds to different choices of the lengths of the basis vectors.

Now if we add one more output signal, thereby the dimensions of $C$ becoming $2\times N$, we can notice that we still have exactly the same $N$ degrees of freedom. If we attempt to change any of $b_{km}$ we need to compensate this in both of $c_{nk}$ for the same $k$ by dividing these $c_{nk}$ by ${\alpha }_k$, the latter being the ratio of the new and old values of $b_{km}$. Respectively, if we change any of $c_{nk}$ in one of the two rows of $C$, this immediately requires the compensating change of $b_{km}$, which in turn requires that the same change occurs not just in one but in both rows of $C$. Adding more rows to $C$ and/or more columns to $B$ we see that the available freedom degrees are still the same and correspond to the freedom of choice of the basis vector lengths.

Thus, aside from the free choice of the basis vector lengths (and of their ordering) the transfer matrix uniquely defines the diagonal form of the state- space system. Respectively, for a non-diagonal form, if the matrix $A$ is given, then the transformation $T$ to the diagonal form is uniquely defined (up to the lengths and the ordering of the basis vectors), and, since the transfer function uniquely defines the matrices $B^{\prime }$, $C^{\prime }$ and $D^{\prime }$ of the diagonal form, the matrices $B=T^{-1}{B}^{\prime }$, $C=C^{\prime }T$ and $D=D^{\prime }$ are also uniquely defined.

Steady-state response

Apparently, there is the usual freedom in regards to the choice of the steady-state response arising out of evaluating the inverse Laplace transform of $H(s)\mathbf{X}(s)$ to the left or to the right of the poles of $H(s)$. The change of the steady- state response (7.15) depending on the choice of the inverse Laplace transform’s integration path in (7.15) to the left or to the right (or in between) the poles of $H(s)$ poses no fundamentally new questions compared to the previous discussion in the analysis of 1- and 2-pole transient responses and results simply in the changes of the amplitudes of transient response partials.

Diagonalization in case of coinciding poles

Even if two or more poles of the system coincide, it still might be diagonalizable, if the eigenvectors corresponding to these poles are distinct. It might seem that this is the most probable situation, after all, what are the changes of two vectors coinciding, or at least being collinear? Without trying to engage ourselves into an analysis of the respective probabilities, we are going to look at this fact from a different angle.

Namely, given a diagonal state space form with some of the eigenvalues coinciding, we are going to have identical entires in the matrix $(s-A{)}^{-1}$, as one can easily see from (7.17). This means that the order of the common denominator of the elements of $(s-A{)}^{-1}$ will be less than $N$ and respectively the order of the denominator of the transfer matrix $H(s)$ will also be less than $N$. This means that the effective order of the system is less than $N$ and the system is degenerate.

Thus, a non-degenerate system with coinciding poles cannot be diagonalized. In such cases we will have to use the Jordan normal form, which we discuss a bit later.

7.9 Real diagonal form

Given a state-space system we could decide to implement it in a diagonal form by first performing a diagonalizing change of basis and then implementing the obtained diagonal state space form. However, if the system has complex poles, the underlying Jordan 1-poles of the system will become complex too, respec- tively generating complex signals $u_n$. So, while the system has real input and real output, internally it would need to deal with complex signals. Of course, in a digital world using complex signals internally in a system shouldn’t be a big problem. But, for one, this is simply unusual and complicates the implementa- tion structure. More importantly, operations on complex numbers are at least twice as expensible as the same operations on real numbers. We therefore wish to convert a diagonal form containing complex poles to a purely real system, while retaining as much of the diagonalization as possible.

Since the system itself and the matrix $A$ in the original basis are real, the complex poles need to come in conjugate pairs. Without loss of generality we can order the poles in such a way that complex-conjugate pairs come first, followed by purely real poles: $p_1,p_1^{\ast },p_3,p_3^{\ast },\dots ,p_N$ (where $p_2=p_1^{\ast }$, $p_4=p_3^{\ast }$, etc.) We will refer to the complex poles $p_1,p_3,\dots$ as the odd poles and to $p_1^{\ast }$, $p_3^{\ast },\dots$ as the even poles. When referring to odd/even poles we will mean only the essentially complex poles, the purely real poles being excluded. Since the poles $p_n$ are eigenvalues of $A$, we will be referring to even/odd eigenvalues and respectively to even/odd eigenvectors.

Let ${\mathbf{v}}_1$ be the eigenvector corresponding to $p_1$, that is $A{\mathbf{v}}_1=p_1{\mathbf{v}}_1$. Then, since $A$ has purely real coefficients, $A\overline{{\mathbf{v}}_1}=\overline{A{\mathbf{v}}_1}=\overline{p_1{\mathbf{v}}_1}=p_1^{\ast }\overline{{\mathbf{v}}_1}$, (where $\overline{\mathbf{v}}$ denotes conjugation of vector’s components). Thus $\overline{{\mathbf{v}}_1}$ is the eigenvector corre- sponding to $p_1^{\ast }$. Obviously, the same applies to any other even/odd eigenvector. Therefore we can choose a set of eigenvectors such that even eigenvectors are component conjugates of odd eigenvectors: ${\mathbf{v}}_1,\overline{{\mathbf{v}}_1},{\mathbf{v}}_3,\overline{{\mathbf{v}}_3},\dots ,{\mathbf{v}}_N$.

If ${\mathbf{u}}^{\prime }=T\mathbf{u}$ is the diagonalizing transformation of the system, the new basis must consist of eigenvectors of $A$. Respectively, since $\mathbf{u}=T^{-1}{\mathbf{u}}^{\prime }$, the columns of $T^{-1}$ must consist of the new basis vectors, that is of the eigenvectors of $A$ (or, more precisely, consist of coordinates of these eigenvectors in the original basis). We will choose

$$T^{-1}=\left(\begin{array}{cccccc}{\mathbf{v}}_1& \overline{{\mathbf{v}}_1}& {\mathbf{v}}_3& \overline{{\mathbf{v}}_3}& \cdots & {\mathbf{v}}_N\end{array}\right)$$

This means that applying component conjugation to $T^{-1}$ swaps the even and the odd columns of $T^{-1}$, which can be expressed as

$$\overline{T^{-1}}=T^{-1}S$$

where

$$S=\left(\begin{array}{cccccc}0& 1& 0& 0& \cdots & 0\\ 1& 0& 0& 0& \cdots & 0\\ 0& 0& 0& 1& \cdots & 0\\ 0& 0& 1& 0& \cdots & 0\\ & & & & \ddots & \\ 0& 0& 0& 0& \cdots & 1\end{array}\right)$$

is the “swapping matrix”. Note that elements of $S$ are purely real and that $S^{-1}=S$.

Since

$$\overline{T}\cdot \overline{T^{-1}}=\overline{T{T}^{-1}}=1^{\ast }=1$$

component conjugation and matrix inversion commute:

$${\overline{T}}^{-1}=\overline{T^{-1}}=T^{-1}S$$

Reciprocating the leftmost and the rightmost expressions we have

$$\overline{T}={\left(T^{-1}S\right)}^{-1}=S^{-1}T=ST$$

That is, component conjugation of $T$ swaps its even and odd rows. Or, put in a slightly different way, the even/odd rows of $T$ are component conjugates of each other, and so are the even/odd columns of $T^{-1}$.

Let’s now concentrate on the first conjugate pair of poles. Taking the diag- onalized form equations (7.16) we extract those specifically concerning the first two poles:

$${\dot{u}}_1^{\prime }=p_1{u}_1^{\prime }+{\mathbf{b}}_1^{\prime \mathsf{T}}\cdot \mathbf{x}\text{\hspace{1em}(7.19a)}$$

$${\dot{u}}_2^{\prime }=p_1^{\ast }{u}_2^{\prime }+{\mathbf{b}}_2^{\prime \mathsf{T}}\cdot \mathbf{x}\text{\hspace{1em}(7.19b)}$$

$$\mathbf{y}={\mathbf{c}}_1^{\prime }{u}_1^{\prime }+{\mathbf{c}}_2^{\prime }{u}_2^{\prime }+\sum _{n=3}^N{\mathbf{c}}_n^{\prime }{u}_n^{\prime }+D\mathbf{x}\text{\hspace{1em}(7.19c)}$$

(where we need to employ the prime notation (7.10) for the diagonalized form, since we explicitly used the diagonalizing transformation ${\mathbf{u}}^{\prime }=T\mathbf{u}$, thus the non-primed state $\mathbf{u}$ referring to the non-diagonalized form).

Using (7.9) and recalling that the first two rows of $T$ are component con- jugates of each other, we must conclude that so are the first two rows of $B^{\prime }$, that is ${\mathbf{b}}_2^{\prime \mathsf{T}}=\overline{{\mathbf{b}}_1^{\prime \mathsf{T}}}$. Recalling that the first two colums of $T^{-1}$ are component conjugates of each other, we conclude that ${\mathbf{c}}_2^{\prime }=\overline{{\mathbf{c}}_1^{\prime }}$. On the other hand, writing out the first two rows of (7.13) in the diagonal case we have

$$u_1^{\prime }(t)=e^{p_{1}t}{u}_1^{\prime }(0)+{\int }_0^t{e}^{p_1(t-\tau )}{\mathbf{b}}_1^{\prime \mathsf{T}}\mathbf{x}dt$$

$$u_2^{\prime }(t)=e^{p_1^{\ast }t}{u}_2^{\prime }(0)+{\int }_0^t{e}^{p_1^{\ast }(t-\tau )}\overline{{\mathbf{b}}_1^{\prime \mathsf{T}}}\mathbf{x}dt$$

Except for the initial state term, the right-hand side of the second equation is a complex conjugate of the right-hand side of the first one. Regarding the initial state term, practically seen, we would have the following situations

• the initial state would be either zero, in which case $u_2^{\prime }(t)=u_1^{\prime \ast }(t)$,
• or it would be a result of some previous signal processing by the system, where previously to that processing the initial state would be zero, in which case $u_2^{\prime }(0)=u_1^{\prime \ast }(0)$ and respectively $u_2^{\prime }(t)=u_1^{\prime \ast }(t)$.

Therefore, we can simply require that $u_2^{\prime }(0)=u_1^{\prime \ast }(0)$, and thus the output signals $u_1^{\prime }(t)$ and $u_2^{\prime }(t)$ of the first two Jordan 1-poles are mutually conjugate. Respectively, the contribution of $u_1^{\prime }(t)$ and $u_2^{\prime }(t)$ into $\mathbf{y}$ in (7.19c), being equal to ${\mathbf{c}}_1^{\prime }{u}_1^{\prime }+{\mathbf{c}}_2^{\prime }{u}_2^{\prime }$, turns out to be a sum of two conjugate values and is therefore purely real. Obviously, the same applies to all other complex conjugate pole pairs.

Thus, even equations of (7.16a) do not contribute any new information about the system and we could drop them, simply computing even state signals as conjugates of odd state signals: $u_2^{\prime }=u_1^{\prime \ast }$, $u_4^{\prime }=u_3^{\prime \ast }$, etc. At the same time we could rewrite the odd equations of (7.16a) explicitly using real and imaginary parts of the signals $u^{\prime }$:

$$\frac{d}{dt}\mathrm{Re}{u}_n^{\prime }=\mathrm{Re}{p}_n\mathrm{Re}{u}_n^{\prime }-\mathrm{Im}{p}_n\mathrm{Im}{u}_n^{\prime }+\left(\mathrm{Re}{\mathbf{b}}_n^{\prime \mathsf{T}}\right)\cdot \mathbf{x}\text{\hspace{1em}(7.20a)}$$

$$\frac{d}{dt}\mathrm{Im}{u}_n^{\prime }=\mathrm{Im}{p}_n\mathrm{Re}{u}_n^{\prime }+\mathrm{Re}{p}_n\mathrm{Im}{u}_n^{\prime }+\left(\mathrm{Im}{\mathbf{b}}_n^{\prime \mathsf{T}}\right)\cdot \mathbf{x}\text{\hspace{1em}(7.20b)}$$

Therefore we can introduce the new state variables, taking purely real values:

$$\begin{array}{rl}{u}_n^{\prime \prime }& =\mathrm{Re}{u}_n^{\prime }=\frac{u_n^{\prime }+u_{n+1}^{\prime }}{2}\\ u_{n+1}^{\prime \prime }& =\mathrm{Im}{u}_n^{\prime }=\frac{u_n^{\prime }-u_{n+1}^{\prime }}{2j}\end{array}\}\text{for odd }{p}_n\text{\hspace{1em}(7.21a)}$$

and

$$u_n^{\prime \prime }=u_n^{\prime }\text{for purely real }{p}_n\text{\hspace{1em}(7.21b)}$$

Then (7.20) turn into

$${\dot{u}}_n^{\prime \prime }=\mathrm{Re}{p}_n\cdot u_n^{\prime \prime }-\mathrm{Im}{p}_n\cdot u_{n+1}^{\prime \prime }+\left(\mathrm{Re}{\mathbf{b}}_n^{\prime \mathsf{T}}\right)\cdot \mathbf{x}\text{\hspace{1em}(7.22a)}$$

$${\dot{u}}_{n+1}^{\prime \prime }=\mathrm{Im}{p}_n\cdot u_n^{\prime \prime }+\mathrm{Re}{p}_n\cdot u_{n+1}^{\prime \prime }+\left(\mathrm{Im}{\mathbf{b}}_n^{\prime \mathsf{T}}\right)\cdot \mathbf{x}\text{\hspace{1em}(7.22b)}$$

and the respective terms in (7.19c) turn into

$${\mathbf{c}}_n^{\prime }{u}_n^{\prime }+{\mathbf{c}}_{n+1}^{\prime }{u}_{n+1}^{\prime }={\mathbf{c}}_n^{\prime }{u}_n^{\prime }+{\mathbf{c}}_n^{\prime \ast }{u}_n^{\prime \ast }={\mathbf{c}}_n^{\prime }{u}_n^{\prime }+{\left({\mathbf{c}}_n^{\prime }{u}_n^{\prime }\right)}^{\ast }=2\mathrm{Re}\left({\mathbf{c}}_n^{\prime }{u}_n^{\prime }\right)=$$

$$=2\mathrm{Re}{\mathbf{c}}_n^{\prime }\cdot u_n^{\prime \prime }-2\mathrm{Im}{\mathbf{c}}_n^{\prime }\cdot u_{n+1}^{\prime \prime }$$

Thus we have obtained a purely real system

$${\dot{\mathbf{u}}}^{\prime \prime }=\left(\begin{array}{cccccc}\mathrm{Re}{p}_1& -\mathrm{Im}{p}_1& 0& 0& \cdots & 0\\ \mathrm{Im}{p}_1& \mathrm{Re}{p}_1& 0& 0& \cdots & 0\\ 0& 0& \mathrm{Re}{p}_3& -\mathrm{Im}{p}_3& \cdots & 0\\ 0& 0& \mathrm{Im}{p}_3& \mathrm{Re}{p}_3& \cdots & 0\\ 0& 0& 0& 0& \ddots & 0\\ 0& 0& 0& 0& \cdots & p_N\end{array}\right){\mathbf{u}}^{\prime \prime }+\left(\begin{array}{c}\mathrm{Re}{\mathbf{b}}_1^{\prime \mathsf{T}}\\ \mathrm{Im}{\mathbf{b}}_1^{\prime \mathsf{T}}\\ \mathrm{Re}{\mathbf{b}}_3^{\prime \mathsf{T}}\\ \mathrm{Im}{\mathbf{b}}_3^{\prime \mathsf{T}}\\ ⋮\\ {\mathbf{b}}_N^{\prime \mathsf{T}}\end{array}\right)\mathbf{x}\text{\hspace{1em}(7.23a)}$$

$$\mathbf{y}=\left(\begin{array}{cccccc}2\mathrm{Re}{c}_1^{\prime }& -2\mathrm{Im}{c}_n^{\prime }& 2\mathrm{Re}{c}_3^{\prime }& -2\mathrm{Im}{c}_3^{\prime }& \cdots & c_N^{\prime }\end{array}\right){\mathbf{u}}^{\prime \prime }+D\mathbf{x}\text{\hspace{1em}(7.23b)}$$

We will refer to (7.23) as the real diagonal form. It represents the system as a set of parallel 2-poles (7.22) (and optionally additional parallel 1-poles if some of the system poles are real).

Note that the substitutions (7.21) are expressible as another linear transfor- mation ${\mathbf{u}}^{\prime \prime }=T^{\prime }{\mathbf{u}}^{\prime }$ where

$$T^{\prime }=\left(\begin{array}{cccccc}\frac{1}{2}& \frac{1}{2}& 0& 0& \cdots & 0\\ \frac{1}{2j}& -\frac{1}{2j}& 0& 0& \cdots & 0\\ 0& 0& \frac{1}{2}& \frac{1}{2}& \cdots & 0\\ 0& 0& \frac{1}{2j}& -\frac{1}{2j}& \cdots & 0\\ & & & & \ddots & \\ 0& 0& 0& 0& \cdots & 1\end{array}\right)$$

Therefore the real diagonal form of the system is related to the original form by a change of basis, where the respective transformation matrix is $T^{\prime }T$.

Jordan 2-poles

The 2-poles (7.22) in the real diagonal form are fully analogous to the 1-poles occuring in the diagonal form. They will also occur in the real Jordan normal form. For that reason we will refer to them as Jordan 2-poles. They are also sometimes (especially in their discrete-time counterpart form) referred to as coupled-form resonators.

The key feature of the Jordan 2-pole topology is that in the absence of the input signal, the system state is spiralling in a circle of an exponentially decaying (or growing) radius. Indeed, recalling that equations (7.22) are simply separate equations for the real and imaginary components of a complex signal $u_n^{\prime }$, we can return to using the equation (7.19a), which by letting $\mathbf{x}=0$ and turns into

$$\dot{u}=pu$$

where we also dropped the indices and the prime notation for simplicity. Re- spectively

$$\frac{d}{dt}\log u=\frac{\dot{u}}{u}=p=\mathrm{Re}p+j\mathrm{Im}p\text{\hspace{1em}(7.24)}$$

On the other hand

$$\log u=\ln |u|+j\arg u$$

therefore

$$\frac{d}{dt}\log u=\frac{d}{dt}\ln |u|+j\frac{d}{dt}\arg u\text{\hspace{1em}(7.25)}$$

Equating the right-hand sides of (7.24) and (7.25), we obtain

$$\frac{d}{dt}\ln |u|+j\frac{d}{dt}\arg u=\mathrm{Re}p+j\mathrm{Im}p$$