State-Space Representation of Systems

Twenty-six chapters have been built on a single description of a system: a relation between one input and one output. Chapter 7 wrote it as a convolution, Chapter 19 as a transfer function \(H(s)\), Chapter 25 as \(H(z)\). It is a powerful description and it is not a complete one, and three separate cracks in it are worth setting out plainly before the repair is offered.

The first crack is initial conditions. The transfer function is defined as the ratio \(Y(s)/U(s)\) with all initial conditions zero — it is a zero-state description by construction. Chapter 19 had to reach for the unilateral Laplace transform, dragging \(y(0^-)\) and \(\dot y(0^-)\) into the algebra by hand, precisely because \(H(s)\) has no room for them. For a system that starts from a charged capacitor or a spinning flywheel, the transfer function is telling only part of the story.

The second crack is internal behaviour. A transfer function describes the terminals and nothing else. If an amplifier inside a feedback loop is saturating, or a capacitor voltage is exceeding its rating, or an internal signal is growing without bound while the output stays polite, \(H(s)\) will not say so. Section 27-9 constructs a system that is perfectly well behaved at its output while a state inside it grows like \(e^{3t}\).

The third crack is scale. Real systems have several inputs and several outputs. A power converter has a duty ratio and a load current going in, an output voltage and an inductor current coming out. With transfer functions this means a matrix of them, one per input–output pair, each with its own algebra; and a system of \(p\) inputs and \(q\) outputs needs \(pq\) of them, all sharing the same denominator and all recomputed from scratch whenever the plant changes.

State space repairs all three at once, and it does so by a change of viewpoint that is almost embarrassingly simple: instead of one equation of order \(n\), write \(n\) equations of order one.

Chapter 5 defined a system with memory as one whose output depends on more than the present input. That definition raises a question it does not answer: how much memory, and stored in what? The state is the answer.

The word smallest is doing real work. A record of the entire past input would certainly determine the future, but it is not a state — it is not minimal, and it is not finite. The state is the minimal sufficient summary, and for a lumped system of order \(n\) it consists of exactly \(n\) numbers.

In physical systems the state variables are almost always the energy-storage variables, and this is a reliable way to find them. A capacitor stores \(\tfrac12 Cv^2\), so its voltage is a state; an inductor stores \(\tfrac12 Li^2\), so its current is a state. A mass stores kinetic energy, so its velocity is a state, and a spring stores potential energy, so its extension is a state. The reason is physical: energy cannot change instantaneously without infinite power, so these variables are continuous, and their values at any instant summarise everything the past has deposited in the system.

Counting them gives the order for free. A circuit with two capacitors and one inductor is third order and needs three state variables; a mass–spring–damper is second order and needs two. That agrees with the order of the differential equation, as it must, and it is often much easier to count storage elements than to derive and inspect the equation.

Collect the \(n\) state variables into a column vector \(\mathbf{x}(t)\), the \(p\) inputs into \(\mathbf{u}(t)\) and the \(q\) outputs into \(\mathbf{y}(t)\). For a linear time-invariant system, the derivative of each state variable is a linear combination of the states and the inputs, and each output is likewise. Writing all of those linear combinations at once gives two matrix equations.

The four matrices have fixed sizes and fixed meanings, and it is worth learning them as roles rather than as symbols.

Only \(A\) contains the system's natural behaviour; everything else concerns how the world is connected to it. That separation is what makes state space so convenient for design, and it is why Section 27-8 finds the poles inside \(A\) alone.

The matrix \(D\) is non-zero only when a change in the input reaches the output with no delay whatever. For a strictly proper transfer function — one whose numerator degree is less than its denominator degree — \(D\) is zero, and this is the usual case for physical systems, which cannot respond instantaneously. It is non-zero for something like a resistive divider in parallel with a filter, where part of the signal takes a direct path.

The discrete-time form replaces the derivative by a one-step advance, and is otherwise identical.

Here the state has an unusually concrete meaning: it is exactly the contents of the delay registers in an implementation. The state equation says what the registers will hold on the next clock edge, and the output equation says what to present at the terminals now. A digital filter's state vector is, quite literally, its memory.

There are two routes into the standard form. The physical route is to pick the energy-storage variables and write one equation per storage element, which is what Worked Example 1 does for a series RLC circuit. The algebraic route starts from a differential equation, and it is worth doing in general because the resulting pattern is used everywhere.

Start with the simplest case, an \(n\)-th order equation with no derivatives of the input:

Choose as state variables the output and its first \(n-1\) derivatives: \(x_1 = y\), \(x_2 = \dot y\), up to \(x_n = y^{(n-1)}\). These are called phase variables. Then \(n-1\) of the state equations are free — each state's derivative is simply the next state — and the last one is the original differential equation rearranged for the highest derivative.

Assembled into matrices this is the controllable canonical form, sometimes called the phase-variable form. Every row of \(A\) but the last is a shifted 1; the last row carries the coefficients of the differential equation, negated and in reverse order.

Now allow derivatives of the input on the right-hand side. The naive extension fails immediately: if \(x_1 = y\) then \(\dot x_n\) would contain \(u^{(n-1)}\), and derivatives of the input are not permitted in the standard form. The way round is a device that recurs throughout linear systems — solve an auxiliary problem and then superpose.

Consider

and define an intermediate signal \(w(t)\) that solves the same equation driven by \(u\) alone:

Because the system is linear and time-invariant, if \(u\) produces \(w\) then \(\dot u\) produces \(\dot w\), and any linear combination of derivatives of \(u\) produces the same combination of derivatives of \(w\). The right-hand side we actually want is such a combination, so

Now take the phase variables of \(w\), namely \(x_i = w^{(i-1)}\). The state equation is exactly the canonical form above, because \(w\) obeys the undifferentiated equation. And the output equation is just the linear combination written out, with no derivatives left in it:

The denominator coefficients live in the bottom row of \(A\); the numerator coefficients live in \(C\). Both can be read straight off a transfer function, which makes this the standard way of turning \(H(s)\) into a state model. If the numerator and denominator have the same degree the transfer function is not strictly proper; divide out first to leave a constant plus a strictly proper remainder, and that constant is \(D\).

The standard form has a picture, and the picture is close to being a wiring diagram. Every state variable is the output of one integrator; the state equation says what to feed into that integrator, and the output equation says what to tap off. Drawing it makes the roles of the four matrices immediately visible.

Note where the feedback is. The only closed loop in the diagram runs from the state, through \(A\), back into the integrator. That loop is the system's own dynamics; \(B\), \(C\) and \(D\) sit outside it. This is the picture behind the claim that the natural behaviour is a property of \(A\) alone.

For the canonical form of the previous section the diagram becomes very concrete. Take the second-order case \(\ddot y + a_1\dot y + a_0 y = b_1\dot u + b_0 u\), with \(x_1 = w\) and \(x_2 = \dot w\). Two integrators in cascade generate \(x_2\) and then \(x_1\); the coefficients \(-a_1\) and \(-a_0\) feed those signals back to the input summer; and \(b_1\), \(b_0\) tap them forward to the output.

Replace each integrator by a unit delay \(z^{-1}\) and the same diagram becomes the direct-form realisation of a digital filter that Chapter 25 drew from the difference equation. This is not a coincidence or an analogy: the state-space form and the direct forms are the same structure described in two notations, and the state vector is the contents of the delay line.

Everything so far has been description. Now solve \(\dot{\mathbf{x}} = A\mathbf{x} + B\mathbf{u}\) with \(\mathbf{x}(0)\) given.

Begin with the unforced case \(\dot{\mathbf{x}} = A\mathbf{x}\). If \(A\) were a scalar \(a\), the answer would be \(x(t) = e^{at}x(0)\), and the reasonable guess is that the matrix case is the same with a matrix exponential. Define it by the series that defines the scalar one:

The series converges for every square \(A\) and every \(t\), because the factorials outrun the growth of \(\|A^k\|\). Two properties follow directly from it. Setting \(t=0\) leaves \(e^{A\cdot 0} = I\). And differentiating term by term gives

So \(\mathbf{x}(t) = e^{At}\mathbf{x}(0)\) satisfies the unforced equation and matches the initial condition, and by uniqueness it is the solution. The matrix \(\Phi(t) = e^{At}\) is called the state transition matrix, because it transports the state from one instant to another: \(\mathbf{x}(t_2) = \Phi(t_2-t_1)\mathbf{x}(t_1)\).

For the forced case use an integrating factor, exactly as for the scalar first-order equation. Multiply through by \(e^{-At}\) and collect:

The left side is an exact derivative because \(\tfrac{d}{dt}e^{-At} = -Ae^{-At}\), by the property just proved. Integrating from \(0\) to \(t\) and multiplying by \(e^{At}\) — which is legitimate since \(e^{At}e^{-At} = I\), so the matrix exponential is always invertible — gives the complete answer.

The zero-state term is a convolution. Reading it against Chapter 7, the kernel \(Ce^{A(t-\tau)}B\) plays the part of \(h(t-\tau)\), so the impulse response of the state-space system is

The impulse in \(D\delta(t)\) is the direct feedthrough, and it is present only when \(D \ne 0\) — a further confirmation that \(D\) represents an instantaneous path.

Computing \(e^{At}\) by summing the series is almost never practical. Three better methods are standard. The first, and the one used in the worked examples, transforms the problem: taking the Laplace transform of \(\dot{\mathbf{x}} = A\mathbf{x}\) with initial state \(\mathbf{x}(0)\) gives \(s\mathbf{X}(s) - \mathbf{x}(0) = A\mathbf{X}(s)\), hence \(\mathbf{X}(s) = (sI-A)^{-1}\mathbf{x}(0)\), and comparing with \(\mathbf{x}(t) = e^{At}\mathbf{x}(0)\) identifies the pair.

The second method diagonalises \(A\), and is developed in Section 27-9. The third uses the Cayley–Hamilton theorem — every matrix satisfies its own characteristic equation, so \(A^n\) is a combination of lower powers, so the infinite series collapses to a polynomial of degree \(n-1\) with time-dependent coefficients. For hand computation on second-order problems the Laplace route is usually quickest, and it has the advantage of reusing machinery already built.

The discrete case needs no calculus at all — just iteration, which is the honest advantage of discrete time noted as far back as Chapter 3. Starting from \(\mathbf{x}[n+1] = A\mathbf{x}[n] + B\mathbf{u}[n]\) and \(\mathbf{x}[0]\), step forward:

The pattern is unmistakable, and an induction on \(n\) confirms it.

Reading the convolution off gives the impulse response directly. Setting \(\mathbf{x}[0] = \mathbf{0}\) and \(u[m] = \delta[m]\) leaves only the \(m=0\) term, so \(\mathbf{x}[n] = A^{n-1}B\) for \(n \ge 1\), and

These coefficients \(D,\ CB,\ CAB,\ CA^2B,\dots\) are called the Markov parameters of the system. They are exactly the impulse response samples, and they show that the entire input–output behaviour is determined by those particular products — a fact that Section 27-9 uses to explain how a mode can be present in \(A\) and yet absent from every one of them.

Building the discrete model from a difference equation follows the continuous recipe with delays in place of integrators. For

with \(D = 0\) whenever there is no \(u[n]\) term on the right. The transfer function follows the same route as in continuous time, with \(z\) replacing \(s\): \(H(z) = C(zI-A)^{-1}B + D\).

State space did not replace the transfer function; it contains it. Transform the state equation with zero initial state, which is the condition under which a transfer function is defined in the first place:

Substituting into \(\mathbf{Y}(s) = C\mathbf{X}(s) + D\mathbf{U}(s)\) gives the bridge between the two descriptions.

Now use the cofactor formula for the inverse, \(M^{-1} = \operatorname{adj}(M)/\det(M)\):

The denominator is \(\det(sI-A)\), which is precisely the characteristic polynomial of \(A\). Its roots are the eigenvalues of \(A\). So the poles of the transfer function are eigenvalues of \(A\) — and the qualification matters enormously: they are among the eigenvalues, not necessarily all of them. If a factor of \(\det(sI-A)\) also divides the numerator, it cancels, and that eigenvalue disappears from \(H(s)\) while remaining perfectly present in the system.

One more property is worth recording because it justifies the remark in Section 27-2 that the choice of state variables does not matter. Suppose new states \(\bar{\mathbf{x}} = T^{-1}\mathbf{x}\) are defined by an invertible matrix \(T\). Substituting throughout gives a new model with

and the transfer function is unchanged, since

Similar matrices have the same characteristic polynomial and therefore the same eigenvalues. The matrices \(A,B,C,D\) are a description; the eigenvalues and the transfer function are the system.

Suppose \(A\) has \(n\) distinct eigenvalues \(\lambda_1,\dots,\lambda_n\) with eigenvectors \(\mathbf{v}_1,\dots,\mathbf{v}_n\). Put the eigenvectors side by side as the columns of \(M\); then \(AM = M\Lambda\) with \(\Lambda\) diagonal, so \(A = M\Lambda M^{-1}\). Since \(A^k = M\Lambda^kM^{-1}\), every term of the exponential series carries the same \(M\), and the sum simplifies dramatically:

The zero-input response is therefore a combination of exponentials in the eigenvalues:

Each term is a natural mode. Its time behaviour is fixed by the eigenvalue \(\lambda_i\); the eigenvector \(\mathbf{v}_i\) fixes the pattern across the state variables — the relative proportions in which they move when that mode alone is excited. If the initial state happens to be a pure eigenvector, only one mode is present, and the state decays or grows along a fixed direction without ever changing shape.

This is the same conclusion Chapter 9 reached from the characteristic equation of a differential equation, and it is now visible as a statement about a matrix. Complex eigenvalues occur in conjugate pairs for a real \(A\), and their pair of terms combines into a real damped sinusoid, exactly as in Chapter 19.

Stability follows immediately. Every mode decays if and only if every eigenvalue lies strictly in the left half-plane; a single eigenvalue with positive real part guarantees that some initial state grows without bound.

Which brings us back to the crack identified in Section 27-1. Asymptotic stability is a statement about the eigenvalues of \(A\); BIBO stability, as defined in Chapter 8 and tested in Chapters 19 and 25, is a statement about the poles of \(H(s)\). Section 27-8 showed that the poles are a subset of the eigenvalues. Asymptotic stability therefore implies BIBO stability, and the converse fails exactly when a cancellation removes a bad eigenvalue from the transfer function.

Such an eigenvalue is called a hidden mode, and there are two ways for a mode to hide. It is uncontrollable if the input cannot excite it — no choice of \(u\) puts any energy into it. It is unobservable if it never reaches the output — it may be raging inside, but \(C\) is blind to it. Either failure removes the mode from every Markov parameter \(CA^{k}B\), hence from the impulse response, hence from \(H(s)\).

The standard tests are matrix rank conditions, and for an \(n\)-th order system they take a compact form.

The system is controllable when \(\mathcal{C}\) has rank \(n\), and observable when \(\mathcal{O}\) has rank \(n\); for a single-input single-output system both are square, so the test is simply whether the determinant is non-zero. A realisation that is both controllable and observable is minimal: it uses the fewest possible state variables, no cancellation occurs, and the eigenvalues and the poles coincide exactly. Worked Example 6 builds a system that fails the first test and shows what goes wrong.

Problems 1 to 3 build state models; 4 and 5 solve them; 6 and 7 return to the transfer function; 8 concerns what the transfer function cannot tell you. In every numerical problem, check \(e^{A\cdot 0} = I\) and check the DC gain against \(H(0)\) before believing the answer.