System Analysis Using the Laplace Transform

Chapter 9 wrote down the equation that governs almost every lumped physical system an undergraduate meets — a linear constant-coefficient differential equation relating an input \(x(t)\) to an output \(y(t)\) — and solved it the classical way, with a homogeneous solution, a particular solution and a set of constants fixed at the end. Chapter 18 supplied the property that makes all of that unnecessary: differentiation becomes multiplication by \(s\).

Take the general equation and transform it term by term, with the system initially at rest so that no boundary terms appear:

Both sides are now polynomials multiplying transforms, and the ratio \(Y(s)/X(s)\) can simply be divided out. The result depends only on the coefficients of the equation — that is, only on the system.

Two names carry over from Chapter 9. The denominator \(A(s)\) is the characteristic polynomial, and its roots — the poles of \(H(s)\) — are precisely the characteristic roots that determined the natural response there. Nothing new has been discovered; what has changed is that finding them no longer requires guessing \(y(t)=e^{\lambda t}\) and substituting, because the transform performed the substitution automatically. The numerator roots, the zeros, played no role at all in the classical method, and that is exactly why the classical method never explained why two systems with the same natural frequencies can behave so differently.

Real systems are rarely at rest when the input arrives. The capacitor holds charge, the inductor carries current, the shaft is already turning. Section 18-9 built the tool for exactly this: the unilateral transform carries the initial conditions into the algebra as additive terms.

Consider a second-order system, which is enough to show the whole pattern:

Collect the terms in \(Y\) on the left and everything else on the right. The polynomial multiplying \(Y\) is \(A(s)=s^2+a_1s+a_0\) — the characteristic polynomial again — and the right-hand side splits into an input part and an initial-condition part:

That last observation matters more than it looks. Because the two pieces share a denominator, they share poles, and their natural-mode terms can reinforce or cancel one another. Worked Example 2 contains a case where a mode present in both halves disappears entirely from the sum — a reminder that "zero-input response" and "natural response" are not synonyms.

It is worth keeping three different decompositions of the same \(y(t)\) apart, because examiners enjoy the distinction:

They coincide only in special cases. The zero-state response of a stable system driven by a step, for instance, contains natural-mode terms (which are transient) as well as a constant (which is steady-state), so it is neither purely forced nor purely steady-state.

Expand a rational \(H(s)\) with distinct poles into partial fractions and invert term by term. For a causal system every term is right-sided, so

Each pole contributes one exponential, and the pole's position in the plane dictates that exponential's character completely. Write \(p = \sigma + j\omega_d\). The real part \(\sigma\) is the exponent of the envelope \(e^{\sigma t}\), and the imaginary part \(\omega_d\) is the rate at which the term oscillates. Everything else follows from those two sentences.

A pole on the negative real axis at \(s=-a\) gives a plain decay \(e^{-at}\) with time constant \(1/a\): the further left the pole, the faster the mode dies. A conjugate pair at \(\sigma\pm j\omega_d\) combines, because the residues are conjugates for a real system, into a single real term \(2|A|e^{\sigma t}\cos(\omega_d t + \angle A)\) — a ringing that decays if \(\sigma \lt 0\). A pair exactly on the imaginary axis gives an oscillation that never decays. A pole at the origin gives a constant, which is why an integrator has one. And a pole anywhere in the right half-plane gives a mode that grows without bound.

Repeated poles change the picture only slightly. A pole of multiplicity \(m\) at \(s=p\) produces terms \(t^{k}e^{pt}\) for \(k=0,\dots,m-1\), by the differentiation-in-\(s\) result of Section 18-5. The polynomial factor grows, but any exponential with \(\sigma \lt 0\) beats it eventually, so a repeated pole in the left half-plane still decays. On the imaginary axis it is a different story: a double pole at \(s=\pm j\omega_0\) gives \(t\cos\omega_0 t\), which grows. That is why "simple poles" appears in the definition of marginal stability.

What, then, do the zeros do? They create no modes at all — the exponentials in \(h(t)\) are fixed entirely by the poles. What zeros control is the residues: how much of each mode appears. A zero placed close to a pole makes that pole's residue small, because the residue \(A_i = B(p_i)/A'(p_i)\) contains \(B(p_i)\), and \(B\) is nearly zero there. Place the zero exactly on the pole and the mode vanishes from the response altogether.

Chapter 17 established the two facts this section uses. A causal system has a right-sided impulse response, so its ROC is the half-plane to the right of the rightmost pole. A BIBO-stable system has an absolutely integrable impulse response, and \(\int|h(t)|dt \lt \infty\) is exactly the statement that the defining integral converges at \(s=j\omega\) — so its ROC contains the imaginary axis. Put the two together and the standard test drops out.

Every word in that statement is load-bearing, and dropping "causal" is the usual mistake. A given \(H(s)\) is not one system. Take \(H(s)=\dfrac{1}{(s+1)(s-2)}\), whose poles are at \(-1\) and \(+2\). Three regions of convergence are compatible with those poles, and each describes a genuinely different system:

The algebra is identical in all three rows. Only the ROC differs, and it decides whether each pole contributes a right-sided or a left-sided piece. A physical system is causal, so the first row is the one that describes a circuit — and it is unstable, which no amount of rewriting the algebra will cure.

Between stable and unstable sits marginal stability: simple poles on the imaginary axis and none to the right. Such a system does not blow up on its own, but it is not BIBO stable either. An undamped resonator \(H(s)=\omega_0^2/(s^2+\omega_0^2)\) driven by the bounded input \(\cos\omega_0 t\) produces an output growing like \(t\sin\omega_0 t\); the input frequency lands exactly on the pole, and the response is unbounded. A bounded input has produced an unbounded output, which is the definition of instability from Chapter 8.

For low-order systems the poles need not be computed at all. Write the characteristic polynomial as \(s^2+a_1s+a_0=(s-p_1)(s-p_2)\), so that \(a_1=-(p_1+p_2)\) and \(a_0=p_1p_2\). If the roots are a complex pair \(\sigma\pm j\omega\), then \(a_1=-2\sigma\) and \(a_0=\sigma^2+\omega^2 \gt 0\) automatically, so \(a_1 \gt 0\) is exactly the condition \(\sigma \lt 0\). If they are real, then both are negative precisely when their sum is negative and their product positive — again \(a_1 \gt 0\) and \(a_0 \gt 0\). Either way:

For higher orders, all coefficients being positive remains necessary — a polynomial with all roots in the left half-plane factors into terms \((s+a)\) and \((s^2+bs+c)\) with \(a,b,c \gt 0\), and a product of such factors cannot produce a negative or missing coefficient — but it is no longer sufficient. Deciding the general case is the business of the Routh–Hurwitz criterion, which belongs to a control course; for this subject, factor the polynomial or trust the second-order rule.

Section 18-6 showed that a cascade multiplies system functions and a parallel connection adds them. Feedback is the third arrangement, and it is the one that changes the poles. Let the forward path be \(G(s)\), the feedback path \(F(s)\), and let the comparator subtract. Then \(E=X-FY\) and \(Y=GE\), so \(Y = G(X - FY)\), and collecting \(Y\):

The poles of the closed-loop system are the roots of \(1+G(s)F(s)=0\), and in general they lie nowhere near the poles of \(G\). That single sentence is why feedback is used at all: it is the one interconnection that can move a pole. Cascading and adding blocks merely collects the poles that were already there, and neither can stabilise an unstable plant. Worked Example 5 takes a system with a pole stuck at the origin and moves it into the left half-plane by closing a loop around it.

The reverse question — given \(H(s)\), what circuit produces it? — has an equally systematic answer. Because \(1/s\) is an integrator (Section 18-7), any rational \(H(s)\) can be drawn as integrators, constant gains and adders. Write the second-order case as a ratio and introduce an intermediate signal \(W(s)\) defined by \(X(s)=A(s)W(s)\):

Read more plainly: define \(w(t)\) so that \(\ddot w + a_1\dot w + a_2 w = x(t)\); then \(y = b_0\ddot w + b_1\dot w + b_2 w\) reproduces \(H(s)\) exactly, because dividing the second relation by the first gives back the ratio of polynomials. Now build \(\ddot w\) at a summing junction, integrate it twice to obtain \(\dot w\) and \(w\), feed those back with gains \(-a_1\) and \(-a_2\), and tap all three signals with gains \(b_0,b_1,b_2\) to form the output. Two integrators serve both the feedback and the feed-forward paths, which is why this arrangement is the economical one.

The same \(H(s)\) admits other realisations. Expanding it into partial fractions and building each term separately gives the parallel form; factoring it into quadratics and cascading them gives the cascade form. All three are identical on paper and differ in practice, because real components have tolerances and real arithmetic has finite precision, and a high-order direct form is notoriously sensitive to small coefficient errors. Chapter 25 revisits exactly this question for discrete-time systems, where the coefficients are numbers stored in a register and the sensitivity is a design constraint rather than a curiosity.

Two special cases account for most of what engineers actually design with, partly because they are common and partly because a high-order system is usually dominated by the one or two poles nearest the imaginary axis — the others decay too quickly to matter.

The first-order system \(H(s)=1/(\tau s+1)\) has a single pole at \(s=-1/\tau\). Its impulse response is \((1/\tau)e^{-t/\tau}u(t)\) and its step response, from \(Y=H/s\),

which reaches 63.2% of its final value at \(t=\tau\) and is within 2% of it after about \(4\tau\). There is no overshoot and no oscillation; a single real pole cannot produce either.

The second-order system is where the interesting behaviour starts. It is universally written in terms of two parameters, the undamped natural frequency \(\omega_n\) and the damping ratio \(\zeta\):

Four regimes follow directly from the discriminant. With \(\zeta=0\) the poles sit on the imaginary axis and the system oscillates for ever at \(\omega_n\). With \(0 \lt \zeta \lt 1\) they are a complex pair and the response is underdamped — a decaying ring at the damped frequency \(\omega_d=\omega_n\sqrt{1-\zeta^2}\). At \(\zeta=1\) they coincide on the real axis at \(-\omega_n\): critically damped, the fastest approach without overshoot. For \(\zeta \gt 1\) they separate along the real axis and the response is overdamped, dominated by the slower of the two.

The underdamped step response is worth deriving once. Split \(Y(s)=\omega_n^2/\big[s(s^2+2\zeta\omega_ns+\omega_n^2)\big]\) by completing the square in the denominator:

The two damped-sinusoid pairs came straight from Chapter 17's table, and the last step is the standard collapse of \(a\cos+b\sin\) into a single sinusoid. Now differentiate. Everything except the sine term cancels, leaving \(\dot y(t)=\dfrac{\omega_n}{\sqrt{1-\zeta^2}}e^{-\zeta\omega_nt}\sin\omega_d t\), which is zero when \(\omega_d t=k\pi\). The first maximum is therefore at \(t_p=\pi/\omega_d\), and substituting it into \(y\) with \(\sin(\pi+\theta)=-\sin\theta=-\sqrt{1-\zeta^2}\) gives the peak value directly.

Those three sentences are the whole design vocabulary. Want less overshoot? Swing the poles towards the real axis. Want a faster settle? Push them left. Want a faster first peak without changing the overshoot? Move them outwards along the same ray, increasing \(\omega_n\) at fixed \(\zeta\).

For a stable system every mode arising from a system pole decays, so whatever survives as \(t\to\infty\) must have come from the input's own poles. That single observation organises the whole of steady-state analysis.

Take a step input first. Then \(Y(s)=H(s)/s\), and provided \(H\) is stable the only pole of \(sY(s)=H(s)\) that could sit on or right of the axis has been removed, so the final value theorem of Section 18-8 applies:

The steady-state response to a unit step is simply \(H(0)\), the system function evaluated at the origin — the DC gain. For the standard second-order system \(H(0)=\omega_n^2/\omega_n^2=1\), which is why its step response settles at exactly 1 no matter how it gets there.

A sinusoidal input is more interesting, and it closes a loop with Chapter 15. Let \(x(t)=\cos(\omega_0 t)u(t)\), so \(X(s)=s/(s^2+\omega_0^2)\) with poles at \(\pm j\omega_0\). Expanding \(Y(s)=H(s)X(s)\) in partial fractions produces terms at the poles of \(H\) — all of which decay — plus two terms at \(\pm j\omega_0\). The residue at \(s=j\omega_0\) is

and the residue at \(s=-j\omega_0\) is its conjugate, since \(h(t)\) is real. Those two terms give \(\tfrac12 H(j\omega_0)e^{j\omega_0t}+\tfrac12H^{*}(j\omega_0)e^{-j\omega_0t}\), which is the real part of \(H(j\omega_0)e^{j\omega_0t}\).

Since \(H(j\omega)\) is just \(H(s)\) restricted to the imaginary axis, and a rational \(H(s)\) is a product of factors \((s-z_i)\) over factors \((s-p_k)\), the frequency response can be read off the plot with no algebra at all. Each factor \((j\omega - z_i)\) is a vector in the plane, running from the zero \(z_i\) to the test point \(j\omega\); its length is the magnitude of the factor and its angle is the phase.

Two consequences dominate filter design. When the test point passes close to a pole, the distance \(d_k\) in the denominator becomes small and \(\big|H(j\omega)\big|\) rises sharply: that is resonance, and the closer the pole is to the axis — the smaller \(\zeta\) — the taller and narrower the peak. When the test point passes close to a zero, a numerator distance becomes small and the magnitude drops: a zero on the axis produces an exact null, which is how a notch filter kills a mains-hum component at 50 Hz.

The plot also explains the gross shape at the two extremes. As \(\omega\to 0\) the vectors reduce to distances from the origin, giving the DC gain \(H(0)\); as \(\omega\to\infty\) every vector becomes essentially \(j\omega\) itself, so \(\big|H(j\omega)\big|\to |K|\,\omega^{M-N}\) — the response rolls off at \(20(N-M)\) dB per decade, decided by nothing more than the excess of poles over zeros. The example in the figure has one zero at the origin and two poles: it climbs from zero at DC, peaks near the poles and falls away afterwards, which is a bandpass.

If a channel has distorted a signal according to \(H(s)\), the obvious repair is to cascade an inverse system \(H_{\text{inv}}(s)=1/H(s)\), because the convolution property then makes the overall system function 1 and the cascade transparent. Forming \(1/H\) is trivial on paper — swap numerator and denominator, so the poles of the inverse are the zeros of the original and vice versa. Whether the result is usable is another matter.

For the inverse to be causal and stable, all of its poles must lie in the open left half-plane. Those poles are the zeros of \(H(s)\). So a stable causal inverse exists exactly when \(H(s)\) has all its zeros in the left half-plane as well as all its poles.

The extreme case of a right-half-plane zero is the all-pass system, whose zeros are the exact mirror images of its poles: \(H_{\text{ap}}(s)=\dfrac{s-a}{s+a}\) with \(a \gt 0\). Evaluate it on the axis. The point \(j\omega\) is the same distance from \(+a\) as it is from \(-a\), by symmetry, so the numerator and denominator vectors have equal length and \(\big|H_{\text{ap}}(j\omega)\big|=1\) at every frequency. The system passes all frequencies with unchanged amplitude and alters only phase — hence the name.

This gives the useful decomposition. Any stable rational \(H(s)\) with a right-half-plane zero at \(s=z_0\) can be written as

The first factor has the offending zero reflected into the left half-plane and is therefore minimum phase; the second is all-pass. Since the all-pass factor has unit magnitude everywhere, the two systems have identical magnitude responses and differ only in phase — and of the two, the minimum-phase one has the smaller phase lag at every frequency, which is where the name comes from. A right-half-plane zero also leaves a visible fingerprint in the time domain: the step response of such a system typically sets off in the wrong direction before recovering, an initial undershoot that no amount of gain can remove.

Problems 1 to 3 go from equation to response; 4 and 5 test stability and the role of the ROC; 6 to 8 use the second-order measures, feedback and the pole–zero geometry. Sketch the pole–zero plot before computing anything — it usually tells you the answer's shape in advance.