The Laplace Transform and Region of Convergence

Part 4 was a considerable success. Chapter 13 gave every reasonable aperiodic signal a spectrum, Chapter 14 turned convolution into multiplication, and Chapter 15 made filtering a matter of choosing a curve. It is worth asking what the word "reasonable" was doing in that sentence.

The Fourier integral \(\int x(t)e^{-j\omega t}dt\) converges only if \(x(t)\) decays fast enough for the integral to exist, since \(|e^{-j\omega t}|=1\) contributes no decay of its own. Absolute integrability is the usual sufficient condition, and it fails for a whole class of perfectly ordinary engineering signals.

The second and fifth rows are the serious ones. An unstable system has an impulse response that grows without bound, and the entire question of stability is a question about such responses. A theory that can only describe systems already known to be stable cannot be used to decide whether a system is stable, and cannot describe what happens when feedback is applied to something that is not. The apparatus of Part 4 is silent exactly where the engineering is most interesting.

The third and fourth rows are a milder annoyance. A ramp and a step are not exotic; they are the standard test inputs. Chapter 14 could handle the step only by admitting an impulse at \(\omega=0\), a device that works but that sits uneasily inside every subsequent calculation.

Take a signal \(x(t)\) whose Fourier integral diverges because \(x\) grows. Before transforming, tame it: multiply by a real decaying exponential \(e^{-\sigma t}\) chosen to beat the growth. If \(\sigma\) is large enough, the product \(x(t)e^{-\sigma t}\) does decay, the Fourier integral of the product converges, and we have a legitimate spectrum — not of \(x\), but of a modified \(x\) that we can keep track of.

The two exponentials have merged into one, whose exponent is a complex number. Name that number \(s=\sigma+j\omega\) and the result is a function of \(s\) rather than of \(\omega\) — which is to say a function on a plane rather than on a line.

The picture makes the price obvious. Whether the integral converges now depends on \(\sigma\), and different values of \(\sigma\) succeed for different signals. For \(x(t)=e^{2t}u(t)\), any \(\sigma \gt 2\) works and no \(\sigma \le 2\) does. The transform is therefore not defined on the whole \(s\)-plane, and stating which part of the plane it is defined on is not a footnote — it is part of the answer, as Section 17-3 will show.

One practical remark before going on. Most textbooks and every circuits course use the unilateral transform \(X(s)=\int_{0^-}^{\infty}x(t)e^{-st}dt\), whose lower limit is the origin. For causal signals the two definitions agree, and the unilateral version has the advantage of absorbing initial conditions, which Chapter 19 will exploit when it solves differential equations. This chapter works with the bilateral transform because it is the version in which the region of convergence has anything interesting to say: a unilateral transform's ROC is always a right half-plane, so nobody has to think about it, and nobody learns what it means.

The region of convergence, universally abbreviated ROC, is the set of complex numbers \(s\) for which the defining integral converges absolutely. Everything in this chapter turns on one observation about it: convergence depends on \(s\) only through \(\sigma=\operatorname{Re}\{s\}\).

The reason is that \(|e^{-st}| = |e^{-\sigma t}||e^{-j\omega t}| = e^{-\sigma t}\), so the magnitude of the integrand, which is what decides absolute convergence, never involves \(\omega\) at all. If the integral converges for one value of \(s\), it converges for every \(s\) with the same real part.

Work the basic example carefully, because every later result is a variation on it. Let \(x(t)=e^{-at}u(t)\) with \(a\) real. The step confines the integral to \(t \ge 0\):

At the upper limit, \(|e^{-(s+a)t}| = e^{-(\sigma+a)t}\), which tends to zero if and only if \(\sigma+a \gt 0\). Under that condition, and only under it, the bracket evaluates to \(0-\big(-1/(s+a)\big)\):

Notice that the algebra is valid for any real \(a\), positive or negative. If \(a \gt 0\) the signal decays and the ROC includes the \(j\omega\) axis; if \(a \lt 0\) the signal grows and the ROC lies strictly to the right of the axis. The Laplace transform did not flinch at the growing signal — it simply reported a region that excludes \(\sigma=0\), which is precisely the statement that no Fourier transform exists.

That is the first appearance of the central theme of this chapter. The ROC is not bookkeeping. It carries physical information that the expression \(1/(s+a)\) does not.

Almost every transform met in this course is a rational function — a ratio of polynomials in \(s\). That is not a coincidence: Chapter 9 described continuous-time systems by linear constant-coefficient differential equations, and Chapter 18 will show that transforming such an equation always produces a ratio of polynomials.

The roots \(z_i\) of the numerator are the zeros, where \(X(s)=0\); the roots \(p_i\) of the denominator are the poles, where \(|X(s)|\) becomes infinite. Plotting them in the complex plane — zeros as circles, poles as crosses — gives the pole–zero plot, which together with the gain \(K\) and the ROC determines the signal completely.

Poles are where the transform blows up, so no pole can lie inside the ROC: if the integral converged at that \(s\), \(X(s)\) would be a finite number there. This single observation, developed in Section 17-5, generates almost the whole list of ROC properties.

Two conventions of the \(s\)-plane are worth fixing now, because Part 5 and Part 7 both lean on them. The vertical axis \(\sigma=0\) is the \(j\omega\) axis, and it is the home of the Fourier transform: a point on it corresponds to a pure oscillation with no growth. The left half-plane \(\sigma \lt 0\) is the home of decaying exponentials; the right half-plane \(\sigma \gt 0\) of growing ones. A pole's real part is the decay rate of the term it contributes, and its imaginary part is that term's oscillation frequency — so a complex-conjugate pair at \(-\alpha \pm j\beta\) contributes \(e^{-\alpha t}\cos(\beta t+\phi)\), the damped sinusoid of Chapter 3.

Six statements cover every case that arises. None needs to be memorised if the convergence integral of Section 17-3 is kept in view.

1. The ROC consists of strips parallel to the \(j\omega\) axis. Proved already: convergence depends only on \(\sigma\).

2. The ROC contains no poles. Proved already: \(X\) is finite wherever the integral converges.

3. If \(x(t)\) has finite duration and is absolutely integrable, the ROC is the entire \(s\)-plane. The integral runs over a finite interval \([T_1,T_2]\), on which the factor \(e^{-\sigma t}\) is bounded above by \(e^{-\sigma T_1}\) or \(e^{-\sigma T_2}\) whichever is larger, for every real \(\sigma\). A bounded factor times an integrable function over a finite interval always integrates, so no value of \(s\) is excluded.

4. If \(x(t)\) is right-sided — zero before some \(T_1\) — the ROC is a right half-plane, \(\operatorname{Re}\{s\} \gt \sigma_0\). Here is the argument. Suppose the integral converges for some \(\sigma_1\), so \(\int_{T_1}^{\infty}|x(t)|e^{-\sigma_1 t}dt\) is finite. Now take any \(\sigma_2 \gt \sigma_1\). For \(t \ge T_1\) we have \(e^{-\sigma_2 t} = e^{-\sigma_1 t}e^{-(\sigma_2-\sigma_1)t} \le e^{-\sigma_1 t}e^{-(\sigma_2-\sigma_1)T_1}\), a constant multiple of the integrand that already converged. So convergence at \(\sigma_1\) forces convergence at every larger \(\sigma\), and the ROC extends rightward without limit.

5. If \(x(t)\) is left-sided, the ROC is a left half-plane, \(\operatorname{Re}\{s\} \lt \sigma_0\). The same inequality, run with the signs reversed: on \(t \le T_2\) a more negative \(\sigma\) only helps.

6. If \(x(t)\) is two-sided, the ROC is a vertical strip, or is empty. Split the signal at any convenient instant into a left-sided piece and a right-sided piece. The right-sided piece converges on some right half-plane \(\sigma \gt \sigma_R\), the left-sided piece on some left half-plane \(\sigma \lt \sigma_L\), and the transform of the sum exists exactly where both do — their intersection. If \(\sigma_R \lt \sigma_L\) that intersection is the strip \(\sigma_R \lt \sigma \lt \sigma_L\). If \(\sigma_R \ge \sigma_L\) it is empty, and the signal has no Laplace transform at all.

For rational transforms these six statements sharpen into a rule that is quick to apply. The poles cut the plane into vertical strips; the ROC is one of those strips, and which one is decided by the sidedness of the signal. A right-sided signal takes the strip to the right of the rightmost pole; a left-sided signal the strip to the left of the leftmost pole; a two-sided signal one of the strips in between.

Now the point that makes the ROC unavoidable. Take the left-sided signal \(x(t)=-e^{-at}u(-t)\), which is zero for positive time and, for negative time, the mirror of the exponential met in Section 17-3. The step \(u(-t)\) confines the integral to \(t \le 0\):

As \(t \to -\infty\) the magnitude \(e^{-(\sigma+a)t}\) tends to zero if and only if \(\sigma+a \lt 0\). Under that condition the limit vanishes and

This is the single most examinable idea in the chapter, and it generalises: a rational \(X(s)\) whose \(N\) poles have distinct real parts admits \(N+1\) different ROCs, hence \(N+1\) different signals, all sharing the same algebra. Worked Example 2 works all three cases of a two-pole transform, and they turn out to be a decaying causal signal, a growing anticausal one, and a two-sided signal that grows in both directions.

A genuinely two-sided example is worth having in hand. Let \(x(t)=e^{-b|t|}\) with \(b \gt 0\). Split at the origin. The right-hand piece \(e^{-bt}u(t)\) transforms to \(1/(s+b)\) for \(\sigma \gt -b\). The left-hand piece is \(e^{bt}\) for \(t\lt 0\), so \(\int_{-\infty}^{0}e^{(b-s)t}dt = 1/(b-s)\) provided \(\operatorname{Re}\{b-s\} \gt 0\), that is \(\sigma \lt b\). Adding:

The strip contains the \(j\omega\) axis, and setting \(s=j\omega\) gives \(2b/(b^2+\omega^2)\), exactly the Fourier transform Chapter 13 computed for the same signal. That agreement is not a coincidence but a theorem, and Section 17-8 states it.

Four short derivations produce most of the table. The impulse is immediate from the sifting property of Chapter 3: \(\int\delta(t)e^{-st}dt = e^{0}=1\), and since nothing was assumed about \(s\), the ROC is the entire plane. The step is the case \(a=0\) of the exponential, giving \(1/s\) with \(\operatorname{Re}\{s\} \gt 0\) — the pole sits exactly on the \(j\omega\) axis, which is why the Fourier transform of \(u(t)\) needed an impulse to exist at all.

Repeated poles come from differentiating the exponential pair with respect to \(a\). Since \(\frac{\partial}{\partial a}e^{-at}u(t) = -t\,e^{-at}u(t)\) and \(\frac{\partial}{\partial a}\frac{1}{s+a} = -\frac{1}{(s+a)^2}\), the pair \(t\,e^{-at}u(t) \leftrightarrow 1/(s+a)^2\) follows, with the ROC unchanged; repeating the trick gives \(t^{n-1}e^{-at}u(t)/(n-1)! \leftrightarrow 1/(s+a)^n\).

Sinusoids come from Euler's formula and linearity. Writing \(\cos\omega_0 t = \tfrac12(e^{j\omega_0 t}+e^{-j\omega_0 t})\) and applying the exponential pair to each term with \(a = \mp j\omega_0\):

Both component ROCs are \(\operatorname{Re}\{s\} \gt 0\), since \(\operatorname{Re}\{\mp j\omega_0\}=0\), so the sum converges there. The poles at \(s=\pm j\omega_0\) sit on the imaginary axis — an undamped oscillation neither grows nor decays, and its poles mark the boundary between the two behaviours.

Apply all of this to a system, with \(h(t)\) the impulse response of Chapter 7 and \(H(s)\) its Laplace transform — the transfer function. Three properties that Chapter 8 defined in the time domain now become statements about a picture.

Fourier existence. Setting \(s=j\omega\) in the Laplace integral gives the Fourier integral, so \(X(j\omega)\) exists precisely when the ROC contains the \(j\omega\) axis. When it does, the Fourier transform is obtained by evaluating \(X(s)\) on that axis, as the two-sided example of Section 17-6 demonstrated. When it does not, no amount of algebra will produce one.

Causality. A causal system has \(h(t)=0\) for \(t \lt 0\), which makes \(h\) right-sided, so by property 4 its ROC is a right half-plane. The converse needs care — a right-half-plane ROC guarantees right-sidedness only when \(H(s)\) is rational, which in this course it always is. For rational transfer functions the working rule is exact: causal if and only if the ROC lies to the right of the rightmost pole.

Stability. Chapter 8 proved that an LTI system is BIBO stable exactly when \(\int|h(t)|dt \lt \infty\). That integral is the convergence integral of Section 17-3 evaluated at \(\sigma=0\). So the system is stable if and only if the ROC includes the \(j\omega\) axis — the same condition as Fourier existence, which is why a stable system always has a frequency response and an unstable one never does.

The two conditions can also conflict, and that is instructive rather than pathological. A transfer function with a pole in the right half-plane can be given an ROC that includes the \(j\omega\) axis, making it stable — but that ROC is then a strip, not a right half-plane, so the system is not causal. Such a system is realisable only with delay, as a processor working on recorded data, which is exactly how non-causal filters are used in offline signal processing. Worked Example 6 constructs one.

Recovering \(x(t)\) from \(X(s)\) is, in principle, a contour integral. Undo the reasoning of Section 17-2: \(X(\sigma+j\omega)\) is the Fourier transform of \(x(t)e^{-\sigma t}\), so invert that Fourier transform and multiply the decaying factor back out.

The vertical line of integration must lie inside the ROC — that is the only place where \(X\) is defined — and this is where the ROC earns its keep a second time: different admissible lines give different signals, exactly as Section 17-6 promised. In practice the contour integral is almost never evaluated. For rational transforms, partial fractions plus the table of Section 17-7 does the job in three steps.

Then invert each term separately, and here is the step students skip: the ROC decides each term individually. Compare the pole \(p_i\) with the ROC. If the ROC lies to the right of \(p_i\), that term is right-sided and contributes \(A_i e^{p_i t}u(t)\). If the ROC lies to the left of \(p_i\), the term is left-sided and contributes \(-A_i e^{p_i t}u(-t)\). A single transform can therefore produce a signal with pieces on both sides of the origin, one per pole, and Worked Example 2 does exactly that.

Two housekeeping notes. If the numerator degree equals or exceeds the denominator degree, divide first; the quotient contributes impulses and their derivatives, since \(1 \leftrightarrow \delta(t)\) and \(s \leftrightarrow \delta'(t)\). And complex poles arrive in conjugate pairs whose coefficients are also conjugate, so the pair is best recombined into the damped-sinusoid entries of the table rather than inverted separately.

Problems 1 to 3 are direct computations from the definition; 4 to 6 turn on the ROC; 7 and 8 connect the transform to system behaviour. In every answer, state the region — an expression alone earns no marks and, more importantly, identifies no signal.