- Why every Laplace property is really two statements — one about \(X(s)\) and one about the ROC — and how to spot the cases where the ROC grows.
- The two shifts: a delay in time becomes the factor \(e^{-st_0}\), while multiplying by \(e^{s_0t}\) translates the whole \(s\)-plane picture, ROC and all.
- Scaling, reversal and conjugation, and the reason a real signal must have its poles and zeros in conjugate pairs.
- Differentiation in time \(\big(\,\to sX(s)\,\big)\) — the single property that turns a differential equation into a polynomial — and its mirror, differentiation in \(s\).
- The convolution property \(y = x * h \Leftrightarrow Y(s)=X(s)H(s)\), proved from the definition, and why the ROC can be larger than the intersection.
- The initial and final value theorems, both derived from one identity, together with the conditions under which each is allowed.
- The unilateral transform and the differentiation rule that carries \(y(0^-)\) and \(\dot y(0^-)\) into the algebra — the tool Chapter 19 uses to solve circuits with stored energy.
From Definition to Toolkit
Chapter 17 built a short table of transform pairs, and every entry in it cost an integral. That is an honest way to start and a hopeless way to continue. The signals an engineer actually meets are not the entries of a table; they are entries that have been delayed, scaled, differentiated, windowed, multiplied by exponentials and convolved with one another. If each such manipulation forced a return to \(\int x(t)e^{-st}\,dt\), the transform would be no faster than the differential equation it was meant to replace.
The escape is the same one Chapter 14 provided for the Fourier transform and Chapter 24 will provide for the \(z\)-transform. A handful of structural results say what a given operation on \(x(t)\) does to \(X(s)\), once and for all. Combine three of them and the transform of a complicated signal falls out in two lines. The table of pairs then shrinks to about five genuinely independent entries, and everything else is derived on demand.
There is one difference from the Fourier case, and it is the difference that gives this chapter its character. A Fourier transform is a function of the single real variable \(\omega\), so a property is one equation. A Laplace transform is a function on a plane together with the region of that plane where the defining integral converges, so a property is two statements: what happens to the algebra, and what happens to the ROC. Quoting the first without the second is the commonest way to produce a confidently wrong answer in this subject.
Linearity and What It Does to the ROC
Linearity is immediate. The defining integral is linear in \(x\), so a weighted sum of signals transforms to the same weighted sum of transforms. The only question worth asking is where the result converges.
If the integral for \(x_1\) converges on \(R_1\) and the integral for \(x_2\) converges on \(R_2\), then for any \(s\) lying in both, the integral for \(ax_1+bx_2\) certainly converges. So the ROC of the sum contains the intersection. It can be strictly larger, and understanding when is the first real lesson of this chapter.
Containment, not equality. The ROC is larger than the intersection precisely when the addition cancels a pole that was limiting one of the two regions.
Here is the cancellation in the flesh. Take
Each transform has a pole at \(s=-1\), and each ROC is the half-plane to the right of it. Add the two signals, however, and the \(e^{-t}u(t)\) terms annihilate one another: what is left is \(x_1(t)+x_2(t) = e^{-2t}u(t)\), whose transform is \(1/(s+2)\) with ROC \(\operatorname{Re}\{s\} \gt -2\). The pole at \(-1\) has vanished from the algebra, and with it the obstruction that was holding the ROC back. The region of convergence has grown.
The Two Shifts
Two properties are called shifting, one in each domain, and they are not symmetrical in their effect on the ROC. Take the time shift first. Delay \(x(t)\) by \(t_0\) and transform:
The substitution is a change of variable in a convergent integral, and \(e^{-st_0}\) is a finite non-zero number for every finite \(s\). Nothing about convergence has changed, so the ROC is untouched.
The shift in \(s\) runs the other way. Multiply the signal by \(e^{s_0t}\) and the exponent in the integrand becomes \(-(s-s_0)t\), so the transform is the old one evaluated at \(s-s_0\) — the entire \(s\)-plane picture, poles, zeros and ROC together, is translated by \(s_0\).
A delay changes the transform but not where it converges. Multiplying by an exponential changes both, and the ROC slides by \(\operatorname{Re}\{s_0\}\) — right for a growing exponential, left for a decaying one.
Two special cases deserve names. Setting \(s_0 = -a\) with \(a\) real and positive gives the damping rule: multiplying a signal by \(e^{-at}\) shifts its transform \(a\) units to the left. This is why every damped entry in Chapter 17's table is the undamped entry with \(s\) replaced by \(s+a\); the cosine pair \(s/(s^2+\omega_0^2)\) becomes \((s+a)/\big[(s+a)^2+\omega_0^2\big]\) without a single new integral. Setting \(s_0 = j\omega_0\) instead gives modulation, which slides the picture vertically and leaves the ROC — a vertical strip — exactly where it was.
The time-shift property also carries a warning about what \(H(s)\) can look like. A pure delay of \(T\) seconds is an LTI system with impulse response \(\delta(t-T)\), so its system function is \(e^{-sT}\). That is not a ratio of polynomials, and no finite differential equation produces it. Transport lag in a pipeline, propagation delay on a line, computation delay in a controller — all of them push a system outside the rational world, which is why they are usually approximated by a rational function before analysis proceeds.
Scaling, Reversal and Conjugation
Compress the time axis and the transform must stretch, exactly as it did for the Fourier transform in Chapter 14 — but now the stretching happens in a plane, and it drags the ROC with it. Substituting \(\tau = at\) in the defining integral, with \(a \gt 0\) so that the limits keep their order,
For \(a \lt 0\) the substitution also reverses the limits of integration, and restoring their order supplies a second minus sign; the two combine into \(1/|a|\). The convergence condition transforms with the variable: \(s/a\) must lie in \(R\), so the new ROC is \(aR\), the old region scaled — and flipped about the imaginary axis when \(a\) is negative.
Reversal is the case \(a=-1\), and it is worth checking against a pair you already know. From Chapter 17, \(e^{-t}u(t) \leftrightarrow 1/(s+1)\) with \(\operatorname{Re}\{s\} \gt -1\). Reversing the signal gives \(e^{t}u(-t)\), and the property predicts \(1/(-s+1) = -1/(s-1)\) on \(\operatorname{Re}\{s\} \lt 1\). Confirm it directly: \(\int_{-\infty}^{0}e^{t}e^{-st}dt = \big[e^{(1-s)t}/(1-s)\big]_{-\infty}^{0} = 1/(1-s)\), and the lower limit vanishes only when \(\operatorname{Re}\{1-s\} \gt 0\), that is \(\operatorname{Re}\{s\} \lt 1\). The right-sided signal has become left-sided and its half-plane has flipped, precisely as the ROC rules of Chapter 17 demand.
Conjugation looks like bookkeeping and delivers a structural fact. Conjugating the whole defining integral gives \(\int x^{*}(t)e^{-s^{*}t}dt = X^{*}(s^{*})\), so \(x^{*}(t) \leftrightarrow X^{*}(s^{*})\). If \(x(t)\) is real then \(x^{*}=x\), and therefore \(X(s) = X^{*}(s^{*})\) identically. Evaluate that at a pole: if \(X\) blows up at \(s=p\), it blows up at \(s=p^{*}\) as well.
Differentiation in Time and in the s-Domain
This is the property the whole transform was invented for. Integrate \(\int x'(t)e^{-st}dt\) by parts, taking \(u = e^{-st}\) and \(dv = x'(t)dt\):
The boundary term is where the ROC earns its keep. For \(s\) inside the region of convergence the product \(x(t)e^{-st}\) is absolutely integrable, and an absolutely integrable function that is being driven by an exponential must tend to zero at both ends — otherwise the integral would diverge. So the bracket vanishes and only the second term survives.
Differentiation becomes multiplication by \(s\). The ROC can only grow, and it grows exactly when the factor \(s\) cancels a pole of \(X(s)\) sitting at the origin.
Read that key result twice, because it is the hinge of Part 5. An \(n\)th-order linear constant-coefficient differential equation is a statement about \(x, \dot x, \ddot x, \dots\); apply the transform and every derivative becomes a power of \(s\), and the equation becomes a polynomial identity. Calculus has turned into algebra. Chapter 9 solved such equations by guessing homogeneous solutions and matching coefficients; Chapter 19 will solve them by dividing two polynomials.
The property also reproduces a fact from Chapter 3 for free. Since \(u(t) \leftrightarrow 1/s\) and \(\delta(t) = du/dt\), the transform of the impulse must be \(s \cdot (1/s) = 1\) — and here the ROC really does grow, from \(\operatorname{Re}\{s\} \gt 0\) to the whole plane, because the factor \(s\) has cancelled the pole at the origin.
Differentiating in the other domain is easier still. Differentiate \(X(s)=\int x(t)e^{-st}dt\) with respect to \(s\), moving the derivative inside the integral, and each integrand picks up a factor \(-t\).
Multiplying by \(t\) does not change how fast the signal grows or decays exponentially, so the region of convergence is exactly the same.
One application generates a whole family of table entries. Starting from \(e^{-at}u(t)\leftrightarrow 1/(s+a)\) and applying the rule once,
Setting \(a=0\) recovers \(t\,u(t)\leftrightarrow 1/s^2\) and, more generally, \(t^{n-1}u(t)/(n-1)! \leftrightarrow 1/s^{n}\). Every repeated-pole entry in the table is one differentiation away from the entry above it, and the ROC never moves — which is the transform's way of saying that a factor of \(t\) is powerless against an exponential.
The Convolution Property
Chapter 7 established that an LTI system's output is the convolution of its input with its impulse response, and Chapter 8 showed how awkward that integral is to evaluate. Chapter 14 gave the Fourier remedy. The Laplace version is the same statement with a wider licence, because it applies to signals whose Fourier transforms do not exist.
Write the transform of \(y(t)=\int x(\tau)h(t-\tau)d\tau\) and exchange the order of integration:
The inner integral was evaluated by the time-shift property of Section 18-3, and \(H(s)\) then came out of the outer integral because it does not depend on \(\tau\). The exchange of order is legitimate wherever both integrals converge absolutely, which is exactly the intersection of the two regions.
The single most useful property in Part 5. Every cascade of LTI systems becomes a product; every convolution integral becomes a multiplication followed by a partial-fraction expansion.
Once again the containment can be strict, and for the same reason as in Section 18-2: if a zero of \(H(s)\) lands on a pole of \(X(s)\), the pole disappears from the product and the ROC opens out. The extreme case is the inverse system, where \(H_{\text{inv}}(s)=1/H(s)\) so that the product is 1 — every pole cancelled, ROC the entire plane, and the cascade reduced to a wire.
The mirror property, multiplication in time, is far less convenient. It corresponds to a convolution in the \(s\)-domain along a vertical contour, \(x_1(t)x_2(t) \leftrightarrow \frac{1}{2\pi j}\int_{c-j\infty}^{c+j\infty}X_1(p)X_2(s-p)\,dp\), and in practice one avoids it by rewriting the product as a sum of exponentials first. The single case worth memorising is multiplication by \(e^{-at}\), which is not a general product at all but the \(s\)-shift of Section 18-3.
Integration in Time
The running integral of a signal is a convolution in disguise. Since \(u(t-\tau)\) is 1 exactly when \(\tau \lt t\),
Now apply the convolution property with \(u(t) \leftrightarrow 1/s\), whose ROC is \(\operatorname{Re}\{s\} \gt 0\). The transform is \(X(s)/s\), and the ROC contains the intersection of \(R\) with the right half-plane.
Integration divides by \(s\) and introduces a pole at the origin — unless \(X(0)=0\), in which case a zero of \(X\) cancels it and the ROC is not restricted at all.
The exception in that last sentence has a physical reading. A pole at the origin is the transform's way of recording a non-zero final value: the integral of a signal with non-zero net area does not settle to zero. If \(X(0)=\int x(t)dt = 0\), the signal has no net area, its running integral returns to zero, and no pole at the origin appears. Compare Chapter 14, where the Fourier integration property carried an extra impulse \(\pi X(0)\delta(\omega)\) for exactly this reason; the Laplace version needs no such patch, because the \(s\)-plane can hold a pole at the origin where the \(\omega\) axis could only hold an impulse.
Read the other way, the property says that a block labelled \(1/s\) is an integrator. That reading is what makes the block diagrams of Chapter 19 possible: any differential equation can be redrawn as integrators, gains and adders, and the resulting picture is a circuit you could build.
The Initial and Final Value Theorems
Sometimes only the two ends of a response matter: how it starts and where it settles. Both can be read off \(X(s)\) without inverting it at all, and remarkably both come from the same identity. Take a signal that is zero for \(t \lt 0\), and apply the differentiation property with the lower limit written as \(0^-\) so that any jump at the origin is included:
Now push \(s\) to each extreme. As \(s \to \infty\) along the real axis, \(e^{-st}\) collapses onto the instant \(t=0\), so the integral retains only what the derivative does there — namely the jump \(x(0^+)-x(0^-)\) — and everything later is crushed to nothing. Hence \(\lim_{s\to\infty}sX(s) = x(0^-) + \big[x(0^+)-x(0^-)\big] = x(0^+)\).
As \(s \to 0\) the exponential becomes 1 and the integral becomes the total change in \(x\), which is \(x(\infty)-x(0^-)\) provided that limit exists. Hence \(\lim_{s\to 0}sX(s) = x(0^-) + x(\infty) - x(0^-) = x(\infty)\).
The initial value theorem requires \(x(t)=0\) for \(t \lt 0\) and no impulse or higher singularity at the origin. The final value theorem additionally requires that \(x(\infty)\) exist, which holds exactly when every pole of \(sX(s)\) lies strictly inside the left half-plane.
The condition on the final value theorem is not a technicality, and it is where most marks are lost. The derivation assumed the total change \(x(\infty)-x(0^-)\) was a finite number. If \(X(s)\) has a pole in the right half-plane, \(x(t)\) grows without bound and there is no final value; if it has a pole on the \(j\omega\) axis away from the origin, \(x(t)\) oscillates for ever and never settles. In both cases \(\lim_{s\to 0}sX(s)\) may still produce a perfectly innocent-looking number, and that number is meaningless.
| \(X(s)\) | \(\lim_{s\to 0}sX(s)\) | Actual \(x(\infty)\) | Verdict |
|---|---|---|---|
| \(\dfrac{5}{s(s+2)}\) | \(5/2\) | \(5/2\) | Valid — poles at \(-2\) only |
| \(\dfrac{\omega_0}{s^2+\omega_0^2}\) | \(0\) | does not exist | Invalid — poles at \(\pm j\omega_0\) |
| \(\dfrac{1}{s-2}\) | \(0\) | \(\infty\) | Invalid — pole at \(+2\) |
| \(\dfrac{1}{s^2}\) | \(1\) | \(\infty\) | Invalid — double pole at the origin |
The Unilateral Transform and the Property Table
Everything so far has been bilateral: the integral runs over the whole time axis, and the transform describes a signal that has always existed. Practical problems are not like that. A switch is closed at \(t=0\), the capacitor already holds charge, the flywheel is already turning, and the question is what happens next. For those problems the natural object is the unilateral Laplace transform, which begins at \(0^-\) — just before the switch — so that any impulse applied at the origin is captured.
For a signal that is already zero before the origin, the two transforms coincide, so the whole table of pairs from Chapter 17 carries over unchanged. The ROC also stops being an issue: the integral starts at a finite time, so the signal is right-sided by construction and the ROC is always a right half-plane. It is customary to leave it unmentioned.
What changes is differentiation, and the change is the entire point. Integrating by parts over \([0^-,\infty)\) leaves a boundary term at the lower limit that no longer vanishes:
Applying the same rule to \(\dot x\) in place of \(x\) gives the second derivative, and induction gives the rest:
The initial conditions enter the algebra as ordinary additive terms. Solve the resulting equation for \(\mathcal{X}(s)\) and the complete response — forced and natural together — falls out in one step, with no arbitrary constants to determine afterwards.
That is a genuine advance over Chapter 9, where the natural response was written with unknown constants and the initial conditions were imposed at the end by solving simultaneous equations. Here they are built in from the start. Chapter 19 does nothing else for four sections.
Two cautions. First, the time-shift property survives only for delays: \(x(t-t_0)u(t-t_0) \to e^{-st_0}\mathcal{X}(s)\) for \(t_0 \gt 0\), because a delay pushes signal to the right, away from the lower limit. An advance drags material across \(t=0\) and out of the integral, and needs a correction term. Second, \(0^-\) and \(0^+\) are not interchangeable when the input contains an impulse: a current impulse into a capacitor changes its voltage instantaneously, so \(v(0^+) \ne v(0^-)\), and it is \(v(0^-)\) — the value before the event — that belongs in the formula.
| Property | Signal | Transform | ROC |
|---|---|---|---|
| Linearity | \(ax_1(t)+bx_2(t)\) | \(aX_1(s)+bX_2(s)\) | \(\supseteq R_1\cap R_2\) |
| Time shift | \(x(t-t_0)\) | \(e^{-st_0}X(s)\) | \(R\) |
| Shift in \(s\) | \(e^{s_0t}x(t)\) | \(X(s-s_0)\) | \(R+\operatorname{Re}\{s_0\}\) |
| Time scaling | \(x(at)\) | \(\dfrac{1}{|a|}X(s/a)\) | \(aR\) |
| Reversal | \(x(-t)\) | \(X(-s)\) | \(-R\) |
| Conjugation | \(x^{*}(t)\) | \(X^{*}(s^{*})\) | \(R\) |
| Differentiation in \(t\) | \(\dfrac{d^{\,n}x}{dt^{\,n}}\) | \(s^{n}X(s)\) | \(\supseteq R\) |
| Differentiation in \(s\) | \(t^{\,n}x(t)\) | \((-1)^{n}\dfrac{d^{\,n}X}{ds^{\,n}}\) | \(R\) |
| Integration | \(\displaystyle\int_{-\infty}^{t}x(\tau)d\tau\) | \(\dfrac{X(s)}{s}\) | \(\supseteq R\cap\{\operatorname{Re}\{s\} \gt 0\}\) |
| Convolution | \(x_1(t)*x_2(t)\) | \(X_1(s)X_2(s)\) | \(\supseteq R_1\cap R_2\) |
| Initial value | \(x(0^+)\) | \(\lim_{s\to\infty}sX(s)\) | causal, no impulse at \(0\) |
| Final value | \(x(\infty)\) | \(\lim_{s\to 0}sX(s)\) | poles of \(sX(s)\) in LHP |
Worked Examples
Problem. Find the Laplace transform of \(x(t) = t\,e^{-2t}\cos(3t)\,u(t)\) and state its ROC.
Solution. Peel the signal apart in the order the properties were built. Start from the pair established in Chapter 17,
Multiplying by \(e^{-2t}\) is the \(s\)-shift with \(s_0=-2\): replace \(s\) by \(s+2\) everywhere and slide the ROC two units left.
Multiplying by \(t\) is differentiation in \(s\) with a sign change. Write \(w=s+2\), so the transform above is \(w/(w^2+9)\), and use the quotient rule:
so \(X(s) = -\,\dfrac{d}{ds}\left[\dfrac{s+2}{(s+2)^2+9}\right] = \dfrac{(s+2)^2-9}{\big[(s+2)^2+9\big]^{2}}\), with ROC \(\operatorname{Re}\{s\} \gt -2\) — unchanged by the factor \(t\).
A quick sanity check: the denominator has a double pole at \(s=-2\pm j3\), which is right, since \(t\,e^{-2t}\cos 3t\) is a damped oscillation whose envelope decays more slowly than \(e^{-2t}\) alone. Repeated poles are the signature of a polynomial-times-exponential signal, exactly as Section 18-5 predicted.
Problem. Find \(X(s)\) for \(x(t) = t\big[u(t)-u(t-2)\big]\), and comment on the ROC.
Solution. The temptation is to write \(1/s^2 - e^{-2s}/s^2\) and stop. That is wrong, because the time-shift property applies to \(x(t-t_0)\) — the whole signal delayed — and the second term here is \(t\,u(t-2)\), which is not the ramp \(r(t)=t\,u(t)\) delayed by 2. Force it into the right form by adding and subtracting:
Now every term is a standard signal delayed by 2, and the property applies term by term with \(r(t)\leftrightarrow 1/s^2\) and \(u(t)\leftrightarrow 1/s\):
As for the ROC: \(x(t)\) is non-zero only on \(0 \lt t \lt 2\), and Chapter 17 established that a finite-duration bounded signal converges for every \(s\). The apparent pole at the origin must therefore be an illusion, and it is — expanding for small \(s\) with \(e^{-2s}\approx 1-2s+2s^2\) gives \(X(s) \approx \big[1-(1-2s+2s^2)\big]/s^2 - 2(1-2s)/s = (2s-2s^2)/s^2 - 2/s + 4 = 2/s - 2 - 2/s + 4 = 2\), a finite value. The singular terms cancel exactly, as they must.
Problem. Evaluate \(y(t) = e^{-t}u(t) * e^{-3t}u(t)\).
Solution. In the \(s\)-domain the convolution is a product, and both factors are right-sided so the ROC of the product is \(\operatorname{Re}\{s\} \gt -1\), the more restrictive of the two:
Both poles lie to the left of the ROC, so both terms are right-sided, and
Compare the direct route. The convolution integral is \(\int_{0}^{t}e^{-\tau}e^{-3(t-\tau)}d\tau = e^{-3t}\int_{0}^{t}e^{2\tau}d\tau = e^{-3t}\frac{e^{2t}-1}{2} = \tfrac12(e^{-t}-e^{-3t})\) for \(t \gt 0\) — the same answer, but only after getting the limits of the sliding integral right, which is the part of Chapter 7 that costs students the most time.
Two structural checks are worth forming into a habit. At \(t=0\) the answer gives \(y(0)=0\), which is correct: the convolution of two causal signals starts at zero. And \(\lim_{s\to\infty}sY(s) = \lim s/\big[(s+1)(s+3)\big] = 0\), which the initial value theorem confirms.
Problem. Let \(x(t)=e^{-2t}u(t)\) and \(g(t)=\int_{-\infty}^{t}x(\tau)\,d\tau\). Find \(G(s)\) by the property, then verify by integrating \(x\) explicitly.
Solution. By the property, \(G(s) = X(s)/s = \dfrac{1}{s(s+2)}\), with ROC \(\operatorname{Re}\{s\} \gt 0\): the region has been cut back from \(\operatorname{Re}\{s\} \gt -2\) because a new pole has appeared at the origin.
Directly, for \(t \gt 0\), \(g(t) = \int_{0}^{t}e^{-2\tau}d\tau = \tfrac12\big(1-e^{-2t}\big)\), and \(g(t)=0\) for \(t \lt 0\). Transforming term by term,
The two agree. The pole at the origin is doing real work: \(g(t)\) tends to \(1/2\) rather than to zero, and the final value theorem reads that off at once as \(\lim_{s\to 0}sG(s) = \lim_{s\to 0}1/(s+2) = 1/2\). Had \(x(t)\) carried zero net area, \(X(0)\) would have been zero, the numerator would have cancelled the new pole, and \(g(t)\) would have returned to zero.
Problem. For each of \(X_1(s)=\dfrac{2s+5}{s(s^2+4s+3)}\), \(X_2(s)=\dfrac{4}{s^2+4}\) and \(X_3(s)=\dfrac{1}{s-2}\), decide whether the final value theorem may be used, and find \(x(\infty)\) where it may.
Solution. Always factor first. \(s^2+4s+3=(s+1)(s+3)\), so
The theorem applies, and \(x_1(\infty) = \dfrac{5}{(1)(3)} = \dfrac53\). Confirm by partial fractions: the residue of \(X_1\) at the origin is \(5/[(1)(3)]=5/3\), so \(x_1(t) = \tfrac53 + (\text{terms in } e^{-t}, e^{-3t})\), which does indeed settle to \(5/3\).
For \(X_2\), \(sX_2(s)=4s/(s^2+4)\) tends to 0 as \(s\to 0\), which looks like an answer. It is not: the poles of \(sX_2\) sit at \(s=\pm j2\), on the imaginary axis, and \(x_2(t)=2\sin(2t)u(t)\) oscillates for ever with no final value at all. The theorem was never applicable.
For \(X_3\), \(sX_3(s)=s/(s-2)\) also tends to 0, and again the number is worthless: the pole at \(s=+2\) puts \(x_3(t)=e^{2t}u(t)\) in the right half-plane, growing without bound. Both failures produce a plausible finite limit, which is exactly why the pole check must come first and the limit second.
Problem. Solve \(\dfrac{dy}{dt} + 3y(t) = e^{-t}u(t)\) with \(y(0^-)=2\), using the unilateral transform.
Solution. Transform both sides. The derivative contributes \(sY(s)-y(0^-)=sY(s)-2\), and the right-hand side is \(1/(s+1)\):
The split is automatic and it is worth naming: the first term is what the stored energy alone would produce, the second what the input alone would produce. Expanding the second term with the residues \(1/(-1+3)=\tfrac12\) and \(1/(-3+1)=-\tfrac12\),
Check it two ways. At \(t=0\): \(\tfrac32+\tfrac12=2=y(0^-)\), so the initial condition is satisfied without any constant having been fitted afterwards. And substituting into the equation, \(\dot y = -\tfrac92 e^{-3t}-\tfrac12 e^{-t}\) while \(3y = \tfrac92 e^{-3t}+\tfrac32 e^{-t}\); the \(e^{-3t}\) terms cancel and the sum is \(e^{-t}\), as required. Chapter 19 takes this method and applies it to systems of any order.
Chapter Summary
An equation for \(X(s)\) and a statement about the ROC. The ROC contains the intersection and grows when a pole cancels.
\(x(t-t_0)\leftrightarrow e^{-st_0}X(s)\), ROC unchanged. \(e^{s_0t}x(t)\leftrightarrow X(s-s_0)\), ROC translated by \(\operatorname{Re}\{s_0\}\).
\(dx/dt \leftrightarrow sX(s)\) turns differential equations into polynomials; \(t\,x(t)\leftrightarrow -dX/ds\) generates every repeated-pole pair.
\(x*h \leftrightarrow XH\). Cascades multiply, parallel branches add, and \(1/s\) is an integrator.
\(x(0^+)=\lim_{s\to\infty}sX(s)\), \(x(\infty)=\lim_{s\to0}sX(s)\) — the second only when every pole of \(sX(s)\) is in the open LHP.
\(\dot x \to s\mathcal{X}(s)-x(0^-)\). Initial conditions become additive terms, so the complete response emerges in one step.
Problems
Problems 1 to 3 exercise the shifts and scalings; 4 to 6 use differentiation, integration and convolution; 7 and 8 test whether you know when a theorem may be quoted. State the ROC every time — an answer without one is incomplete.
- Find the Laplace transform and ROC of \(x(t) = e^{-3(t-2)}u(t-2)\), and of \(y(t)=e^{-3t}u(t-2)\). Explain in one sentence why the two answers differ by a constant factor.
- Given \(x(t)\leftrightarrow X(s)\) with ROC \(-1 \lt \operatorname{Re}\{s\} \lt 2\), state the transform and ROC of (a) \(x(2t)\), (b) \(x(-t)\), (c) \(e^{-3t}x(t)\), (d) \(x(t-4)\).
- Use the \(s\)-shift together with the cosine pair to obtain \(e^{-at}\cos(\omega_0 t)u(t) \leftrightarrow (s+a)/\big[(s+a)^2+\omega_0^2\big]\), and sketch the pole locations for \(a=1,\ \omega_0=2\).
- Find the transform of \(t^{2}e^{-4t}u(t)\) by applying differentiation in \(s\) twice, and confirm the result against the general repeated-pole formula.
- A signal has \(X(s)=\dfrac{s+1}{s^2+5s+6}\) with ROC \(\operatorname{Re}\{s\} \gt -2\). Find the transform of \(\int_{-\infty}^{t}x(\tau)d\tau\), state its ROC, and find its final value.
- Evaluate \(e^{-2t}u(t) * u(t)\) using the convolution property, and check your answer by direct integration.
- For each of \(\dfrac{3}{s(s+1)}\), \(\dfrac{s}{s^2+9}\), \(\dfrac{s+2}{s(s-1)}\) and \(\dfrac{10}{(s+2)^2}\), say whether the final value theorem may be applied, and give \(x(\infty)\) where it may.
- Solve \(\ddot y + 4\dot y + 3y = 0\) with \(y(0^-)=1\) and \(\dot y(0^-)=0\) using the unilateral transform, and verify both initial conditions from your answer.