CHAPTER 1
Laplace Transform
The big idea: turn a differential equation in time \(t\) into an algebra problem in \(s\). Solve the algebra, then transform back. Every topic below has a picture you can play with.
1.0 What the Laplace transform actually does
Definition 1.1 — the "weighted area" machine
Definition
\[F(s)=\mathcal{L}\{f(t)\}=\int_0^{\infty} f(t)\,e^{-st}\,dt\]
for every \(s\) for which the improper integral converges.
Read it as a recipe. Take your signal \(f(t)\). Multiply it by a decaying exponential \(e^{-st}\). Measure the total area under the product. That single number is \(F(s)\). Change \(s\) and you get a different number, so \(F\) is a new function of \(s\).
Laplace machine: watch the area become F(s)
f(t)e−stf(t)e−st (area = F(s))
See itDrag \(s\) to the left until \(s\) is close to \(a\) with \(f=e^{at}\). The yellow area explodes. That is why \(\mathcal{L}\{e^{at}\}=\frac{1}{s-a}\) needs \(\operatorname{Re}(s)>a\). The integral only converges when the exponential beats the growth of \(f\).
1.1 The must-know table (Theorem 1.2)
Tap each card to flip it. Try to say the answer before you flip.
\(\mathcal{L}\{1\}\)tap
\(\dfrac1s,\ \ s>0\)
\(\mathcal{L}\{t\}\)tap
\(\dfrac1{s^2}\)
\(\mathcal{L}\{t^n\}\)tap
\(\dfrac{n!}{s^{n+1}}\)
\(\mathcal{L}\{e^{at}\}\)tap
\(\dfrac1{s-a},\ \ s>a\)
\(\mathcal{L}\{\sin at\}\)tap
\(\dfrac{a}{s^2+a^2}\)
\(\mathcal{L}\{\cos at\}\)tap
\(\dfrac{s}{s^2+a^2}\)
\(\mathcal{L}\{\sinh at\}\)tap
\(\dfrac{a}{s^2-a^2}\)
\(\mathcal{L}\{\cosh at\}\)tap
\(\dfrac{s}{s^2-a^2}\)
\(\mathcal{L}\{\delta(t-a)\}\)tap
\(e^{-as}\)
\(\mathcal{L}\{u(t-a)\}\)tap
\(\dfrac{e^{-as}}{s}\)
Memory tricksin has the a on top, cos has the s on top. The hyperbolic versions are identical except \(+a^2\) becomes \(-a^2\).
Exam proof: L{sin at} and L{cos at} (2025 Q1(a)(i), 6 marks)
Use Euler: \(e^{iat}=\cos at+i\sin at\). Compute one integral instead of two.
\[\mathcal{L}\{e^{iat}\}=\int_0^\infty e^{-(s-ia)t}\,dt=\left[\frac{-e^{-(s-ia)t}}{s-ia}\right]_0^\infty=\frac{1}{s-ia}\quad(\operatorname{Re}s>0)\]
The upper limit vanishes because \(|e^{-(s-ia)t}|=e^{-\operatorname{Re}(s)t}\to0\).
Rationalise the denominator:
\[\frac{1}{s-ia}=\frac{s+ia}{(s-ia)(s+ia)}=\frac{s}{s^2+a^2}+i\,\frac{a}{s^2+a^2}\]
By linearity \(\mathcal{L}\{e^{iat}\}=\mathcal{L}\{\cos at\}+i\,\mathcal{L}\{\sin at\}\). Match real and imaginary parts:
\[\boxed{\mathcal{L}\{\cos at\}=\frac{s}{s^2+a^2}},\qquad \boxed{\mathcal{L}\{\sin at\}=\frac{a}{s^2+a^2}}\]
Exercise 1.4 · piecewise f
For \(f=-1\) on \([0,2)\) and \(f=1\) for \(t\ge2\), split the integral: \(\int_0^2 -e^{-st}dt+\int_2^\infty e^{-st}dt=\dfrac{2e^{-2s}-1}{s}\).
1.2 Linearity and the first shifting theorem
Multiplying by \(e^{at}\) in time = sliding \(F\) to the right by \(a\) in \(s\)
Theorem 1.5 · Linearity\[\mathcal{L}\{f+g\}=\mathcal{L}\{f\}+\mathcal{L}\{g\},\quad \mathcal{L}\{cf\}=c\,\mathcal{L}\{f\}\]
Theorem 1.6 · s-shifting\[\mathcal{L}\{e^{at}f(t)\}=F(s-a)\]Replace every \(s\) in \(F\) with \(s-a\).
Slide a and see both worlds move
originalwith eat
Exercise 1.7 results (all from s-shifting)
\[\mathcal{L}\{e^{at}t^n\}=\frac{n!}{(s-a)^{n+1}},\quad \mathcal{L}\{e^{at}\cos bt\}=\frac{s-a}{(s-a)^2+b^2},\quad \mathcal{L}\{e^{at}\sin bt\}=\frac{b}{(s-a)^2+b^2}\]
\[\mathcal{L}\{e^{at}\cosh bt\}=\frac{s-a}{(s-a)^2-b^2},\qquad \mathcal{L}\{e^{at}\sinh bt\}=\frac{b}{(s-a)^2-b^2}\]
Exercise 1.9 · \(\mathcal{L}\{e^{-3t}\cos^2 t\}\)
Write \(\cos^2t=\tfrac12+\tfrac12\cos2t\), so \(\mathcal{L}\{\cos^2t\}=\frac1{2s}+\frac{s}{2(s^2+4)}\). Then shift \(s\to s+3\): \(\ \dfrac{1}{2(s+3)}+\dfrac{s+3}{2((s+3)^2+4)}\).
Time scaling (Theorem 1.10)
Theorem\[\mathcal{L}\{f(\alpha t)\}=\frac1\alpha F\!\left(\frac s\alpha\right),\quad\alpha>0\]
Check with Example 1.11: \(\mathcal{L}\{\sin t\}=\frac1{s^2+1}\), so \(\mathcal{L}\{\sin5t\}=\frac15\cdot\frac1{(s/5)^2+1}=\frac{5}{s^2+25}\).
1.3 Transforms of derivatives and integrals
This is the reason Laplace solves ODEs: derivatives turn into multiplication by \(s\)
\(y'\)
→
\(sY-y(0)\)
\(y''\)
→
\(s^2Y-sy(0)-y'(0)\)
\(\int_0^t f\)
→
\(F(s)/s\)
General formula (1.4)
\[\mathcal{L}\{y^{(n)}\}=s^nY(s)-s^{n-1}y(0)-s^{n-2}y'(0)-\cdots-y^{(n-1)}(0)\]
The powers of \(s\) count down as the derivatives at 0 count up.
Exam proof: L{f''} = s²F − s f(0) − f'(0) (2025 Q1(a)(ii), 6 marks)
Start with the first derivative. Integrate by parts with \(u=e^{-st}\), \(dv=f'(t)dt\):
\[\mathcal{L}\{f'\}=\int_0^\infty e^{-st}f'(t)\,dt=\Big[e^{-st}f(t)\Big]_0^\infty+s\int_0^\infty e^{-st}f(t)\,dt\]
Because \(f\) is of exponential order \(a\), \(|f(t)|\le Me^{at}\). So for \(\operatorname{Re}s>a\), \(e^{-st}f(t)\to0\) as \(t\to\infty\). The bracket gives \(0-f(0)\):
\[\mathcal{L}\{f'\}=sF(s)-f(0)\]
Apply the same result to \(g=f'\):
\[\mathcal{L}\{f''\}=s\,\mathcal{L}\{f'\}-f'(0)=s\big(sF(s)-f(0)\big)-f'(0)\]
\[\boxed{\mathcal{L}\{f''(t)\}=s^2F(s)-sf(0)-f'(0)},\qquad \operatorname{Re}(s)>a\]
Exam proof: L{∫₀ᵗ f(u)du} = F(s)/s (2021 Q1(a)(ii), 5 marks)
Let \(g(t)=\int_0^t f(u)\,du\). Then \(g(0)=0\) and by the fundamental theorem of calculus \(g'(t)=f(t)\).
Apply the derivative rule to \(g\): \(\mathcal{L}\{g'\}=s\,\mathcal{L}\{g\}-g(0)=s\,\mathcal{L}\{g\}\).
But \(\mathcal{L}\{g'\}=\mathcal{L}\{f\}=F(s)\). So \(F(s)=s\,\mathcal{L}\{g\}\), which gives \(\boxed{\mathcal{L}\left\{\int_0^t f(u)du\right\}=\dfrac{F(s)}{s}}\).
1.4 Multiplying and dividing by t
Multiplying by \(t\) means differentiate in \(s\). Dividing by \(t\) means integrate in \(s\).
Theorem 1.13\[\mathcal{L}\{t^nf(t)\}=(-1)^n\frac{d^n}{ds^n}F(s)\]
Theorem 1.16\[\mathcal{L}\left\{\frac{f(t)}t\right\}=\int_s^\infty F(p)\,dp\]
Exam proof: division by t (2021 Q1(a)(i), 5 marks)
Write the right-hand side using the definition of \(F\):
\[\int_s^\infty F(p)\,dp=\int_s^\infty\!\!\int_0^\infty f(t)e^{-pt}\,dt\,dp\]
Swap the order of integration (allowed for convergent integrals):
\[=\int_0^\infty f(t)\left(\int_s^\infty e^{-pt}dp\right)dt\]
The inner integral is \(\left[\frac{-e^{-pt}}{t}\right]_{p=s}^{\infty}=\frac{e^{-st}}{t}\).
\[\int_0^\infty \frac{f(t)}{t}e^{-st}\,dt=\mathcal{L}\left\{\frac{f(t)}{t}\right\}\quad\blacksquare\]
Using both rules (2021 Q1(b), 10 marks)
Spot the pattern: \(\int_0^\infty e^{-t}g(t)\,dt\) is just \(\mathcal{L}\{g\}\) at \(s=1\). Here \(g(t)=\int_0^t\frac{\sin u}{u}du\).
Division rule: \(\mathcal{L}\{\frac{\sin t}t\}=\int_s^\infty\frac{dp}{p^2+1}=\frac\pi2-\tan^{-1}s=\tan^{-1}\frac1s\).
Integral rule: \(\mathcal{L}\{g\}=\frac1s\tan^{-1}\frac1s\). Put \(s=1\): \(\ \tan^{-1}1=\boxed{\pi/4}\).
For \(\mathcal{L}\left\{\int_0^{t/2}\frac{1-e^{-2x}}{x}dx\right\}\): first \(\mathcal{L}\{\frac{1-e^{-2t}}{t}\}=\int_s^\infty\left(\frac1p-\frac1{p+2}\right)dp=\ln\frac{s+2}{s}\).
Integral rule: \(G(s)=\frac1s\ln\left(1+\frac2s\right)\) for \(g(t)=\int_0^t\). Then the time-scaling rule with \(\alpha=\frac12\): \(\mathcal{L}\{g(t/2)\}=2G(2s)=\frac{2}{2s}\ln\left(1+\frac1s\right)=\boxed{\frac{\ln(1+1/s)}{s}}\).
Note on the paperThe identity \(\ln(1+1/s)/s\) only works if the integrand is \(\frac{1-e^{-2x}}{x}\). The scanned paper looks like \(e^{2x}\); with \(e^{+2x}\) you would get \(\ln(1-1/s)/s\). State your reading in the script if this appears.
Examples 1.15 and 1.18
\(\mathcal{L}\{t\sinh at\}=-\frac{d}{ds}\frac{a}{s^2-a^2}=\frac{2as}{(s^2-a^2)^2}\) · \(\mathcal{L}\{t^2\cos at\}=\frac{d^2}{ds^2}\frac{s}{s^2+a^2}=\frac{2s(s^2-3a^2)}{(s^2+a^2)^3}\)
\(\mathcal{L}\{\frac{\sin3t}{t}\}=\tan^{-1}\frac3s\) · \(\mathcal{L}\{e^{5t}\frac{\sin2t}{t}\}=\tan^{-1}\frac{2}{s-5}\)
\(\mathcal{L}\{\frac{\sin3t}{t}\}=\tan^{-1}\frac3s\) · \(\mathcal{L}\{e^{5t}\frac{\sin2t}{t}\}=\tan^{-1}\frac{2}{s-5}\)
1.5 Unit step and the second shifting theorem
\(u(t-a)\) is a switch that turns on at \(t=a\)
Definition 1.19\[u(t)=\begin{cases}0&t<0\\1&t\ge0\end{cases}\]
Theorem 1.21 · t-shifting\[\mathcal{L}\{u(t-a)f(t-a)\}=e^{-as}F(s)\]
Delay a signal: f(t−a)u(t−a) versus the common mistake f(t)u(t−a)
f(t)u(t−a)f(t−a) ✔ shifted copyu(t−a)f(t) ✘ just switched on
Recipe for piecewise functions
Each jump in value adds a step. Example 1.20: \(f=8\) for \(t<2\) then \(6\) gives \(f=8-2u(t-2)\), so \(F=\frac8s-\frac{2e^{-2s}}{s}\).
For \(\mathcal{L}\{u(t-a)g(t)\}\), rewrite \(g(t)\) in terms of \((t-a)\) first. Example 1.23: \(t^2=(t-2)^2+4(t-2)+4\), so \(\mathcal{L}\{u(t-2)t^2\}=e^{-2s}\left(\frac2{s^3}+\frac4{s^2}+\frac4s\right)\).
For \(\mathcal{L}\{u(t-a)g(t)\}\), rewrite \(g(t)\) in terms of \((t-a)\) first. Example 1.23: \(t^2=(t-2)^2+4(t-2)+4\), so \(\mathcal{L}\{u(t-2)t^2\}=e^{-2s}\left(\frac2{s^3}+\frac4{s^2}+\frac4s\right)\).
1.6 Dirac delta: an infinitely tall, infinitely thin spike of area 1
The 2022 paper builds δ as the limit of rectangles δₙ. Watch it happen.
Definition 1.25
\[\int_{-\infty}^{\infty}\delta(x)\,dx=1,\qquad \int_{-\infty}^{\infty}f(x)\delta(x-a)\,dx=f(a)\]
It is a distribution (a generalized function), not an ordinary function.
δₙ(t) = n on [a − 1/2n, a + 1/2n]: area stays 1, the integral of f·δₙ tends to f(a)
δₙ(t)f(t)f·δₙ area
Exam proof (2022 Q1(a)–(b), 12 marks)
\(\delta_n\) is zero outside \([a-\frac1{2n},a+\frac1{2n}]\). So
\[\int_0^\infty f(t)\delta_n(t)\,dt=n\int_{a-1/2n}^{a+1/2n}f(t)\,dt\quad(n\text{ large so the interval is inside }(0,\infty)).\]
Mean value theorem for integrals (\(f\) continuous): there is \(c_n\) in the interval with \(\int_{a-1/2n}^{a+1/2n}f\,dt=f(c_n)\cdot\frac1n\). So the integral equals \(f(c_n)\).
As \(n\to\infty\), the interval shrinks to \(a\), so \(c_n\to a\) and by continuity \(f(c_n)\to f(a)\):
\[\int_0^\infty f(t)\delta(t-a)\,dt=\lim_{n\to\infty}f(c_n)=f(a)\]
Take \(f(t)=e^{-st}\), which is continuous at \(a\):
\[\mathcal{L}\{\delta(t-a)\}=\int_0^\infty e^{-st}\delta(t-a)\,dt=\boxed{e^{-as}}\]
With \(a=0\) this gives \(\mathcal{L}\{\delta(t)\}=1\). Example 1.30: \(\mathcal{L}\{\delta(t-5)e^{-3t}\}=e^{-15}e^{-5s}\).
1.7 Periodic functions
Integrate over one period only, then divide by \(1-e^{-sT}\)
Theorem 1.31\[\mathcal{L}\{f(t)\}=\frac{\int_0^T e^{-st}f(t)\,dt}{1-e^{-sT}}\]
Only the highlighted period matters
1.8 Inverse Laplace transform
Go backwards: partial fractions, complete the square, then read the table
Shift in s\(\mathcal{L}^{-1}\{F(s+a)\}=e^{-at}f(t)\)
Delay\(\mathcal{L}^{-1}\{e^{-as}F(s)\}=u(t-a)f(t-a)\)
Scale\(\mathcal{L}^{-1}\{F(\alpha s)\}=\frac1\alpha f(t/\alpha)\)
Decision flowchart
Is there \(e^{-as}\)?
pull it out, shift at the end
pull it out, shift at the end
→
Denominator factors?
partial fractions
partial fractions
→
Irreducible quadratic?
complete the square \((s-a)^2+b^2\)
complete the square \((s-a)^2+b^2\)
→
Match numerator
to \((s-a)\) and \(b\)
to \((s-a)\) and \(b\)
Worked: Example 1.38 (4) and (6)
\(\frac{1}{s(s+3)}=\frac{A}{s}+\frac{B}{s+3}\). Cover-up: \(A=\frac{1}{0+3}=\frac13\), \(B=\frac{1}{-3}\). So \(f(t)=\frac13\left(1-e^{-3t}\right)\).
\(F=\frac{2(s-1)e^{-4s}}{(s-1)^2+1}\). Ignore \(e^{-4s}\) first: \(\mathcal{L}^{-1}\left\{\frac{2(s-1)}{(s-1)^2+1}\right\}=2e^{t}\cos t\).
Now apply the delay of 4: \(f(t)=2e^{t-4}\cos(t-4)\,u(t-4)\).
\(\frac{1+e^{-\pi s}}{s^2+1}\to\sin t+u(t-\pi)\sin(t-\pi)=\sin t\,(1-u(t-\pi))\). The second term cancels the first after \(t=\pi\).
1.9 Convolution: flip, slide, multiply, add up
\((f*g)(t)=\int_0^t f(\tau)g(t-\tau)\,d\tau\) and \(\mathcal{L}\{f*g\}=F(s)G(s)\)
Convolution animator
f(τ)g(t−τ) flipped and slidoverlap area(f*g)(t) built so far
Remark 1.45\(\mathcal{L}\{f\,g\}\ne\mathcal{L}\{f\}\cdot\mathcal{L}\{g\}\). The product rule belongs to convolution, not ordinary multiplication.
Exercise 1.48 (2)\(\frac{s}{(s^2+a^2)^2}=\frac{s}{s^2+a^2}\cdot\frac1{s^2+a^2}\), so \(f=\cos at * \frac{\sin at}{a}=\frac{1}{a}\int_0^t\cos a\tau\sin a(t-\tau)d\tau=\frac{t\sin at}{2a}\).
Example 1.47: \(\int_0^te^{2(t-\tau)}\tau^3d\tau=\tau^3*e^{2t}\), so its transform is \(\frac{6}{s^4}\cdot\frac{1}{s-2}\).
Example 1.47: \(\int_0^te^{2(t-\tau)}\tau^3d\tau=\tau^3*e^{2t}\), so its transform is \(\frac{6}{s^4}\cdot\frac{1}{s-2}\).
1.10 Solving ODEs: the 4-box pipeline
Figure 1.1 from the notes. Click through the stages.
Differential equation
in t
in t
→ 𝓛
Algebraic equation
in s, ICs included
in s, ICs included
→
Solve for Y(s)
partial fractions
partial fractions
→ 𝓛⁻¹
Solution y(t)
Example 1.52: \(y''+5y'+6y=0,\ y(0)=2,\ y'(0)=3\).
Transform every term: \(\big(s^2Y-2s-3\big)+5\big(sY-2\big)+6Y=0\).
\((s^2+5s+6)Y=2s+13\Rightarrow Y=\dfrac{2s+13}{(s+2)(s+3)}=\dfrac{9}{s+2}-\dfrac{7}{s+3}\).
\(\boxed{y(t)=9e^{-2t}-7e^{-3t}}\). Check: \(y(0)=2\) ✔, \(y'(0)=-18+21=3\) ✔.
Spring–mass hit by a hammer at t = 3 (2022 Q1(c), 6 marks)
\(y''+y=\delta(t-3),\ y(0)=y'(0)=0\). Transform: \(s^2Y+Y=e^{-3s}\).
\(Y=\dfrac{e^{-3s}}{s^2+1}\). Without the exponential this is \(\sin t\).
Delay by 3: \(\boxed{y(t)=\sin(t-3)\,u(t-3)}\). The mass sits still until the hit, then oscillates with amplitude 1.
t =
LR circuit (2021 Q1(c), 5 marks)
\(I'+I=V(t)\), \(I(0)=0\), with \(V(t)=\int_0^te^{2u}\sin3u\,du\).
s-shift then integral rule: \(\mathcal{L}\{e^{2t}\sin3t\}=\frac{3}{(s-2)^2+9}\), so \(\mathcal{L}\{V\}=\dfrac{3}{s\left((s-2)^2+9\right)}\).
\((s+1)I(s)=\mathcal{L}\{V\}\Rightarrow I(s)=\dfrac{3}{s(s+1)(s^2-4s+13)}\).
Partial fractions: \(\ \dfrac{3/13}{s}-\dfrac{1/6}{s+1}+\dfrac{-5(s-2)-3}{78\left((s-2)^2+9\right)}\).
\[\boxed{I(t)=\frac{3}{13}-\frac{1}{6}e^{-t}-\frac{e^{2t}}{78}\left(5\cos3t+\sin3t\right)}\]Check: \(I(0)=\frac{18-13-5}{78}=0\) ✔.
Series RLC circuit lab (Example 1.56)
Kirchhoff gives \(Li'+Ri+\frac1C\int_0^ti\,d\tau=v(t)\). With \(v=\delta(t)\): \(\left(Ls+R+\frac{1}{Cs}\right)I(s)=1\). For \(L=1,R=4,C=\frac13\): \(I=\frac{s}{(s+1)(s+3)}\), so \(i(t)=\frac32e^{-3t}-\frac12e^{-t}\).
Impulse response of a series RLC: change R, L, C
System of ODEs: two masses, three springs (2025 Q1(b), 13 marks)
With \(m=1,k=2\): \(y_1''=-4y_1+2y_2\), \(y_2''=2y_1-4y_2\), \(y_i(0)=0\), \(y_i'(0)=1\).
Transform: \((s^2+4)Y_1-2Y_2=1\) and \(-2Y_1+(s^2+4)Y_2=1\).
Subtract the equations: \((s^2+6)(Y_1-Y_2)=0\Rightarrow Y_1=Y_2\).
Then \((s^2+2)Y_1=1\Rightarrow Y_1=Y_2=\dfrac1{s^2+2}\).
\[\boxed{y_1(t)=y_2(t)=\tfrac{1}{\sqrt2}\sin\sqrt2\,t}\]Both masses move together, so the middle spring never stretches.
Example 1.54 · integro-differential\(y'=1-\sin t-\int_0^ty\,d\tau,\ y(0)=0\Rightarrow sY=\frac1s-\frac{1}{s^2+1}-\frac Ys\), so \(Y=\frac{1}{s^2+1}-\frac{s}{(s^2+1)^2}\) and \(y=\sin t-\frac{t}{2}\sin t\).
1.11 LTI systems and transfer functions
Output = input convolved with the impulse response
x(t)
→
h(t)
impulse response
impulse response
→
y = h * x
X(s)
→
H(s)
transfer function
transfer function
→
Y = H·X
DefinitionsLinear: \(c_1x_1+c_2x_2\mapsto c_1y_1+c_2y_2\). Time invariant: delaying the input delays the output by the same amount. With zero initial conditions, \(H(s)=\dfrac{Y(s)}{X(s)}\).
Input \(x=9u(t)\Rightarrow X=\frac9s\). Output \(y=\sin(9t)u(t)\Rightarrow Y=\frac{9}{s^2+81}\).
\(H(s)=\frac{Y}{X}=\frac{9}{s^2+81}\cdot\frac{s}{9}=\boxed{\frac{s}{s^2+81}}\)
Impulse response \(h(t)=\mathcal{L}^{-1}\{H\}=\boxed{\cos9t}\) for \(t\ge0\).