The Central Idea
For a function \(f(t)\) defined for \(t\ge0\), its Laplace transform is
\[ \boxed{ \mathcal L\{f(t)\}(s) =F(s) =\int_0^\infty e^{-st}f(t)\,dt }. \]The transform combines \(f(t)\) with the exponential weight \(e^{-st}\). The result is no longer a function of time. It is a function of \(s\), which may be real or complex.
The conceptual move is
\[ \boxed{ f(t) \longrightarrow F(s) }. \]The inverse Laplace transform reverses the process:
\[ \boxed{ \mathcal L^{-1}\{F(s)\}=f(t) }. \]Why the Exponential Weight Matters
The factor \(e^{-st}\) can suppress growth as \(t\) becomes large. If \(s=\sigma+i\omega\), then
\[ e^{-st}=e^{-\sigma t}e^{-i\omega t}. \]The real part \(\sigma\) controls exponential decay or growth in the weighting, while the imaginary part \(\omega\) contributes oscillation. This gives the Laplace transform a connection to Fourier analysis while also allowing it to handle many functions for which an ordinary Fourier transform is inconvenient or does not converge in the same way.
The Transform Variable \(s\)
In applications, the transform variable is commonly written
\[ s=\sigma+i\omega. \]It is therefore useful to think of the Laplace transform as living on the complex \(s\)-plane. The horizontal coordinate represents \(\sigma\), and the vertical coordinate represents \(\omega\).
Much of introductory work treats \(s\) algebraically. Later analysis studies poles, zeros, regions of convergence, stability, and complex inversion more systematically.
When Does the Transform Exist?
The defining integral must converge:
\[ \int_0^\infty e^{-st}f(t)\,dt. \]A common sufficient condition is that \(f\) be piecewise continuous on every finite interval and of exponential order. This means that for sufficiently large \(t\), there are constants \(M>0\) and \(a\) such that
\[ |f(t)|\le M e^{at}. \]Under such conditions, the factor \(e^{-st}\) can dominate the growth when the real part of \(s\) is sufficiently large.
A First Transform: The Constant Function
Let
\[ f(t)=1. \]Then
\[ \mathcal L\{1\} =\int_0^\infty e^{-st}\,dt. \]For \(\operatorname{Re}(s)>0\),
\[ \int_0^\infty e^{-st}\,dt =\left[-\frac1s e^{-st}\right]_0^\infty =\frac1s. \]Therefore,
\[ \boxed{\mathcal L\{1\}=\frac1s}. \]Transform of an Exponential
Let
\[ f(t)=e^{at}. \]Then
\[ \mathcal L\{e^{at}\} =\int_0^\infty e^{-st}e^{at}\,dt =\int_0^\infty e^{-(s-a)t}\,dt. \]When \(\operatorname{Re}(s)>a\),
\[ \boxed{ \mathcal L\{e^{at}\}=\frac1{s-a} }. \]The denominator shows a basic structural rule: multiplication by an exponential in time shifts the transform in the \(s\)-domain.
Transform of a Power
For a nonnegative integer \(n\),
\[ \boxed{ \mathcal L\{t^n\}=\frac{n!}{s^{n+1}} }, \qquad \operatorname{Re}(s)>0. \]For example,
\[ \mathcal L\{t\}=\frac1{s^2}, \qquad \mathcal L\{t^2\}=\frac2{s^3}, \qquad \mathcal L\{t^3\}=\frac6{s^4}. \]These formulas are especially useful when polynomial forcing terms appear in differential equations.
Sine and Cosine
Two essential transform pairs are
\[ \boxed{ \mathcal L\{\cos bt\} =\frac{s}{s^2+b^2} } \]and
\[ \boxed{ \mathcal L\{\sin bt\} =\frac{b}{s^2+b^2} }. \]These formulas connect oscillatory motion in time with rational expressions in \(s\). They appear constantly in vibration, circuits, control systems, and forced differential equations.
Linearity
Like the Fourier transform, the Laplace transform is linear. If \(a\) and \(b\) are constants, then
\[ \boxed{ \mathcal L\{af(t)+bg(t)\} =aF(s)+bG(s) }. \]This allows complicated functions to be transformed term by term whenever the required transforms exist.
For example,
\[ \mathcal L\{3t^2-4\sin 2t\} =3\frac{2}{s^3}-4\frac{2}{s^2+4}. \]Thus,
\[ \boxed{ \mathcal L\{3t^2-4\sin 2t\} =\frac6{s^3}-\frac8{s^2+4} }. \]A Small Transform Table
The following pairs form a basic working toolkit.
- \(1 \longleftrightarrow 1/s\)
- \(t^n \longleftrightarrow n!/s^{n+1}\)
- \(e^{at} \longleftrightarrow 1/(s-a)\)
- \(\sin bt \longleftrightarrow b/(s^2+b^2)\)
- \(\cos bt \longleftrightarrow s/(s^2+b^2)\)
- \(\sinh bt \longleftrightarrow b/(s^2-b^2)\)
- \(\cosh bt \longleftrightarrow s/(s^2-b^2)\)
In practice, transform methods combine this table with algebraic identities, shifting rules, differentiation formulas, and partial fractions.
Inverse Laplace Transforms
Solving a transformed problem eventually requires returning from \(F(s)\) to \(f(t)\). We write
\[ \boxed{ \mathcal L^{-1}\{F(s)\}=f(t) }. \]At an introductory level, inversion is often performed by recognizing a known transform pair and rewriting the expression into a familiar form.
For example,
\[ F(s)=\frac{5}{s^2+25}. \]Since
\[ \mathcal L\{\sin 5t\}=\frac5{s^2+25}, \]we obtain
\[ \boxed{ \mathcal L^{-1}\left\{\frac5{s^2+25}\right\} =\sin 5t }. \]Partial Fractions and Inversion
Rational transform expressions are often inverted by partial fraction decomposition. Consider
\[ F(s)=\frac{5}{s(s+2)}. \]Write
\[ \frac{5}{s(s+2)} =\frac{A}{s}+\frac{B}{s+2}. \]Solving gives
\[ A=\frac52, \qquad B=-\frac52. \]Therefore,
\[ F(s)=\frac52\frac1s-\frac52\frac1{s+2}. \]Taking inverse transforms,
\[ \boxed{ f(t)=\frac52\left(1-e^{-2t}\right) }. \]First Shifting Theorem
If
\[ \mathcal L\{f(t)\}=F(s), \]then multiplication by an exponential shifts the transform:
\[ \boxed{ \mathcal L\{e^{at}f(t)\}=F(s-a) }. \]For example, because
\[ \mathcal L\{\cos bt\}=\frac{s}{s^2+b^2}, \]we have
\[ \boxed{ \mathcal L\{e^{at}\cos bt\} =\frac{s-a}{(s-a)^2+b^2} }. \]Completing the Square for Inverse Transforms
Quadratic denominators often need to be rewritten before a transform pair is visible. Suppose
\[ F(s)=\frac{s+3}{s^2+6s+13}. \]Complete the square:
\[ s^2+6s+13=(s+3)^2+4. \]Hence,
\[ F(s)=\frac{s+3}{(s+3)^2+2^2}. \]Therefore,
\[ \boxed{ \mathcal L^{-1}\{F(s)\} =e^{-3t}\cos 2t }. \]The Transform of a Derivative
The most important operational property for differential equations is
\[ \boxed{ \mathcal L\{f'(t)\} =sF(s)-f(0) }. \]This formula converts differentiation into multiplication by \(s\), while also carrying the initial value \(f(0)\) into the transformed equation.
For the second derivative,
\[ \boxed{ \mathcal L\{f''(t)\} =s^2F(s)-sf(0)-f'(0) }. \]Higher Derivatives
The pattern continues. For the \(n\)th derivative,
\[ \boxed{ \mathcal L\{f^{(n)}(t)\} =s^nF(s) -s^{n-1}f(0) -s^{n-2}f'(0) -\cdots -f^{(n-1)}(0) }. \]This is why the Laplace transform is particularly well matched to initial-value problems. The initial conditions become part of the algebra automatically.
Solving a First-Order Initial-Value Problem
Solve
\[ y'+2y=6, \qquad y(0)=1. \]Let
\[ Y(s)=\mathcal L\{y(t)\}. \]Transform both sides:
\[ sY(s)-1+2Y(s)=\frac6s. \]Therefore,
\[ (s+2)Y(s)=1+\frac6s. \]So
\[ Y(s)=\frac{s+6}{s(s+2)}. \]Decompose:
\[ \frac{s+6}{s(s+2)} =\frac3s-\frac2{s+2}. \]Hence,
\[ \boxed{ y(t)=3-2e^{-2t} }. \]A quick check gives \(y(0)=1\), as required.
Solving a Second-Order Initial-Value Problem
Consider
\[ y''+4y=0, \qquad y(0)=2, \qquad y'(0)=0. \]Transforming gives
\[ s^2Y(s)-2s+4Y(s)=0. \]Thus,
\[ (s^2+4)Y(s)=2s, \]so
\[ Y(s)=\frac{2s}{s^2+4}. \]Recognizing the cosine transform,
\[ \boxed{ y(t)=2\cos 2t }. \]Why Initial Conditions Fit So Naturally
In a direct differential-equation method, one often first finds a family of solutions and then uses initial conditions to determine constants. In the Laplace method, the initial data enter during the transformation step.
The transformed equation is therefore usually an algebraic equation for \(Y(s)\). Once \(Y(s)\) is isolated and inverted, the resulting function is already tailored to the specified initial conditions.
Multiplication by \(t\)
If
\[ \mathcal L\{f(t)\}=F(s), \]then
\[ \boxed{ \mathcal L\{t f(t)\} =-F'(s) }. \]More generally,
\[ \boxed{ \mathcal L\{t^n f(t)\} =(-1)^n F^{(n)}(s) }. \]This property provides another way to generate transform pairs from known ones.
Example: Using Differentiation in the \(s\)-Domain
We know
\[ \mathcal L\{\sin bt\}=\frac{b}{s^2+b^2}. \]Therefore,
\[ \mathcal L\{t\sin bt\} =-\frac{d}{ds}\left(\frac{b}{s^2+b^2}\right). \]Differentiating,
\[ \boxed{ \mathcal L\{t\sin bt\} =\frac{2bs}{(s^2+b^2)^2} }. \]Integration in Time
Define
\[ g(t)=\int_0^t f(\tau)\,d\tau. \]Then
\[ \boxed{ \mathcal L\left\{\int_0^t f(\tau)\,d\tau\right\} =\frac{F(s)}{s} }. \]Thus integration in time corresponds to division by \(s\) in the transform domain.
Unit Step Functions
A function that changes formula at a particular time can be represented with a unit step function. For a shift time \(a\), define
\[ u(t-a)= \begin{cases} 0, & tSecond Shifting Theorem
If
\[ \mathcal L\{f(t)\}=F(s), \]then a delayed copy satisfies
\[ \boxed{ \mathcal L\{u(t-a)f(t-a)\} =e^{-as}F(s) }. \]The factor \(e^{-as}\) therefore represents a delay of \(a\) units in time. This is one of the most useful transform-domain signatures in systems and signal analysis.
Example: A Delayed Input
Consider
\[ g(t)=u(t-3)(t-3). \]Since
\[ \mathcal L\{t\}=\frac1{s^2}, \]the second shifting theorem gives
\[ \boxed{ \mathcal L\{u(t-3)(t-3)\} =\frac{e^{-3s}}{s^2} }. \]The Dirac Impulse
In idealized models, an instantaneous input at time \(a\) is represented by the Dirac delta \(\delta(t-a)\). It is not an ordinary function in the usual elementary sense, but it is an important generalized object in transform and systems theory.
Its Laplace transform is
\[ \boxed{ \mathcal L\{\delta(t-a)\}=e^{-as} }. \]Thus a pure time delay and an exponential factor in the \(s\)-domain are again closely linked.
Convolution
For functions defined on \(t\ge0\), their convolution is
\[ \boxed{ (f*g)(t) =\int_0^t f(\tau)g(t-\tau)\,d\tau }. \]The Laplace transform converts convolution into multiplication:
\[ \boxed{ \mathcal L\{f*g\} =F(s)G(s) }. \]Equivalently,
\[ \boxed{ \mathcal L^{-1}\{F(s)G(s)\} =(f*g)(t) }. \]This property is fundamental in linear systems. It will be developed in much greater depth in the next lesson.
Example: Inversion by Convolution
Suppose
\[ F(s)=\frac{1}{s(s^2+1)}. \]Write
\[ F(s)=\frac1s\cdot\frac1{s^2+1}. \]Since
\[ \mathcal L^{-1}\left\{\frac1s\right\}=1, \qquad \mathcal L^{-1}\left\{\frac1{s^2+1}\right\}=\sin t, \]the inverse transform is the convolution
\[ f(t)=\int_0^t 1\cdot\sin(t-\tau)\,d\tau. \]Evaluating,
\[ \boxed{ f(t)=1-\cos t }. \]Transfer Functions
In a linear time-invariant system with zero initial conditions, a transformed input \(X(s)\) and output \(Y(s)\) are often related by
\[ \boxed{ Y(s)=H(s)X(s) }. \]The function
\[ \boxed{ H(s)=\frac{Y(s)}{X(s)} } \]is called the transfer function. It describes how the system modifies inputs in the transform domain.
Poles and zeros of \(H(s)\) encode important information about system behavior, resonance, transients, and stability.
Example: A First-Order System
Consider the zero-initial-condition model
\[ y'+ay=bx. \]Transforming gives
\[ sY(s)+aY(s)=bX(s). \]Thus,
\[ Y(s)=\frac{b}{s+a}X(s), \]so the transfer function is
\[ \boxed{ H(s)=\frac{b}{s+a} }. \]Poles
A pole is a value of \(s\) at which a transform expression becomes unbounded because its denominator vanishes without cancellation.
For
\[ F(s)=\frac1{s+3}, \]there is a pole at
\[ \boxed{s=-3}. \]This corresponds to the time-domain function \(e^{-3t}\). In many system models, the locations of poles are closely related to exponential modes in the time response.
Zeros
A zero is a value of \(s\) at which the numerator of a transform or transfer function vanishes without cancellation.
For example,
\[ H(s)=\frac{s+1}{(s+2)(s+4)} \]has a zero at
\[ \boxed{s=-1} \]and poles at
\[ \boxed{s=-2,\,-4}. \]Pole-zero structure becomes especially important in control, filtering, and spectral system analysis.
Region of Convergence
A transform formula is incomplete if we ignore where the defining integral converges. The region of convergence, or ROC, is the set of complex \(s\)-values for which the Laplace integral exists.
For the one-sided transform
\[ \mathcal L\{e^{at}\}=\frac1{s-a}, \]the integral converges when
\[ \boxed{ \operatorname{Re}(s)>a }. \]Introductory transform tables often suppress the ROC for convenience, but in more advanced analysis it carries essential information.
One-Sided and Two-Sided Laplace Transforms
This lesson uses the one-sided Laplace transform
\[ \int_0^\infty e^{-st}f(t)\,dt, \]which is especially suited to initial-value problems beginning at \(t=0\).
A two-sided transform instead uses
\[ \int_{-\infty}^{\infty}e^{-st}f(t)\,dt. \]The two-sided form is useful in broader signal and systems theory because it treats negative and positive time together. Its region of convergence must be tracked carefully.
Connection with the Fourier Transform
Write
\[ s=\sigma+i\omega. \]Then the Laplace kernel becomes
\[ e^{-st}=e^{-\sigma t}e^{-i\omega t}. \]The factor \(e^{-i\omega t}\) is the Fourier oscillation, while \(e^{-\sigma t}\) adds exponential weighting. In settings where the imaginary axis lies inside the Laplace transform's region of convergence, evaluating on \(s=i\omega\) connects the Laplace transform to a Fourier transform.
This relationship explains why the two transforms share many structural properties while serving somewhat different purposes.
Initial-Value and Frequency-Domain Viewpoints
Fourier methods emphasize how a signal is distributed among oscillatory frequencies. Laplace methods add a second dimension: exponential growth and decay. This makes the \(s\)-domain particularly effective for systems whose behavior includes both oscillation and transient response.
The two viewpoints are complementary rather than competing.
A Reliable Procedure for Solving Initial-Value Problems
- Write the differential equation and initial conditions clearly.
- Let \(Y(s)=\mathcal L\{y(t)\}\).
- Transform every term.
- Insert the initial conditions immediately.
- Solve algebraically for \(Y(s)\).
- Simplify \(Y(s)\), often using factoring or partial fractions.
- Take the inverse Laplace transform.
- Check the initial conditions.
- Substitute into the original differential equation when practical.
Example: Forced Second-Order System
Solve
\[ y''+y=1, \qquad y(0)=0, \qquad y'(0)=0. \]Transform:
\[ s^2Y(s)+Y(s)=\frac1s. \]Therefore,
\[ Y(s)=\frac1{s(s^2+1)}. \]Use partial fractions:
\[ \frac1{s(s^2+1)} =\frac1s-\frac{s}{s^2+1}. \]Hence,
\[ \boxed{ y(t)=1-\cos t }. \]Example: Exponential Forcing
Solve
\[ y'+y=e^{2t}, \qquad y(0)=0. \]Transform:
\[ sY(s)+Y(s)=\frac1{s-2}. \]Thus,
\[ Y(s)=\frac1{(s+1)(s-2)}. \]Partial fractions give
\[ \frac1{(s+1)(s-2)} =-\frac1{3(s+1)}+\frac1{3(s-2)}. \]Therefore,
\[ \boxed{ y(t)=\frac13e^{2t}-\frac13e^{-t} }. \]Example: Delayed Forcing
Suppose an input is zero until time \(t=4\) and then becomes a constant of magnitude \(3\):
\[ f(t)=3u(t-4). \]Since \(\mathcal L\{1\}=1/s\), the delayed form gives
\[ \boxed{ F(s)=\frac{3e^{-4s}}{s} }. \]The factor \(e^{-4s}\) records the four-unit delay directly in the transform domain.
Checking a Laplace-Transform Solution
Transform methods involve several stages, so checking is important. For an initial-value problem, verify at least two things:
- the proposed solution satisfies the initial conditions;
- substitution into the differential equation reproduces the required forcing.
A transform-domain check can also be useful: re-transform the final answer and compare it with the derived expression for \(Y(s)\).
Common Error: Forgetting Initial Terms
It is incorrect to write
\[ \mathcal L\{y'\}=sY(s) \]unless \(y(0)=0\). In general,
\[ \boxed{ \mathcal L\{y'\}=sY(s)-y(0) }. \]Likewise, a second derivative includes both \(y(0)\) and \(y'(0)\).
Common Error: Confusing the Two Shift Rules
The first shifting theorem concerns multiplication by an exponential:
\[ e^{at}f(t) \longleftrightarrow F(s-a). \]The second shifting theorem concerns a delay in time:
\[ u(t-a)f(t-a) \longleftrightarrow e^{-as}F(s). \]A shift in the \(s\)-variable and a shift in time are different operations and should not be interchanged.
Common Error: Ignoring Algebra Before Inversion
An inverse transform is often not visible until the expression is simplified. Useful algebraic steps include
- factoring denominators;
- partial fraction decomposition;
- completing the square;
- splitting numerators;
- matching shifted transform forms.
Transform methods are therefore partly an exercise in recognizing algebraic structure.
Common Error: Treating \(s\) as Time
After transformation, \(s\) is the independent variable and \(t\) has been integrated out. Expressions such as \(Y(s)\) belong to the transform domain.
Only after applying the inverse transform do we return to a function of time.
Common Error: Ignoring the Region of Convergence
A symbolic expression such as
\[ \frac1{s-a} \]is not the whole story. For the one-sided transform of \(e^{at}\), convergence requires
\[ \operatorname{Re}(s)>a. \]In elementary calculations the ROC is often implicit, but in advanced work it must be stated and analyzed explicitly.
Common Error: Cancelling Meaningful Structure Too Early
Algebraic cancellation can sometimes hide important information about poles, zeros, or system realizations. In basic inverse-transform calculations, simplification is usually helpful, but in systems analysis it can matter whether a factor represents a physical mode before cancellation.
The appropriate level of simplification depends on the question being asked.
Laplace Transforms and Piecewise Models
One practical advantage of the Laplace transform is that piecewise inputs can be encoded with unit steps. Instead of solving a differential equation separately on many intervals and matching solutions manually, the forcing can often be written as one step-function expression and transformed at once.
This is particularly useful for switched circuits, delayed controls, impacts, staged forcing, and finite-duration pulses.
Finite-Duration Pulses
A pulse of height \(A\) that turns on at \(t=a\) and turns off at \(t=b\) can be written
\[ f(t)=A\bigl[u(t-a)-u(t-b)\bigr]. \]Its Laplace transform is
\[ \boxed{ F(s)=\frac{A}{s}\left(e^{-as}-e^{-bs}\right) }. \]This gives a compact transform-domain representation of an input active only during a finite time interval.
Laplace Transforms in Engineering and Applied Mathematics
Laplace methods appear whenever linear differential systems, transients, and initial conditions are central. Common applications include
- electrical circuits;
- mechanical vibration;
- control systems;
- signal processing;
- heat and diffusion models;
- population and compartment models;
- chemical and process systems;
- probability and renewal models;
- integral equations.
The method is especially valuable when the model contains switching, impulses, or piecewise forcing that would be cumbersome to manage entirely in the time domain.
Laplace Transforms as an Operator Method
It is useful to view the Laplace transform as changing the representation of a problem. In the time domain, differentiation is an operator. In the transform domain, that operator becomes multiplication by powers of \(s\), together with initial-value corrections.
This is a recurring theme throughout mathematics: a difficult operation in one representation can become a simpler operation in another.
Practice 1: Basic Transform
Find
\[ \mathcal L\{4t^3\}. \]Practice 2: Exponential Shift
Find
\[ \mathcal L\{e^{2t}\sin 3t\}. \]Practice 3: Inverse Transform
Find
\[ \mathcal L^{-1}\left\{\frac{s}{s^2+16}\right\}. \]Practice 4: Partial Fractions
Find
\[ \mathcal L^{-1}\left\{\frac{6}{s(s+3)}\right\}. \]Practice 5: Transform of a Derivative
If \(Y(s)=\mathcal L\{y(t)\}\), \(y(0)=2\), and \(y'(0)=-1\), write
\[ \mathcal L\{y''(t)\} \]in terms of \(Y(s)\).
Practice 6: First-Order Initial-Value Problem
Solve
\[ y'+3y=0, \qquad y(0)=5. \]Practice 7: Second-Order Initial-Value Problem
Solve
\[ y''+9y=0, \qquad y(0)=0, \qquad y'(0)=6. \]Practice 8: Delayed Function
Find the Laplace transform of
\[ u(t-2)(t-2)^2. \]Practice 9: Transfer Function
For the zero-initial-condition system
\[ y''+4y'+5y=2x, \]find the transfer function
\[ H(s)=\frac{Y(s)}{X(s)}. \]Practice 10: Conceptual Check
Explain why the factor \(e^{-as}\) in a Laplace transform is naturally associated with a delay of \(a\) units in time.
Practice Answers
Practice 1
Since
\[ \mathcal L\{t^3\}=\frac{3!}{s^4}=\frac6{s^4}, \]linearity gives
\[ \boxed{ \mathcal L\{4t^3\}=\frac{24}{s^4} }. \]Practice 2
Start with
\[ \mathcal L\{\sin 3t\}=\frac3{s^2+9}. \]Replace \(s\) by \(s-2\):
\[ \boxed{ \mathcal L\{e^{2t}\sin 3t\} =\frac3{(s-2)^2+9} }. \]Practice 3
Since
\[ \mathcal L\{\cos 4t\}=\frac{s}{s^2+16}, \]we have
\[ \boxed{\cos 4t}. \]Practice 4
Decompose:
\[ \frac6{s(s+3)} =\frac2s-\frac2{s+3}. \]Therefore,
\[ \boxed{ 2-2e^{-3t} }. \]Practice 5
Use
\[ \mathcal L\{y''\} =s^2Y(s)-sy(0)-y'(0). \]Substituting \(y(0)=2\) and \(y'(0)=-1\),
\[ \boxed{ \mathcal L\{y''\}=s^2Y(s)-2s+1 }. \]Practice 6
Transforming,
\[ sY(s)-5+3Y(s)=0. \]Hence,
\[ Y(s)=\frac5{s+3}, \]so
\[ \boxed{ y(t)=5e^{-3t} }. \]Practice 7
Transform:
\[ s^2Y(s)-6+9Y(s)=0. \]Therefore,
\[ Y(s)=\frac6{s^2+9} =2\frac3{s^2+9}. \]Hence,
\[ \boxed{ y(t)=2\sin 3t }. \]Practice 8
Since
\[ \mathcal L\{t^2\}=\frac2{s^3}, \]the delay by \(2\) gives
\[ \boxed{ \mathcal L\{u(t-2)(t-2)^2\} =\frac{2e^{-2s}}{s^3} }. \]Practice 9
With zero initial conditions,
\[ (s^2+4s+5)Y(s)=2X(s). \]Thus,
\[ \boxed{ H(s)=\frac{2}{s^2+4s+5} }. \]Practice 10
The second shifting theorem states
\[ u(t-a)f(t-a) \longleftrightarrow e^{-as}F(s). \]Multiplication by \(e^{-as}\) therefore marks that the original time-domain behavior has been postponed until \(t=a\). The transform factor records the delay algebraically without changing the underlying shape \(f\).
What to Remember
The one-sided Laplace transform is
\[ \boxed{ \mathcal L\{f(t)\}(s) =\int_0^\infty e^{-st}f(t)\,dt }. \]- the transform converts a time-domain function into an \(s\)-domain function;
- linearity allows sums to be transformed term by term;
- exponential multiplication shifts the transform in \(s\);
- time delays produce factors of \(e^{-as}\);
- differentiation becomes multiplication by \(s\) plus initial-value terms;
- integration in time corresponds to division by \(s\);
- partial fractions are a major tool for inverse transforms;
- convolution in time becomes multiplication in the transform domain;
- transfer functions describe linear system behavior algebraically;
- poles, zeros, and regions of convergence add structural information;
- Laplace transforms are especially effective for initial-value problems and switched inputs.
The central working habit is to move deliberately between representations: transform the problem, simplify it in the \(s\)-domain, and then invert while checking that the resulting time-domain function satisfies the original model.
Where We Go Next
One of the most important properties of both Fourier and Laplace transforms is that convolution in the original domain becomes multiplication in the transform domain. That fact is more than a computational shortcut: it is the mathematical foundation of how linear systems combine inputs with system responses.
The next lesson, Convolution, develops this operation directly. It will examine convolution integrals, discrete convolution, impulse responses, commutativity and associativity, support, smoothing, filtering, and the convolution theorem across Fourier, Laplace, and computational settings.