Unit 3: Laplace Transform

I. Orientation

The Laplace transform converts a time-domain function into an algebraic function of a complex variable. For a function (f(t)) defined for (t\geq 0), it is defined by

[
\mathcal{L}{f(t)}=F(s)=\int_{0}^{\infty}e^{-st}f(t)\,dt
]

where (t) is time, (s) is the transform variable, (e^{-st}) is the kernel, and (F(s)) is the Laplace transform of (f(t)). This method is especially useful in robotics because motor, actuator, sensor, and mechanical-system models are commonly expressed as linear differential equations with initial conditions.

• Domain: Functions are normally considered for (t\geq 0), representing the response after a system starts.
• Existence: The transform generally exists when (f(t)) is piecewise continuous on every finite interval and of exponential order.
• Exponential order: There are constants (M,a,T>0) such that (|f(t)|\leq Me^{at}) for (t>T).
• Initial conditions: Values such as (f(0)), (f'(0)), and (f''(0)) enter directly when derivatives are transformed.
• Linearity: The transform changes differentiation and integration into algebraic operations involving (s).
• Convergence: The integral converges in a suitable region of the (s)-plane, commonly (\operatorname{Re}(s)>a).

II. Laplace Transforms of Various Standard Functions — Basic Transform Pairs

A standard transform pair provides a direct link between a time-domain function and its (s)-domain representation. These pairs are the main computational foundation of the method.

A. Laplace transforms of various standard functions

The following standard functions occur frequently in control and robotic-system models.

• Constant function: For (f(t)=1),

L{1} = 1/s,       s > 0

• Power function: For (f(t)=t^n), where (n) is a nonnegative integer,

L{t^n} = n! / s^(n+1)

For example, (\mathcal{L}{t^2}=2/s^3).

• Exponential function: For (f(t)=e^{at}), where (a) is a constant,

L{e^(at)} = 1/(s-a)

The transform converges for (s>a) in the real-variable interpretation.

• Trigonometric functions: For angular frequency (\omega),

L{sin(ωt)} = ω/(s^2+ω^2)
L{cos(ωt)} = s/(s^2+ω^2)

Here (\omega) is measured in radians per unit time.

• Hyperbolic functions:

L{sinh(at)} = a/(s^2-a^2)
L{cosh(at)} = s/(s^2-a^2)

• Unit-step function: The unit step (u(t-a)) changes from (0) to (1) at (t=a). Its transform is

L{u(t-a)} = e^(-as)/s

This factor is important for modeling a command or force applied after a delay (a).

• Impulse function: The Dirac impulse (\delta(t)), concentrated at (t=0), satisfies

L{δ(t)} = 1

An impulse input is useful for identifying the dynamic response of a robotic system.

III. Properties of Laplace Transforms — Algebraic Rules

Properties reduce lengthy integrations to short algebraic calculations. They also explain how changes in a signal affect its transform.

A. Properties of Laplace transforms

For functions (f(t)) and (g(t)), with transforms (F(s)) and (G(s)), the main properties are as follows.

• Linearity: For constants (a) and (b),

L{a f(t) + b g(t)} = aF(s) + bG(s)

For example, (\mathcal{L}{3t+2}=3/s^2+2/s).

• First shifting theorem: Multiplication by (e^{at}) shifts (s):

L{e^(at) f(t)} = F(s-a)

Thus, (\mathcal{L}{e^{2t}\sin t}=1/((s-2)^2+1)).

• Second shifting theorem: A delay of (a) time units produces a factor (e^{-as}):

L{u(t-a) f(t-a)} = e^(-as)F(s)

The function (u(t-a)) ensures that the delayed signal is zero before (t=a).

• Change of scale: For (a>0),

L{f(at)} = (1/a) F(s/a)

This relates a time-scaled signal to its transform.

• Multiplication by (t):

L{t f(t)} = -dF(s)/ds

Consequently, (\mathcal{L}{t\sin(at)}) can be found by differentiating (a/(s^2+a^2)) with respect to (s).

• Initial-value theorem:

f(0+) = lim(sF(s)) as s → ∞

This applies when the required limit exists and no impulse occurs at (t=0).

• Final-value theorem:

lim(f(t)) as t → ∞ = lim(sF(s)) as s → 0

This requires that all poles of (sF(s)) lie in the left half of the (s)-plane, except possibly a simple pole at (s=0).

IV. Inverse Laplace Transforms — Returning to the Time Domain

The inverse Laplace transform recovers (f(t)) from its transform (F(s)). It is written as (f(t)=\mathcal{L}^{-1}{F(s)}).

A. Inverse Laplace transforms

The usual procedure is to simplify (F(s)), decompose it into recognizable terms, and use standard transform pairs.

• Partial fractions: Rational functions are split into simpler fractions. For

F(s) = (2s+5)/((s+1)(s+2))

write (F(s)=A/(s+1)+B/(s+2)), then determine (A) and (B).

• Completing the square: A quadratic denominator is arranged in the form

(s-a)^2 + ω^2

This identifies shifted sine and cosine transforms.

• Worked example: Let

F(s) = (s+3)/((s+3)^2+4)

Since (\mathcal{L}{e^{-3t}\cos(2t)}=(s+3)/((s+3)^2+4)),

L^(-1){F(s)} = e^(-3t) cos(2t)

• Repeated poles: Terms such as (1/(s+a)^2) correspond to (te^{-at}), because

L{t e^(-at)} = 1/(s+a)^2

• Convolution: If (F(s)=G(s)H(s)), then

L^(-1){G(s)H(s)}
= ∫[0 to t] g(τ)h(t-τ)dτ

where (g) and (h) are the inverse transforms of (G) and (H). Convolution represents the accumulated effect of an input on a system.

V. Transform of Derivatives and Integrals — Handling Dynamics

The transform of derivatives converts differential equations into algebraic equations. This is the central reason Laplace methods are effective for dynamic systems.

A. Transform of derivatives and integrals

For (f(t)) with initial values (f(0)), (f'(0)), and so on:

• First derivative:

L{f'(t)} = sF(s) - f(0)

• Second derivative:

L{f''(t)} = s^2F(s) - s f(0) - f'(0)

• Nth derivative:

L{f^(n)(t)}
= s^nF(s) - s^(n-1)f(0) - ... - f^(n-1)(0)

The initial conditions appear as subtraction terms.

• Integral from zero to time (t):

L{∫[0 to t] f(τ)dτ} = F(s)/s

Here (\tau) is a dummy integration variable.

• Worked example: For

y'' + 3y' + 2y = 0,
y(0)=1, y'(0)=0

the transformed equation is

(s^2Y-s) + 3(sY-1) + 2Y = 0

Therefore,

Y(s) = (s+3)/(s^2+3s+2)

The initial values have been incorporated before solving for (Y(s)).

VI. Applications to Solution of Ordinary Linear Differential Equations with Constant Coefficients — System Response

Laplace transforms solve initial-value problems by replacing derivatives with powers of (s). The method applies directly to models whose coefficients are constant.

A. Applications to solution of ordinary linear differential equations with constant coefficients

A general linear equation has the form

a_n y^(n) + a_(n-1)y^(n-1) + ... + a_1 y' + a_0 y = r(t)

where (a_0,\ldots,a_n) are constants, (y(t)) is the unknown response, and (r(t)) is the input.

• Transform the equation: Replace (y(t)) by (Y(s)), and transform (r(t)). Initial conditions are inserted using derivative formulas.
• Solve algebraically: Collect all terms containing (Y(s)), giving

Y(s) = R(s) / (a_n s^n + ... + a_1 s + a_0)

with additional numerator terms caused by initial conditions.

• Factor interpretation: The denominator polynomial is the characteristic polynomial. Its roots determine natural modes such as (e^{-2t}) or (e^{-t}\sin t).
• Input and response: Zero-input response comes from initial conditions; zero-state response comes from (r(t)). Their sum is the complete response.
• Robotics application: A mass-damper system,

m x'' + c x' + kx = F(t)

uses displacement (x(t)), mass (m), damping (c), stiffness (k), and force (F(t)). Its transform gives the transfer relation between force and displacement.

• Transfer function: With zero initial conditions,

G(s) = X(s)/F(s) = 1/(m s^2 + c s + k)

This describes the plant dynamics independently of a particular input.

VII. Simultaneous Differential Equations — Coupled System Models

Simultaneous differential equations describe several interacting variables. Laplace transformation converts the coupled differential equations into simultaneous algebraic equations.

A. Simultaneous differential equations

For two unknown functions (x(t)) and (y(t)), a typical system is

x' = a x + b y + f(t)
y' = c x + d y + g(t)

where (a,b,c,d) are constants and (f(t),g(t)) are external inputs.

• Transform each equation: If (X(s)=\mathcal{L}{x(t)}) and (Y(s)=\mathcal{L}{y(t)}),

sX - x(0) = aX + bY + F(s)
sY - y(0) = cX + dY + G(s)

• Arrange algebraically: Move unknown transforms to the left:

(s-a)X - bY = x(0) + F(s)
-cX + (s-d)Y = y(0) + G(s)

• Eliminate or use matrices: The system can be written as

[s-a   -b ] [X] = [x(0)+F(s)]
[-c    s-d] [Y]   [y(0)+G(s)]

Solving this matrix equation gives (X(s)) and (Y(s)).

• Worked example: For (x'=y), (y'=-x), (x(0)=1), (y(0)=0),

sX - 1 = Y
sY = -X

Using (Y=sX-1), substitution gives ((s^2+1)X=s). Hence

X = s/(s^2+1),     Y = 1/(s^2+1)

and therefore (x(t)=\cos t), (y(t)=\sin t).

• Robotic significance: Coupled joint motion, translational-rotational dynamics, and interacting state variables naturally produce simultaneous equations. Laplace methods expose coupling through off-diagonal matrix terms and support systematic state-space analysis.