# Laplace Transform

• ## Laplace Transform

• ### Introduction

Consider following equations:

${5t^2 + 6t - 1 = 0 \ and \ 3 \frac {d^2 y}{dx^2} - 2 \frac {dy}{dx} + 9y = x^4 sin (x)}$

Clearly, first equation is algebraic and second equation is differential. It takes a lot of efforts to solve the differential equation, when compared to algebraic equation.

The Laplace transform is a tool, named after French mathematician Pierre-Simon Laplace, which converts a differential equation into an algebraic equation.

Note that the solution will be a function instead of a number.

• ### Definition

Let ${f(t)}$ be a real-valued function, defined for ${t > 0}$. The Laplace transform of ${f(t)}$ is defined as

${\mathcal {L} [f(t)] = \int \limits_0^{\infty} e^{-st} f(t) dt = F(s)}$

Thus, it transforms ${f(t)}$ into another function (of a complex variable) ${F(s)}$.

• ### When Does Laplace Transform Exist?

Note : Important from a math. perspective, can be left out by engineering students. Nonetheless, it’s necessary to know this.

Let there be a constant ${\alpha}$ such that ${e^{- \alpha t} |f(t)|}$ is bounded as ${t \to \infty}$. In other words,

${ \lim \limits_{t \to \infty} e^{- \alpha t} |f(t)| \le M , \ M \ is \ finite}$

Then, ${f(t)}$ is said to be of exponential order ${\alpha}$.

If ${f(t)}$ is piecewise continuous in every finite interval of ${[t, \infty)}$ and is of exponential order ${\alpha}$, then Laplace transform of ${f(t)}$, i.e. ${F(s)}$ exists for all ${s > \alpha}$.

• ### Linearity (Superposition Principle) of Laplace Transform

If ${a}$ and ${b}$ are constants and ${f(t)}$ and ${g(t)}$ are 2 functions with existence of Laplace transforms, then,

${\mathcal {L} [a f(t) + b g(t)] = a \mathcal {L}[f(t)] + b \mathcal {L} [g(t)]}$

• ### Laplace Transforms of Standard Functions

${\mathcal {L} [1] = \frac {1}{s}, \ s>0}$

${\mathcal {L} [e^{at}] = \frac {1}{s-a} , \ s > a}$

${\mathcal {L} [sin (at)] = \frac {a}{s^2 + a^2}, \ s> 0}$

${\mathcal {L} [cos (at)] = \frac {s}{s^2 + a^2}, \ s> 0}$

${\mathcal {L} [sinh (at)] = \frac {a}{s^2 - a^2}, \ s> |a|}$

${\mathcal {L} [cosh (at)] = \frac {s}{s^2 - a^2}, \ s> |a|}$

${\mathcal {L} [t^n], \ t > n -1 \ is \frac {\Gamma (n+1)}{s^{n+1}}, s > 0}$

• ### First Shifting Theorem

If ${\mathcal {L} [f(t)] = F(s)}$, then,

${\mathcal {L} [e^{-at}f(t)] = F(s+a)}$

• ### Second Shifting Theorem

If ${\mathcal {L} [f(t)] = F(s)}$ and ${F(t) = f(t-a), t > a}$ and ${F(t) = 0 , t < a}$, then,

${\mathcal {L} [F(t)] = e^{-as} F(s)}$

• ### Change of Scale Theorem

If ${\mathcal {L} [f(t)] = F(s)}$, then,

${\mathcal {L} [f(at)] = \frac {1}{a}F \big (\frac {s}{a} \big)}$

• ### Laplace Transform of Derivatives of Functions, Roughly corresponds to multiplication

If ${\mathcal {L} [f(t)] = F(s)}$, then,

$latex{\mathcal {L} [f'(t)] =s F(s) – f(0)}&s=2$

${\mathcal {L} [f''(t)] =s^2 F(s) - s f(0) - f'(0)}$

${\mathcal {L} [f'''(t)] =s^3 F(s) - s^2 f(0) - s f'(0) - f''(0)}$

Form an expression for ${\mathcal {L}[f^n(t)]}$.

• ### Laplace Transform of Derivatives of Functions, Roughly corresponds to division

If ${\mathcal {L} [f(t)] = F(s)}$, then,

${\mathcal {L} \Big [ \int \limits_0^{t} f(u)du \Big ] = \frac {1}{s} F(s)}$

${\mathcal {L} \Big [ \int \limits_0^{t} \int \limits_0^{t} \int \limits_0^{t} \cdots \int \limits_0^{t} f(u)du \Big ] = \frac {1}{s^n} F(s)}$

• ### Laplace Transform of ${t^n f(t)}$ and ${\frac {f(t)}{t^n}}$, Roughly corresponds to differentiation and integration

${\mathcal {L} [t^n f(t)] = (-1)^n \frac {d^n}{ds^n} F(s)}$

${\mathcal {L} \Big[ \frac {f(t)}{t^n} \Big] = \int \limits_s^{\infty} \int \limits_s^{\infty} \cdots \int \limits_s^{\infty} F(s) ds^n}$

• ### Convolution

The convolution of 2 functions ${f(t)}$ and ${g(t)}$ is given by

${f(t) * g(t) = \int \limits_0^{t} f(u) g(t-u) du}$

It is commutative, associative and distributive over addition.

Laplace transform of convolution is given by

${\mathcal {L}[f(t) * g(t)] = F(s) G(s)}$

It can be shown that ${1 * 1 * 1 * \cdots * 1 = \frac {t^{n-1}}{n-1}}$.

• ### Initial Value Theorem, ${t \to 0}$

If ${\mathcal {L} [f(t)] = F(s)}$, then,

${\lim \limits_{t \to 0} f(t) = \lim \limits_{s \to \infty} s F(s)}$

• ### Final Value Theorem, ${t \to \infty}$

If ${\mathcal {L} [f(t)] = F(s)}$, then,

${\lim \limits_{t \to \infty} f(t) = \lim \limits_{s \to 0} s F(s)}$

• ## Inverse Laplace Transform

This is an operation, which gets us back to the original function. In other words, if ${F(s)}$ is the Laplace transform of ${f(t)}$, then the inverse Laplace transform is defined as

${\mathcal {L}^{-1} [ F(s)] = f(t)}$

It is a linear transform.

• ### Formulas

Having developed the formulas for Laplace transform, we can easily get the following :

${\mathcal {L}^{-1} \frac {1}{s} = 1}$

${\mathcal {L}^{-1} \frac {1}{s-a} = e^{at}}$

${\mathcal {L}^{-1} \frac {1}{s^2 + a^2} = \frac {1}{a} sin (at)}$

${\mathcal {L}^{-1} \frac {s}{s^2 + a^2} = cos (at)}$

${\mathcal {L}^{-1} \frac {1}{s^2 - a^2} = \frac {1}{a} sinh (at)}$

${\mathcal {L}^{-1} \frac {s}{s^2 - a^2} = cosh (at)}$

${\mathcal {L}^{-1} \frac {1}{s^{n}} = \frac {t^{n-1}}{\Gamma (n)}}$

• ### Theorems

The theorems discussed for Laplace transform are presented here in terms of inverse Laplace transform.

${\mathcal {L}^{-1} [F(s+a)] = e^{-at} f(t)}$

${\mathcal {L}^{-1} [e^{-as} F(s)] = f(t-a), when \ t> a \ and \ 0 \ when \ t < a}$

${\mathcal {L}^{-1} [F(ks)] = \frac {1}{k} f \Big(\frac {t}{k} \Big)}$

Useful when ${F(s)}$ has logarithmic or inverse trigonometric functions :

${\mathcal {L}^{-1} \Big (\frac {d^n}{ds^n} F(s) \Big) = (-1)^n t^n f(t)}$

${\mathcal {L}^{-1} \Big ( \int \limits_{s}^{\infty} \int \limits_{s}^{\infty} \cdots \int \limits_{s}^{\infty} F(s)ds^n \Big) = \frac {f(t)}{t^n}}$

${\mathcal {L}^{-1} [s F(s)] = f'(t), \ when \ f(0)=0}$

Useful when ${F(s)}$ has a power of ${s}$ in the denominator :

${\mathcal {L}^{-1} \Big ( \frac {F(s)}{s} \Big ) = \int \limits_0^t f(t)dt}$

Convolution Theorem :

${\mathcal {L}^{-1} [F(s) G(s)] = f(t) * g(t)}$

• ### Partial Fractions

The method of partial fractions greatly simplifies ${F(s)}$ by splitting the function into fractions, whose denominators are components of the denominator of ${F(s)}$. Getting the coefficients of the fractions can take time, so try if the shifting theorem can be applied.

• ### Solving ODEs using Laplace transform

We solved linear ODEs of higher order in 1st unit. They can also be solved using Laplace transforms. We first take the Laplace transform of the differential equation by using the properties above. This converts the ODE into an algebraic equation. We solve it to get ${F(s)}$. Taking the inverse Laplace transform of ${F(s)}$ gives us the required function or the solution of the ODE.