# Laplace Transforms Calculations Examples with Solutions

Laplace transforms calculations with examples including step by step explanations are presented.

## Definition of Laplace Transform

If $f(t)$ is a one sided function such that $f(t) = 0$ for $t \lt 0$ then the Laplace transform $F(s)$ is defined by the improper integral $\mathscr{L}\{f(t)\} = F(s) = \int_{0}^{+\infty} f(t) e^{-st} dt$ or the more precise definition to accommodate functions such as the delta function $\delta (t)$ as we will see in example 5 below. $\mathscr{L}\{f(t)\} = F(s) = \int_{0^{-}}^{+\infty} f(t) e^{-st} dt$ where $s$ is allowed to be a complex number for which the improper integral above converges.
In what follows, $j$ is the imaginary unit defined by $j = \sqrt{-1}$

Example 1
Find the Laplace transform of function $f(t)$ defined by $f(t) = 1$ Solution to Example 1
Use the definition of the Laplace transform given above
$\displaystyle F(s) = \int_{0}^{+\infty} f(t) e^{-st} dt$
$f(t) = 1$ Over the interval of integration $[0, \infty )$, hence $F(s)$ simplifies to
$\displaystyle F(s) = \int_{0}^{+\infty} e^{-st} dt$
Calculate the above improper integral as follows
$\displaystyle F(S) = \lim_{T \to +\infty} \left[ -\dfrac{1}{s} e^{-st} \right]_{0}^{T}$

$\quad \quad \displaystyle = \lim_{T \to +\infty} - \dfrac{e^{-sT} - e^{0}}{s}$
If the real part of $s$ is greater than zero, $\lim_{T \to +\infty} e^{-sT} = 0$ and therefore the integral converges and $F(S)$ is given by
$F(S) = \dfrac{1}{s}$

Example 2
Find the Laplace transform of function $f(t)$ defined by $f(t) = e^{at}$ Solution to Example 2
Use the definition given above
$\displaystyle F(s) = \int_{0}^{+\infty} e^{at} e^{-st} dt$
Simplify exponents
$\displaystyle \quad \quad = \int_{0}^{+\infty} e^{(a-s)t} dt$
Calculate the above improper integral
$F(S) = \lim_{T \to +\infty} \left[ \dfrac{1}{a - s} e^{(a-s)t} \right]_{0}^{T}$

$\displaystyle \quad \quad = \lim_{T \to +\infty} \dfrac{e^{(a-s)T} - e^{0}}{a-s}$
For the real part of $s$ greater than the real part of $a$ , $\lim_{T \to +\infty} e^{(a-s)T} = 0$ and therefore the integral converges and $F(S)$ is given by
$F(S) = \dfrac{1}{s - a}$

Example 3
Find the Laplace transform of function $f(t)$ defined by $f(t) = \sin(\omega t)$ Solution to Example 3
Use the definition given above
$\displaystyle F(s) = \int_{0}^{+\infty} \sin(\omega t) e^{-st} dt$
Express $\sin(\omega t)$ in terms of exponentials as follows
$\sin(\omega t) = \dfrac{e^{j \omega t } - e^{ - j \omega t }}{2 j}$
Substitute and calculate the integral

$\displaystyle F(s) = \int_{0}^{+\infty} \dfrac{e^{j \omega t } - e^{ - j \omega t }}{2 j} e^{-st} dt$
Split the integrand and rewrite the integral as a sum/difference of integrals
$\quad \quad \displaystyle = \int_{0}^{+\infty} \dfrac{e^{j \omega t} e^{- s t}}{2 j} dt - \int_{0}^{+\infty} \dfrac{ e^{-j \omega t}e^{ - st}}{2 j} dt$
Group the exponents and factor $t$ out
$\quad \quad \displaystyle = \int_{0}^{+\infty} \dfrac{e^{ (j \omega - s) t}}{2 j} dt - \int_{0}^{+\infty} \dfrac{ e^{ -(j \omega + s) t}}{2 j} dt$
Evaluate the integral
$\quad \quad \displaystyle = \lim_{T \to +\infty} \left[ \dfrac{1}{2j( j \omega - s)} e^{(j\omega-s)t} \right]_{0}^{T} - \lim_{T \to +\infty} \left[ \dfrac{1}{-2j( j \omega + s)} e^{ - (j\omega+s)t} \right]_{0}^{T}$

$\quad \quad \displaystyle = \lim_{T \to +\infty} \dfrac{e^{(j\omega-s)T} - e^0}{2j( j \omega - s)} - \lim_{T \to +\infty} \dfrac{e^{ - (j\omega-s)T }- e^0}{-2j( j \omega + s)}$

If the real part of $s$ is greater than zero , $\lim_{T \to +\infty} \dfrac{e^{(j\omega-s)T}}{2j( j \omega - s)} = 0$ and $\lim_{T \to +\infty} \dfrac{e^{ - (j\omega + s)T }}{-2j( j \omega + s)} = 0$ therefore the integral converges and $F(S)$ is given by
$\displaystyle F(s) = - \dfrac {1}{2j( j \omega - s)} - \dfrac {1}{2j( j \omega + s)}$
Set to the common denominator and simplify to obtain

$\displaystyle F(s) = \dfrac{\omega}{\omega^2+s^2}$

Example 4
Find the Laplace transform of function $f(t) = \cosh(\omega t)$.
Solution to Example 4
Use the definition of the Laplace transform
$\displaystyle F(s) = \int_{0}^{+\infty} \cosh(\omega t) e^{-st} dt$
Express $\cosh(\omega t)$ in terms of exponentials as follows
$\cosh(\omega t) = \dfrac{e^{\omega t}+e^{-\omega t}}{2}$
Substitute and calculate the integral

$\displaystyle F(s) = \int_{0}^{+\infty} \dfrac{e^{\omega t } + e^{-\omega t }}{2 } e^{-st} dt$
Split integrand
$\quad \quad \displaystyle = \int_{0}^{+\infty} \dfrac{e^{ \omega t} e^{ - s t }}{2} dt + \int_{0}^{+\infty} \dfrac{ e^{ -\omega t}e^{ - s t}}{2} dt$
Group exponents and factor $t$ out
$\quad \quad \displaystyle = \int_{0}^{+\infty} \dfrac{e^{ (\omega - s)t }}{2} dt + \int_{0}^{+\infty} \dfrac{ e^{ -(\omega + s)t}}{2} dt$
Evaluate the integrals
$\quad \quad \displaystyle = \lim_{T \to +\infty} \left[ \dfrac{1}{2( \omega - s)} e^{(\omega-s)t} \right]_{0}^{T} + \lim_{T \to +\infty} \left[ \dfrac{1}{ - 2( \omega + s)} e^{ - (\omega+s)t} \right]_{0}^{T}$

$\quad \quad \displaystyle = \lim_{T \to +\infty} \dfrac{e^{(\omega-s)T}-e^0}{2( \omega - s)} + \lim_{T \to +\infty} \dfrac{ e^{ - (\omega+s)T} - e^0 }{ - 2( \omega + s)}$

For real part of $s$ greater that $\omega$ , $\lim_{T \to +\infty} e^{(\omega-s)T} = 0$ and $\lim_{T \to +\infty} e^{-(\omega + s)T} = 0$, hence the integral converges and is given by

$\quad \quad \displaystyle F(s) = \dfrac{-1}{2(\omega - s)} + \dfrac{-1}{-2(\omega + s)}$
and simplifies to
$\displaystyle F(s) = \dfrac{s}{s^2 - \omega^2}$

Example 5 Laplace transform of Dirac Delta Functions.
Find the Laplace transform of the delta functions: a) $\delta (t)$ and b) $\delta (t - a) , a \gt 0$
Solution to Example 5
We first recall that that integrals involving delta functions are evaluated as follows
$\displaystyle \int_{A}^{B} f(t) \delta(t - a) dt = \begin{cases} 1 & \text{for} A \lt a \lt B \\ 0 & \text{otherwise} \\ \end{cases}$
a)
To find the Laplace transform of $\delta (t)$, we need to precise definition of the Laplace transform given by
$\displaystyle \mathscr{L}\{\delta(t)\} = \int_{0^{-}}^{+\infty} \delta(t) e^{-st} dt$
The interval of integration starts from $0^{-}$ to accommodate the detla function $\delta(t)$ in the integration as shown above.
Evaluate the above integral
$\displaystyle \mathscr{L}\{\delta(t)\} = \int_{0^{-}}^{+\infty} \delta(t) e^{-st} dt = e^0 = 1$
b)
For the function $\delta (t - a) , a \gt 0$,
Because $a \gt 0$, the definition of the Laplace transform gives
$\displaystyle \mathscr{L}\{\delta(t - a)\} = \int_{0}^{+\infty} \delta(t - a) e^{-st} dt = e^{-as}$

More Formulas and Properties of Laplace Transform are included.