# 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.