Laplace transforms calculations with examples including step by step explanations are presented.
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
\( \displaystyle \mathscr{L}\{\delta(t - a)\} = \int_{0}^{+\infty} \delta(t - a) e^{-st} dt = e^{-as} \)