

If f(t) is a function of t, defined for all t > 0, the Laplace transform of f(t), denote by the symbol
L{ f (t) } is defined by:
where s may be real or complex. In circuit applications we assume s = σ + jω.
The operation L{f(t)} transforms a function f(t) in the time domain into a
function F(s) in the complex frequency domain or simply the s domain.

Solved examples:  At 
e^{−at} 
te^{−at} 
sin ωt 
Acosωt 
e^{−at}cosωt 
sin(ωt + θ) 
cosh ωt 
df(t) / dt 
e^{−t}(t^{2} + 12t + 10) 




Example 1a: Obtain the Laplace transform of f(t) = At where A is a constant.


Example 2a: Obtain the Laplace transform of f(t) = e^{−at} where a is a constant.  
Example 3a: Obtain the Laplace transform of f(t) = te^{−at} where a is a constant.


Example 4a: Find the Laplace transform of f(t) = sin(ωt)
Apply integration by parts second time on the cos term on the right side of the integral:


Example 5a: Find the Laplace transform of f(t) = A cos ωt
Apply integration by parts second time on the cos term on the right side of the integral:


Example 6a: Find the Laplace transform of f(t) = e^{−at} cos ωt
The Laplace trensform is defined by the integral:
The integrals on both sides of the equation are the same and can be added.


Example 7a: Find the Laplace transform of f(t) = sin (ωt + θ)
Apply integration by parts second time on the integral at the right side to get:


Example 8a: Find the Laplace transform of f(t) = cosh ωt
The second term reach zero at infinity, the first term is zero only when s>ω


Example 9a: Find the Laplace transform of the derivative of f^{'}(t) = df(t) / dt.


Example 10a: Find the Laplace transform of: f(t) = e^{−t} (t^{2} + 12t + 10)









Laplace transform by partial fraction technique.


After aranging terms we get the following form:
By comparing the coefficients we get the following set of linear equations:
Solving the set of linear equations we get: A = 1, B = 5, C =− 8, D = 0.75


Find the denominator roots (also called poles) and rewrite the polinomials as followes:
s^{3} + 2s^{2} + s + 1 = A(s^{3} + 10s^{2} + 29s + 20) + B(s + 4)(s + 5) + C(s + 1)(s + 5) + D(s + 1)(s + 4)
After arranging the powers of s we get the values:
s^{3} + 2s^{2} + s + 1 = (A)s^{3} + (10A + B + C + D)s^{2} + (29A + 9B + 6C + 5D)s + (20A + 20B + 5C + 4D)
The coefficients of s in the right and left side of the equation must be the same, now we have to solve the system of linear equations of 4 unknowns:
And the solutions of the coefficients are: A = 1 B = 0.0833 C = 11.667 D = − 19.75


