LAPLACE TRANSFORMS LT
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Complex Fourier transform is also called as Bilateral Laplace Transform. This is used to solve
differential equations. Consider an LTI system exited by a complex exponential signal of the form x
t = Gest.
Where s = any complex number = σ + jω ,
σ = real of s, and
ω = imaginary of s
The response of LTI can be obtained by the convolution of input with its impulse response i.e.
∞
y(t) = x(t) × h(t) = ∫−∞ h(τ) x(t − τ)dτ
∞
= ∫−∞ h(τ) Ges(t−τ) dτ
∞
= Gest . ∫−∞ h(τ) e(−sτ) dτ
y(t) = Gest . H(S) = x(t). H(S)
∞
Where H S = Laplace transform of h(τ) = ∫−∞ h(τ)e−sτ dτ
∞
Similarly, Laplace transform of x(t) = X(S) = ∫−∞ x(t)e−st dt . . . . . . (1)
Relation between Laplace and Fourier transforms
∞
Laplace transform of x(t) = X(S) = ∫−∞ x(t)e−st dt
Substitute s= σ + jω in above equation.
∞
→ X(σ + jω) = ∫−∞ x(t)e−(σ+jω)t dt
∞
= ∫−∞ [x(t)e−σt ]e−jωt dt
∴ X(S) = F. T [x(t)e−σt ] . . . . . . (2)
X(S) = X(ω) for s = jω
Inverse Laplace Transform
You know that X(S) = F. T [x(t)e−σt ]
→ x(t)e−σt = F. T −1 [X(S)] = F. T −1 [X(σ + jω)]
∞
= 12 π ∫−∞ X(σ + jω)ejωt dω
1 ∞
x(t) = eσt 2π ∫−∞ X(σ + jω)ejωt dω
1 ∞
= 2π
∫−∞ X(σ + jω)e(σ+jω)t dω . . . . . . (3)
Here, σ + jω =s
jdω = ds → dω = ds/j
∴ ()= 1
( ) st . . . . . . (4)
∞
∴ x(t) = 1
2πj
∫−∞ X(s)est ds . . . . . . (4)
Equations 1 and 4 represent Laplace and Inverse Laplace Transform of a signal xt .
Conditions for Existence of Laplace Transform
Dirichlet's conditions are used to define the existence of Laplace transform. i.e.
The function ft has finite number of maxima and minima.
There must be finite number of discontinuities in the signal ft ,in the given interval of time.
It must be absolutely integrable in the given interval of time. i.e.
∞
∫−∞ | f(t)| dt < ∞
Initial and Final Value Theorems
If the Laplace transform of an unknown function xt is known, then it is possible to determine the
initial and the final values of that unknown signal i.e. xt at t=0+ and t=∞.
Initial Value Theorem
Statement: if xt and its 1st derivative is Laplace transformable, then the initial value of xt is given
by
x(0+ ) = lim SX(S)
s→∞
Final Value Theorem
Statement: if xt and its 1st derivative is Laplace transformable, then the final value of xt is given
by
x(∞) = lim SX(S)
s→∞
