Unit-2: Laplace Transform - Complete Formulas & Examples
Unit-2: Laplace Transform (Complete Theory + Formulas + Solved Examples)
1. Definition of Laplace Transform
\[ \mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) \, dt \quad (s > \sigma) \]2. Existence Theorem
Laplace transform exists if:- f(t) is piecewise continuous on [0, ∞)
- f(t) is of exponential order: |f(t)| ≤ M e^{αt} for t > T
3. Standard Laplace Transforms (Important Table)
| f(t) | \(\mathcal{L}\{f(t)\}\) |
|---|---|
| 1 | \(\frac{1}{s}\) |
| t | \(\frac{1}{s^2}\) |
| t^n | \(\frac{n!}{s^{n+1}}\) |
| e^{at} | \(\frac{1}{s - a}\) |
| \sin at | \(\frac{a}{s^2 + a^2}\) |
| \cos at | \(\frac{s}{s^2 + a^2}\) |
| \sinh at | \(\frac{a}{s^2 - a^2}\) |
| \cosh at | \(\frac{s}{s^2 - a^2}\) |
| t f(t) | \(-\frac{d}{ds} F(s)\) |
4. Properties of Laplace Transform
| Property | Time Domain | Laplace Domain |
|---|---|---|
| First Shifting | e^{at} f(t) | F(s - a) |
| Second Shifting | f(t - a) u(t - a) | e^{-as} F(s) |
| Multiplication by t | t f(t) | -F'(s) |
| Multiplication by t^n | t^n f(t) | (-1)^n F^{(n)}(s) |
| Division by t | \(\frac{f(t)}{t}\) | \(\int_s^\infty F(s) \, ds\) |
| Differentiation | f'(t) | s F(s) - f(0) |
| f''(t) | s^2 F(s) - s f(0) - f'(0) | |
| f^{(n)}(t) | s^n F(s) - s^{n-1} f(0) - \cdots - f^{(n-1)}(0) | |
| Integration | \(\int_0^t f(\tau) d\tau\) | \(\frac{F(s)}{s}\) |
5. Unit Step Function (Heaviside Function)
\[ u(t - a) = \begin{cases} 0 & t < a \\ 1 & t \geq a \end{cases} \] \[ \mathcal{L}\{u(t - a)\} = \frac{e^{-as}}{s} \] \[ \mathcal{L}\{f(t - a) u(t - a)\} = e^{-as} F(s) \]6. Laplace Transform of Periodic Function
If f(t) has period T, then: \[ \mathcal{L}\{f(t)\} = \frac{1}{1 - e^{-sT}} \int_0^T e^{-st} f(t) \, dt \]7. Inverse Laplace Transform
\[ f(t) = \mathcal{L}^{-1}\{F(s)\} \] Using partial fractions, standard forms, properties.8. Convolution Theorem
\[ \mathcal{L}\{f(t) * g(t)\} = F(s) G(s) \] where \( (f * g)(t) = \int_0^t f(\tau) g(t - \tau) d\tau \)\[ \mathcal{L}^{-1}\{F(s) G(s)\} = f(t) * g(t) \]
Example 1: Solve y'' + 4y = sin 2t, y(0)=1, y'(0)=0
Take LT:
\( s^2 Y - s + 4Y = \frac{2}{s^2 + 4} \)
\( Y(s) = \frac{s}{s^2 + 4} + \frac{2}{(s^2 + 4)^2} \)
\( \mathcal{L}^{-1}\left\{\frac{s}{s^2 + 4}\right\} = \cos 2t \)
\( \mathcal{L}^{-1}\left\{\frac{2}{(s^2 + 4)^2}\right\} = \sin 2t - 2t \cos 2t \) (using known form)
y(t) = \cos 2t + \frac{1}{2} (\sin 2t - 2t \cos 2t)
Take LT:
\( s^2 Y - s + 4Y = \frac{2}{s^2 + 4} \)
\( Y(s) = \frac{s}{s^2 + 4} + \frac{2}{(s^2 + 4)^2} \)
\( \mathcal{L}^{-1}\left\{\frac{s}{s^2 + 4}\right\} = \cos 2t \)
\( \mathcal{L}^{-1}\left\{\frac{2}{(s^2 + 4)^2}\right\} = \sin 2t - 2t \cos 2t \) (using known form)
y(t) = \cos 2t + \frac{1}{2} (\sin 2t - 2t \cos 2t)
Example 2 (With Unit Step): Solve y' + y = f(t), y(0)=0 where f(t) = { 1 (02) }
f(t) = u(t) - u(t-2)
LT: sY + Y = \frac{1 - e^{-2s}}{s}
Y(s) = \frac{1 - e^{-2s}}{s(s+1)} = \frac{1}{s(s+1)} - \frac{e^{-2s}}{s(s+1)}
Partial: \frac{1}{s(s+1)} = 1 - \frac{1}{s+1}
y(t) = (1 - e^{-t}) - (1 - e^{-(t-2)}) u(t-2)
f(t) = u(t) - u(t-2)
LT: sY + Y = \frac{1 - e^{-2s}}{s}
Y(s) = \frac{1 - e^{-2s}}{s(s+1)} = \frac{1}{s(s+1)} - \frac{e^{-2s}}{s(s+1)}
Partial: \frac{1}{s(s+1)} = 1 - \frac{1}{s+1}
y(t) = (1 - e^{-t}) - (1 - e^{-(t-2)}) u(t-2)
Example 3 (Convolution): Solve y' - 2y = sin t, y(0)=0
Y(s)(s - 2) = \frac{1}{s^2 + 1}
Y(s) = \frac{1}{(s-2)(s^2 + 1)}
Using convolution: solution is ∫ e^{2τ} sin(t - τ) dτ from 0 to t
Or partial fractions:
y(t) = \frac{1}{5} (2 \sin t - \cos t + e^{2t})
Y(s)(s - 2) = \frac{1}{s^2 + 1}
Y(s) = \frac{1}{(s-2)(s^2 + 1)}
Using convolution: solution is ∫ e^{2τ} sin(t - τ) dτ from 0 to t
Or partial fractions:
y(t) = \frac{1}{5} (2 \sin t - \cos t + e^{2t})
Example 4 (Simultaneous DEs):
x' = 3x + y
y' = x + 3y, x(0)=4, y(0)=1
LT:
sX - 4 = 3X + Y
sY - 1 = X + 3Y
(s-3)X - Y = 4
-X + (s-3)Y = 1
Solve: X(s) = \frac{4s - 13}{(s-3)^2 - 1} = \frac{4s-13}{s^2 - 6s + 8}
= \frac{4s-13}{(s-2)(s-4)} → partial → X(s) = \frac{5}{s-2} - \frac{1}{s-4}
x(t) = 5e^{2t} - e^{4t}, y(t) = -5e^{2t} + 6e^{4t}
x' = 3x + y
y' = x + 3y, x(0)=4, y(0)=1
LT:
sX - 4 = 3X + Y
sY - 1 = X + 3Y
(s-3)X - Y = 4
-X + (s-3)Y = 1
Solve: X(s) = \frac{4s - 13}{(s-3)^2 - 1} = \frac{4s-13}{s^2 - 6s + 8}
= \frac{4s-13}{(s-2)(s-4)} → partial → X(s) = \frac{5}{s-2} - \frac{1}{s-4}
x(t) = 5e^{2t} - e^{4t}, y(t) = -5e^{2t} + 6e^{4t}
Summary of Key Inverse Transforms
| F(s) | f(t) |
|---|---|
| \(\frac{1}{(s+a)^n}\) | \(\frac{t^{n-1} e^{-at}}{(n-1)!}\) |
| \(\frac{s}{(s+a)^2 + b^2}\) | e^{-at} \cos bt |
| \(\frac{b}{(s+a)^2 + b^2}\) | e^{-at} \sin bt |
| \(\frac{1}{s(s^2 + k^2)}\) | \(\frac{1}{k^2} (1 - \cos kt)\) |
| \(\frac{1}{(s^2 + k^2)^2}\) | \(\frac{\sin kt - kt \cos kt}{2k^3}\) |