7.1 Laplace Transform

Detailed Theory: Laplace Transform

1. Introduction to Laplace Transform

1.1 What is Laplace Transform?

The Laplace transform is an integral transform that converts a function of time (t) into a function of complex frequency (s). It's a powerful tool for solving differential equations and analyzing linear systems.

Key Idea: Transforms time-domain problems into simpler algebraic problems in s-domain.

1.2 Why Use Laplace Transform?

1. Converts differential equations to algebraic equations

2. Handles initial conditions automatically

3. Useful for solving linear ordinary differential equations (ODEs)

4. Essential in control theory, signal processing, and circuit analysis

5. Can handle discontinuous functions (like step functions)

1.3 Basic Definition

The Laplace transform of a function $f(t)$, defined for $t \geq 0$, is:

$F(s) = \mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \, dt$

where:

• $f(t)$ is the time-domain function

• $F(s)$ is the Laplace transform (s-domain function)

• $s = \sigma + i\omega$ is a complex frequency variable

• The integral converges for certain values of $s$

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2. Existence Conditions and Region of Convergence

2.1 Conditions for Existence

For the Laplace transform to exist, the function $f(t)$ must satisfy:

1. Piecewise continuous on every finite interval $[0, T]$

2. Of exponential order as $t \to \infty$

2.2 Exponential Order

A function $f(t)$ is of exponential order if there exist constants $M > 0$, $c > 0$, and $T > 0$ such that:

$|f(t)| \leq Me^{ct} \quad \text{for all } t \geq T$

This means $f(t)$ doesn't grow faster than an exponential function.

2.3 Region of Convergence (ROC)

The set of complex numbers $s$ for which the Laplace integral converges.

If $f(t)$ is of exponential order with constant $c$, then the Laplace transform converges for $\text{Re}(s) > c$.

Example: For $f(t) = e^{at}$, ROC is $\text{Re}(s) > a$

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3. Basic Laplace Transforms

3.1 Elementary Functions

a) Unit Step Function (Heaviside Function)

$u(t) = \begin{cases} 0 & t < 0 \\ 1 & t \geq 0 \end{cases}$

$\mathcal{L}\{u(t)\} = \frac{1}{s}, \quad \text{Re}(s) > 0$

b) Exponential Function

$\mathcal{L}\{e^{at}\} = \frac{1}{s-a}, \quad \text{Re}(s) > a$

Special cases:

• $\mathcal{L}\{1\} = \mathcal{L}\{e^{0t}\} = \frac{1}{s}, \quad \text{Re}(s) > 0$

• $\mathcal{L}\{e^{-at}\} = \frac{1}{s+a}, \quad \text{Re}(s) > -a$

c) Sine and Cosine Functions

$\mathcal{L}\{\sin(\omega t)\} = \frac{\omega}{s^2 + \omega^2}, \quad \text{Re}(s) > 0$

$\mathcal{L}\{\cos(\omega t)\} = \frac{s}{s^2 + \omega^2}, \quad \text{Re}(s) > 0$

d) Hyperbolic Sine and Cosine

$\mathcal{L}\{\sinh(at)\} = \frac{a}{s^2 - a^2}, \quad \text{Re}(s) > |a|$

$\mathcal{L}\{\cosh(at)\} = \frac{s}{s^2 - a^2}, \quad \text{Re}(s) > |a|$

e) Power Functions (tⁿ)

For $n = 0, 1, 2, \ldots$:

$\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}, \quad \text{Re}(s) > 0$

Special cases:

• $\mathcal{L}\{t^0\} = \mathcal{L}\{1\} = \frac{1}{s}$

• $\mathcal{L}\{t\} = \frac{1}{s^2}$

• $\mathcal{L}\{t^2\} = \frac{2}{s^3}$

• $\mathcal{L}\{t^3\} = \frac{6}{s^4}$

3.2 Basic Laplace Transform Table

Here's a useful reference table:

f(t) (t ≥ 0)	F(s) = 𝓛{f(t)}	ROC
1 (unit step)	$\frac{1}{s}$	$\text{Re}(s) > 0$
$e^{at}$	$\frac{1}{s-a}$	$\text{Re}(s) > a$
$t^n$ (n = 0,1,2,...)	$\frac{n!}{s^{n+1}}$	$\text{Re}(s) > 0$
$\sin(\omega t)$	$\frac{\omega}{s^2 + \omega^2}$	$\text{Re}(s) > 0$
$\cos(\omega t)$	$\frac{s}{s^2 + \omega^2}$	$\text{Re}(s) > 0$
$\sinh(at)$	$\frac{a}{s^2 - a^2}$	$$\text{Re}(s) >
$\cosh(at)$	$\frac{s}{s^2 - a^2}$	$$\text{Re}(s) >
$t^n e^{at}$	$\frac{n!}{(s-a)^{n+1}}$	$\text{Re}(s) > a$
$e^{at} \sin(\omega t)$	$\frac{\omega}{(s-a)^2 + \omega^2}$	$\text{Re}(s) > a$
$e^{at} \cos(\omega t)$	$\frac{s-a}{(s-a)^2 + \omega^2}$	$\text{Re}(s) > a$

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4. Properties of Laplace Transform

4.1 Linearity Property

The Laplace transform is linear:

$\mathcal{L}\{a f(t) + b g(t)\} = a F(s) + b G(s)$

where $a$ and $b$ are constants.

Proof:

$\mathcal{L}\{a f(t) + b g(t)\} = \int_{0}^{\infty} e^{-st}[a f(t) + b g(t)] \, dt$

$= a \int_{0}^{\infty} e^{-st} f(t) \, dt + b \int_{0}^{\infty} e^{-st} g(t) \, dt$

$= a F(s) + b G(s)$

Example: Find $\mathcal{L}\{3t^2 + 2\sin(4t) - 5e^{-t}\}$

$= 3\mathcal{L}\{t^2\} + 2\mathcal{L}\{\sin(4t)\} - 5\mathcal{L}\{e^{-t}\}$

$= 3 \times \frac{2}{s^3} + 2 \times \frac{4}{s^2 + 16} - 5 \times \frac{1}{s+1}$

$= \frac{6}{s^3} + \frac{8}{s^2 + 16} - \frac{5}{s+1}$

4.2 First Shifting Theorem (s-Shift)

If $\mathcal{L}\{f(t)\} = F(s)$, then:

$\mathcal{L}\{e^{at} f(t)\} = F(s-a)$

Proof:

$\mathcal{L}\{e^{at} f(t)\} = \int_{0}^{\infty} e^{-st} e^{at} f(t) \, dt$

$= \int_{0}^{\infty} e^{-(s-a)t} f(t) \, dt = F(s-a)$

Example 1: Find $\mathcal{L}\{e^{2t} \sin(3t)\}$

We know: $\mathcal{L}\{\sin(3t)\} = \frac{3}{s^2 + 9}$

Using s-shift with $a=2$: $\mathcal{L}\{e^{2t} \sin(3t)\} = \frac{3}{(s-2)^2 + 9}$

Example 2: Find $\mathcal{L}\{e^{-t} t^3\}$

We know: $\mathcal{L}\{t^3\} = \frac{3!}{s^4} = \frac{6}{s^4}$

Using s-shift with $a=-1$: $\mathcal{L}\{e^{-t} t^3\} = \frac{6}{(s+1)^4}$

4.3 Second Shifting Theorem (t-Shift)

If $\mathcal{L}\{f(t)\} = F(s)$, then:

$\mathcal{L}\{f(t-a) u(t-a)\} = e^{-as} F(s)$

where $u(t-a)$ is the unit step function:

$u(t-a) = \begin{cases} 0 & t < a \\ 1 & t \geq a \end{cases}$

This is also called the Time Delay property.

Example: Find $\mathcal{L}\{(t-2)^3 u(t-2)\}$

We know: $\mathcal{L}\{t^3\} = \frac{6}{s^4}$

Using t-shift with $a=2$: $\mathcal{L}\{(t-2)^3 u(t-2)\} = e^{-2s} \times \frac{6}{s^4} = \frac{6e^{-2s}}{s^4}$

4.4 Change of Scale Property

If $\mathcal{L}\{f(t)\} = F(s)$, then:

$\mathcal{L}\{f(at)\} = \frac{1}{a} F\left(\frac{s}{a}\right), \quad a > 0$

Example: Find $\mathcal{L}\{\sin(4t)\}$ using $\mathcal{L}\{\sin(t)\} = \frac{1}{s^2 + 1}$

$\mathcal{L}\{\sin(4t)\} = \frac{1}{4} \times \frac{1}{(s/4)^2 + 1} = \frac{1}{4} \times \frac{1}{s^2/16 + 1}$

$= \frac{1}{4} \times \frac{16}{s^2 + 16} = \frac{4}{s^2 + 16}$

4.5 Differentiation in Time Domain

a) First Derivative

If $\mathcal{L}\{f(t)\} = F(s)$, then:

$\mathcal{L}\{f'(t)\} = sF(s) - f(0)$

b) Second Derivative

$\mathcal{L}\{f''(t)\} = s^2 F(s) - sf(0) - f'(0)$

c) n-th Derivative

$\mathcal{L}\{f^{(n)}(t)\} = s^n F(s) - s^{n-1} f(0) - s^{n-2} f'(0) - \cdots - f^{(n-1)}(0)$

Important: These formulas incorporate initial conditions automatically!

Example: Find $\mathcal{L}\{f'(t)\}$ if $f(t) = t^2$ and $f(0)=1$

$\mathcal{L}\{t^2\} = \frac{2}{s^3}$

$f'(t) = 2t$

$\mathcal{L}\{f'(t)\} = \mathcal{L}\{2t\} = 2 \times \frac{1}{s^2} = \frac{2}{s^2}$

Using formula: $\mathcal{L}\{f'(t)\} = sF(s) - f(0) = s \times \frac{2}{s^3} - 1 = \frac{2}{s^2} - 1$

Wait, we get $\frac{2}{s^2} - 1$ vs $\frac{2}{s^2}$. This is because $f(0)=1$, but for $f(t)=t^2$, $f(0)=0$, not 1. Let me correct:

For $f(t)=t^2$, $f(0)=0^2=0$, so:

$\mathcal{L}\{f'(t)\} = s \times \frac{2}{s^3} - 0 = \frac{2}{s^2}$

Correct.

4.6 Integration in Time Domain

If $\mathcal{L}\{f(t)\} = F(s)$, then:

$\mathcal{L}\left\{\int_{0}^{t} f(\tau) \, d\tau\right\} = \frac{F(s)}{s}$

Example: Find $\mathcal{L}\left\{\int_{0}^{t} \sin(2\tau) \, d\tau\right\}$

First, $\int_{0}^{t} \sin(2\tau) \, d\tau = \left[-\frac{1}{2}\cos(2\tau)\right]_{0}^{t} = -\frac{1}{2}\cos(2t) + \frac{1}{2}$

Using formula: $\mathcal{L}\{\sin(2t)\} = \frac{2}{s^2 + 4}$

So $\mathcal{L}\left\{\int_{0}^{t} \sin(2\tau) \, d\tau\right\} = \frac{1}{s} \times \frac{2}{s^2 + 4} = \frac{2}{s(s^2 + 4)}$

4.7 Differentiation in s-Domain

If $\mathcal{L}\{f(t)\} = F(s)$, then:

$\mathcal{L}\{t^n f(t)\} = (-1)^n \frac{d^n}{ds^n} F(s)$

Special case (n=1): $\mathcal{L}\{t f(t)\} = -\frac{d}{ds} F(s)$

Example: Find $\mathcal{L}\{t \sin(3t)\}$

We know: $\mathcal{L}\{\sin(3t)\} = \frac{3}{s^2 + 9} = F(s)$

Then: $\mathcal{L}\{t \sin(3t)\} = -\frac{d}{ds} \left(\frac{3}{s^2 + 9}\right)$

$= -\frac{d}{ds} [3(s^2 + 9)^{-1}] = -3 \times (-1)(s^2 + 9)^{-2} \times 2s$

$= \frac{6s}{(s^2 + 9)^2}$

4.8 Integration in s-Domain

If $\mathcal{L}\{f(t)\} = F(s)$, then:

$\mathcal{L}\left\{\frac{f(t)}{t}\right\} = \int_{s}^{\infty} F(u) \, du$

provided the integral converges.

Example: Find $\mathcal{L}\left\{\frac{\sin t}{t}\right\}$

We know: $\mathcal{L}\{\sin t\} = \frac{1}{s^2 + 1}$

So: $\mathcal{L}\left\{\frac{\sin t}{t}\right\} = \int_{s}^{\infty} \frac{1}{u^2 + 1} \, du$

$= \left[\tan^{-1} u\right]_{s}^{\infty} = \frac{\pi}{2} - \tan^{-1} s = \cot^{-1} s$

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5. Inverse Laplace Transform

5.1 Definition

The inverse Laplace transform converts from s-domain back to time domain:

$f(t) = \mathcal{L}^{-1}\{F(s)\}$

If $\mathcal{L}\{f(t)\} = F(s)$, then $\mathcal{L}^{-1}\{F(s)\} = f(t)$.

5.2 Linearity of Inverse Transform

$\mathcal{L}^{-1}\{a F(s) + b G(s)\} = a \mathcal{L}^{-1}\{F(s)\} + b \mathcal{L}^{-1}\{G(s)\}$

5.3 Common Inverse Transforms

F(s)	f(t) = 𝓛⁻¹{F(s)}
$\frac{1}{s}$	1
$\frac{1}{s-a}$	$e^{at}$
$\frac{1}{s^2}$	t
$\frac{1}{s^{n+1}}$	$\frac{t^n}{n!}$
$\frac{\omega}{s^2 + \omega^2}$	$\sin(\omega t)$
$\frac{s}{s^2 + \omega^2}$	$\cos(\omega t)$
$\frac{1}{(s-a)^2 + \omega^2}$	$\frac{1}{\omega} e^{at} \sin(\omega t)$
$\frac{s-a}{(s-a)^2 + \omega^2}$	$e^{at} \cos(\omega t)$

5.4 Method of Partial Fractions

Most common method for finding inverse transforms. Express $F(s)$ as sum of simpler fractions.

Case 1: Distinct Linear Factors

If $F(s) = \frac{P(s)}{(s-a_1)(s-a_2)\cdots(s-a_n)}$, then:

$F(s) = \frac{A_1}{s-a_1} + \frac{A_2}{s-a_2} + \cdots + \frac{A_n}{s-a_n}$

Example: Find $\mathcal{L}^{-1}\left\{\frac{s+1}{s^2 + 5s + 6}\right\}$

Factor denominator: $s^2 + 5s + 6 = (s+2)(s+3)$

Write: $\frac{s+1}{(s+2)(s+3)} = \frac{A}{s+2} + \frac{B}{s+3}$

Multiply both sides by $(s+2)(s+3)$:

$s+1 = A(s+3) + B(s+2)$

Solve for A and B:

• Set $s=-2$: $-2+1 = A(1) \Rightarrow -1 = A \Rightarrow A=-1$

• Set $s=-3$: $-3+1 = B(-1) \Rightarrow -2 = -B \Rightarrow B=2$

So: $\frac{s+1}{(s+2)(s+3)} = \frac{-1}{s+2} + \frac{2}{s+3}$

Now inverse transform:

$\mathcal{L}^{-1}\left\{\frac{s+1}{s^2+5s+6}\right\} = -\mathcal{L}^{-1}\left\{\frac{1}{s+2}\right\} + 2\mathcal{L}^{-1}\left\{\frac{1}{s+3}\right\}$

$= -e^{-2t} + 2e^{-3t}$

Case 2: Repeated Linear Factors

If $F(s) = \frac{P(s)}{(s-a)^n}$, then:

$F(s) = \frac{A_1}{s-a} + \frac{A_2}{(s-a)^2} + \cdots + \frac{A_n}{(s-a)^n}$

Example: Find $\mathcal{L}^{-1}\left\{\frac{s}{(s+1)^3}\right\}$

Write: $\frac{s}{(s+1)^3} = \frac{A}{s+1} + \frac{B}{(s+1)^2} + \frac{C}{(s+1)^3}$

Multiply by $(s+1)^3$: $s = A(s+1)^2 + B(s+1) + C$

Expand: $s = A(s^2 + 2s + 1) + B(s+1) + C = As^2 + (2A+B)s + (A+B+C)$

Compare coefficients:

• $s^2$: $A = 0$

• $s^1$: $2A + B = 1 \Rightarrow 0 + B = 1 \Rightarrow B=1$

• $s^0$: $A + B + C = 0 \Rightarrow 0 + 1 + C = 0 \Rightarrow C=-1$

So: $\frac{s}{(s+1)^3} = \frac{0}{s+1} + \frac{1}{(s+1)^2} + \frac{-1}{(s+1)^3}$

Inverse transform:

$\mathcal{L}^{-1}\left\{\frac{s}{(s+1)^3}\right\} = \mathcal{L}^{-1}\left\{\frac{1}{(s+1)^2}\right\} - \mathcal{L}^{-1}\left\{\frac{1}{(s+1)^3}\right\}$

We know: $\mathcal{L}^{-1}\left\{\frac{1}{(s-a)^{n+1}}\right\} = \frac{t^n e^{at}}{n!}$

So: $= te^{-t} - \frac{t^2}{2} e^{-t} = e^{-t} \left(t - \frac{t^2}{2}\right)$

Case 3: Quadratic Factors

For $F(s) = \frac{Ps + Q}{(s-a)^2 + b^2}$, complete to match sine/cosine forms.

Example: Find $\mathcal{L}^{-1}\left\{\frac{3s+7}{s^2+2s+5}\right\}$

Complete square in denominator: $s^2+2s+5 = (s+1)^2 + 4$

Write numerator to match forms: $3s+7 = 3(s+1) + 4$

So: $\frac{3s+7}{s^2+2s+5} = \frac{3(s+1)}{(s+1)^2+4} + \frac{4}{(s+1)^2+4}$

Inverse transform:

$= 3\mathcal{L}^{-1}\left\{\frac{s+1}{(s+1)^2+4}\right\} + 2\mathcal{L}^{-1}\left\{\frac{2}{(s+1)^2+4}\right\}$

$= 3e^{-t} \cos(2t) + 2e^{-t} \sin(2t) = e^{-t}[3\cos(2t) + 2\sin(2t)]$

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6. Solving Differential Equations using Laplace Transform

6.1 General Procedure

1. Take Laplace transform of both sides of ODE

2. Use derivative formulas to incorporate initial conditions

3. Solve algebraic equation for $Y(s)$ (Laplace transform of solution)

4. Use inverse Laplace transform to find $y(t)$

6.2 Example 1: First Order ODE

Solve: $y' + 2y = 0$, with $y(0) = 3$

Step 1: Take Laplace transform of both sides:

$\mathcal{L}\{y' + 2y\} = \mathcal{L}\{0\}$

$\mathcal{L}\{y'\} + 2\mathcal{L}\{y\} = 0$

$[sY(s) - y(0)] + 2Y(s) = 0$

Step 2: Substitute $y(0)=3$:

$sY(s) - 3 + 2Y(s) = 0$

Step 3: Solve for $Y(s)$:

$(s+2)Y(s) = 3$

$Y(s) = \frac{3}{s+2}$

Step 4: Inverse transform:

$y(t) = \mathcal{L}^{-1}\left\{\frac{3}{s+2}\right\} = 3e^{-2t}$

Check: $y' = -6e^{-2t}$, $2y = 6e^{-2t}$, sum = 0 ✓, $y(0)=3$ ✓

6.3 Example 2: Second Order ODE

Solve: $y'' - 3y' + 2y = 4e^{2t}$, with $y(0)=1$, $y'(0)=0$

Step 1: Laplace transform:

$\mathcal{L}\{y'' - 3y' + 2y\} = \mathcal{L}\{4e^{2t}\}$

$[s^2Y(s) - sy(0) - y'(0)] - 3[sY(s) - y(0)] + 2Y(s) = \frac{4}{s-2}$

Step 2: Substitute initial conditions:

$[s^2Y(s) - s - 0] - 3[sY(s) - 1] + 2Y(s) = \frac{4}{s-2}$

Step 3: Simplify:

$s^2Y(s) - s - 3sY(s) + 3 + 2Y(s) = \frac{4}{s-2}$

$(s^2 - 3s + 2)Y(s) - s + 3 = \frac{4}{s-2}$

Note: $s^2 - 3s + 2 = (s-1)(s-2)$

So: $(s-1)(s-2)Y(s) = \frac{4}{s-2} + s - 3$

Step 4: Solve for $Y(s)$:

$Y(s) = \frac{4}{(s-2)^2(s-1)} + \frac{s-3}{(s-1)(s-2)}$

Step 5: Partial fractions for first term:

$\frac{4}{(s-2)^2(s-1)} = \frac{A}{s-1} + \frac{B}{s-2} + \frac{C}{(s-2)^2}$

Multiply by $(s-2)^2(s-1)$: $4 = A(s-2)^2 + B(s-1)(s-2) + C(s-1)$

• Set $s=1$: $4 = A(1)^2 \Rightarrow A=4$

• Set $s=2$: $4 = C(1) \Rightarrow C=4$

• Set $s=0$: $4 = 4A + 2B - C = 16 + 2B - 4 \Rightarrow 4 = 12 + 2B \Rightarrow 2B = -8 \Rightarrow B=-4$

So: $\frac{4}{(s-2)^2(s-1)} = \frac{4}{s-1} - \frac{4}{s-2} + \frac{4}{(s-2)^2}$

Step 6: Partial fractions for second term:

$\frac{s-3}{(s-1)(s-2)} = \frac{D}{s-1} + \frac{E}{s-2}$

Multiply: $s-3 = D(s-2) + E(s-1)$

• Set $s=1$: $-2 = -D \Rightarrow D=2$

• Set $s=2$: $-1 = E \Rightarrow E=-1$

So: $\frac{s-3}{(s-1)(s-2)} = \frac{2}{s-1} - \frac{1}{s-2}$

Step 7: Combine:

$Y(s) = \left[\frac{4}{s-1} - \frac{4}{s-2} + \frac{4}{(s-2)^2}\right] + \left[\frac{2}{s-1} - \frac{1}{s-2}\right]$

$= \frac{6}{s-1} - \frac{5}{s-2} + \frac{4}{(s-2)^2}$

Step 8: Inverse transform:

$y(t) = 6\mathcal{L}^{-1}\left\{\frac{1}{s-1}\right\} - 5\mathcal{L}^{-1}\left\{\frac{1}{s-2}\right\} + 4\mathcal{L}^{-1}\left\{\frac{1}{(s-2)^2}\right\}$

$= 6e^t - 5e^{2t} + 4te^{2t}$

Final answer: $y(t) = 6e^t + e^{2t}(4t - 5)$

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7. Convolution Theorem

7.1 Definition of Convolution

The convolution of two functions $f(t)$ and $g(t)$ is:

$(f * g)(t) = \int_{0}^{t} f(\tau) g(t-\tau) \, d\tau$

Convolution is commutative: $f * g = g * f$

7.2 Convolution Theorem

$\mathcal{L}\{f(t) * g(t)\} = F(s) G(s)$

Equivalently: $\mathcal{L}^{-1}\{F(s) G(s)\} = f(t) * g(t)$

7.3 Application Example

Find $\mathcal{L}^{-1}\left\{\frac{1}{s(s^2+1)}\right\}$

We can write: $\frac{1}{s(s^2+1)} = \frac{1}{s} \times \frac{1}{s^2+1}$

Let $F(s) = \frac{1}{s}$, $G(s) = \frac{1}{s^2+1}$

Then $f(t) = \mathcal{L}^{-1}\{F(s)\} = 1$

$g(t) = \mathcal{L}^{-1}\{G(s)\} = \sin t$

By convolution theorem:

$\mathcal{L}^{-1}\left\{\frac{1}{s(s^2+1)}\right\} = f(t) * g(t) = 1 * \sin t$

$= \int_{0}^{t} 1 \cdot \sin(t-\tau) \, d\tau = \int_{0}^{t} \sin(t-\tau) \, d\tau$

Let $u = t-\tau$, then $du = -d\tau$:

$= \int_{\tau=0}^{\tau=t} \sin u \, (-du) = \int_{u=t}^{u=0} \sin u \, (-du) = \int_{0}^{t} \sin u \, du$

$= [-\cos u]_{0}^{t} = -\cos t + \cos 0 = 1 - \cos t$

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8. Unit Step Function and Periodic Functions

8.1 Unit Step Function (Heaviside Function)

$u(t-a) = \begin{cases} 0 & t < a \\ 1 & t \geq a \end{cases}$

Useful for representing functions that "turn on" at time $t=a$.

Laplace transform: $\mathcal{L}\{u(t-a)\} = \frac{e^{-as}}{s}$

8.2 Representing Piecewise Functions

Any piecewise function can be written using unit step functions.

Example: $f(t) = \begin{cases} 0 & t < 2 \\ t & 2 \leq t < 4 \\ 4 & t \geq 4 \end{cases}$

Can be written as: $f(t) = t[u(t-2) - u(t-4)] + 4u(t-4)$

8.3 Periodic Functions

If $f(t)$ is periodic with period $T$ (i.e., $f(t+T) = f(t)$ for all $t$), then:

$\mathcal{L}\{f(t)\} = \frac{1}{1 - e^{-sT}} \int_{0}^{T} e^{-st} f(t) \, dt$

Example: Find Laplace transform of square wave with period 2:

$f(t) = \begin{cases} 1 & 0 \leq t < 1 \\ 0 & 1 \leq t < 2 \end{cases}$, and $f(t+2) = f(t)$

Here $T=2$:

$\int_{0}^{T} e^{-st} f(t) \, dt = \int_{0}^{1} e^{-st} \cdot 1 \, dt + \int_{1}^{2} e^{-st} \cdot 0 \, dt$

$= \left[-\frac{1}{s} e^{-st}\right]_{0}^{1} = -\frac{1}{s}(e^{-s} - 1) = \frac{1 - e^{-s}}{s}$

So: $\mathcal{L}\{f(t)\} = \frac{1}{1 - e^{-2s}} \times \frac{1 - e^{-s}}{s} = \frac{1 - e^{-s}}{s(1 - e^{-2s})}$

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9. Dirac Delta Function (Impulse Function)

9.1 Definition

The Dirac delta function $\delta(t-a)$ is not a function in the usual sense, but a "generalized function" with:

1. $\delta(t-a) = 0$ for $t \neq a$

2. $\int_{-\infty}^{\infty} \delta(t-a) \, dt = 1$

3. $\int_{-\infty}^{\infty} f(t) \delta(t-a) \, dt = f(a)$ (sifting property)

9.2 Laplace Transform of Delta Function

$\mathcal{L}\{\delta(t-a)\} = e^{-as}$ for $a \geq 0$

Special case: $\mathcal{L}\{\delta(t)\} = 1$

9.3 Physical Interpretation

Represents an instantaneous impulse (like a hammer blow) at time $t=a$.

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10. Applications of Laplace Transform

10.1 Electrical Circuits

• Solving RLC circuit equations

• Analyzing transient responses

• Finding transfer functions

10.2 Control Systems

• Analyzing system stability

• Designing controllers

• Finding system responses

10.3 Mechanical Systems

• Solving spring-mass-damper systems

• Vibration analysis

• Heat transfer problems

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11. Important Formulas Summary

11.1 Basic Transforms

• $\mathcal{L}\{1\} = \frac{1}{s}$

• $\mathcal{L}\{e^{at}\} = \frac{1}{s-a}$

• $\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}$

• $\mathcal{L}\{\sin(\omega t)\} = \frac{\omega}{s^2 + \omega^2}$

• $\mathcal{L}\{\cos(\omega t)\} = \frac{s}{s^2 + \omega^2}$

11.2 Properties

• Linearity: $\mathcal{L}\{af(t)+bg(t)\} = aF(s)+bG(s)$

• s-Shift: $\mathcal{L}\{e^{at}f(t)\} = F(s-a)$

• t-Shift: $\mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s)$

• Derivative: $\mathcal{L}\{f'(t)\} = sF(s)-f(0)$

• Integral: $\mathcal{L}\left\{\int_0^t f(\tau)d\tau\right\} = \frac{F(s)}{s}$

11.3 Convolution Theorem

$\mathcal{L}\{f(t)*g(t)\} = F(s)G(s)$

11.4 Inverse Transform Tips

1. Use partial fractions for rational functions

2. Complete squares for quadratic denominators

3. Use convolution theorem for products

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12. Exam Tips and Common Mistakes

12.1 Common Mistakes to Avoid

1. Forgetting initial conditions in derivative formulas

2. Misapplying t-shift theorem: Must have $u(t-a)$

3. Incorrect partial fractions: Check degree of numerator vs denominator

4. Forgetting ROC when it matters

5. Algebra errors when solving for constants in partial fractions

12.2 Problem-Solving Strategy

1. Identify type of problem: ODE, inverse transform, etc.

2. Choose appropriate method: Partial fractions, properties, etc.

3. Check initial conditions for ODE problems

4. Verify answer by plugging back or using different method

5. Simplify final answer

12.3 Quick Checks

1. Linearity: $\mathcal{L}\{af+bg\} = a\mathcal{L}\{f\}+b\mathcal{L}\{g\}$

2. Derivative: $\mathcal{L}\{f'\} = sF(s)-f(0)$

3. Final Value Theorem: $\lim_{t\to\infty} f(t) = \lim_{s\to 0} sF(s)$

4. Initial Value Theorem: $\lim_{t\to 0^+} f(t) = \lim_{s\to\infty} sF(s)$

This comprehensive theory covers all aspects of Laplace transform with detailed explanations and examples, making it easy to understand while being thorough enough for exam preparation.