Laplace Transform

Nerd Cafe

1. What is the Laplace Transform?

The Laplace Transform is a powerful tool used in engineering, physics, and mathematics to convert time-domain functions (typically functions of t) into frequency-domain functions (functions of s).

Definition:

The Laplace Transform of a function f(t) is defined as:

$$L\left\{ f(t) \right\}=\int_{0}^{\infty }e^{-st}f(t)dt$$

Where:

• f(t): time-domain function

• F(s): Laplace-transformed function

• s: complex frequency variable (can be s=σ+jω)

2. Why Use the Laplace Transform?

• Converts differential equations into algebraic equations.

• Helps in solving initial value problems.

• Very useful in control systems, circuits, and signal processing.

3. Common Laplace Transform Pairs

Example 1

$$f(t)=1$$

A. Step-by-Step Solution

The Laplace Transform of a function f(t) is defined as:

$$L\left\{ f(t) \right\}=\int_{0}^{\infty }e^{-st}f(t)dt$$

So, for 𝑓 (𝑡) = 1 :

$$L\left\{ 1 \right\}=\int_{0}^{\infty }e^{-st}dt$$

This is a basic exponential integral. Let's evaluate it:

$$\int_{0}^{\infty }e^{-st}dt=\left\lfloor \frac{e^{-st}}{-s} \right\rfloor^{\infty }_{0}$$

As we know:

• At 𝑡 → ∞ , 𝑒 ^{− 𝑠 𝑡} → 0 (since 𝑠 > 0)

• At 𝑡=0 , 𝑒^−𝑠𝑡=1

So:

$$L\left\{ 1 \right\}=(-\frac{1}{s}\times 0)-(-\frac{1}{s}\times 1)=\frac{1}{s}$$

B. Python Verification

from sympy import symbols, laplace_transform

# Define variables
t, s = symbols('t s')

# Define the function
f_t = 1

# Compute Laplace Transform
F_s = laplace_transform(f_t, t, s)
print("Laplace Transform of f(t)=1:", F_s)

C. Output:

Laplace Transform of f(t)=1: (1/s, 0, True)

Example 2

$$f(t)=t$$

A. Step-by-Step Solution

Substitute 𝑓(𝑡)=𝑡:

$$L\left\{ t \right\}=\int_{0}^{\infty }e^{-st}tdt$$

Let’s choose:

$$\begin{matrix} u=t⇒du=dt \\ \\ dv=e^{-st}dt\Rightarrow v=-\frac{1}{s}e^{-st} \end{matrix}$$

Apply the formula:

$$\int_{0}^{\infty }te^{-st}dt=\left[ -\frac{t}{s}e^{-st} \right]_{0}^{\infty }+\int_{0}^{\infty }\frac{1}{s}e^{-st}dt$$

First term:

$$\left[ -\frac{t}{s}e^{-st} \right]_{0}^{\infty }=0$$

Second term:

$$\int_{0}^{\infty }\frac{1}{s}e^{-st}dt=\frac{1}{s}\left\lfloor \frac{e^{-st}}{-s} \right\rfloor_{0}^{\infty }=\frac{1}{s}(0-(-\frac{1}{s})) =\frac{1}{s^{2}}$$

B. Python Code Verification

from sympy import symbols, laplace_transform

# Define variables
t, s = symbols('t s')

# Define the function
f_t = t

# Compute Laplace Transform
F_s = laplace_transform(f_t, t, s)
print("Laplace Transform of f(t)=t:", F_s)

C. Output

Laplace Transform of f(t)=1: (s**(-2), 0, True)

Example 3

Let’s now mathematically calculate the Laplace Transform of:

$$f(t)=t^{2}$$

From the definition of Laplace Transform:

$$L\left\{ f(t) \right\}=\int_{0}^{\infty }e^{-st}f(t)dt$$

Substitute 𝑓(𝑡)=𝑡²:

$$L\left\{ f(t) \right\}=\int_{0}^{\infty }e^{-st}t^{2}dt$$

A. Step-by-Step Integration

We use integration by parts, or more efficiently, we use the gamma function property for powers of ttt.

But first, we’ll do it manually.

Let:

$$\begin{matrix} u=t^{2}\Rightarrow du=2tdt \\ \\ dv=e^{-st}dt\Rightarrow v=-\frac{1}{s}e^{-st} \end{matrix}$$

So:

$$\int_{}^{}t^{2}e^{-st}dt=-\frac{1}{s}t^{2}e^{-st}+\int_{}^{}\frac{2t}{s}e^{-st}dt$$

Now let:

$$\begin{matrix} u=t\Rightarrow du=dt \\ \\ dv=e^{-st}dt\Rightarrow v=-\frac{1}{s}e^{-st} \end{matrix}$$

So:

$$\int_{}^{}te^{-st}dt=-\frac{t}{s}e^{-st}+\frac{1}{s}\int_{}^{}e^{-st}dt=-\frac{t}{s}e^{-st}-\frac{1}{s^{2}}e^{-st}$$

Now plug everything back in:

$$\int_{0}^{\infty }t^{2}e^{-st}dt=\left[ -\frac{t^{2}}{s} e^{-st}\right]_{0}^{\infty }+\left[ \frac{2}{s}(-\frac{t}{s}e^{-st}-\frac{1}{s}e^{-st}) \right]_{0}^{\infty }$$

All terms vanish at t→∞, and at t=0, all t-dependent terms are zero, but the constant exponential part survives. So we are left with:

$$L\left\{ t^{2} \right\}=\frac{2}{t^{3}}\;\;, \;\; for\;\;s>0$$

B. General Rule (for future use):

$$L\left\{ t^{n} \right\}=\frac{n!}{t^{n+1}}\;\;, \;\; for\;\;s>0$$

So for 𝑛 = 2:

$$L\left\{ t^{2} \right\}=\frac{2}{t^{3}}\;\;, \;\; for\;\;s>0$$

C. Python Verification (SymPy)

from sympy import symbols, laplace_transform

# Define variables
t, s = symbols('t s')

# Define the function
f_t = t**2

# Compute Laplace Transform
F_s = laplace_transform(f_t, t, s)
print("Laplace Transform of f(t)=t^2:", F_s)

D. Output:

Laplace Transform of f(t)=1: (2/s**3, 0, True)

Example 4

Let's now mathematically compute the Laplace Transform of:

$$f(t)=e^{a.t}$$

From the definition of Laplace Transform:

$$L\left\{ f(t) \right\}=\int_{0}^{\infty }e^{-st}f(t)dt$$

Substitute 𝑓(𝑡)=e^at:

$$L\left\{ e^{a.t} \right\}=\int_{0}^{\infty }e^{-st}e^{at}dt$$

A. Step-by-Step Integration

We now compute:

$$\int_{0}^{\infty }e^{-st}e^{at}dt=\int_{0}^{\infty }e^{(a-s)t}dt$$

B. Condition:

• Converges if s>a, because we need (a−s)<0 so the exponential decays.

C. Integrate

$$\int_{0}^{\infty }e^{(a-s)t}dt=\frac{1}{a-s}e^{(a-s)t}$$

D. Apply limits from 0 to ∞ .

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

E. Final Result:

$$L\left\{ e^{at} \right\}=\frac{1}{s-a}$$

F. Python Verification

from sympy import symbols, laplace_transform, exp

# Define variables
t, s, a = symbols('t s a')

# Define the function
f_t = exp(a * t)

# Compute Laplace Transform
F_s = laplace_transform(f_t, t, s)
print("Laplace Transform of f(t)=e^(at):", F_s)

G. Output:

Laplace Transform of f(t)=e^(at): (1/(-a + s), re(a), True)

Keywords

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