Problem 35 - Differential Equations and Linear Algebra

Question

Recall that according to Euler's formula $$ e^{i b t}=\cos b t+i \sin b t $$ since the Laplace transform is linear, it follows that $$ \begin{aligned} L[\cos b t] &=\operatorname{Re}\left(L\left[e^{i b t}\right]\right) \\ L[\sin b t] &=\operatorname{Im}\left(L\left[e^{i b t}\right]\right) \end{aligned} $$ Find \(\left.L\left[e^{i b t}\right], \text { and hence, derive Equations ( } 10.1 .4\right)\) and \((10.1 .5)\)

Step-by-Step Solution

Answer

In summary, we found the Laplace transform of \(e^{i b t}\), which is \(L[e^{i b t}] = \frac{-1}{s - i b}\). After splitting it into real and imaginary parts, we derived Equations (10.1.4) and (10.1.5) as follows: \(L[\cos b t] = \frac{-s}{s^2 + b^2}\) and \(L[\sin b t] = \frac{-b}{s^2 + b^2}\).

Step 1: Recall the definition of Laplace transform

The Laplace transform of a function f(t), denoted \(L[f(t)]\) or \(\mathcal{L}\{f(t)\}\), is defined as follows: \[ L[f(t)] = \int_{0}^{\infty} e^{-st} f(t) dt \] Where s is a complex variable.

Step 2: Apply the Laplace transform to \(e^{i b t}\)

Now, let's apply the Laplace transform to the function \(e^{i b t}\) and evaluate the integral. \[ L[e^{i b t}] = \int_{0}^{\infty} e^{-st} e^{i b t} dt = \int_{0}^{\infty} e^{(-s + i b)t} dt \]

Step 3: Evaluate the integral

To find the integral, we can now use integration with respect to t: \[ \begin{aligned} L[e^{i b t}] &= \int_{0}^{\infty} e^{(-s + i b)t} dt \\ &= \left[\frac{1}{-s+i b} e^{(-s+i b)t}\right]_0^{\infty} \\ &= \frac{1}{-s+i b} \lim_{T\to\infty} \left( e^{(-s+i b)T} - e^{0} \right) \\ &= \frac{1}{-s+i b} \left( 0 -1\right) \\ &= \frac{-1}{s - i b} \end{aligned} \] So, we have found that: \[ L[e^{i b t}] = \frac{-1}{s - i b} \]

Step 4: Find the real and imaginary parts

Now, we need to split the Laplace transform of \(e^{i b t}\) into its real and imaginary parts. In order to do this, let's multiply the numerator and the denominator by the complex conjugate of the denominator, which is \(s + i b\): \[ L[e^{i b t}] = \frac{-1}{s - i b} \cdot \frac{s+i b}{s+i b} = \frac{-(s+i b)}{s^2 + b^2} \] Now we observe that the real and imaginary parts are as follows: \[ \operatorname{Re}\left(L\left[e^{i b t}\right]\right) = \frac{-s}{s^2 + b^2} \quad \text{and} \quad \operatorname{Im}\left(L\left[e^{i b t}\right]\right) = \frac{-b}{s^2 + b^2} \]

Step 5: Find Laplace transforms of \(\cos(bt)\) and \(\sin(bt)\)

We know that: \[ L[\cos b t] = \operatorname{Re}\left(L\left[e^{i b t}\right]\right) \quad \text{and} \quad L[\sin b t] = \operatorname{Im}\left(L\left[e^{i b t}\right]\right) \] So, we can now write down equations (10.1.4) and (10.1.5): \[ L[\cos b t] = \frac{-s}{s^2 + b^2} \quad \text{and} \quad L[\sin b t] = \frac{-b}{s^2 + b^2} \]

Key Concepts

Euler's Formula

Understanding Euler's Formula is crucial when dealing with problems involving complex numbers and trigonometric functions. This elegant formula is given by

• \( e^{i \theta} = \cos \theta + i \sin \theta \)

This expression relates the complex exponential function with trigonometric functions \( \cos \theta \) and \( \sin \theta \). Here, \( i \) is the imaginary unit, and \( \theta \) is an angle in radians.Euler's Formula is incredibly powerful because it provides a bridge between complex exponentials and oscillatory functions such as sine and cosine.
This means we can use exponentials to simplify calculations involving these oscillating functions. In exercises involving the Laplace transform, Euler's formula helps break down complex exponential expressions into real and imaginary parts, which represent cosine and sine respectively.

Complex Numbers

Complex numbers are numbers that include both a real part and an imaginary part. They are usually expressed as

• \( a + bi \), where \( a \) is the real part and \( bi \) is the imaginary part.

The imaginary unit \( i \) is defined such that \( i^2 = -1 \). Complex numbers allow for the representation of oscillatory behavior using Euler's Formula by incorporating imaginary components.When solving problems that involve complex numbers, such as in the Laplace transform, you often need to separate the expression into real and imaginary components.
This distinguishes between the cosine and sine components within the context of complex exponentials. Multiplying complex numbers involves distributing the terms and utilizing the property \( i^2 = -1 \).

Integration Techniques

Integration techniques are fundamental in calculus and are necessary to perform the Laplace transform. The Laplace transform itself involves integrating a function over an interval from 0 to \( \infty \).
This is executed with precision using exponential functions.In integration, certain techniques such as u-substitution, integration by parts, and recognizing standard integral forms are often employed to solve complex integrals. In the context of the Laplace transform:

• The variable \( s \) is introduced, which is typically a complex variable, making the integration a bit more involved with complex arithmetic.
• Exponential integrals are evaluated meticulously to simplify expressions.

Understanding these techniques allows for effective evaluation of integrals necessary to compute transforms as observed with

• \( L[e^{abt}] \).

Fourier Transform

Fourier Transform is a critical concept that, like the Laplace Transform, converts time-domain signals into frequency-domain representations. While the Laplace Transform is primarily used for linear systems analysis, the Fourier Transform focusses on frequency analysis.Both transforms share similarities, such as using complex exponential functions to decompose signals. Fourier Transform analysis breaks down functions into constituent frequencies.

• It uses the form: \( F(s) = \int_{-\infty}^{\infty} f(t)e^{-i2\pi st} dt \).

The Fourier Transform is essential in signal processing, allowing us to understand the frequency components within a signal.
While the Laplace Transform deals with transcendental expressions over the real axis, the Fourier Transform covers the entire frequency spectrum. This makes it invaluable in fields such as engineering and physics for analyzing waves and physical phenomena.