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Magazine Chapter 24: Problem 3
If \(f(t)= \begin{cases}1 & |t| \leqslant 2 \\ 0 & \text { otherwise }\end{cases}\) and \(g(t)=\mathrm{e}^{\mathrm{j} t}\), find \(\left.\mathcal{F} \mid f(t) g(t)\right\\}\).
Short Answer
Expert verified
The Fourier transform is \( H(\omega) = \frac{2 \sin(2(1-\omega))}{1-\omega} \).
Step by step solution
01
Understanding the Functions
The function \( f(t) \) is a piecewise function that equals 1 when \( |t| \leq 2 \) and 0 elsewhere. The function \( g(t) = e^{jt} \) represents a complex exponential.
02
Formulate f(t)g(t)
Since \( f(t) \) is 1 for \( |t| \leq 2 \) and 0 otherwise, the product \( f(t)g(t) \) is given by \( f(t)g(t) = e^{jt} \) for \( |t| \leq 2 \) and 0 otherwise.
03
Fourier Transform Definition
The Fourier transform \( \mathcal{F}\{ h(t) \} \) of a function \( h(t) \) is defined as \( H(\omega) = \int_{-\infty}^{\infty} h(t) e^{-j\omega t} \, dt \). We apply this to \( h(t) = f(t)g(t) \).
04
Compute the Fourier Transform
Since \( f(t)g(t) = e^{jt} \) for \( |t| \leq 2 \) and 0 otherwise, we compute \[ H(\omega) = \int_{-2}^{2} e^{jt} e^{-j\omega t} \, dt. \]Substituting, we have \[ H(\omega) = \int_{-2}^{2} e^{jt-j\omega t} \, dt = \int_{-2}^{2} e^{jt(1-\omega)} \, dt. \]
05
Evaluate the Integral
Evaluate the integral:\[ H(\omega) = \int_{-2}^{2} e^{jt(1-\omega)} \, dt = \left[ \frac{1}{j(1-\omega)} e^{jt(1-\omega)} \right]_{-2}^{2}. \]The evaluation gives \[ H(\omega) = \frac{1}{j(1-\omega)} \left[ e^{j2(1-\omega)} - e^{-j2(1-\omega)} \right]. \]
06
Simplify Using Euler's Formula
Using Euler's formula, which states that \( e^{jx} - e^{-jx} = 2j \sin(x) \), we simplify:\[ H(\omega) = \frac{1}{j(1-\omega)} (2j \sin(2(1-\omega))). \] Simplifying further, we find:\[ H(\omega) = \frac{2 \sin(2(1-\omega))}{1-\omega}. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Piecewise Function
Piecewise functions are a type of function which have different expressions based on the value of the input variable. In this context, the function \( f(t) = \begin{cases} 1 & \text{if } |t| \leq 2 \ 0 & \text{otherwise} \end{cases} \) is a simple example of a piecewise function. This means that for any value of \( t \) within the interval \( -2 \leq t \leq 2 \), the function outputs 1. For any other value of \( t \), outside this interval, the function outputs 0.
- This makes the function value depend on the domain of \( t \).
- Piecewise functions are helpful to model scenarios where characteristics change over different intervals.
Complex Exponential Function
The complex exponential function can be represented as \( g(t) = e^{jt} \), where \( j \) is the imaginary unit, equivalent to \( \sqrt{-1} \). This form is foundational in the analysis of signals, especially when dealing with harmonic and oscillatory behavior. Complex exponentials are significant in engineering and physics for several reasons:
- They simplify the analysis of periodic functions.
- These functions encapsulate rotating phases and amplitudes.
- Managing these with Euler's formula bridges real and imaginary components.
Euler's Formula
Euler's formula is a key relationship in mathematics, expressed as \( e^{jx} = \cos(x) + j\sin(x) \). This formula is extremely useful as it connects complex exponentials with trigonometric functions. This linkage helps in converting expressions back and forth between exponential and trigonometric forms easily.
- It enables transformation equations like \( e^{jx} - e^{-jx} = 2j \sin(x) \).
- In Fourier analysis, Euler's formula is vital for simplifying expressions involving complex numbers.
- When evaluating integrals with complex exponents, Euler鈥檚 formula is regularly applied to simplify calculations into manageable trigonometric functions.
Integral Evaluation
Integral evaluation is a fundamental concept in calculus used to find the total value or area under a curve represented by a function. In solving problems involving Fourier transforms, evaluating integrals enables the determination of a function鈥檚 frequency spectrum.
- The Fourier transform integral \( H(\omega) = \int_{-\infty}^{\infty} h(t) e^{-j\omega t} \, dt \) is calculate to obtain the frequency representation.
- This involves integrating within specific boundaries, namely wherever our piecewise function is non-zero. Here, it鈥檚 between \( t = -2 \) and \( t = 2 \).
- After setting up the integral for our product \( f(t)g(t) \), the evaluation involves careful substitution and simplification.
