Problem 12
Question
Evaluate Fourier transforms of the following distributional functions: (a) \(\delta(x-a)\). (b) \(\delta^{\prime}(x-a)\). (c) \(\delta^{(n)}(x-a)\). (d) \(\delta\left(x^{2}-a^{2}\right)\) (c) \(\delta^{\prime}\left(x^{2}-a^{2}\right)\)
Step-by-Step Solution
Verified Answer
(a) \( e^{-j\omega a} \), (b) \( j\omega e^{-j\omega a} \), (c) \((j\omega)^n e^{-j\omega a} \), (d) undefined, (e) undefined.
1Step 1: Fourier Transform of Delta function
Fourier transform \(F(\omega)\) of a function f(t) is given by \(F(\omega) = \int_{-\infty}^{+\infty}f(t) e^{-j\omega t} dt \). If function f(t) = \(\delta (t-a)\) ,we get \(F(\omega) = e^{-j\omega a} \) .
2Step 2: Fourier Transform of Delta function derivatives
The Fourier Transform of the nth derivative of the delta function \(\delta^{(n)}(t-a)\) is given by \( (j\omega)^n e^{-j\omega a} \). This is obtained by differentiating the Fourier transform of the delta function n times.
3Step 3: Fourier Transform of Zero function delta
The Fourier transform of a zero function delta, \(\delta(x^2-a^2)\) is zero as it's undefined. Delta function is only defined for linear functions, not for quadratic or higher degree functions.
4Step 4: Fourier Transform of Zero function delta derivative
The derivative of a zero function delta, \(\delta'(x^2-a^2)\) is also not defined, hence its Fourier Transform is also zero.
Key Concepts
Delta functionDistributional functionsDerivatives of delta functionZero function delta
Delta function
The delta function, often denoted as \( \delta(x) \), is a special type of function used in mathematical analysis. It is not a function in the traditional sense but rather a "distribution" that assigns values to other functions through integration. The delta function is defined such that it is zero everywhere except at a single point, where it is theoretically infinitely high, and its integral over the entire real line is 1.
The delta function is useful in various fields such as physics and engineering because it can model an ideal impulse, often used to simplify calculations. In the context of Fourier transforms, the delta function results in a constant function over frequency, specifically \( e^{-j\omega a} \), making it particularly simple to work with. The application of the delta function in Fourier transforms provides a powerful tool for analyzing signals and systems.
The delta function is useful in various fields such as physics and engineering because it can model an ideal impulse, often used to simplify calculations. In the context of Fourier transforms, the delta function results in a constant function over frequency, specifically \( e^{-j\omega a} \), making it particularly simple to work with. The application of the delta function in Fourier transforms provides a powerful tool for analyzing signals and systems.
Distributional functions
Distributional functions, like the delta function, are used to describe mathematical objects that traditional functions cannot handle. These are generalized functions which extend the concept of functions and are crucial in areas such as signal processing and quantum mechanics.
Key elements of distributional functions include:
Key elements of distributional functions include:
- They are not functions in the typical sense. Instead, they act on functions through integration.
- Distributions can describe complex phenomena and singular measurements.
- They extend the rigor and applications of mathematical analysis.
Derivatives of delta function
The derivatives of the delta function are an extension of the distributional concept. They help address situations requiring derivatives of non-standard mathematical objects. When dealing with the nth derivative of the delta function, denoted as \( \delta^{(n)}(x-a) \), the Fourier transform yields a very specific result: \( (j\omega)^n e^{-j\omega a} \).
Important factors to consider:
Important factors to consider:
- The mathematical operation involves the multiplication of powers of \( j\omega \), reflecting how frequency components influence the derivative.
- Derivatives of the delta function are significant in accurately modeling impulses and systems with sudden changes.
Zero function delta
The zero function delta, represented as \( \delta(x^2-a^2) \), poses challenges in the realm of distributional functions. This term is considered undefined within the concept of the delta function since it involves a polynomial expression rather than a linear one.
Key aspects include:
Key aspects include:
- The delta function's properties are not applicable to non-linear expressions like \( x^2-a^2 \).
- Consequently, its Fourier transform result is zero, signifying that the application is invalid for quadratic or higher degree functions.
- This highlights the limitations when using delta functions in traditional mathematical contexts that are non-linear.
Other exercises in this chapter
Problem 9
Show the identities $$ \frac{d}{d x}(\delta(f(x)))=f^{\prime}(x) \delta^{\prime}(f(x)) $$ and $$ \delta(f(x))+f(x) \delta^{\prime}(f(x))=0 $$ Hence show that \(
View solution Problem 10
Find the Fourier transforms of the functions $$ f(x)= \begin{cases}1 & \text { if }-a \leq x \leq a \\ 0 & \text { otherwise }\end{cases} $$ and $$ g(x)= \begin
View solution Problem 13
Prove that $$ x^{m} \delta^{(n)}(x)=(-1)^{m} \frac{n !}{(n-m) !} \delta^{(n-m)}(x) \quad \text { for } n \geq m $$ Hence show that the Fourier trarsform of the
View solution Problem 14
Show that the Fourier transform of the distribution $$ \delta_{0}+\delta_{d i}+\delta_{2 a}+\cdots+\delta_{(2 n-1)} $$ is a distribution with density $$ \frac{1
View solution