A signal x(t) has a Fourier transform X(ω). If x(t) is a real and odd function of t, then X(ω) is

Solution (By Examveda Team)

Here's the explanation for the given Fourier Transform question:

The question asks about the properties of the Fourier Transform, X(ω), of a signal x(t) when x(t) is both real and odd.

Here's a breakdown to understand the answer:

Key Concepts:

* Fourier Transform: A mathematical tool that decomposes a signal into its constituent frequencies.
* Real Function: A function whose values are real numbers for all real inputs.
* Odd Function: A function x(t) is odd if x(-t) = -x(t). Odd functions are symmetrical about the origin.
* Even Function: A function x(t) is even if x(-t) = x(t). Even functions are symmetrical about the y-axis.

Properties of Fourier Transform with Symmetry:

* If x(t) is real, then X(ω) has Hermitian symmetry, meaning X(-ω) = X^*(ω) where * denotes complex conjugate. In simpler terms, the real part of X(ω) is even, and the imaginary part is odd.
* If x(t) is odd, the Fourier transform X(ω) is imaginary. This means it only contains imaginary components and no real components.

Applying to the question:

Since x(t) is both real and odd:

1. Because x(t) is real, we know X(ω) has Hermitian symmetry.
2. Because x(t) is odd, we know X(ω) is purely imaginary.

Combining these two properties, if X(ω) is purely imaginary (say, X(ω) = j*Y(ω), where Y(ω) is a real-valued function) and has Hermitian symmetry, it means:

X(-ω) = X^*(ω)
j*Y(-ω) = (j*Y(ω))^*
j*Y(-ω) = -j*Y(ω) (because the conjugate of j is -j)
Y(-ω) = -Y(ω)

This means Y(ω) is an odd function. Since X(ω) = j*Y(ω), and Y(ω) is odd, then X(ω) itself is an imaginary and odd function.

Therefore, the correct answer is Option B: An imaginary and odd function of ω