GATE PH 2022 • Q26
MSQ • 1 Mark
Section 1: Mathematical Physics
Fourier and Laplace transforms
Subtopic: Fourier Series & Integral Transforms
Question Statement:
If g(k) is the Fourier transform of f(x), then which of the following are true?
Options:
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(A) g(-k) = +g∗(k) implies f(x) is real
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(B) g(-k) = -g∗(k) implies f(x) is purely imaginary
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(C) g(-k) = +g∗(k) implies f(x) is purely imaginary
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(D) g(-k) = -g∗(k) implies f(x) is real
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.26)**
**1. Fourier Transform Symmetry Properties**:
- Fourier transform definition:
$g(k) = \int_{-\infty}^{\infty} f(x) e^{-ikx}\, dx$
- Complex conjugate:
$g^*(k) = \int_{-\infty}^{\infty} f^*(x) e^{ikx}\, dx$
- Evaluating $g(-k)$:
$g(-k) = \int_{-\infty}^{\infty} f(x) e^{ikx}\, dx$
**2. Step-by-Step Derivation**:
- **Condition 1**: $g(-k) = +g^*(k)$
$\int_{-\infty}^{\infty} f(x) e^{ikx}\, dx = \int_{-\infty}^{\infty} f^*(x) e^{ikx}\, dx$
Taking inverse Fourier transform implies $f(x) = f^*(x)$, meaning **$f(x)$ is purely real**. **Statement (A) is TRUE**.
- **Condition 2**: $g(-k) = -g^*(k)$
$\int_{-\infty}^{\infty} f(x) e^{ikx}\, dx = -\int_{-\infty}^{\infty} f^*(x) e^{ikx}\, dx$
$\implies f(x) = -f^*(x)$
This is the precise definition of a **purely imaginary** function. **Statement (B) is TRUE**.
**3. Official Verification**:
The verified official IIT Master Answer is **Options (A, B)**.
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