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In the Fourier transform, if the time domain signal x(t) is real and even, then the frequency domain signal X(jΩ) will be:
- a)imaginary and even
- b)imaginary and odd
- c)real and even
- d)real and odd
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
In the Fourier transform, if the time domain signal x(t) is real and e...
Understanding the Fourier Transform of Real Even Functions
When analyzing the Fourier transform of a time-domain signal \(x(t)\) that is real and even, we can derive important properties about its frequency-domain counterpart \(X(j\Omega)\).
Properties of Real and Even Functions
- A real and even function satisfies the condition:
\[
x(t) = x(-t)
\]
- This symmetry in the time domain implies specific characteristics in the frequency domain.
Fourier Transform Definition
The Fourier transform of \(x(t)\) is defined as:
\[
X(j\Omega) = \int_{-\infty}^{\infty} x(t) e^{-j\Omega t} dt
\]
Evaluating the Transform
1. Substituting for Even Function:
- Since \(x(t)\) is even, we can express the integral as:
\[
X(j\Omega) = \int_{-\infty}^{\infty} x(t) \cos(\Omega t) dt - j \int_{-\infty}^{\infty} x(t) \sin(\Omega t) dt
\]
2. Properties of Cosine and Sine:
- The cosine term:
- \( \int_{-\infty}^{\infty} x(t) \cos(\Omega t) dt \) is real and even.
- The sine term:
- \( \int_{-\infty}^{\infty} x(t) \sin(\Omega t) dt = 0 \) (since \(x(t)\) is even and \(\sin(\Omega t)\) is odd).
3. Conclusion:
- Thus, the Fourier transform simplifies to:
\[
X(j\Omega) = \text{Real (even)}
\]
Final Result
Since \(X(j\Omega)\) is derived as a real function with even symmetry, the correct answer is:
- Option C: Real and Even.
This property is essential in signal processing and communication systems, as it helps in understanding the behavior of linear time-invariant systems.
When analyzing the Fourier transform of a time-domain signal \(x(t)\) that is real and even, we can derive important properties about its frequency-domain counterpart \(X(j\Omega)\).
Properties of Real and Even Functions
- A real and even function satisfies the condition:
\[
x(t) = x(-t)
\]
- This symmetry in the time domain implies specific characteristics in the frequency domain.
Fourier Transform Definition
The Fourier transform of \(x(t)\) is defined as:
\[
X(j\Omega) = \int_{-\infty}^{\infty} x(t) e^{-j\Omega t} dt
\]
Evaluating the Transform
1. Substituting for Even Function:
- Since \(x(t)\) is even, we can express the integral as:
\[
X(j\Omega) = \int_{-\infty}^{\infty} x(t) \cos(\Omega t) dt - j \int_{-\infty}^{\infty} x(t) \sin(\Omega t) dt
\]
2. Properties of Cosine and Sine:
- The cosine term:
- \( \int_{-\infty}^{\infty} x(t) \cos(\Omega t) dt \) is real and even.
- The sine term:
- \( \int_{-\infty}^{\infty} x(t) \sin(\Omega t) dt = 0 \) (since \(x(t)\) is even and \(\sin(\Omega t)\) is odd).
3. Conclusion:
- Thus, the Fourier transform simplifies to:
\[
X(j\Omega) = \text{Real (even)}
\]
Final Result
Since \(X(j\Omega)\) is derived as a real function with even symmetry, the correct answer is:
- Option C: Real and Even.
This property is essential in signal processing and communication systems, as it helps in understanding the behavior of linear time-invariant systems.
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In the Fourier transform, if the time domain signal x(t) is real and e...
Let F(ω) is the Fourier transform of f(t).
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In the Fourier transform, if the time domain signal x(t) is real and even, then the frequency domain signal X(jΩ) will be:a)imaginary and evenb)imaginary and oddc)real and evend)real and oddCorrect answer is option 'C'. Can you explain this answer?
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