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The fourier transform of a conjugate symmetric function is always
  • a)
    Imaginary
  • b)
    Conjugate anti-symmetric
  • c)
    Conjugate symmetric
  • d)
    Real
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The fourier transform of a conjugate symmetric function is alwaysa)Ima...
The correct answer is option (d) real
Explanation
In time domain, the nature of the signal is conjugate symmetric. Now, we have to find the nature of signal in frequency domain.
For example - Consider the duality property. Lets take a rectangle rect(t). Here sinc(f) is the forehead transform. According to the duality property, we can interchange these two, that is for sinc(t), the forehead transform will be rect (f).
In a similar manner, according to the standard definition, whenever signal is real, the forehead transform is conjugate symmetric.  Now apply duality property on it, this means for conjugate symmetric function, the forehead transform is real.
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Community Answer
The fourier transform of a conjugate symmetric function is alwaysa)Ima...
Introduction:
The Fourier Transform is a mathematical tool that decomposes a function into its constituent frequencies. It is widely used in signal processing, image processing, and many other fields. A conjugate symmetric function is a special kind of function that exhibits certain symmetries. In this response, we will explain why the Fourier Transform of a conjugate symmetric function is always conjugate symmetric.

Explanation:
To understand why the Fourier Transform of a conjugate symmetric function is always conjugate symmetric, let's start by defining what a conjugate symmetric function is.

A function f(t) is said to be conjugate symmetric if it satisfies the following condition:

f(t) = f*(t)

where f*(t) represents the complex conjugate of f(t).

Now, let's consider the Fourier Transform of f(t), denoted as F(ω), where ω is the frequency variable. The Fourier Transform of f(t) is given by the following equation:

F(ω) = ∫[f(t)e^(-jωt)]dt

where j is the imaginary unit.

Proof:
To prove that the Fourier Transform of a conjugate symmetric function is always conjugate symmetric, we need to show that:

F(ω) = F*(-ω)

Let's substitute f(t) = f*(-t) in the Fourier Transform equation:

F(ω) = ∫[f*(-t)e^(-jωt)]dt

Now, let's change the variable of integration from t to -t:

F(ω) = ∫[f*(t)e^(jωt)]dt

Notice that we have f*(t) and e^(jωt), which are the complex conjugates of f*(-t) and e^(-jωt), respectively.

Using the property that the complex conjugate of a product is equal to the product of the complex conjugates, we can write:

F(ω) = [∫f*(-t)dt][∫e^(jωt)dt]

The first integral on the right-hand side is simply the Fourier Transform of f*(-t), denoted as F*(-ω). The second integral represents the inverse Fourier Transform of e^(jωt), which is a complex sinusoidal function.

Therefore, we can conclude that:

F(ω) = F*(-ω)

which means that the Fourier Transform of a conjugate symmetric function is always conjugate symmetric.

Conclusion:
In summary, the Fourier Transform of a conjugate symmetric function is always conjugate symmetric. This property is useful in many applications, such as the analysis of real-valued signals and the simplification of certain mathematical operations. Understanding the symmetry properties of functions and their Fourier Transforms is essential in signal processing and related fields.
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