Correct Answer - Option 2 : X*(e
jω)
Concept:
Fourier Transform:
\(F.T\left[ {x\left( t \right)} \right] = X\left( ω \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - jω t}}dt\)
\(I.F.T\left[ {X\left( ω \right)} \right] = x\left( t \right)\)
\(= \frac{1}{{2π }}\mathop \smallint \limits_{ - \infty }^\infty X\left( ω \right){e^{jω t}}dt\)
Some properties of fourier transform:
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Properties
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X(f) form
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X(ω) form
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Time scaling x(at)
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\(\frac{1}{{\left| a \right|}} X \left( {\frac{f}{a}} \right)\)
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\(\frac{1}{{\left| a \right|}}X\left( {\frac{ω}{a}} \right)\)
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Time reversal x(-t)
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X(-f)
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X(-ω)
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Time shift x(t ± t0)
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\({e^{ \pm 2π f{t_0}}} X\left( t \right)\)
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\({e^{ \pm jω {t_0}}} X\left( ω \right)\)
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Frequency shift \(x\left( t \right){e^{ \pm j{ω _0}t}}\)
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X(f ± f0)
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X(ω ± ω0)
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Differentiation in time
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\(\frac{d}{{dt}}x\left( t \right)↔ j2π f ~X\left( f \right)\)
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\(\frac{d}{{dt}}x\left( t \right)↔ jω ~X\left( ω \right)\)
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Conjugation x[n]
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X*(e-j2πf)
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X*(e-jω)
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Time reversal x[-n]
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X(e-j2πf)
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X(e-jω)
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Duality
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x(t) ↔ X(f)
x(t) ↔ -X(f)
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x(t) ↔ X(ω)
x(t) ↔2π X(-ω)
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Analysis:
We know that:
\({x^*}\left[ { + n} \right]\mathop \leftrightarrow \limits^{F.T} {X^*}\left( {{e^{ - j\omega }}} \right) = {X_1}\left( \omega \right)\)
Then,
\({x^*}\left[ { + n} \right]\mathop \leftrightarrow \limits^{F.T} {X^*}\left( { - \omega } \right) = {X^*}\left( {{e^{ - j\left( { - \omega } \right)}}} \right)\)
\(= {X^*}\left( {{e^{j\omega }}} \right)\)
Option (2) correct.
More information:
- Fourier transform of the real signal is always even conjugate in nature.
- F.T [Real & even signal] = purely real and even.
- F.T [Real & odd signal] = purely imaginary and odd.
- Shifting in the time domain only changes the phase spectrum of the signal.
