Which one of the following is the Fourier transform of the signal given in Fig. II, if the Fourier transform of the signal in Fig. I is \(=2\dfrac{\sin\omega T_1}{\omega}\) ?
\(2\dfrac{\sin\omega T_1}{\omega}e^{-j\omega T_1}\)
Compare the two waveforms. Fig. I is a rectangular pulse of width 2T1 centred on the origin. Fig. II is the same pulse — same height, same width — but occupying 0 to 2T1. Its centre has moved from 0 to T1, so
\(f_2(t)=f_1(t-T_1)\)
a pure delay of T1, with no change of shape or amplitude.
Apply the time-shifting property.
\(f(t-t_0)\ \longleftrightarrow\ F(\omega)e^{-j\omega t_0}\)
With \(t_0=T_1\):
\(F_2(\omega)=2\dfrac{\sin\omega T_1}{\omega}\,e^{-j\omega T_1}\)
which is option 2.
What the result means physically. The exponential has unit magnitude, so
\(|F_2(\omega)|=|F_1(\omega)|\)
— delaying a signal cannot change its magnitude spectrum, only its phase, and the added phase \(-\omega T_1\) is linear in frequency. That is the definition of distortionless delay: every frequency component is held back by the same time.
Ruling out the others. A positive exponent (option 1) would mean a time advance, shifting the pulse the wrong way. Options 3 and 4 have lost the factor 2, so their transforms would not reduce to the given F1(ω) when the shift is removed — and option 4's exponent mixes T1 with a bare constant 2, which is dimensionally impossible inside \(e^{j\omega t}\).
A quick zero-frequency check. At ω = 0 the transform must equal the area under the pulse, which is \(1\times2T_1=2T_1\). Both the given F1 and option 2 give \(2\lim_{\omega\to0}\frac{\sin\omega T_1}{\omega}=2T_1\) ✓, while options 3 and 4 give only T1.
Hence, the Fourier transform of Fig. II is \(2\dfrac{\sin\omega T_1}{\omega}e^{-j\omega T_1}\).
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