Discussion
If \(\tilde{f}(s)\) is the Fourier transform of \(f(t)\), then Fourier transform of \(f(t - a)\) is :
Answer: D
The Fourier transform of \(f(t - a)\) is \(e^{isa}\tilde{f}(s)\).
With the definition \(\tilde{f}(s) = \int f(t)e^{ist}\, dt\) (the sign convention the key follows), substitute \(u = t - a\):
\[ \begin{aligned} \int f(t - a)e^{ist}\, dt &= e^{isa}\int f(u)e^{isu}\, du \\ &= e^{isa}\tilde{f}(s) \end{aligned} \]
This is the shifting theorem: a shift in \(t\) multiplies the transform by a phase factor without changing its argument. With the opposite sign convention the factor would be \(e^{-isa}\), but the argument would still be \(s\), which rules out \(\tilde{f}(s + a)\). The real exponential \(e^{sa}\) is a Laplace-type factor, not a Fourier one.
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