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i have a channel impulse response \$h(\tau,t)=\delta(\tau-1/4t)+b\delta(\tau-T)\$

if I would like to take Fourier transform, I can take FT for each delta function and then sum up a result.

\$FT(h(\tau,t))=FT(\delta(\tau-1/4t))+b FT(\delta(\tau-T))\$

How to find FT for the first delta function is clear, but I don't understand how to find ft of the second delta function, \$FT(\delta(\tau-T))\$.

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You can use the property for time shifts:

$$ \mathcal{F}\{f(t-\tau)\} = e^{-j\omega\tau}\mathcal{F}\{f(t)\}\ $$

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