# Fourier transform of a sum of 2 delta function

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))\$$.

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