Obtaining fourier transform for Root Raised Cosine pulse

I have been trying to obtain the Fourier transform for time domain root-raised-cosine pulse using Mathematica. The answer I'm getting from Mathematica is not the correct answer. When I input the frequency domain response of the RRC pulse and take the inverse Fourier transform, I don't get the correct answer. Can someone please explain this? I'm trying to get the correct answer as a sanity check before getting modified pulse shapes. The code I have used is given below.

FourierTransform[
(4*a*t*Cos[((1 + a)*Pi*t)/Ts] + Ts*Sin[((1 - a)*Pi*t)/Ts])/
(Pi*t*(Ts^(-1))^(3/2)*(-16*a^2*t^2 + Ts^2)), t, ω
]

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Change pi to Pi and maybe add Assumptions->a>0. Whether this will give you the result you want I cannot say, but it will at least improve the odds. –  Daniel Lichtblau Apr 4 '13 at 13:30
Your expression differs by a factor of Sqrt[1/Ts] from that given on Wikipedia. Are you sure it's correct and that you're using a consistent normalization? –  Oleksandr R. Apr 4 '13 at 13:38
Also have a look at the option FourierParameters in the documentation. Depending on the field that you come from (electrical engineering, physics, mathematics) there are different conventions regarding the constant factors in the forward and backward transforms. –  Thies Heidecke Apr 4 '13 at 17:55
Thanks very much for the replys. @Daniel Lichtblau I made the changes as you have said. But I'm still getting the same error. THe normalizing factor should also come in front of the function, but it doesn't change the Mathematica output much. $FourierTransform[ Sqrt[1/Ts](4*a*t*Cos[((1 + a)*Pi*t)/Ts] + Ts*Sin[((1 - a)*Pi*t)/Ts])/ (Pi*t*(Ts^(-1))^(3/2)*(-16*a^2*t^2 + Ts^2)), t, ω ]$ –  James Apr 5 '13 at 3:15
Expected output is $\sqrt(Ts)$ for $0<|f| \leq \frac{(1-\alpha)}{2 Ts}$ and $\sqrt(Ts) cos (\frac{\pi |f| Ts} {2 \alpha} - \frac{\pi (1-\alpha)}{4 \alpha})$ for $\frac{1-\alpha)}{2 Ts} \leq |f| \leq \frac{(1+\alpha)}{2 Ts}$ and $0$ for $|f| > \frac{(1+\alpha)}{2 Ts}$ –  James Apr 5 '13 at 3:23