I would like some help clearing up some confusion.
The Fourier transform of a triangular pulse with base $2T_b$ is given as,
$$T_b \mbox{sinc}^2(\pi f T_b)$$
You can also get this result by realizing that 2 square pulses each with width $T_b$ convolved together make a triangular pulse with width $2T_b$.
The Fourier transform of a square pulse with width $T_b$ is given as,
$$T_b \mbox{sinc}(\pi f T_b)$$
Using the property,
$$\text{Convolution in time} \longleftrightarrow \text{Multiplication in frequency}$$
You can find that the convolution of two square pulses together, is just their fourier transforms multiplied together. Thus, a triangular pulse of width $2T_b$ is just the fourier transforms of two square pulses multiplied together, right?
Using this result I find that the Fourier transform of the triangular pulse is,
$$T_b^2 \mbox{sinc}^2(\pi f T_b)$$
Though this does not seem correct as it has an extra factor of $T_b$. Could someone please clear my confusion?
Best Answer
Both are correct. There is no contradiction. It's just that you're not talking of the Fourier transform of the same function in both cases. You forgot to consider the height of the triangle when convolving the square pulses.
If you take two square pulses of base $T_b$ and of height 1, their convolution is indeed a triangular pulse of width $2T_b$, but of height $T_b$, not 1. So the Fourier transform of the triangular pulse of width $2T_b$ and height $T_b$ is indeed $$T_b^2\mathrm{sinc}^2(\pi T_b f)$$
Now when you speak of "a triangular pulse with base $T_b$", you seem to be implying "of height 1". If that's what you mean then it follows from above that the Fourier transform for that triangular pulse is $$T_b\mathrm{sinc}^2(\pi T_b f)$$