# Convolution of two square pulses and the fourier transform of a triangular pulse

I would like some help clearing up some confusion.

The fourier transform of a triangular pulse with base $$2T_b$$ is given as,

$$T_bsinc^2(\pi f T_b)$$

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,

$$T_bsinc(\pi f T_b)$$

Using the property,

Convolution in time $$<=======>$$ 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,

$$T_b^2 sinc^2(\pi f T_b)$$

Though this does not seem correct as it has an extra factor of $$T_b$$,

Could someone please clear my confusion?

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 $$T_b\mathrm{sinc}^2(\pi T_b f)$$