I have this function:
$$f(t)=\left | \cos\Bigl(\frac{2\pi t }{T}\Bigr) \right |$$, for $$\left | t \right |\leq \frac{5T}{4}$$ and zero otherwise.
I have to find its Fourier transform and plot its amplitude and phase specter.
My initial idea was to look at it as a multiple of a square impulse and cosine, so I could do convolution in the frequency domain, but I'm not sure how to proceed with that.
Best Answer
Using the definition we have $$F(f) = \int_{-\infty}^{\infty} f(t) e^{- 2 \pi i t f} \ dt = \int_{-\frac{5T}{4}}^{\frac{5T}{4}} \vert \cos(\frac{2\pi t }{T}) \vert e^{- 2 \pi i t f} \ dt $$ Noting that $\cos(x) \geq 0$ when $- \frac{\pi}{2} \leq x \leq \frac{\pi}{2}$ and negative elsewhere on $[-\pi,\pi]$, with periodicity on neighbouring intervals, and using the fact that $\cos(x)$ is an even function, we get $$F(f) = 2 \Big( \int_{0}^{\frac{T}{4}} \cos(\frac{2\pi t }{T}) e^{- 2 \pi i t f} \ dt - \int_{\frac{T}{4}}^{\frac{3T}{4}} \cos(\frac{2\pi t }{T}) e^{- 2 \pi i t f} \ dt + \int_{\frac{3T}{4}}^{\frac{5T}{4}} \cos(\frac{2\pi t }{T}) e^{- 2 \pi i t f} \ dt \Big)$$ Now to fully find $F(f)$ we need an expression for $\int \cos(\frac{2\pi t }{T}) e^{- 2 \pi i t f}$. There are many ways to do it. One way is to use Euler's formula, $$\int \cos(\frac{2\pi t }{T}) e^{- 2 \pi i t f} = \int \cos(\frac{2\pi t }{T}) \cos( 2 \pi i t f) - i \int \cos(\frac{2\pi t }{T}) \sin( 2 \pi i t f)$$ and solve each integral found in the real and imaginary parts above. I'll leave the math part for you ;)