Find the Fourier transform of the given function.

Question

We have to find the Fourier transform of the given function:

The answer to the above question is:

However, I get a slightly different answer. I think that a negative sign, which ought to have been added is missing. Please help me with solving this question. I am enclosing herewith my answer sheet.

Answer 1

Writing the integral as,

$$I = \int_{-2}^0-te^{-j2\pi ft}dt+\int_{0}^2te^{-j2\pi ft}dt$$ Replace $t$ by $-t$ in the 1st integral

$$I = \int_{0}^2t\left(e^{-j2\pi ft}+e^{j2\pi ft}\right)dt=\int_0^22t\cos(2\pi ft)dt$$ $$I = \left[2t\sin\frac{(2\pi ft)}{2\pi f} - 2\cos(2\pi ft)\frac{-1}{(2\pi f)^2}\right]^2_0 = 4\frac{\sin(4\pi f)}{2\pi f} +2\frac{\cos(4\pi f)-1}{(2\pi f)^2}$$

$$I = 8\frac{\sin(4\pi f)}{4\pi f} -4\frac{\sin^2(2\pi f)}{(2\pi f)^2}$$

$$I = 8\ \text{sinc}(4 f) - 4\ \text{sinc}^2(2f)$$

Answer 2

The Fourier transform of $x(t)$ evaluated at $f=0$ is the integral of $x(t)$ (times some positive number depending on the domain and the convention used). As the integral is positive, you can use this to check for minus sign errors - if the answer is either $8\operatorname{sinc}(4f) - 4 \operatorname{sinc}^2(2f) $ or its negative, then it has to be $8\operatorname{sinc}(4f) - 4 \operatorname{sinc}^2(2f) $ because $$8\operatorname{sinc}(4\times 0) - 4 \operatorname{sinc}^2(2\times 0) >0.$$