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Given plots of magnitude and phase of a Fourier transform, how can I obtain the analytical formula for the signal?

I know that $$X_1(j\omega) = 3j\omega[H(\omega+3\pi)-H(\omega-3\pi)]$$ but I can't figure out a way from this point to the answer, which is $$x_1(t)=\frac{3}{\pi t^2}[3\pi t\cos(3\pi t)-\sin(3\pi t)]$$

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  • $\begingroup$ What is $H(\omega)$? Did you take $\angle X_1(j\omega)$ into account when writing $X_1(j\omega)$? Are you sure about the correctness of $x_1(t)$? What happens when $t=0$? $\endgroup$ – MBaz Dec 30 '16 at 23:42
  • $\begingroup$ It is the Heaviside step function, it is so natural for me that I forgot to define it here, I am really sorry!! $\endgroup$ – Vinícius Lopes Simões Dec 31 '16 at 0:04
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In this case, it is quite simple. In general, it may be more difficult. I'm not sure what you are getting at with $H(j \omega)$ since you don't define it anywhere, but you are trying to get the analytical time domain expression for the signal from the plots. This is possible by noticing that $$X_1(j \omega) = |X_1(j \omega)| \exp\left( j \angle X_1(j \omega)\right).$$ In this case, due to the plots' simplicity, we can generate a formula by reading directly from the axes. In this case, $$X_1(j \omega) = 3 j \omega \left[ u(\omega + 3 \pi) - u(\omega - 3 \pi) \right],$$ where $u(t)$ is the heaviside step function.

Now, we can calculate the Inverse Fourier Transform with the formula, $$ x_1(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} X_1(j \omega) \exp\left(j \omega t \right) d \omega. $$ Plugging in our $X_1(j \omega)$ and simplifying leads us to the expression $$ x_1(t) = \frac{1}{2 \pi} \int_{-3 \pi}^{3 \pi} 3 j \omega \exp\left(j \omega t \right) d \omega.$$

I assume that you can evaluate this integral. If you need more assistance, let me know.

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You already have a correct answer, but I'd like to show you a little trick that allows you to derive the solution without solving any integrals. You've already figured out the analytical expression for $X_1(j\omega)$, which is an important step. Now you could observe that this expression contains a factor $j\omega$. Remember that multiplication by $j\omega$ in the frequency domain corresponds to differentiation in the time domain:

$$\mathcal{F}\left\{\frac{dx(t)}{dt}\right\}\Longleftrightarrow j\omega X(j\omega)\tag{1}$$

This means that you could obtain the function $x_1(t)$ by first deriving the inverse Fourier transform of

$$X(j\omega)=\frac{X_1(j\omega)}{j\omega}=3[u(\omega+3\pi)-u(\omega-3\pi)]\tag{2}$$

which is a simple rectangular function. It's corresponding time domain function is

$$x(t)=\frac{3\sin(3\pi t)}{\pi t}\tag{3}$$

This is a basic Fourier transform relationship, which you should know by heart. The desired function $x_1(t)$ is now given by the derivative of $(3)$:

$$\begin{align}x_1(t)=\frac{dx(t)}{dt}&=\frac{(3\pi)^2 t\cos(3\pi t))-3\pi\sin(3\pi t))}{(\pi t)^2}\\&=\frac{3}{\pi t^2}[3\pi t\cos(3\pi t)-\sin(3\pi t)]\tag{4}\end{align}$$

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