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I want to find the pre envelope of

$$x(t) = {\Pi}_{a}(t)\cos(2{\pi}f_{0}t)$$ where

I found the Fourier transform to be
$$ \begin{align} X(f) & \triangleq \mathfrak{F}[{\Pi}_{a}(t)\cos(2{\pi}f_{0}t)] \\ & = a\text{ sinc}(2a(f-f_{0})) + a\text{ sinc}(2a(f+f_{0})) \\ \end{align} $$
So $$X_{+}(f) = 2u(f)X(f)$$
But I don't know how to continue from here. Is there an easier way to find the pre envelope with another method?

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  • $\begingroup$ uhm, you sure that $$ X_+(f) = 2u(f)X(f) \quad ?$$ looks like some of $a \ \operatorname{sinc}(2 a (f-f_0))$ could leak into where $f < 0$. i think you might have assume that $|f| \ll f_0$ to say that. $\endgroup$ – robert bristow-johnson Jan 18 '16 at 3:53
  • $\begingroup$ s-mat-pcs.oulu.fi/~ssa/ESignals/sig4_5.htm $\endgroup$ – temp8jfhfhf Jan 18 '16 at 4:19
  • $\begingroup$ i know about the Hilbert transform and the pre-envelope (what i like to call the "complex envelope)". anyway, if $\frac1a \ll f_0$, then you can say that $$ X_+(f) \ \approx \ 2a \ \operatorname{sinc}(2a(f-f_0)) \quad . $$ then what is $x_+(t)$ ? which is, i think the pre-envelope. $\endgroup$ – robert bristow-johnson Jan 18 '16 at 6:38
  • $\begingroup$ also, i asked this at your other question: can you be clear about the meaning of $$ \Pi_a(t) $$ ? how wide is this rectangular function? is it as wide as $a$ or is it $2a$? $\endgroup$ – robert bristow-johnson Jan 18 '16 at 6:43
  • $\begingroup$ It is from $-a$ till $a$. The reverse Fourier of $X_{+}(f)$ seems difficult since it also has $u(f)$, so I will need to make a convolution later, and I would like to avoid that. $x_{+}(t)$ is the pre-envelope $\endgroup$ – temp8jfhfhf Jan 18 '16 at 6:54
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The pre-envelope is also called analytic signal. Its Fourier transform is indeed given by the expression in your question:

$$X_+(f)=2X(f)u(f)\tag{1}$$

where $X(f)$ is the Fourier transform of the original signal, and $u(f)$ is the unit step function. Obviously, $X_+(f)$ has only positive frequency components. The analytic signal $x_+(t)$ with its Fourier transform given by $(1)$ is necessarily a complex-valued signal:

$$x_+(t)=x(t)+j\mathcal{H}\{x(t)\}\tag{2}$$

where $\mathcal{H}\{\cdot\}$ denotes the Hilbert transform. Note that the analytic signal is not the same as the complex envelope. For a band pass signal $x(t)$, the complex envelope is a low pass signal, whereas the analytic signal is a band pass signal.

From $(2)$, in order to compute the analytic signal you need to compute the Hilbert transform of $x(t)$:

$$\hat{x}(t)=\mathcal{H}\{x(t)\}=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{x(\tau)}{t-\tau}d\tau\tag{3}$$

For the given signal you get

$$\hat{x}(t)=\frac{1}{\pi}\int_{-a}^a\frac{\cos(2\pi f_0\tau)}{t-\tau}d\tau\tag{4}$$

The result of $(4)$ can be written in terms of the cosine integral $\text{Ci}(x)$ and the sine integral $\text{Si}(x)$:

$$\hat{x}(t)=\frac{1}{\pi}\left\{\cos(2\pi f_0t)\left[\text{Ci}(2\pi f_0(t+a))-\text{Ci}(2\pi f_0(t-a))\right]+\\\sin(2\pi f_0)\left[\text{Si}(2\pi f_0(t+a))-\text{Si}(2\pi f_0(t-a))\right]\right\}\tag{5}$$

I don't think that anybody expected you to come up with that awful expression. Anyway, for large values of $f_0$, the first term in $(5)$ becomes very small, and the second term converges to $\sin(2\pi f_0)\Pi_a(t)$. So for large $f_0$ you get the expected result

$$x_+(t)\approx x(t)+j\sin(2\pi f_0)\Pi_a(t)=\Pi_a(t)e^{j2\pi f_0 t}\tag{6}$$

Obviously, the respresentation

$$x(t)=\text{Re}\left\{\Pi_a(t)e^{j2\pi f_0 t}\right\}\tag{7}$$

is always valid, but the complex-valued signal $\Pi_a(t)e^{j2\pi f_0 t}$ is no analytic signal, it's just a good approximation of the analytic signal for large values of $f_0$.

Note that for $x(t)=m(t)\cos(2\pi f_0t)$ with $m(t)$ a band-limited function, i.e. $M(f)=0$ for $|f|>B$, the complex-valued signal $m(t)e^{j2\pi f_0 t}$ is an analytic signal, as long as $f_0>B$. The problem with the function given in your question is that $\Pi_a(t)$ is not band-limited.

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