# Pre-envelope of ${\Pi}_{a}(t)\cos(2{\pi}f_{0}t)$

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?

• 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. – robert bristow-johnson Jan 18 '16 at 3:53
• s-mat-pcs.oulu.fi/~ssa/ESignals/sig4_5.htm – temp8jfhfhf Jan 18 '16 at 4:19
• 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. – robert bristow-johnson Jan 18 '16 at 6:38
• 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 $2a$? – robert bristow-johnson Jan 18 '16 at 6:43
• 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 – temp8jfhfhf Jan 18 '16 at 6:54

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.