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Let $x_\mathrm{a}(t)$ be the analytic signal for real signal $x(t)$. I want to find an expression for $\mathscr{F}\{|x_\mathrm{a}(t)|^2\}(f)$ in terms of $x(t)$. The analytic signal can be written as $x_\mathrm{a}(t) = x(t)+j\hat{x}(t)$ where $\hat{x}(t)$ denotes the Hilbert transform of $x(t)$. So we have:

$$\begin{align} |x_\mathrm{a}(t)|^2 &= (x(t)+j\hat{x}(t))(x(t)-j\hat{x}(t)) \\ &=x^2(t)+\hat{x}^2(t) \\ \end{align}$$

Taking Fourier transform and using its properties (Fourier transform of product is convolution of Fourier transforms), we get:

$$\begin{align} \mathscr{F}\{|x_\mathrm{a}(t)|^2\}(f) &= (X(\lambda)\star X(\lambda))(f)+(\hat{X}(\lambda)\star\hat{X}(\lambda))(f) \\ &= (X(\lambda)\star X(\lambda))(f)+(j\ \text{sgn}(\lambda)X(\lambda)\star j\ \text{sgn}(\lambda)X(\lambda))(f) \\ &= (X(\lambda)\star X(\lambda))(f)-(\text{sgn}(\lambda)X(\lambda)\star\text{sgn}(\lambda)X(\lambda))(f) \\ \end{align}$$

By definition of convolution, we have:

$$ \mathscr{F}\{|x_\mathrm{a}(t)|^2\}(f) = \int_{-\infty}^{+\infty}X(\lambda)X(f-\lambda)d\lambda - \int_{-\infty}^{+\infty}\text{sgn}(\lambda)\text{sgn}(f-\lambda)X(\lambda)X(f-\lambda)d\lambda $$

Consider $f>0$ and split the integrals:

$$\begin{align} \mathscr{F}\{|x_\mathrm{a}(t)|^2\}(f) &= \int_{-\infty}^{0}X(\lambda)X(f-\lambda)d\lambda + \int_{0}^{f}X(\lambda)X(f-\lambda)d\lambda + \int_{f}^{+\infty}X(\lambda)X(f-\lambda)d\lambda \\ &\qquad - \left( -\int_{-\infty}^{0}X(\lambda)X(f-\lambda)d\lambda + \int_{0}^{f}X(\lambda)X(f-\lambda)d\lambda - \int_{f}^{+\infty}X(\lambda)X(f-\lambda)d\lambda \right) \\ &= 2\int_{-\infty}^{0}X(\lambda)X(f-\lambda)d\lambda + 2\int_{f}^{+\infty}X(\lambda)X(f-\lambda)d\lambda \\ \end{align}$$

Is this computation correct? Also I don't know whether it can be simplified further.

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  • $\begingroup$ If I were you, since you're in continuous-time domain and that's where the Laplace Transform sometimes exists which uses "$s$", I would turn every instance of "$s(t)$" to $x(t)$ and "$S(f)$" to $X(f)$. $\endgroup$ Nov 6, 2023 at 17:50
  • $\begingroup$ You need to learn \begin{align} and \end{align} in $\LaTeX$. And change = to &= in those aligned multi-equation statements. Might I show you how to do that? $\endgroup$ Nov 6, 2023 at 17:54
  • $\begingroup$ @robertbristow-johnson Thanks, I changed the notation. Yes please, I would appreciate it. $\endgroup$
    – S.H.W
    Nov 6, 2023 at 18:04

1 Answer 1

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Your result is correct, but it can be simplified by noticing that the two integrals in your solution are actually identical, which for $f>0$ leads to

$$\mathscr{F}\left\{|x_\mathrm{a}(t)|^2\right\}=4\int_f^{\infty}X(\nu)X(f-\nu)d\nu,\qquad f>0$$

This can be seen by substituting $\nu=f-\lambda$ in the first integral of your solution:

\begin{align*} \int_{-\infty}^0X(\lambda)X(f-\lambda)d\lambda &= \int_f^{\infty}X(f-\nu)X(\nu)d\nu \end{align*}

A more straightforward way to arrive at this result is to write the spectrum of the analytic signal as

$$X_\mathrm{a}(f)=2X(f)u(f)$$

where $u(f)$ is the unit step function. With

$$|x_\mathrm{a}(t)|^2=x_\mathrm{a}(t)x_\mathrm{a}^*(t)\Longleftrightarrow X_\mathrm{a}(f)\star X_\mathrm{a}^*(-f)$$

we obtain

\begin{align*} \mathscr{F}\left\{|x_\mathrm{a}(t)|^2\right\} &= 4\int_{-\infty}^{\infty}X(\nu)u(\nu)X^*(\nu-f)u(\nu-f)d\nu \\ &= 4\int_{\max\{0,f\}}^{\infty}X(\nu)X^*(\nu-f)d\nu \\ &= 4\int_{\max\{0,f\}}^{\infty}X(\nu)X(f-\nu)d\nu\tag{1} \end{align*}

where I've used $X(f)=X^*(-f)$ since $x(t)$ is real. Note that $(1)$ is valid for positive as well as negative $f$.

EXAMPLE:

Let's look at a very simple example to see how Eq. $(1)$ gives a correct result. Let $x(t)=\cos(2\pi f_0t)$. We have $x_\mathrm{a}(t)=e^{j2\pi f_0t}$ and $|x_\mathrm{a}(t)|^2=1$. Consequently, $\mathscr{F}\{|x_\mathrm{a}(t)|^2\}=\delta(f)$.

Now we use Eq. $(1)$ to obtain the same result. We have

$$X(f)=\frac12\big[\delta(f-f_0)+\delta(f+f_0)\big]$$

From Eq. $(1)$ we have

\begin{align*} \require{cancel} \mathscr{F}\left\{|x_\mathrm{a}(t)|^2\right\} &= 4\frac14\int_{\max\{0,f\}}^{\infty}\big[\delta(\nu-f_0)+\cancel{\delta(\nu+f_0)}\big]\\&\qquad\qquad\big[\cancel{\delta(f-\nu-f_0)}+\delta(f-\nu+f_0)\big]d\nu \\ &= \int_{\max\{0,f\}}^{\infty}\delta(\nu-f_0)\delta(f-\nu+f_0)d\nu \\ &= \delta(f) \end{align*}

where I've cancelled all Dirac impulses whose argument will not become zero inside the integration interval.

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  • $\begingroup$ Thanks. If we consider $x(t) = \cos(2\pi f_1t) + \cos(2\pi f_2t)$, by direct computation of $|x_a(t)|^2$, it can be seen that $\mathscr{F}\left\{|x_a(t)|^2\right\}$ is a sum of $\delta(f)$, $\delta(f_1 - f_2)$ and $\delta(f_2 - f_1)$ (each term has a weight). How can we obtain this result using $(1)$? $\endgroup$
    – S.H.W
    Nov 7, 2023 at 8:59
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    $\begingroup$ @S.H.W: In that case, $X(f)$ has $4$ Dirac impulses. In the integral $(1)$ you can remove all Dirac impulses whose argument will not become zero inside the integration interval. E.g., you'll have a term $X(\nu+f_1)$, and since $\nu$ (and $f_1$) will always be positive, you can remove that term. You'll be left with the sum of two Diracs times the sum of two other Diracs, which after integration will give you the desired result. $\endgroup$
    – Matt L.
    Nov 7, 2023 at 12:52
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    $\begingroup$ @S.H.W: I've added a simple example to my answer, just to show how it works. $\endgroup$
    – Matt L.
    Nov 7, 2023 at 13:22
  • $\begingroup$ Thanks a lot for great answer. $\endgroup$
    – S.H.W
    Nov 7, 2023 at 16:04

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