Do all phase quadrature signals form orthogonal pairs with their in-phase versions

Question

I have been studying analog modulation, in that i found a concept of quadrature carrier multiplexing which says that two baseband signals with possibly different bandwidths can be modulated to occupy the same transmission bandwidth in the passband by modulating them with carrier waves that are in phase quadrature.

We typically use cosines as carrier waves and their quadrature version is the sine, which are orthogonal

Now I am thinking that, by this argument, do all signals form orthogonal pairs to their phase quadrature versions ?

$$\int_{-\infty}^\infty x(t)x^\wedge(t)dt=0$$ where $x^\wedge(t)$ is the hilbert transform of x(t)

If this is true, does this concept has anything to do with the idea of canonical representation of passband signals in terms of their in-phase and quadrature phase representation. Could anyone enlighten me on this?

Answer 1

It is true that a (real-valued) signal $x(t)$ and its Hilbert transform $\hat{x}(t)$ are orthogonal, i.e.

$$\int_{-\infty}^{\infty}x(t)\hat{x}(t)dt=0\tag{1}$$

This can be shown as follows. First, note that according to Parseval's theorem we have

$$\int_{-\infty}^{\infty}x(t)y^*(t)dt=\int_{-\infty}^{\infty}X(f)Y^*(f)df\tag{2}$$

where $^*$ denotes complex conjugation, and $X(f)$ and $Y(f)$ are the Fourier transforms of $x(t)$ and $y(t)$, respectively.

If we want to apply $(2)$ to $(1)$, we first need the Fourier transform of $\hat{x}(t)$, which is given by

$$\mathcal{F}\left\{\hat{x}(t)\right\}=-j\operatorname{sgn}(f)X(f)\tag{3}$$

With $(2)$ and $(3)$, the integral in $(1)$ can be written as

\begin{align*} \int_{-\infty}^{\infty}x(t)\hat{x}(t)dt &= \int_{-\infty}^{\infty}X(f)j\operatorname{sgn}(f)X^*(f)df\\ &= j\int_{-\infty}^{\infty}|X(f)|^2\operatorname{sgn}(f)df\tag{4} \end{align*}

Since $x(t)$ is real-valued, $|X(f)|^2$ is an even function, and, consequently, the integrand in $(4)$ is an odd function. Hence, the integral $(4)$ equals zero (in the Cauchy principal value sense), and $(1)$ is satisfied.