# Demonstrate that $x(t)$ and $\hat{x}(t)$ are orthogonal

I have to demonstrate that a function $x(t)$ and his Hilbert's transform are orthogonal, it is said:

$$\int^{\infty}_{-\infty} x(t) \cdot \hat{x}(t) dt = 0$$

I have tried the exercise using Parseval's theorem:

$$\int^{\infty}_{-\infty} x^2(t) \cdot \hat{x}^2(t) dt = \int^{\infty}_{-\infty} |X(f) \ast \hat{X}(f)|^2 df = \int^{\infty}_{-\infty} |X(f) \ast (-jsgn(f)X(f))|^2 df =$$

$$= \int^{0}_{-\infty} |X(f) \ast jX(f)|^2 df + \int^{\infty}_{0} |X(f) \ast (-jX(f))|^2 df$$

Using properties of convoution ($ax(t) \ast y(t)) = a(x(t) \ast y(t))$: $$\int^{\infty}_{-\infty} x^2(t) \cdot \hat{x}^2(t) dt = \int^{0}_{-\infty} | j(X(f) \ast X(f))|^2 df + \int^{\infty}_{0} |-j(X(f) \ast X(f))|^2 df =$$ $$=\int^{0}_{-\infty} |X(f) \ast X(f)|^2 df + \int^{\infty}_{0} |X(f) \ast X(f)|^2 df =$$ $$= -\int^{\infty}_{0} |X(f) \ast X(f)|^2 df + \int^{\infty}_{0} |X(f) \ast X(f)|^2 df = 0$$

But I don't know how to relationate both integrals. Maybe that isn't the way to demonstrate the exercise.

• See page 14 here. – MBaz Mar 28 '18 at 15:43

You are making this too complicated. Parseval's theorem in a more general form is $$\int_{t}^{ } u(t)^* \cdot v(t)\cdot dt= \int_{f}^{ } U(f)^* \cdot V(f)\cdot df$$

$$\int_{t}^{ } x(t)^* \cdot {\hat{x}}(t)\cdot dt= \int_{f}^{ } X(f)^* \cdot X(f)\cdot (-j \cdot sign(f))\cdot df$$

$$= -j \cdot \int_{f}^{ } |X(f)|^2 \cdot sign(f) \cdot df = 0$$

The final integral is zero only if $x(t)$ is real. This implies that $|X(f)| = |X(-f)|$ and hence the integral part can be split in two (below 0 and above 0) and because of the signum() function both parts cancel.

You have to use the following form of Parseval's identity:

$$\int_{-\infty}^{\infty}x(t)y^*(t)dt=\int_{-\infty}^{\infty}X(f)Y^*(f)df\tag{1}$$

where $^*$ denotes complex conjugation, and where $X(f)$ and $Y(f)$ are the Fourier transforms of $x(t)$ and $y(t)$, respectively. With $y(t)=\hat{x}(t)$ and $Y(f)=-j\,\textrm{sgn}(f)X(f)$, we get from $(1)$

$$\int_{-\infty}^{\infty}x(t)\hat{x}^*(t)dt=j\int_{-\infty}^{\infty}|X(f)|^2\textrm{sgn}(f)df\tag{2}$$

For real-valued $x(t)$, $|X(f)|^2$ is an even function, and, consequently, the integrand on the right-hand side of $(2)$ is odd. This implies that the integral on the right-hand side of $(2)$ equals zero if it is interpreted as a Cauchy principal value.

Note that we've used the fact that $x(t)$ is real-valued. For complex-valued $x(t)$, the integrals in $(2)$ are generally not zero.

The Hilbert transform is a complicated operator. Its precise definition, and that of its Fourier transform, require a lot of care: this is a singular integral, and one should prove results in the proper function spaces (are functions integrable, in which $L_1$, $L_2$ spaces, etc.) There are two volumes by F. W. King on the Hilbert transform: Hilbert transforms.

The previous answers rely on Fourier/Parseval versions, and did not (so far) give the proper mathematical conditions. The good thing is that Fourier offers a much more intuitive vision. However, in the spirit of the law of parsimony, I would draw another version without Fourier (whose 250th anniversary is being celebrated in France), based on the symmetry/anti-symmetry properties in the time domain for real signals.

So, as a complement, I propose a pure time version, and as well without taking care about integration condition (bad bad me), but this can provide a different insight. Let us start from:

$$\hat{x}(t)={\frac {1}{\pi }}\int _{-\infty }^{\infty }{\frac {x(\tau )}{t-\tau }}\,d\tau$$ then $$P_x=\pi\int_{-\infty}^{\infty} x(t)\hat{x}(t)dt=\int_{-\infty}^{\infty} x(t)\int_{-\infty }^{\infty }{\frac {x(\tau )}{t-\tau }}\,d\tau dt =\int_{-\infty}^{\infty} \int_{-\infty }^{\infty } x(t){\frac {x(\tau )}{t-\tau }}\,d\tau dt$$

thus, exchanging integral signs (Fubini-without-a-care)

$$P_x= \int_{-\infty}^{\infty} x(\tau )\int_{-\infty }^{\infty } \left({\frac {x(t)}{t-\tau }}dt \right) d\tau$$

then (switching the mute variables, not necessary, but to make notations clearer):

$$P_x= \int_{-\infty}^{\infty} x(t ) \left(\int_{-\infty }^{\infty }{\frac {x(\tau)}{\tau-t }}d\tau\right) dt$$ thus $$P_x= \int_{-\infty}^{\infty} x(t ) \left(-\hat{x}(t)\right) dt =-P_x\,.$$

Thus, $P_x=0$. In other words, via basic manipulations, you turn a function into its opposite.

Let us finish with a more limited, but intuitive insight: if $x(t)$ is even (resp. odd), then $\hat{x}(t)$ is odd (resp. even ). So, the product is odd, with a zero integral. An example is the cosine, whose Hilbert transform (in some well-defined sense) is a sine. And cosine and sine are orthogonal. But combined, they yield the complex exponential $\cos t + i \sin t$.