6
$\begingroup$

In Signals and Systems by A. V. Oppenheim, A. S. Willsky, S. Hamid Nawab, 2nd Edition, and Signals and Systems, Simon Haykins, Barry Van Veen, 2nd Edition there is a problem related to energy of real-valued even and odd signal.

Energy of an arbitrary real-valued signal $x(t)$ is equal to the sum of the energy of the even component $x_{e}(t)$ and energy of the odd component $x_{o}(t)$ i.e. \begin{equation} \int_{-\infty}^{\infty}x^2(t)dt = \int_{-\infty}^{\infty}x_{e}^2(t)dt + \int_{-\infty}^{\infty}x_{o}^2(t)dt\notag \end{equation} Which can be proved very easily by expressing $x(t) = x_{e}(t) + x_{o}(t)$ and evaluating its energy as \begin{align} \int_{-\infty}^{\infty}x^2(t)dt &= \int_{-\infty}^{\infty}\big(x_{e}(t) + x_{o}(t)\big)^2dt\notag\\ &= \int_{-\infty}^{\infty}x_{e}^{2}(t)dt + \int_{-\infty}^{\infty}x_{o}^{2}(t)dt +2\int_{-\infty}^{\infty}x_{e}(t)x_{o}(t)dt\notag. \end{align} The term $x_{e}(t)x_{o}(t)$ in the last integration corresponds to an odd signal (multiplication of even and odd signal results into odd signal) and its integration (joint energy) is equal to zero (area under an odd signal over symmetrical limits is zero). Hence \begin{align} \int_{-\infty}^{\infty}x^2(t)dt = \int_{-\infty}^{\infty}x_{e}^{2}(t)dt + \int_{-\infty}^{\infty}x_{o}^{2}(t)dt\notag \end{align}

I was wondering, what if I consider $x(t)$ to be a complex signal, will above property still be valid. I tried as \begin{equation} \int_{-\infty}^{\infty}|x(t)|^2dt = \int_{-\infty}^{\infty}|x_{e}(t)|^2dt + \int_{-\infty}^{\infty}|x_{o}(t)|^2dt\notag \end{equation} Expressing $x(t) = x_{e}(t) + x_{o}(t)$ and evaluating its energy as \begin{align} \int_{-\infty}^{\infty}|x(t)|^2dt &= \int_{-\infty}^{\infty}\big|x_{e}(t) + x_{o}(t)\big|^2dt\notag\\ &= \int_{-\infty}^{\infty}|x_{e}(t)|^{2}dt + \int_{-\infty}^{\infty}|x_{o}(t)|^{2}dt +\int_{-\infty}^{\infty}[x_{e}(t)x_{o}^{*}(t) + x_{o}(t)x_{e}^{*}(t)]dt\notag \end{align}

On expanding the term \begin{align} x_{e}(t)x_{o}^{*}(t) &= \big(x_{er}(t) + jx_{ei}(t)\big)\big(x_{or}(t) - jx_{oi}(t)\big)\\ &= x_{er}(t)x_{or}(t) + x_{ei}(t)x_{oi}(t) - jx_{er}(t)x_{oi}(t) + jx_{ei}(t)x_{or}(t)\\ x_{o}(t)x_{e}^{*}(t) &= \big(x_{or}(t) + jx_{oi}(t)\big)\big(x_{er}(t) - jx_{ei}(t)\big)\\ &= x_{er}(t)x_{or}(t) + x_{ei}(t)x_{oi}(t) - jx_{ei}(t)x_{or}(t) + jx_{er}(t)x_{oi}(t) \end{align} Further on adding this terms results into \begin{align} \int_{-\infty}^{\infty}|x(t)|^2dt &= \int_{-\infty}^{\infty}|x_{e}(t)|^{2}dt + \int_{-\infty}^{\infty}|x_{o}(t)|^{2}dt +2\int_{-\infty}^{\infty}\big(x_{er}(t)x_{or}(t) + x_{oi}(t)x_{ei}(t)\big)dt\notag \end{align} The terms $x_{er}(t)x_{or}(t)$ and $x_{oi}(t)x_{ei}(t)$ in the last integration corresponds to an odd signal and its integration (joint energy) is equal to zero. Hence \begin{equation} \int_{-\infty}^{\infty}|x(t)|^2dt = \int_{-\infty}^{\infty}|x_{e}(t)|^2dt + \int_{-\infty}^{\infty}|x_{o}(t)|^2dt\notag \end{equation} I would appreciate if anyone here can validate the proof. Is it correct to say that the same property too holds for complex signal.

Improve this question
$\endgroup$
  • $\begingroup$ deconstruct $x_e$ and $x_o$ into real and imaginary part; fully execute the the $x_ex_o^*$ and $x_e^*x_o$ products :) $\endgroup$ – Marcus Müller Mar 13 '18 at 14:15
  • $\begingroup$ @MarcusMüller, I have expanded them... $\endgroup$ – Meet Mar 13 '18 at 14:57
  • $\begingroup$ I appreciate you remind me this property $\endgroup$ – Laurent Duval Mar 13 '18 at 17:27
3
$\begingroup$

Your result is correct but note that for complex signals, the even and odd parts are defined by

$$x_e(t)=\frac12\left[x(t)+x^*(-t)\right]\tag{1}$$ and $$x_o(t)=\frac12\left[x(t)-x^*(-t)\right]\tag{2}$$

where $^*$ denotes complex conjugation.

From $(1)$ and $(2)$ it follows that the real part of $x_e(t)$ is even and its imaginary part is odd, whereas $x_o(t)$ has an odd real part and an even imaginary part.

We get

$$|x(t)|^2=|x_e(t)+x_o(t)|^2=|x_e(t)|^2+|x_o(t)|^2+2\,\text{Re}\{x_e(t)x_o^*(t)\}\tag{3}$$

From

$$\begin{align}x_e(t)x_o^*(t)&=\frac14\left[x(t)+x^*(-t)\right]\left[x^*(t)-x(-t)\right]\\&=\frac14\left[|x(t)|^2-x(t)x(-t)+x^*(t)x^*(-t)-|x(-t)|^2\right]\end{align}\tag{4}$$

we see that $x_e(t)x_o^*(t)$ is an odd signal according to definition $(2)$ (it is the odd part of $\frac12 \left[|x(t)|^2-x(t)x(-t)\right]$), and, consequently, its real part is odd. This implies that the integral over the last term in $(3)$ is zero, and the stated property holds.

Improve this answer
$\endgroup$
  • $\begingroup$ Good explanation. $\endgroup$ – QMC Mar 16 '18 at 4:50
  • $\begingroup$ Please correct me. The equation in $(1)$ and $(2)$ are known as conjugate-symmetric (CS) and conjugate-antisymmetric (CA) part of the complex signal. Where in general by the definition of CS i.e. $\mathbf{x}(-t) = \mathbf{x}^{*}(t)$, where $\operatorname{\mathbb{R}e}\{\mathbf{x}(t)\}$ has even symmetry and $\operatorname{\mathbb{I}m}\{\mathbf{x}(t)\}$ has odd symmetry. Whereas, CA is defined as $\mathbf{x}(-t) = -\mathbf{x}^{*}(t)$, where $\operatorname{\mathbb{R}e}\{\mathbf{x}(t)\}$ has odd symmetry and $\operatorname{\mathbb{I}m}\{\mathbf{x}(t)\}$ has even symmetry. $\endgroup$ – Meet Mar 18 '18 at 11:53
  • $\begingroup$ @Meet: Yes, also know as Hermitian (anti-)symmetry. $\endgroup$ – Matt L. Mar 18 '18 at 12:05
  • $\begingroup$ @MattL.: How can you say its even and odd signals decomposition are given as in $(1)$ and $(2)$. Or there is no difference in Hermitian and even. And skew-Hermitian and odd regarding complex signal. $\endgroup$ – Meet Mar 18 '18 at 12:16
  • 1
    $\begingroup$ @Meet: Because that's how even (symmetric) and odd (anti-symmetric) are usually defined for complex signals. $\endgroup$ – Matt L. Mar 18 '18 at 12:17
0
$\begingroup$

Sometimes, notations can simplify the computations.

Let us write $\underline{x} = \overline{x(-t)}$, where $\overline{\cdot} $ is the complex conjugate. In the complex case, it is more often called Hermitian/skew-Hermitian decomposition

$$\frac{x+\underline{x}}{2} (\textrm{hermitian})+\frac{x-\underline{x}}{2} (\textrm{skew-hermitian})$$

then under the integral, the second term is:

$$\frac{x+\underline{x}}{2}\overline{\frac{x+\underline{x}}{2}}+\frac{x-\underline{x}}{2}\overline{\frac{x-\underline{x}}{2}}$$

Multiplying by 4, one gets:

$$(x+\underline{x})\cdot(\overline{x+\underline{x}})+(x-\underline{x})\cdot(\overline{x-\underline{x}})$$

and cross terms cancel nicely.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.