0
\$\begingroup\$

I would like to understand why \$\int_{-\infty}^{+\infty}g^2(t)cos^2(2\pi f_c t)dt = \mathcal{E}_g/2\$

\$g(t)\$ is a pulse of duration T.

the background is:

reading the book "Digital Communications" 5th Edition from Proakis chapter 3, page 99 he defines PAM signals as:

$$ s_m(t) = Re[A_m g(t)e^{j2\pi f_c t}] $$

where \$g(t)\$ is a real funciton, \$A_m\$ is real and usually \$A_m = \{\pm1, \pm3, \pm5, ... \pm (M-1)\}\$ where \$M\$ is the number of amplitudes so that \$M=2^k\$ and \$k\$ is the number of bits per symbol.

for example, if \$k=2\$, then \$M=4\$ and \$A_m = \{\pm1, \pm3\}\$

The energy of a signal \$f(t)\$ is given by the inner product:

$$\mathcal{E}_f = <f(t), f(t)> = \int_{-\infty}^{+\infty}f(t)f^*(t)dt$$ where \$f^*(t)\$ is the complex conjugate of \$f(t)\$.

For PAM, the energy of the singal \$s_m(t)\$ is: $$\mathcal{E}_s = \int_{-\infty}^{+\infty}s_m^2(t)dt$$

for the baseband signal \$s_m(t)=A_mg(t)\$ we have $$\mathcal{E}_s = A_m^2\int_{-\infty}^{+\infty}g^2(t)dt=A_m^2\mathcal{E}_g$$

then a basis for baseband PAM signals would be

$$\phi (t)=\frac{g(t)}{||g(t)||}= \frac{g(t)}{\sqrt{\mathcal{E}_g}}$$

for the bandpass PAM signals, the basis would be

$$\phi (t)=\frac{g(t)cos(2\pi f_c t)}{||g(t)cos(2\pi f_c t)||}= \frac{g(t)cos(2\pi f_c t)}{\sqrt{\int_{-\infty}^{+\infty}g^2(t)cos^2(2\pi f_c t)dt}}$$

The next step is exacly what I don't understand: the integral in the denominator is \$\int_{-\infty}^{+\infty}g^2(t)cos^2(2\pi f_c t)dt = \mathcal{E}_g/2\$. Why is that?

\$\endgroup\$
2
  • \$\begingroup\$ Are you sure the integral is \$ \int_{-\infty}^{+\infty} \$ and not \$ \int_{0}^{T} \$ ? If it's the former, maybe \$ g(t) = 0 \$ outside of \$ [0,T] \$ with \$ T=\frac{1}{f_c} \$ \$\endgroup\$
    – Rahmany
    Commented Oct 3, 2022 at 13:27
  • \$\begingroup\$ yes I'm sure, that comes simply from the definition of the norm of a function, or the energy of a funciton. \$g(t)\$ is a pulse of duration T, but I can't see how this leads to \$\mathcal{E}_s/2\$. Maybe I should add in the question the info that \$g(t)\$ is a pulse of duration T \$\endgroup\$
    – nanogauss
    Commented Oct 3, 2022 at 20:24

1 Answer 1

0
\$\begingroup\$

Since the band-pass PAM has \$p(t)=g(t)\cdot \cos(\omega t)\$, the energy of \$g(t)\$ is \$E_g\$ (assume).

The energy of a finite cosine signal over an interval 0 to T is 1/2; if we represent a waveform in the frequency domain using Fourier transform it should be BIBO (absolutely integrable).

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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