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?

  • \$\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


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).


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