# Why does norm of the PAM bandpass signal give energy of the pulse over two?

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:

$$\mathcal{E}_f = = \int_{-\infty}^{+\infty}f(t)f^*(t)dt$$ 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?

• 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 $T=\frac{1}{f_c}$ Commented Oct 3, 2022 at 13:27
• 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 Commented Oct 3, 2022 at 20:24

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