I need to calculate the power of the sidebands and the carrier. The carrier is 10*cos(2pi10⁶t) [V] and the modulating is 3*cos(2pi10³t) [V]. The modulated signal is 10*cos(\$w_0\$t)+\$\frac{3}{2}\$*cos(\$w_0\$t+\$w_m\$t)+\$\frac{3}{2}\$*cos(\$w_0\$t-\$w_m\$t).

The furthest I can get is to \$\frac{m²(t)∗u²}{1+m²(t)∗u²}\$ But I can't find what m(t) is.

How can I calculate the percentage of power in the sidebands compared to the total AM power. Thanks in advance.


closed as off-topic by Matt Young, Wesley Lee, uint128_t, Dmitry Grigoryev, laptop2d Apr 4 '17 at 7:16

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    \$\begingroup\$ Homework with no attempt at a solution \$\endgroup\$ – Matt Young Apr 1 '17 at 23:42
  • \$\begingroup\$ yep I have an attemp. I get to \$\frac{m²(t)*u²}{1+m²(t) * u²}\$ \$\endgroup\$ – scottbear Apr 1 '17 at 23:47
  • \$\begingroup\$ @MattYoung but I can't figure out where the m(t) comes from. Not homework but exam study \$\endgroup\$ – scottbear Apr 1 '17 at 23:48
  • \$\begingroup\$ Here is a clue to help you! Think about the Power Spectral Density of the signal. Were you shown how to calculate psd ? \$\endgroup\$ – AugustCrawl Apr 2 '17 at 0:21

Your result consists of three components: One for the carrier and one for each sideband. The multiplier term is the peak voltage of each. The power in each is proportional to the square of the voltages:

10^2 : (3/2)^2 : (3/2)^2.

Then it is trivial to calculate the ratio of the powers.


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