# Power Relation in DSB-FC AM [closed]

So, in my class I have to learn how to derive the power relation in a Double Side Band, Full Carrier Amplitude Modulated Wave However, I have hit a wall somewhere.

The basic equation is this :

$P_{total} = P_{carrier} + P_{lower-side-band} + P_{upper-side-band}$
Then,

$P_{carrier} = \frac{E_c^{2}}{R}$ where $E_c$ is amplitude of carrier.
Now , when converting to R.M.S, I divide the numerator by $\sqrt2$ and I get: $\frac{E_c^2}{2R}$

$P_{lower-side-band} = P_{upper-side-band} = \frac{(\frac{mE_c}{2})^2}{R} = \frac{m^2E_c^2}{4R}$

The problem is that the notes provided to me by my teacher state the numerator is $8R$ and I do not know why. Can someone please tell me if I have made a mistake somewhere.

• This question appears to be off-topic because it is about a specific homework question that isn't likely to be helpful to anyone else. – Jason R Sep 17 '13 at 15:20
• @JasonR yes but shows research effort nonetheless – An SO User Sep 17 '13 at 15:21

Consider your carrier signal as $E_c\cos W_{ct}$ and your message signal is $E_m \cos W_{mt}$.

The double side band plus carrier signal is,

$E_c\cos(W_{ct})+(E_c\cos(W_{ct})E_m\cos(W_{mt}))$

$E_c\cos(W_{ct})+\frac{1}{2}(E_cE_m[\cos(W_{ct}+W_{mt})+\cos(W_{ct}-W_{mt})])$

$Ec*Cos(W_{ct})+\frac{1}{2}E_cE_m\cos(W_{ct}+W_{mt})+\frac{1}{2}*Ec*Em*Cos(W_{ct}-W_{mt})$

Carrier being $E_c\cos(W_{ct})$

Upper Side band being $\frac{1}{2}E_cE_m\cos(W_{ct}+W_{mt})$ and

Lower Side band being $\frac{1}{2}E_cE_m\cos(W_{ct}-W_{mt})$.

For a signal $A\cos W_t$, the average power is $(\frac{A^2}{R}T)*(\int^T_0 \cos^2 W_t) = \frac{{A_c}^2}{2R}.$

Hence,

Average Power of Carrier is $\frac{{E_c}^2}{2R}$

Average Power of Upper Side band is $\frac{{E_cE_m}^2}{8R}$

Average Power of Upper Side band is $\frac{{Ec*Em}^2}{8R}$