Question
The sinusoidal signal \$f(t) = \cos(2 \pi f_m t)\$ is applied to the
input of an FM system. The corresponding modulation signal output for
\$f_m = 1kHz\$, is$$Q(t)=100 \cos(2 \pi 100 \times 10^6 t + 4 \sin(w \pi 1000t)) \space
> V$$across 50 ohm resistive load.
What is the power developed at 100MHz?
My Work
\$Q(t) = 100\cos(2 \pi 100 \times 10^6 t + 4 \sin(w \pi 1000t))\$
\$\space\space\space\space\space\space\space\space
= 100 \cos(2 \pi 100 \times 10^6 t) \cos(4 \sin(2 \pi 1000 t))
– 100 \sin (2 \pi 100 \times 10^6 t) \sin(4 sin(2 \pi 1000 t))\$
Where
\$\space\space\space\space\space\space\space\space\cos(4 \sin(2 \pi 1000 t) \approx 1\$ and;
\$\space\space\space\space\space\space\space\space\sin(4 sin(2 \pi 1000 t)) \approx 4 \sin(2 \pi 1000 t)\$
(These relationships are obtained from my lecture note)
Therefore,
\$\space\space\space\space\space\space\space\space
Q(t) = 100 \cos(2 \pi 100 \times 10^6 t) – 100 \sin (2 \pi 100 \times 10^6 t) \times 4\sin (2 \pi 1000 t)\$
Using equation \$P=\frac{V^2}{2R}\$, \$P\space =\space \frac{100^2}{2\cdot 50}=100\space W\$.
There are two things containing \$100*10^6\$ Hz, I do not know whether it is correct…
Can you explain more about this question to me?
Best Answer
Power developed at 100 MHz can be easily calculated by using Bessel function.
By Bessel function, J0(4) = -0.4
P= 1/2R * Ac^2 * J0(4)^2 = 800/R W