Electronic – Question about FM and PM and Angle Modulation

fmModulation

I am having a bit of trouble understanding how FM works mathematically. I fully understand the conceptual basics regarding how information is transfered by modulating the frequency, but what I don't get (from the equation below), is how this expression inside the cosine term actually changes the frequency. When I think of trying to change the frequency, it makes more sense to me to 2π(fc)(t) a function of Xm [ex. 2π(fc)(1 + Xm)(t) which comes out to be 2π(fc)(t) + 2π(Xm)(t)]. Why do they have the integral in there? And why is the amplitude of the carrier (Ac) specified, when it shouldn't make a difference to the output signal?

FM Equation

Also, would you be able to clarify the difference between FM and PM and angle modulation (a combination of both), and how/when each should be used. Thanks for the help

Best Answer

The instantaneous frequency of a signal \$A\cos(\phi(t))\$ where \$\phi(t)\$ is an arbitrary function of time is defined as the derivative of \$\phi(t)\$ if you want to measure frequency in radians per second and as \$\frac{1}{2\pi}\$ times the derivaive_ of \$\phi(t)\$ if you want to measure frequency in Hertz. Of course, in the common case of a fixed frequency this corresponds to the familiar \$\phi(t) = \omega_c t+\phi_0 = 2\pi f_c t + \phi_0\$.

The standard definition of a frequency-modulated signal is one in which the deviation of the instantaneous frequency (at time \$t_0\$, say), from the carrier frequency is proportional to the value \$x_m(t_0)\$ of the modulating signal \$x(t)\$ at time \$t_0\$. The constant of proportionality is denoted by \$f_{\Delta}\$ in your notation: a \$1\$ volt signal creates a deviation of \$f_{\Delta}\$ Hz. Thus, if \$A\cos(\phi(t))\$ is the FM signal, then we have that $$ \left.\frac{\mathrm d}{\mathrm dt}\phi(t)\right|_{t=t_0} = 2\pi f_c + 2\pi f_{\Delta} x_m(t_0) $$ so that the deviation of the instantaneous frequency \$f_c + f_{\Delta} x_m(t_0)\$ from the carrier frequency \$f_c\$ is \$f_{\Delta} x_m(t_0)\$, just as we want it to be. It then follows from the fundamental theorem of calculus that $$\phi(t_0) = \int_{0}^{t_0}2\pi f_c + 2\pi f_{\Delta} x_m(t_0)\, \mathrm dt = 2\pi f_c t_0 + \int_{0}^{t_0} 2\pi f_{\Delta} x_m(t_0)\, \mathrm dt$$ or, with a slight change in notation, the FM signal can be expressed as $$A\cos\left(2\pi f_c t + \int_{0}^{t} 2\pi f_{\Delta} x_m(\tau)\, \mathrm d\tau\right)$$ the way you have it. Note that \$A\$ is the amplitude of the FM signal and is fixed; it is the frequency that is varying. Surely we need to distinguish between the FM signal when it is created using a voltage-controlled oscillator with an amplitude of \$1\$ volt and when it comes out of the power amplifier and goes to the antenna with a power of 10 kW?