# Is it true that frequency deviation over 2*pi is the phase deviation of an FM/PM signal?

for ∆f frequency deviation if for FM only and PM has Phase deviation. and Kf= ∆f (Hz) of FM? Kf = ∆f (rad/s) for PM? and for th rules:

∆f=(kfmp)/(2p) for FM where mp = mp max-mp min ∆f=(kfmp)/(2p) for PM where mp' =mp'max-mp'min

does this rule apply for frequency deviation of the carrier and the former of Kf and Kp is for the frequency deviation of the message signal?

and do if it is correct that PM has frequency deviation, does FM has phase deviation?

I am so confused and some one clarify me this concept.

Is it true that frequency deviation over 2*pi is the phase deviation of an FM/PM signal?

No. The phase deviation is the integral with respect to time of the frequency deviation. Or, put another way, the frequency deviation is the derivative of the phase deviation with respect to time.

An angle modulated signal can be described as either phase modulated or frequency modulated.

The modulation index is the phase deviation in radians of an angle modulated signal.

The modulation index at a single modulation frequency is $\frac{FrequencyDeviation}{ModulationFrequency}$

Both the modulation index and the frequency deviation are measured as the peak deviation.

Both the frequency deviation and the modulation frequency are measured in the same units, that is they are both in Hz, or radians/s.