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Source 1 defines the modulation index in the expression for FM as follows:

enter image description here

  • As per the expression for frequency modulated wave, isn't the frequency deviation same as Modulation index 'M'?

"The max value for the quantity inside bracket is: {\$(\omega_c + M)t\$} (valid for a short duration around the peak of modulating signal) and hence max frequency deviation ( angular) should be M"

\$\color{red}{\text{EDIT:}}\$ The above (italic) sentence is wrong. The correct max value for the quantity inside bracket is: (\$\omega_c t + M\$)

ie, M is not multiplied by t.


Source 2 defines the modulation index in the expression for FM as follows:

  • This is inline with the expectation listed above

enter image description here

Wiki defines the FM in a different manner, involving the integral of signal, which I think would lead to the expression in Source 1. But still frequency deviation definition seems to be confusing.

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  • \$\begingroup\$ I don't see where you think there is an anomaly (in your title question). \$\endgroup\$
    – Andy aka
    Commented May 3, 2023 at 15:20
  • \$\begingroup\$ @Andyaka: As per Source 1 expression, the maximum frequency is 'wc +M' which means max deviation should be 'M'. This is not inline with the definition of M (which is defined as peak frequency deviation divided by modulation frequency. \$\endgroup\$ Commented May 3, 2023 at 15:46
  • \$\begingroup\$ On second thought , I realized the question is having incorrect statements. I will delete it once the bounty expires. \$\endgroup\$ Commented May 3, 2023 at 16:22
  • \$\begingroup\$ @DivyaK.S It may be useful for you to clarify the incorrect statements and leave the question open. Because: If it's something that has confused you it may confuse others and your in-question comments may be useful to others. \$\endgroup\$
    – Russell McMahon
    Commented May 6, 2023 at 1:39
  • \$\begingroup\$ @DivyaK.S +1. The question now has an answer and may have another soon. Please do not delete it. IF there is content in the question that you wish to alter please do so while retaining the old content - or at very least the sense of it. This is so existing answers are not invalidated by the changes. || Information only: Some people alter questions so the the sense is changed enough so that the new question is quite different to the old and answers are then "wrong". This may ne needed to suit what is really being asked BUT if this happens it's useful to be able to understand what has happened. \$\endgroup\$
    – Russell McMahon
    Commented May 6, 2023 at 4:03

1 Answer 1

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As per the expression for frequency modulated wave, isn't the frequency deviation same as Modulation index 'M'?

Yes and No.

\$\Delta f_p\$ is the peak (maximium) deviation of the frequency transmitted wave from the carrier frequency (in response to modulation).

Normalizing \$\Delta f_p\$ to the maximum modulating frequency \$f_m\$ provides a ratio \$M=\frac{\Delta f_p}{f_m}\$that represents the peak deviation from the carrier relative to the maximum modulation frequency.

So No, M is not the peak deviation from the carrier.

So Yes, M is the peak deviation from the carrier relative to the highest modulating frequency and is called the modulation index.

This is a language problem. When "frequency deviation" is uttered or written the listener or reader is expected to know what is implied.

The max value for the quantity inside bracket is (ωc+M)t (valid for a short duration around the peak of modulating signal) and hence max frequency deviation ( angular) should be M.

This is the "incorrect spot" referred to in the comments. Edit: The correction in the OP is verified here,

The modulation index can be written \$M=\frac{\Delta\omega_p}{\omega_m}\$ and for small values of \$\omega_m t\$, \$\text{sin}(\omega_m t)\approx \omega_m t\$. So the argument of the cosine in the OP is: $$\left(\omega_c t+\frac{\Delta\omega_p}{\omega_m}\text{sin}(\omega_m t)\right)=\left(\omega_c t+\Delta\omega_p t\right)$$

Interestingly, the modulation index stands in as a "peak phase angle" of the transmitted wave frequency. This is called phase modulation and shows that there really is no difference between phase and frequency modulation.

Hope this helps


Reflecting on my answer, I can add some clarification.

Why does the frequency deviation of a FM wave have different definitions?

Actually the definitions are the same. In your question you start with the title question then move to modulation index, then point to a Wikipedia artcle on FM.

In all three references the frequency deviation is the shift in frequency of the carrier to another frequency. The Wikipedia , article is almost good. What is necessary is to see the frequency deviation propagate through the math derivation and to get a visuallization of the reality.

Often the term frequency is used as if it is the actual transmission. What is being transmitted is an electromagnetic wave. The wave shape is only sinusoidal if the input is constant or zero.

The following block diagram outlines a visuallization of an FM modulator.

schematic

simulate this circuit – Schematic created using CircuitLab

THe block diagram shows a typical modulator.

  1. \$V_{power}\$ sets the amplitude of the output.
  2. \$v_m(t)\$ is the modulating frequency.
  3. \$V_{DC}\$ sets the carrier frequency.
  4. \$v_{FM}(t)\$ is the transmitted waveform.

For clarity the symbol \$\omega\$ means angu;ar velocity not frequency. We all know that but often use frequency instead. We also use the phrase "instantaneous frequency" which for me doesn't really make sense.

So the angular velocity is defined as the rate of change of angle (phase) $$\omega=\frac{d\theta}{dt}\tag{Equ 1}$$ which can have an instantaneous value.

The transmitted unmodulated wave, \$v_m(t)=0\$, called the carrier wave, is sinusoidal with a frequency of \$f_c\$.

$$v_{c}(t)=V_{c}\cos\left(\omega_{c}t\right)$$ is the carrier wave. \$\omega_{c}\$ is constant. The carrier frequency is fixed by the modulation constant \$K_\omega\$. $$\omega_c=K_{\omega}v_{DC}\tag{Equ 2}$$.

As \$v_m(t)\$ moves above and below zero volts, (not necessarily sinusoidal), so the frequency of the transmitted wave moves above and below the carrier frequency.

In other words, as the input signal deviates above and below zero, so the frequency of the transmitted wave deviates above and below the carrier frequency.

This is the meaning of "frequency deviation": It is the deviation of the transmitted frequency from the carrier frequency.

This definition should be used. And so I must adjust (italics) my answer above to match.

To verify the equation in the OP, some calculus must be used.

Starting with the transmitted wave: $$v_{FM}(t)=V_{c}\cos\left(\theta(t)\right)$$ If unmodulated, \$\theta(t)=\theta_c=\omega_c t\$

To obtain \$\theta(t)\$ EQU 1 must be rearranged and integrated: $$\theta(t)=\int_{0}^{t}\omega(\tau)d\tau$$

\$\omega(t)\$ is the instantaneous angular velocity (frequency) of \$v_{FM}(t)\$.

So:$$v_{FM}(t)=V_{c}\cos\left(\int_{0}^{t}\omega(\tau)d\tau\right)$$

$$\omega(t)=K_{\omega}\left[V_{DC}+v_{m}(t)\right]$$

$$v_{FM}(t)=V_{c}\cos\left(K_{\omega}V_{DC}t+\int_{0}^{t}K_{\omega}v_{m}(\tau)d\tau\right)=V_{c}\cos\left(\omega_c t+\int_{0}^{t}K_{\omega}v_{m}(\tau)d\tau\right)$$

The integrand in the argument of the cosine is the frequency deviation from the carrier.

\$v_{m}(t)\$ often cannot be defined (audio for example), so a sinusoid is used as a stand-in. $$v_{m}(t)=V_{m}\cos\left(\omega_{m}t\right)$$

So: $$v_{FM}(t)=V_{c}\cos\left(\omega_ct+K_{\omega}\int_{0}^{t}V_{m}\cos\left(\omega_{m}t\right)d\tau\right)$$

Finally by integration: $$v_{FM}(t) =V_{c}\cos\left(\omega_ct+\frac{K_{\omega}V_{m}}{\omega_m}\sin\omega_m t\right)$$

The term \$\frac{K_{\omega}V_{m}}{\omega_m}\sin\omega_m t\$ is the angular (phase if you like) deviation of the transmitted wave from the carrier angle (phase).

So now the form of the equation in the OP appears. First \$K_{\omega}V_{m}\$ is recognized as the maximum deviation for this sinusoid. It can be called:$$\Delta\omega_{cMAX}=K_{\omega}V_m\text{ for this sinusoid.}$$

If \$f_m=\frac{\omega_m}{2\pi}\$ is the highest frequency of the modulation signal, then a modulation index can be defined as: $$M=\frac{\Delta\omega_p}{\omega_m}=\frac{\Delta f_p}{f_m}$$

Where \$\Delta f_p\$ is the peak frequency deviation of the transmitted signal from the carrier frequency. Where \$f_m\$ is the highest modulating frequency used.

The integration delivers the \$f_m\$ that appears in the denominator of the modulation index. It comes from the integration of \$\text{cos}\omega_m t\$.

If your calculus is not a strength then take my word for it.

Frequency deviation is used in the same way for all your sources. It is the deviation of the transmitted frequency from the carrier frequency.

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