I try to implement a frequency modulator.

From what I know, the more I vary the modulation index (which is set by the gain block here,) the more the spectral lines should increase in number.

When I increase the modulation index, the spectrum of my modulated signal does not change. Can someone explain to me why?

My signal (signal modulant) : \$m(t) = 3cos(2 \pi*f_m t ) + 6\$ , where \$f_m\$ = 500Hz

Carrier signal (porteuse) : \$p(t) = cos(2\pi*f_c t )\$ , where \$f_c\$ = 6kHz

Sample time of every signal = 1/1000

Here my schematic diagram:

enter image description here

With a modulation index of 0.2, the spectrum schema gives:

enter image description here

With a modulation index of 0.5, the spectrum schema gives:

enter image description here

Here is the design I have:

Narrow band FM modulation:

enter image description here

The output signal (signal modulé) in the scope gave me :

enter image description here

  • \$\begingroup\$ What is the mathematical operation that the dot product block does? Why do you think that this block diagram creates an FM signal. I think it creates an AM signal. That is why there are only two side bands. What is the integrator block for? \$\endgroup\$
    – AJN
    Oct 1, 2020 at 14:23
  • \$\begingroup\$ In fact this is a example I try to implement and It said that it is a narrow band FM modulation, I'm also very suspicious about the design because the results are false \$\endgroup\$ Oct 1, 2020 at 14:25
  • \$\begingroup\$ Can you show the time domain plot stored in signal modulé? It should be of constant amplitude if it is an FM signal. Can you confirm if it is so. \$\endgroup\$
    – AJN
    Oct 1, 2020 at 14:26
  • \$\begingroup\$ Where is it said that it is a narrow band FM modulation? Can you provide a link to the source? \$\endgroup\$
    – AJN
    Oct 1, 2020 at 14:27
  • \$\begingroup\$ @AJN I updated the post and added to it the necessary ,about the output signal (signal modulé ) and the time domain plot is between 0 and 10 sec \$\endgroup\$ Oct 1, 2020 at 14:32

1 Answer 1


According to Wikipedia, narrow band FM has a modulation index less than 0.3.

This set of slides also mentions a similar circuit (Fig. 4.4 in the link) for narrow band modulator. The modulation index is small there too. Condition \$\beta\int m(t) << 1\$ is not satisfied in your example because the bias added to the message when integrated, produces large values.

The equation for narrow band FM mentioned in your question is derived in the above links. The derivation assumes certain approximation. So it will generate only 2 side bands; one upper and lower. Increasing the modulation index will

  1. Make the assumption made in the derivation invalid.
  2. Have no effect on the narrow band approximation equation.

The "FM" output has varying amplitude since the input message signal has a bias in it. When integrated, the bias integrates to \$kt\$ which is a (linearly) increasing signal. When multiplied with the carrier, this linearly increasing signal results in the output also to increase in amplitude.

The narrowband equation should be used only if the result of the integration of the message signal has small values. Your message signal (with its bias) doesn't have satisfy that condition when it is integrated.

Try after removing the bias. You will still get only two side bands since it is still an approximate formula for FM.

  • \$\begingroup\$ The link I provided says one more thing about narrow band FM. "Amplitude Distortion of Narrow-band FM :The envelope contains a residual amplitude modulation that varies with time". so, even after removing the bias, you will not get perfect constant-amplitude FM. \$\endgroup\$
    – AJN
    Oct 1, 2020 at 15:22
  • \$\begingroup\$ You're perfectly right , after removing the bias , I didn't get the result I want . I think that I will still get only two side bands since it is still an approximate formula for FM . Thank you so much for the time you gave me , I will mark this post solved ,thanks to you @AJN \$\endgroup\$ Oct 1, 2020 at 15:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.