Skip to content
Question Bank (5 marks) · Q5

Q.Derive an expression for the instantaneous value of a FM wave.

Karnataka PUCTextbookLong· 5mImportance★★★★★est
16% · 30/183 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

[!TLDR]

Since FM varies frequency with the message, integrating the instantaneous frequency gives the phase and eFM=Ecsin⁡(ωct+mfsin⁡ωmt)e_{FM}=E_c\sin(\omega_c t+m_f\sin\omega_m t), with mf=Δf/fmm_f=\Delta f/f_m.

In frequency modulation the amplitude of the carrier is kept constant while its instantaneous frequency is varied in proportion to the instantaneous value of the modulating signal em=Emcos⁡ωmte_m=E_m\cos\omega_m t.

The instantaneous frequency is therefore

fi=fc+kfem=fc+kfEmcos⁡ωmt,f_i = f_c + k_f e_m = f_c + k_f E_m\cos\omega_m t,

where fcf_c is the rest (carrier) frequency and kfk_f is a constant of the modulator. The maximum change of frequency, the frequency deviation, is

Δf=kfEm,\Delta f = k_f E_m,

so

fi=fc+Δfcos⁡ωmt.f_i = f_c + \Delta f\cos\omega_m t.

The instantaneous phase angle of the wave is the integral of the instantaneous angular frequency:

θ=∫ωi dt=∫2πfi dt=∫2π(fc+Δfcos⁡ωmt) dt.\theta = \int \omega_i\,dt = \int 2\pi f_i\,dt = \int 2\pi(f_c + \Delta f\cos\omega_m t)\,dt.

Integrating,

θ=2πfct+2πΔf⋅sin⁡ωmtωm.\theta = 2\pi f_c t + 2\pi\Delta f\cdot\frac{\sin\omega_m t}{\omega_m}.

Since ωm=2πfm\omega_m = 2\pi f_m,

θ=ωct+Δffmsin⁡ωmt.\theta = \omega_c t + \frac{\Delta f}{f_m}\sin\omega_m t.

Defining the modulation index of FM as

mf=Δffm,m_f = \frac{\Delta f}{f_m},

we get …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.