Wave Modulation

Part-3

FM Modulation

• Principle

• The phase of the carrier wave is modulated by the modulating wave, as such this becomes non-linear modulation, and thus cannot be expressed using linear superposition principle.

• It is more complex than AM dan DSB, as well as requiring wider bandwidth. But it has better quality of S/N.

Formulation

Suppose the carrier wave is given by:

FM Modulation

• The modulation is on the term ϕ= ϕ(t), where

with ω’ : angular ferequency deviation from the carrier frequency and and ϕ: phase deviation

Two approaches for FM modulation:

• Phase modulation(PM)

• frequency modulation (FM)

FM Modulation

Phase modulation

Modulator wave modulates phase deviation: ϕ(t)= kP Ψm(t)

With kP : phase constant deviation (constraint <= π/Ψm(t)

Frequency modulation

Modulator wave modulate frequency deviation : ω’(t)= kF Ψm(t)

With: thus

Choosing t0=0 and ϕ(0)=0, thus :

Comparions of AM,PM FM modulations

Analog signal Digital Signal

AM

FM

PMSource : Gelombang, MO Tjia, Dabara

Example of PM Modulation

• Given modulated signal of voltage Vm(t) with amplitude kPVm= 1000 rad/s. The carrier is Sinusoidal wave with ωC= 3000 rad/s, its amplitude is A.

• ω(t)=ωC + ω’(t)→

• ω(t)=ωC + kPVm(t)

0<t<2π ms :

ω(t) = 3000+1000 rad/s

2π <t< 3π ms :

ω(t) = 3000-1000 rad/s

3π <t< 4.5π :

ω(t) = 3000+1000 rad/s

Special case: Single tone FM analysis

• Special Case: Pure sinusoidal modulator

Ψm(t)= Ψm cos(ωm t), thus

in angular frequency:

ω’(t)=kF Ψm(t)= kFΨm cos(ωm t)= ω’max cos(ωm t)

With: ω’max= kF Ψm =kF [Ψm(t)]max

FM modulationindex defined as

β = kF Ψm/ωm = ω ‘max /ωm = [ϕ(t)]max

hence the FM modulation becomes:

ΨFM(t)= ΨC cos(ωC t+ βsin(ωm t) )

Special case: Single tone FM analysis

The Spectral characteristics can be studied from the Fourier

expansion:

1. In complex notation:

ΨFM(t)= ΨC cos(ωC t+ βsin(ωm t) )=

ΨC Re { exp [i (ωC t+ βsin(ωm t) )]}

2. The function is periodic as

sin(ωm t) = sin[ωm (t+ 2π/ ωm )], with period T= 2π/ ωm

Special case: Single tone FM analysis

3. The term exp [i (βsin(ωm t) ) expanded using Fourier:

With Cn can be shown as:

For the final step using substitution θ=ωmt.

Jn expresses Bessel function of the first kind of the nth order in real space.

Some properties of Bessel function

a. Nth order Bessel function as a series:

b. For n as integer,

From the plot: only J0 has non zero value at x0

Special case: Single tone FM analysis

4. Using Fourier analysis, the resulting wave from FM

modulation:

5. Frequency spectrum using TF:

Special case: Single tone FM analysis

6. It means, FM signal with pure sinusoidal modulation, contains unlimited number of carrier components and frequency components:

ν= νC± nνm , n=1,2,3,….

Amplitude depends on β, which subsequently is a function of Ψm.

7. Narrowband, β<<1, the contributing amplitude:

J0(β) ≈1 J1(β) ≈ β/2 Jn(β) ≈0, n>1

Which means FM spectrum frequency only has ωC, dan ωC ±ωm

components, similar to AM modulation

Power of FM, Transmission bandwidth

Bandwidth is the range of minimum frequency in the frequency

spectrum which must be transmitted so that sufficient signal

quality is obtained. Empirically it is found from experiment for

FM it is “sufficient” to transmit at 98% power of the FM.

Calculation: ration of the nth component of FM power to the

total power.

an= Nn/NT

Power of FM, Transmission bandwidth

Fourier expansion gives:

Whereas

This last result proves to the fact that the amplitude of FM signal

is constant.

FM Power, Transmission Bandwidth

Given an :

Numerically, if aa>=0.98 and if n >= β+1

Which gives transmission badnwidth B= 2(β+1)ωm = 2(ω’max +ωm)

For Narrow Band FM, β<<1 thus ω’max<< ωm as such B ≈2 ωm

For non sinusoidal modulation, index β is undefined, hence we

use the Carson criteria:

B= 2(D+1)ωm = 2(ω’max +ωm)

with

Example of FM Problem

Sinusoidal carrier FM sinusoidal with frequency 20 MHz,

modulated by sinusoidal wave which causes frequency deviation

of 200 kHz at most. Determine modulation index and the

bandwidth for if the signals are :

a. 1 kHz dan b. 1000 kHz

Answer:

From the problem it is known that

ν’max = 200 kHz = 2x105 Hz = βνm νC= 20 MHz = 2x107 Hz

νC>> ν’max Index β = ν’max / νm bandwidth B= 2(β+1) νm

Answer

a. For νm = 1 kHz = 103Hz

𝛽= 2x105/103=200

this result shows it is a wide band FM with

B = 2(200+1)x103 = 402 kHz

b. For νm = 1000 kHz = 106Hz

𝛽= 2x105/106=0,2

This result shows it is a Narrow Band FM with

B = 2νm = 2000 kHz

FM Demodulation

There are 2 methods:

a. With frequency discriminator :

Convert FM to AM signals and then detected by the envelope

detector.

b. Feedback demodulation

Using PPL (Phase Lock Loop); in principle FM signal is put

through PPL and the error signal output is the intended

demodulated signal.