COMMUNICATION SYSTEM EECB353

Chapter 3(II)ANGLE MODULATION

Dept of Electrical EngineeringUniversiti Tenaga Nasional

Frequency Analysis of Angle-Modulated Waves

)]cos(cos[)( tmtVtm mcc • The output of angle-modulated signal of can be expressed as a

series of sinusoids by using Bessel function of the first kind.

• The table and graph of Bessel function represents normalized voltages for the various freq components of an FM and PM signals. (i.e the number in the table represents actual voltages if the unmodulated carrier has an amplitude of 1V).

Consider the result of expression m(t) above:

)...(...])2cos()2)[cos((])cos())[cos((

cos)()(

2

1

0

mJttmJttmJ

tmJVtm

nmcmc

mcmc

c

c

where m(t) = angle-modulated signal m = modulation index Vc = peak amplitude of the unmodulated carrier J0(m) = carrier component J1(m) = 1st set of side freq displaced from the carrier by m J2(m) = 2nd set of side freq displaced from the carrier by 2m Jn(m) = nth set of side freq displaced from the carrier by nm

Frequency Analysis of Angle-Modulated Waves

Note that amplitude values with –ve sign represent phase shifts of 180 and SB freqs are considered insignificant if their amplitude values less than 0.01 of the unmodulated carrier amplitude.

Frequency Analysis of Angle-Modulated Waves

Figure 1: FM in Frequency Domain

Frequency Analysis of Angle-Modulated Waves

Generally, J0(m) determines the amplitude of the carrier. The nth Bessel function Jn(m) determines the amplitude of the nth pair of sidebands

All the Bessel terms should be multiplied Vc. to find the actual SB amplitude.

If m = 0 zero SB, if m , SB pairs Example of the meaning Jn(m):

J1(2.5)- represent the amplitude of the 1st pair of sidebands for the modulated signal m(t) with m = 2.5.

J1(2.5)- from the table = 0.5, means that 1st SB pairs amplitude are 50% of the unmodulated carrier amplitude (i.e = 0.5Vc).

Jn(m) versus m Figure shows the

curve for the relative amplitudes of the carrier and several sets of side frequencies for values of m up to 10.

It can be seen that the amplitudes of both the carrier and the side frequencies vary at a periodic rate that resembles a damped sine wave.

The negative values of J(m) indicate the relative phase of that side frequency.

1st carrier null

Example 1Consider an angle-modulation system where the message signal m(t)=2 sin(5000t) , frequency deviation sensitivity, Kf = 3000Hz/v and phase deviation sensitivity, Kp = 1.2rad/v. Given the carrier frequency is 1MHz and amplitude of 10V. Determine,

a) The peak frequency and phase deviation.

b) The modulation index for both modulators.

c) Number of sets of significant side frequencies

d) Their amplitudes, then

e) Draw the frequency spectrum showing their relative amplitudes.

Bandwidth Requirement

The actual BW required to pass all the significant SBs for an angle-modulated wave is equal to two times the product of the highest modulating-signal frequency, fm and the number of SBs, n found from the Bessel function.

Carson’s rule approximate the BW necessary to transmit an angle-modulated wave as twice the sum of peak freq deviation, f and the highest modulating-signal freq.fm.

)(2 mfnB

)(2 mffB

Example 2

a) For an FM modulator with a peak frequency deviation f = 10kHz, a modulating-signal frequency fm = 10kHz, Vc = 10V, and a 500kHz carrier, determinea. Actual minimum bandwidth from the Bessel function

tableb. Approximate minimum bandwidth using Carson’s rule,

thenc. Plot the output frequency spectrum for the Bessel

approximation.

b) Compare a percentage saving of a bandwidth in an FM system using Carson’s rule and the one obtained using Bessel Function table. Given a modulating signal frequency is 3 kHz and maximum deviation of 12 kHz

Deviation Ratio (DR)Definition – is defined as the worst-case modulation index and is equal to the maximum peak frequency deviation divided by the maximum modulating-signal frequency.

DR = deviation ratio (unitless) f(max) = maximum peak freq deviation (Hz)

f(max) = maximum modulating-signal freq (Hz)

The worst-case modulation index produces the widest output freq spectrum. Example, for the maximum frequency deviation of 50kHz and the maximum modulating signal frequency of 15kHz, the DR is 3.33.

• This does not mean that whenever a modulation index of 3.33 occurs, the widest BW also occurs at the same time.

• It means that whenever a modulation index of 3.33 occurs for a maximum modulating-signal freq, the widest BW occurs.

(max)

max

mff

DR

Frequency Analysis of Angle-Modulated Waves

Example 3

Determine the deviation ratio and bandwidth for the worse-case (widest-bandwidth) modulation index for an FM broadcast-band transmitter with a maximum frequency deviation of 75kHz and a maximum modulating-signal frequency of 15kHz.

Determine the deviation ratio and maximum bandwidth for an equal modulation index with only half the peak frequency deviation and modulating-signal frequency.

Frequency Analysis of Angle-Modulated Waves

Example 4Given FM and PM modulators with the following parameters:

a. A) Determine the modulation index and sketch the output spectrums for both modulators.

b. B) Change the modulating signal amplitude for both modulators to 4Vp and repeat step (a)

c. C) Change the modulating signal frequency for both modulators to 1kHz and repeat step (a)

FM Modulator PM Modulator Kf = 1.5kHz/v Kp = 0.75rad/v fc = 500kHz fc = 500kHz vm = 2sin(2 2kt) vm = 2sin(2 2kt)

Average Power of an Angle-Modulated Wave

With angle modulation, the power that was originally in the unmodulated carrier is redistributed among the carrier and its sidebands.

It’s independent of modulating signal, modulation index and frequency deviation.

The average power in the unmodulated carrier is :

WR

VP c

c 2

2

where Vc = peak unmodulated carrier voltage (volts)R = load resistance (ohm)

Average Power of an Angle-Modulated Wave

• The modulated carrier power is the sum of the power of the carrier and side frequency components.

• Therefore, total modulated wave power is :

nt PPPPPP ...3210

RV

RV

RV

RV

RVP n

t 22...

22

22

22

2

223

22

21

20

where Pt = total average power P0 = modulated carrier power P1 = power in 1st set of SB P2 = power in 2nd set of SB P3 = power in 3rd set of SB Pn = power in nth set of SB

EXAMPLE 5 :

a. Determine the unmodulated carrier power for the FM modulator with a modulation index, m =1, a modulating signal vm(t)=Vmsin(21000t), and an unmodulated carrier vc(t)=10sin(2500kt), assuming load resistance RL = 50.

b. Determine the total power in the angle-modulated wave.

EXAMPLE 6 :

Consider an FM Modulator with output Xc(t)=100cos[21000t + ), The modulator operates with Kf=8 and has the input message signal vm(t)=5 cos(28t), The modulator is followed by a BPF with a centre frequency of

1000 Hz and a bandwidth 56Hz. Determine :

i) Peak frequency deviationii) Modulation indexiii) Number of sets of significant side frequenciesiv) Their amplitudev) Plot the output frequency spectrum for the Bessel

approximation –show all the amplitude and frequencyvi) Power of the unmodulated carriervii) Power at the filter output viii) Power at modulator output

)(t