TLT-5200/5206 COMMUNICATION THEORY, Exercise 6, Fall 2012
TLT-5200/5206 COMMUNICATION THEORY, Exercise 6, Fall 2012
Problem 1.
» beta = 20/3;
» J = besselj([0:15]',beta)
The instantaneous frequency of an FM modulated signal varies between
fmin = 99.98 MHz and fmax = 100.02 MHz when the modulating signal is a
sine wave with fm = 3 kHz (assume Am = 1).
J =
a) The carrier frequency is in the middle in sine modulation, so
fc 
f max  f min 100.02  99.98

MHz  100 MHz
2
2
b) Carrier swing is the maximum range of instantaneous frequency, so
swing  f max  f min  40kHz
c) Frequency deviation f is the maximum deviation of instantaneous
frequency from the carrier frequency fc
f   f max  f c  100.02MHz  100MHz  20kHz
d) Modulation index  is given by (Am = 1)

0.2817
-0.1053
-0.3133
-0.0827
0.2389
0.3694
0.3151
0.1979
0.1005
0.0432
0.0162
0.0054
0.0016
0.0004
0.0001
0.0000
Based on the Carson's rule, peaks for n = -7…7 ( B = 46 kHz) are to be
considered. The other approximation considers peaks for n = -8…8 (B =
52 kHz) to be "important".
Am
1
f 
20kHz  6.7
fm
3kHz
n=7
e) Theoretically, the bandwidth of an FM signal is always infinite.
Practical approximations for the "important" part of the spectrum are
n=-8
n=8
fm
B  2 f   f m   220kHz  3kHz   46kHz , Carson' s rule
f
fC
B  2 f   2 f m   220kHz  6kHz   52kHz , " low  distortion" rule
f) The spectrum consists of peaks at fc + nfm, n = … -2, -1, 0, 1, 2, … .
The relative amplitudes of the peaks are given by the Bessel functions
Jn(). Now, the modulation index  = 6.7 and the values can be
obtained from tables or from Matlab (next page). Remember also that
J-n() = (-1)nJn().
Figure 1: Spectrum of the FM signal with modulation index  = 6.7 (only
positive frequencies shown).
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n=-7
TLT-5200/5206 COMMUNICATION THEORY, Exercise 6, Fall 2012
Problem 2.
We want to increase the frequency deviation from f = 4kHz to f' =
24kHz. This can be achieved with a frequency multiplier which multiplies
the instantaneous frequency (remember f(t) = fC + fx(t) for FM).

24/4 = 6 times multiplication needed.
TLT-5200/5206 COMMUNICATION THEORY, Exercise 6, Fall 2012
» beta = 3;
» J = besselj([0:10]',beta)
J =
-0.2601
0.3391
0.4861
0.3091
0.1320
0.0430
0.0114
0.0025
0.0005
0.0001
0.0000
» beta = 0.5;
» J = besselj([0:5]',beta)
J =
0.9385
0.2423
0.0306
0.0026
0.0002
0.0000
This, however, increases also the center frequency to 6*5 MHz = 30 MHz

20 MHz extra frequency shift needed to make the final carrier
frequency 50 Mhz (80 MHz or 20 MHz LO signal). Input signal
frequency fm = 8 kHz.
fm = 8 kHz
B = 24 kHz
Block-diagram:
f0 = 5 MHz
f = 4 kHz
x(t)
FM
modulator
 = 0.5
f0' = 30 MHz
f' = 24 kHz
Instantaneous
frequency
multiplier, n=6
fC = 50 MHz
f' = 24 kHz
fm
f
BPF
=3
f0
=3
B (Carson)
LO signal
(80 or 20 MHz)
Figure 2: Block-diagram for processing the signal.
The modulation index is  = Amf/fm, so
 = 4/8 = 0.5 before the instantaneous frequency multiplier
 = 24/8 = 3 after the instantaneous frequency multiplier
Figure 3: FM spectrum before instantaneous frequency multiplier (only
positive frequencies shown).
fm = 8 kHz
B = 64 kHz
f 0'
f
fm
According to the Carson's rule (of thumb), the bandwidth B of the signal is
B = 2(f + fm) meaning that B = 2(4 kHz + 8kHz) = 24 kHz before and
B = 2(24 kHz + 8kHz) = 64 kHz after the instantaneous frequency
multiplier.
Figure 4: FM spectrum after instantaneous frequency multiplier (only
positive frequencies shown).
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B (Carson)
TLT-5200/5206 COMMUNICATION THEORY, Exercise 6, Fall 2012
TLT-5200/5206 COMMUNICATION THEORY, Exercise 6, Fall 2012
Problem 3.
Relative power in 2nmax +1 frequency components around center frequency
The starting point in analyzing FM systems with tone modulation is
always the series expansion
xc (t )  Ac cos c t   (t ) 
1
0.8
 ...

 Ac  J n (  ) cos c  n m  t.
0.6
n  
Based on the first line, the power of the modulated signal xc(t) is always
Ac2/2 in exponential modulations, so

0.4
0.2
 J n ( )2  1.
n  
0
This means that the power in some band W relative to the total power is
given by the sum of squares of Jn() over the given bandwidth W.
Now,  = 2 and we can use Matlab again for calculations. The given band
WFM (see the problem sheet) includes the peaks for n = -2 … 2. We can
also utilize the fact that J-n() = (-1)nJn() , so J-n()2 = Jn()2.
» beta = 2;
» J = besselj([0:1000]',beta);
» Ptot = 2*sum(J(2:1001).^2) + J(1)^2;
Ptot =
1.00000
» P1 = 2*sum(J(2:3).^2) + J(1)^2;
» relative = 100*P1/Ptot
relative =
96.4334
So approximately 96.4 % of the total power is located at the given
bandwidth.
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0
1
2
3
4
5
6
7
8
9
nmax
Figure 5: Relative power as a function of spectral components included
around the center frequency (n = -nmax…nmax). Modulation index  = 2.
The Carson's rule for the bandwidth gives in this case
B = 2(+ 1)fm = 6fm ( nmax = 3 )
whereas the “low-distortion criterion” (this one was emphasized on
lectures) gives
B = 2(+ 2)fm = 8fm ( nmax = 4 ).
With these values, the relative powers are 99.7587 % and 99.9898 %,
respectively.
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