' 2.4 Frequency and Phase Modulation
FIGURE2.12 Bessel Functions
60
TABLE2.1BesselFunctions mJ0J1J2J3J4J5J6J7J8J9J10J11J12J13J14J15J16J17J18 01.00 0.250.980.12 0.50.940.240.03 0.750.860.350.070.01 10.770.440.110.02 1.250.650.510.170.040.01 1.50.510.560.230.060.01 1.750.370.580.290.090.02 20.220.580.350.130.030.01 2.250.080.550.400.170.050.01 2.40.000.520.430.200.060.02 2.5−0.050.500.450.220.070.02 2.75−0.160.430.470.260.100.030.01 3−0.260.340.490.310.130.040.01 3.5−0.380.140.460.390.200.080.030.01 4−0.40−0.070.360.430.280.130.050.01 4.5−0.32−0.230.220.420.350.200.080.030.01 5−0.18−0.330.050.360.390.260.130.050.020.01 5.50.00−0.34−0.120.260.400.320.190.090.030.01 60.15−0.28−0.240.110.360.360.250.130.060.020.01 6.50.26−0.15−0.31−0.030.280.370.300.180.090.040.01 70.30−0.01−0.30−0.170.160.350.340.230.130.060.020.01 7.50.270.14−0.23−0.260.020.280.350.280.170.090.040.010.01 80.170.24−0.11−0.29−0.110.190.340.320.220.130.060.030.01 8.50.040.270.02−0.26−0.210.070.290.340.270.170.090.040.020.01 8.650.000.270.06−0.24−0.230.030.270.340.280.180.100.050.020.01 9−0.090.250.14−0.18−0.27−0.060.200.330.300.210.130.060.030.01 10−0.250.040.260.06−0.22−0.23−0.010.220.320.290.210.120.060.030.01 11−0.17−0.180.140.23−0.01−0.24−0.200.020.230.310.280.200.120.060.030.010.01 120.05−0.22−0.080.200.18−0.07−0.24−0.170.040.230.300.270.200.120.070.030.010.10 130.21−0.07−0.220.000.220.13−0.12−0.24−0.140.070.230.290.260.190.120.070.030.010.01 140.170.13−0.15−0.180.08−0.15−0.23−0.110.080.240.290.250.190.120.070.030.020.01 15−0.010.200.04−0.19−0.120.130.210.03−0.17−0.22−0.090.100.240.280.250.180.120.070.03 16−0.170.090.19−0.04−0.20−0.060.170.18−0.01−0.19−0.21−0.070.110.240.270.240.180.110.07 17−0.17−0.100.160.14−0.11−0.190.000.190.15−0.04−0.20−0.19−0.050.120.240.270.230.170.11
Bessel coefficients are equally valid for peak or RMS voltages, but the user should be careful to keep track of which type of measurement is being used.
When Bessel functions are used, the signal of Equation (2.20) becomes v(t) = A sin (ωct+mƒsin ωmt)
= A {J0(mƒ) sinωct− J1(mƒ)[sin (ωc− ωm)t −sin (ωc+ ωm)t]
+J2(mƒ)[sin (ωc−2ωm)t +sin (ωc+ 2ωm)t]
−J3(mƒ)[sin (ωc−3ωm)t +sin (ωc+ 3ωm)t]
+ ⋅ ⋅ ⋅} (2.21)
With angle modulation, the total signal voltage and power do not change with modulation. Therefore, the appearance of power in the side-bands indicates that the power at the carrier frequency must be reduced be-low its unmodulated value in the presence of modulation. In fact, the carrier-frequency component disappears for certain values of mƒ(for exam-ple, 2.4 and 5.5).
This constant-power aspect of angle modulation can be demonstrated using the table of Bessel functions. For simplicity, normalized values can be used. Let the unmodulated signal have a voltage of one volt RMS across a re-sistance of one ohm. Its power is, of course, one watt. When modulation is applied, the carrier voltage will be reduced and sidebands will appear. J0from FIGURE2.13 FMin the frequency domain
the table will represent the RMS voltage at the carrier frequency and the power at the carrier frequency will be
P V
Similarly, the power in each of the first pair of sidebands will be PSB J
1 1
= 2
The combined power in the first set of sidebands will be twice as much as this, of course. The power in the whole signal will then be
P J J JT = + + + ⋅ ⋅ ⋅0
If the series is carried on far enough, the result will be equal to one watt, regardless of the modulation index.
The bandwidth of an FM or PM signal is to some extent a matter of defi-nition. The Bessel series is infinite, but as can be seen from the table or the graph, the amplitude of the components will gradually diminish until at some point they can be ignored. The process is slower for large values of m, so the number of sets of sidebands that has to be considered is greater for larger modulation indices. A practical rule of thumb is to ignore sidebands with a Bessel coefficient of less than 0.01. The bandwidth, for practical pur-poses, is equal to twice the number of the highest significant Bessel coeffi-cient, multiplied by the modulating frequency.
EXAMPLE2.10
Y
An FM signal has a deviation of 3 kHz and a modulating frequency of 1 kHz.
Its total power is 5 W, developed across a 50Ωresistive load. The carrier fre-quency is 160 MHz.
(a) Calculate the RMS signal voltage.
(b) Calculate the RMS voltage at the carrier frequency and each of the first three sets of sidebands.
(c) Calculate the frequency of each sideband for the first three sideband pairs.
(d) Calculate the power at the carrier frequency, and in each sideband, for the first three pairs.
(e) Determine what percentage of the total signal power is unaccounted for by the components described above.
(f) Sketch the signal in the frequency domain, as it would appear on a spec-trum analyzer. The vertical scale should be power in dBm, and the hori-zontal scale should be frequency.
SOLUTION
(a) The signal power does not change with modulation, and neither does the voltage, which can easily be found from the power equation.
P V
(b) The modulation index must be found in order to use Bessel functions to find the carrier and sideband voltages.
m
From the Bessel function table, the coefficients for the carrier and the first three sideband pairs are:
J0= −0.26 J1=0.34 J2=0.49 J3=0.31
These are normalized voltages, so they will have to be multiplied by the to-tal RMS signal voltage to get the RMS sideband and carrier-frequency voltages.
For the carrier, Vc= J0VT
J0has a negative sign. This simply indicates a phase relationship between the components of the signal. It would be required if we wanted to add together all the components to get the resultant signal. For our present purpose, however, it can be ignored, and we can use
Vc = J VT
= ×
0
0 26. 15 8. V
= 4.11 V
Similarly we can find the voltage for each of the three sideband pairs.
Note that these are voltages for individual components. There will be a lower and an upper sideband with each of these calculated voltages.
V1 = J1VT
(c) The sidebands are separated from the carrier frequency by multiples of the modulating frequency. Here,ƒc=160 MHz andƒm=1 kHz, so there are sidebands at each of the following frequencies.
ƒUSB1 =160 MHz+1 kHz= 160.001 MHz
(d) Since each of the components of the signal is a sinusoid, the usual equa-tion can be used to calculate power. All the components appear across the same 50Ωload.
Similarly, it can be shown that
P1=0.576 W P2=1.2 W P3=0.48 W
(e) To find the total power in the carrier and the first three sets of sidebands, it is only necessary to add the powers calculated above, counting each of the sideband powers twice, because each of the calculated powers represents one of a pair of sidebands. We only count the carrier once, of course.
PT = Pc+2(P1+P2+P3)
= 0.338+2(0.576+1.2+0.48) W
= 4.85 W
This is not quite the total signal power, which was given as 5 W. The re-mainder is in the additional sidebands. To find how much is unaccounted for by the carrier and the first three sets of sidebands, we can subtract. Call the difference Px.
Px= 5−4.85=0.15 W
As a percentage of the total power this is Px(%) .
%
= ×
= 015
5 100
3
All the information we need for the sketch is on hand, except that the power values have to be converted to dBm using the equation
P P
(dBm) log
=10 mW 1 This gives
Pc(dBm) = 10 log 338=25.3 dBm P1(dBm) = 10 log 576=27.6 dBm P2(dBm) = 10 log 1200=30.8 dBm P3(dBm) = 10 log 480=26.8 dBm The sketch is shown in Figure 2.14.
X
FIGURE2.14
Bandwidth For PM, the bandwidth varies directly with the modulating frequency, since doubling the frequency doubles the distance between sidebands. It is also roughly proportional to the maximum phase deviation, since increasing mp increases the number of sidebands. For FM, however, the situation is compli-cated by the fact that
mƒ
= ƒδ
m
For a given amount of frequency deviation, the modulation index is in-versely proportional to the modulating frequency. Recall that the frequency deviation is proportional to the amplitude of the modulating signal. Then, if the amplitude of the modulating signal remains constant, increasing its fre-quency reduces the modulation index. Reducing mƒreduces the number of sidebands with significant amplitude. On the other hand, the increase inƒm
means that the sidebands will be further apart in frequency, since they are separated from each other byƒm. These two effects are in opposite directions.
The result is that the bandwidth does increase somewhat with increasing modulating-signal frequency, but the bandwidth is not directly propor-tional to the frequency. Sometimes FM is called a “constant-bandwidth”
communication mode for this reason, though the bandwidth is not really constant. Figure 2.15 provides a few examples that show the relationship be-tween modulating frequency and bandwidth. For this example the deviation remains constant at 10 kHz as the modulating frequency varies from 2 kHz to 10 kHz.
One other point must be made about the sidebands. With AM, restrict-ing the bandwidth of the receiver has a very simple effect on the signal.
Since the side frequencies farthest from the carrier contain the high-frequency baseband information, restricting the receiver bandwidth reduces its response to high-frequency baseband signals, leaving all else unchanged.
When reception conditions are poor, bandwidth can be restricted to the minimum necessary for intelligibility. For FM the situation is more compli-cated, since even low-frequency modulating signals can generate sidebands that are far removed from the carrier frequency. FM receivers must be de-signed to include all the significant sidebands that are transmitted; other-wise severe distortion, not just limited frequency response, will result.
Carson’s Rule The calculation of the bandwidth of an FM signal from Bessel functions is easy enough, since the functions are available in a table, but it can be a bit te-dious. There is an approximation, known as Carson’s rule, that can be used to find the bandwidth of an FM signal. It is not as accurate as using Bessel functions, but can be applied almost instantly, without using tables or even a calculator.
FIGURE2.15 Variation of FMbandwidth with modulating frequency
Here is Carson’s rule:
B≅ 2[δ(max)+ ƒm(max)] (2.22)
Equation (2.22) assumes that the bandwidth is proportional to the sum of the deviation and the modulating frequency. This is not strictly true. Car-son’s rule also makes the assumption that maximum deviation occurs with the maximum modulating frequency. Sometimes this leads to errors in prac-tical situations, where often the highest baseband frequencies have much less amplitude than lower frequencies, and therefore do not produce as much deviation.
EXAMPLE2.11
Y
Use Carson’s rule to calculate the bandwidth of the signal used in Example 2.10.
SOLUTION
Here there is only one modulating frequency, so B ≅ 2(δ +fm)
= 2(3 kHz+1 kHz)
= 8 kHz
In the previous example we found that 97% of the power was contained in a bandwidth of 6 kHz. An 8-kHz bandwidth would contain more of the signal power. Carson’s rule gives quite reasonable results in this case, with very little work.
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Narrowband and Wideband FM
We mentioned earlier that there are no theoretical limits to the modulation index or the frequency deviation of an FM signal. The limits are practical and result from a compromise between signal-to-noise ratio and bandwidth.
In general, larger values of deviation result in an increased signal-to-noise ratio, while also resulting in greater bandwidth. The former is desirable, but the latter is not, especially in regions of the spectrum where frequency space is in short supply. It is also necessary to have some agreement about devia-tion, since receivers must be designed for a particular signal bandwidth.
For these reasons, the bandwidth of FM transmissions is generally lim-ited by government regulations that specify the maximum frequency devia-tion and the maximum modulating frequency, since both of these affect bandwidth. In general, relatively narrow bandwidth (on the order of 10 to
30 kHz) is used for voice communication, with wider bandwidths for such services as FM broadcasting (about 200 kHz) and satellite television (36 MHz for one system).
FM and Noise The original reason for developing FM was to give improved performance in the presence of noise, and that is still one of its main advantages over AM.
This improved noise performance can actually result in a better signal-to-noise ratio at the output of a receiver than is found at its input.
One way to approach the problem of FM and noise is to think of the noise voltage as a phasor having random amplitude and phase angle. The noise adds to the signal, causing random variations in both the amplitude and phase angle of the signal as seen by the receiver. Figure 2.16 shows this vector addition.
The amplitude component of noise is easily dealt with in a well-designed FM system. Since FM signals do not depend on an envelope for detection, the receiver can employ limiting to remove any amplitude variations from the signal. That is, it can use amplifiers whose output amplitude is the same for a wide variety of input signal levels. The effect of this on the amplitude of a noisy signal is shown in Figure 2.17. As long as the signal amplitude is
FIGURE2.16