TRANSMISSION SYSTEMS: AN INTRODUCTION TO FREQUENCY MODULATION

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1 INTERNATIONAL JOURNAL OF SATELLITE COMMUNICATIONS, VOL. 2, (19x4) TUTORIAL TRANSMISSION SYSTEMS: AN INTRODUCTION TO FREQUENCY MODULATION W. R. CALVERT IiWELSAT, 490 L Enfant Plaza, South West, Washington DC 20024, U.S.A. INTRODUCTION Before we can understand the various carrier configurations, capacities and techniques of frequency modulation an understanding of basic F. M. theory is required. Modulation is the process whereby a signal frequency waveform is impressed in some way upon a higher frequency carrier wave. Frequency modulation is an angle modulation whereby the carrier amplitude is fixed and the frequency is made to vary so that the waveform of frequency variation as a function of time is identical to that of the information. The equation for a frequency modulated wave may be written e = A cos (act+ M sin w,t) where M is the modulation index given by Im Af,, = peak deviation f, = top modulating frequency w, = angular frequency of the modulating signal w, = angular frequency of the carrier The modulation index is directly proportional to the amplitude of the modulating signal and to the deviation, but is inversely proportional to the modulating-signal frequency. Expanding the above expression we have e = A cos w,t cos (Msin w,t) -A sin w,t sin (M sin w,t) This may be written in the form e/a = Jo(M) cos w,t + J,(M)[cos (0, + w,) t - cos (0,- 0,) t] + Jz(M)[cos (w, + 2w,) t + cos (w, - 2w,) t] + J,(M)[cos (w, + 3w,) t - cos (w, - 30,) t] +... in which Jo(M), J,(M), etc. are Bessel functions of This is the first of a series of tutorials for the practising engineer and technician and is released by INTELSAT for the public domain and may be reproduced freely with due acknowledgement to both INTELSAT and John Wiley & Sons, Ltd. the first kind of argument M. The values of these functions can be obtained from tables and a graphical summary can be produced (Figure 1). For example, the value of J, at a modulation index (M) of 1 is Examination of this expression for a frequency-modulated wave shows that it contains a number of components with angular frequencies equal to w,, (w,+w,), (wcurn), (w,+2w,), (wc-2w,), etc. The first of these components is at carrier frequency, the second and third constitute a sideband pair symmetrically disposed about the carrier frequency and displaced from it by a frequency equal to the modulating frequency. The fourth and fifth constitute a pair of sidebands, also symmetrically disposed about the carrier frequency, but displaced from it by twice the modulation frequency. There are a large number of such pairs of sidebands displaced from the carrier frequency by multiples of the modulating frequency. Thus, a single sinusoidal modulating signal can give rise to a large number of pairs of sidebands. The two sidebands constituting each pair in a frequency-modulated wave have the same amplitude, but this is not linearly related to the modulating signal amplitude, rather it is related to a Bessel function of argument M. The amplitude of the carrier-frequency component is also dependent on a Bessel function (.To) of M. The way in which the amplitudes of the carrier and first pair of sidebands depend on M is illustrated in Figure 1. Figure 1 is a graph of carrier and sideband amplitude plotted against modulation index, where modulation index is the ratio of deviation (Af,) to input frequency (f,). So if we have a peak deviation (Af,) of 100 khz and an input frequency of 100 khz, the modulation index (M) would be: 100 khz M= =1 100 khz Figures 2 and 3 show spectral displays of frequency versus amplitude as would be seen on a spectrum analyser. For a modulation index of 1, our Bessel function graph (Figure 1) shows that J3= 0.02, Jz = 0.11, J, = 0.45 and Jo (carrier) = 0.82 (see Figure 2) /84/ $01.00 Received May 1984 Revised May 1984

2 210 W. R. CALVERT a Jlf u z = J2 -E I- 3 0 J3-2 9 JO = CARRIER J1 = IS1 SIDE FREOUENCY 52 = 2ND SIDE FREOUENCY PAIR SIDE FREOUENCIES 53 = 3R0 SIDE 2MD PAIR SIDE FREOUENCIES FREOUENCY PAIR SIDE FREOUENCIES PAIR SIDE FREQUENCIES *.. 5TH PAIR SIDE FREOUENCIES MOOULATIOW INDEX Figure 1. Bessel function. Modulation index = Af/f,,, For a modulation index of 2.4, Jo = 0, i.e. the carrier disappears as shown in Figure 3. What we have done in fact is to derive a relationship between modulation index (fixed by the Bessel functions), deviation and modulating frequency. As can be seen, if we know the deviation and the top modulating frequency we can calculate the modulation index and, by using the Bessel functions, determine the sideband frequency pattern produced. Later on we shall look at the significance of modulation index (M) and Bessel functions when testing an F.M. modulator. In an F.M. system the signal to noise ratio (SIN) improvement increases in direct proportion to the square of the frequency modulation index (M). Therefore the SIN in the top baseband channels will be worse than those at the bottom or, in other words, the noise output level rises linearly through the baseband. The reason for this is that the higher order sidebands of channel 1 appear in the sidebands of channels 2 and 3; the higher order sidebands of channel 2 appear in the sidebands of 3 and 4, etc. Thus, at the output of the discriminator we obtain channel 1 with low noise; channel 2 with higher noise due to the channel 1 sidebands; channel 3 with still higher noise due to the channel 1 plus channel 2 sidebands, etc; thus channel N has noise proportional to N- 1 channels, as shown in Figure 4. A technique known as modulation emphasis is J2 = 0.11 J1 = 0.45 Jo ICxrl = 0.82 J2 J1 used where the lower channel frequencies are attenuated more than the higher ones, and then applied to the modulator. This is pre-emphasis and maintains approximately constant SIN per channel. After the demodulator, the higher frequencies emerge with greater power than the lower ones, and a de-emphasis network is used to attenuate the higher frequencies, and restore all channels to about equal power level. Figure 5 illustrates the normalized pre-emphasis curve. Thus, when testing an F.M. modulator with a test-tone, we need to test at the cross-over point of the network (i.e. no pre-emphasis) which is calculated as ftop, where ftop is the top baseband frequency.* For a 132 channel system where the top modulating frequency is 552 khz, the cross-over frequency would be 552 X or khz, which is a test-tone frequency we can use regardless of the emphasis network. Another useful calculation is to be able to calculate the maximum bandwidth occupied by an F.M. system. This is given by the Carson s rule bandwidth expression which approximates to BW = 2 (Atp + f,> where Af,, = peak deviation f, = top modulating frequency (The above is a useful approximation and holds true JI Jz I 1 J3 1 I JO Figure 3. Disappearing carrier * Note: the latest CC IIT recommendation gives the cross-over frequency as ftod.)

3 TUTORIAL 211 A convert to multichannel r.m.s. deviation (Afmc): Afmc = Afr x NLR = khz X = 529 khz D B I I AMPLITUDE Figures 4A and B. F.M. spectrum Figures 4C and D. Effective noise in channels + NOISE PER CHANNEL INCREASING for values of M greater than unity in which only side frequencies of orders up to n = (M+ 1) make a significant contribution). As we are using a multichannel system the OdBmO test-tone deviation must become a multichannel test-tone deviation, by multiplying it by the noise loading ratio: NLR = log N, for N < 240 channels log N, for N > 240 channels Thus, for a 132 channel system: NLR = log (132) = 7.48 db as a ratio = Using our 132 channel system as our example, the 0 dbmo test-tone deviation is khz. To sl-!j$o P I.[ I I I NORMALISEO FREDUENCY (flfmax) Figure 5. Re-emphasis curve F In most satellite systems there is a bandwidth restriction, so to correct a multiple access satellite system we have to use the multichannel peak factor (G) which is defined as the ratio of the overload voltage to the r.m.s. voltage for a given number of active channels at constant volume. The peak factor used by INTELSAT at the moment is given as 10 db (3.16) and this must be used in the F.M. bandwidth calculation; so to complete the calculations we now have for the occupied bandwidth: Carson Formula Multichannel BW 2(Afmc + fm) 2[ (Afr X NLR) + fml Multichannel BW including peak factor 2[ (Afr x NLR) x G + f, where, G = multichannel peak factor Af, = deviation (single tone) In Table I are reproduced the transmission parameters of regular FDM/FM carriers for the INTELSAT IVA, V and VI satellites. We are now in a position to be able to check and confirm columns 1-6 (columns 7-9 will be covered in a future article). Column 1. Carrier capacity (N) number of channels. Column 2. Top baseband frequency urn). In this case with 4kHz channels as an example using a 132 channel carrier: 132 channels X 4 khz = 528 khz. The lowest channel is 12 khz, and there is a 12 khz guardband between supergroup 1 and supergroup 2. Therefore 528 khz + 12 khz + 12 khz = fm = 552 khz. Column 3. Allocated satellite bandwidth (b,) is the estimated safe bandwidth, (up to 25 per cent extra). Column 4. Occupied bandwidth (b,) can be calculated as 2E(Afr X NLR) X G + fm] = X(Afmc X G 1 + fml = b, The above holds true using Af, or Afmc. Column 5. Deviation for 0 dbmo test-tone. This is dependent on knowing the allocated bandwidth and top modulating frequency.

4 212 W. R. CALVERT Table I. Intelsat IV-A, V, V-A, and VI transmission parameters (regular FDM/FM carriers) Carrier-to-total noise temperature Ratio of unmodulated Deviation ratio at carrier power to Carrier Top Allocated (r.m.s.) operating point Carrier-to- max. carrier power capacity haseband satellite Occupied for 0 drm0 Multichannel ( pw0p noise ratio density under full (number of frequcncy BW unit bandwidth test tone r.m.s. devi- from RF sources) in occupied load conditions channels) (khd (MHz) (MHz) (khz) (khz1 (dbw/k) BW (db) (db/4 khz) N rml ha b, Af, Af,,. C/T ClN 12 t * (X = 1) i *(X= 1) * (X = 1) i3.n * (X = 2) Column 6. Multichannel r.m.s. deviation (Af,,). Knowing the deviation for 0 dbmo test-tone and the number of channels we can calculate the NLR, which in our case is given by the expression NLR = log N where N is the number of channels (n < 240) ; then Afr x NLR = Afmc Using the above information: Multichannel bandwidth = 2[223 X X 3.16) kHz = 2( )kHz = 2(2219) khz = 4.4 (MHz) Having discussed the various parameters of an F.M. system, let us look at one method of testing an F.M. modulator, that is the Bessel zero method of testing deviation sensitivity. Thc variation of output frequency is known as the deviation of the modulator, and the amount of voltage to cause a certain deviation will give a figure known as deviation sensitivity. So, for a known input voltage let us call it level in dbm, the output will deviate by a certain amount and can often be seen quoted -30 dbm/200 khz (r.m.s.) The deviation must always be specified in peak or r.m.s. The input to an F.M. modulator is a varying level which in a single channel system corresponds to the voice signal input. So for a no-input condition the output of a modulator is a single frequency; for a variation in input level, i.e. a different input, it may be carrier plus side frequencies (Figure 6) and for another level may be a pattern such as Figure 7. Now, knowing the above response of an F.M. modulator to an input signal we can identify various patterns, i.e. for a modulation index (M) of 1.4 (using our Bessel functions) a display of equal carrier and first side frequencies can be obtained. For a modulation index of 2.4 a display of disappearing carrier is obtained. (This is first Bessel zero) So, at the instant when we have a modulation index of 2.4 and a top modulating frequency of 552 khz, we have a certain deviation; is this any use to us? Not as it stands, but if we now use the cross-over frequency (0.608 fm) as our new fm we can calculate our deviation (f,.) at this frequency AfP 2.4 = khz Af, = khz (peak) So now we know that at cross-over frequency we have a modulation index (M) of 2.4, which gives us a spectrum analyser display of disappearing carrier (Figure 8). If we now take our original r.m.s. test tone deviation for a 132 channel carrier, which is 223 khz (remembering that this is r.m.s. and our new Af is

5 ~~ FO Figure 6 FO Figure 7 Figures 6 and 7. Spectrum analyser display Table 11. Test-tone deviation and baseband level Deviation r.m.s. level TOP Allocated (r.m.s.) for Test-tone for 1st Carrier baseband satellite 0 dmbo test- frequency carrier capacity frequency BW unit tone 0 *608ftop null (channels) (khz1 (WZ) (kh4 (kw (dbmo) * * * ~~~ t * * _ ~ ~ ~

6 214 W. R. CALVERT J1 I 3. Intelsat Earth Station Technology Manual and Satellite Systems Operation Guide. peak), we have JO Figure 8. Disappearing carrier khz = khz X log = dbmo We now have a correcting level to apply at the input to the modulator (using cross-over frequency) which should give us a carrier null display if the deviation sensitivity of the modulator is correctly set up (Table 11). It is possible to use other combinations such as M= 1.4, or even to derive complex patterns at other modulation index points. Figure 9 gives an example of a typical test arrangement to check the deviation sensitivity of the modulator. For a 132 channel 5MHz carrier we have, from the FDM/FM carrier tables (see Table 11), 1. Test-tone frequency is khz. 2. r.m.s. level for first carrier null is +8-2 dbmo. 3. Insert test-tone of khz from signal generator into baseband input at +8.2 dbmo level. (At a -45 dbr point, this corresponds to an absolute level of = dbm.) FURTHER READING 1. Ledger and Roche, Fundamental Electronics, Blackie and Son Limited. 2.?dub and Schilling, Principles of Communication Systems. McGraw Hill Book Company. PROBLEMS 1. If the modulator sensitivity is quoted as -30 dbm/200 khz r.m.s. what is the deviation when (a) -30 dbm is applied? (b) -33 dbm is applied? (c) -27 dbm is applied? 2. Using the Bessel functions what is the distinctive spectrum obtained for a modulation index (M) of 1*4? 3. What would be the test tone frequency for a hypothetical 144 channel carrier? 4. Given a 0 dbmo test tone (r.m.s.) deviation of 243 khz for the 144 channel carrier (a) what is the multi-channel deviation (r.m.s.)? (b) what is the occupied bandwidth? 5. What would be the correction factor for first carrier null? ANSWERS 1. (a) 200 khz r.m.s. or khz peak. (b) 141 khz r.m.s. or 200 khz peak. (c) 282 khz r.m.s. or 400 khz peak. (Remember deviation is voltage sensitive, i.e deviation obtained). 2. Carrier and equal first side frequencies kHz. 4. (a) khz r.m.s. with NLR of (b) 4-9MHz dB SIGNAL GENERATOR SPECTRUM ANALYSER OISAP. CARRIER Figure 9. Spectrum analyser display

FM AND BESSEL ZEROS TUTORIAL QUESTIONS using the WAVE ANALYSER without a WAVE ANALYSER...137

FM AND BESSEL ZEROS TUTORIAL QUESTIONS using the WAVE ANALYSER without a WAVE ANALYSER...137 FM AND BESSEL ZEROS PREPARATION... 132 introduction... 132 EXPERIMENT... 133 spectral components... 134 locate the carrier... 134 the method of Bessel zeros... 136 looking for a Bessel zero... 136 using