4- Single Side Band (SSB)

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Della Quinn
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1 4- Single Side Band (SSB) It can be shown that: s(t) S.S.B = m(t) cos ω c t ± m h (t) sin ω c t -: USB ; +: LSB m(t) X m(t) cos ω c t -π/ cos ω c t -π/ + s S.S.B m h (t) X m h (t) ± sin ω c t 1

2 Tone Modulation: SSB The Hilbert transform is:

3 Detection of SSB Coherent Detection y(t) = [ m(t) cos ω c t ± m h (t) sin ω c t ] cos ω c t = (1/)m(t) [1+ cos ω c t ] ± (1/) (t) sin ω c t m h After the L.P.F the output equals (1/) m(t) 3

4 SSB Coherent Detection USB Coherent demodulation 4

5 Envelope Detection of SSB s(t) S.S.B = m(t) cos ω c t - m h (t) sin ω c t s S.S.B = R(t) cos [ω c t + Ө(t)] R(t) = m (t) m h (t) Ө(t) = tan -1 m h (t) m(t) The envelope did not express the signal Now, assume a large carrier is added: s S.S.B (t) + A cos ω c t = R(t) cos [ω c t + Ө(t)] 5

6 Envelope Detection of SSB m(t) A ] A m(t) A [1 ] A m(t) A[1 R(t) be ignored will A (t) m and A (t) m terms The (t) m then A m(t) A If ] A (t) m A (t) m A m(t) A[1 (t) m A) (m(t) R(t) 1/ 1/ h h h h 6

7 Envelope Detection of SSB In case of large carrier, the envelope of the S.S.B. signal has the form of m(t) (the base band signal), so the signal can be demodulated by the envelope detector with the condition that A >>> І m(t) І 7

8 5- Vestigial Side Band (V.S.B) Since S.S.B. is difficult to realize, a compromise between S.S.B. and D.S.B. in spectrum can be obtained using V.S.B. [especially when we have important components at low frequency] Most of one side band is passed along with a vestige (remainder) of the other side band. (B.W ~ 1.15 W)(Reproduced by filters) 8

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10 5- Vestigial Side Band (V.S.B) Modulator Demodulator A B X H v (f) C X D L.P.F E cos ω c t cos ω c t What is the shape of H v (f)? 10

11 5- Vestigial Side Band (V.S.B) Spectrum at A prop. to: M(f) Spectrum at B prop. to: (1/)[M(f+f c ) + M(f-f c )] f -f c Spectrum at C prop. to: H v (f)[m(f+f c ) + M(f-f c )] Spectrum at D prop. to : (1/) H v (f+f c ) [ M(f+f c ) + M(f) ] +(1/) H v (f-f c ) [ M(f) + M(f-f c ) ] The central lobe of the spectrum at D must equal M(f) (1/) [ H v (f+f c ) + H v (f-f c ) ] M(f) f c f 11

12 5- Vestigial Side Band (V.S.B) What are the conditions on the vestigial filter? [ H v (f+f c ) + H v (f-f c ) ] = constant in W H v (f) 1 1/ -B-f c -f c f c B+f c f 1

13 5- Vestigial Side Band (V.S.B) What are the conditions on the vestigial filter? [ H v (f+f c ) + H v (f-f c ) ] = constant in W -B B f 13

14 Comparison of Various AM signals AM and AM_SC - Detectors required for AM are simpler - SC requires less power at the transmitter for the same information ( cheaper transmitter) 14

15 Comparison of Various AM signals DSB and SSB - SSB needs only half the bandwidth 15

16 AM Applications 1- Frequency Division Multiplexing (FDM) If we have N signals to be transmitted using AM modulation. All of these are band limited to W. f A1 1 Mod 1 f1 1 Dem 1 A B1 Mod f + Dem B C1 3 Mod 3 f3 3 Dem 3 C 16

17 Examples: radio, TV, telephone backbone, satellite, 17

18 AM Applications 1) Example: Telephone channel multiplexing 1 1 First Level Mux Basic Group 1 second Level Mux 5 Super Group 1 10 Third Level Mux Master Group North American FDM hierarchy 18

19 AM Applications All long-haul telephone channels are multiplexed by FDM using SSB. Basic Group consists of 1 FDM SSB voice signals each of BW = 4 KHz. It uses LSB spectra and occupies 60 to 108 KHz. [ alternate group configuration of 1 USB occupies 148 to 196 KHz] Basic group A (LSB) 19

20 Basic Super group AM Applications consists of 60 channels. It is formed by multiplexing 5 basic groups and it occupies a band of 31 to 55 KHz. [alternate 60 KHz to 300 KHz] Super group 1 (LSB) 0

21 Basic master group AM Applications consiste of 600 channels. It is formed by multiplexing 10 supergroups. 1

23 AM Applications - Super heterodyne receiver The receiver not only has the task of demodulation but it is also required to perform some other tasks such as:- - Tuning : select the desired signal - Filtering : separate the desired signal from other modulated signals. - Amplification: to compensate for the loss in the signal power Super heterodyne receiver fulfills efficiently all the three functions 3

24 Homodyne Vs Heterodyne Direct-conversion problems: LNA and mixers have to be very linear, o.w. the distortion (evenorder distortion) will cause DC offset CMOS circuits cause flicker noise (inverse prop. to freq.) LO leakage causing self-mixing thus DC offset The DC offset can be cancelled out (however simple envelope detection is not then possible; also can cause corruption to digital binary stream) 4

25 Homodyne Vs Heterodyne Direct-conversion problems: LO leakage causing self-mixing thus DC offset B. Razavi, "Design considerations for direct-conversion receivers," in IEEE Transactions on Circuits and Systems II: Analog and Digital Signal Processing, vol. 44, no. 6, pp , Jun

26 Homodyne Vs Heterodyne Heterodyne problems: Image rejection (requires selective filtering especially if IF freq. is small, thus the need to multi-stages filtering (also called super heterodyne) ) 6

27 Homodyne Vs Heterodyne Heterodyne problems: Image rejection Vs adjacent channel power 7

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29 Basic elements of an AM receiver of the superheterodyne type A B C D E R.F. Section Mixer I.F. Section Envelop. detector Audio Amplifier ~ Local oscillator (f L.O ) Common tuning. 9

30 Spectrum at different points I.F. response R.F. response A.F. response 0 f I. f c f at E F at C at A 30

31 at A : [A + m(t)] cos ω c t at B : [A 1 +a 1 m(t)] cos ω c t cos (ω c + ω I.F )t =[A +a m(t)] {cos (ω c + ω I.F )t +cos ω I.F t } at C : [A 3 +a 3 m(t)] cos ω I.F t at D : a 4 m(t) 31

32 The R.F cannot provide adequate selectivity since f c is high. It rejects a lot of adjacent channel interference and amplifies the signal. This is why we translate it to IF frequency to obtain good selectivity. All selectivity is realized in the IF section. The main role of RF section is image frequency rejection. What is the image frequency? 3

33 If f c = 1000 KHz, f L.O. = =1455KHz the image frequency = f c +f I.F. = 1910 KHz. Will also be picked up. Stations that are f I.F apart are called image stations and are rejected by RF filter. Up-conversion: f L.O. = f c +f I.F (1005 to 055 KHz (ratio.045)) Down conversion: f L.O = f c -f I.F (95to 1145KHz (ratio 1.05)) 33

34 F IF = 10.7 MHz 34

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Outline. Communications Engineering 1

Outline. Communications Engineering 1 Outline Introduction Signal, random variable, random process and spectra Analog modulation Analog to digital conversion Digital transmission through baseband channels Signal space representation Optimal