ELG3175 Introduction to Communication Systems. VSB and Introduction to Angle Modulation

Transcription

1 ELG3175 Inoduion o Communiaion Sysems and Inoduion o ngle odulaion

2 oivaion Fo wideband inomaion signals, SSB is diiul o implemen. Fo equeny disiminaion, he ile mus have a shap uo nea he equeny so as o be able o eliminae one band wihou disoing he ohe. When we use phase disiminaion, we equie ilbe ansomes whih ae diiul o implemen i he signal m has a lage bandwidh.

3 odulaion When SSB is diiul o implemen, we use vesigial sideband modulaion. is implemened by equeny disiminaion bu he ileing poess does no ompleely eliminae he unwaned band. In a, some o he desied band is also paially ileed ou.

4 odulao modulao and is ile s equeny esponse ae shown below. The esponse o he ile is denoed as. We noie a ansiion band aound he equeny. m s os - -x +x

5 Speum o a signal -B m B m S DSB-SC - -B m - +B m -B m +B m S - -B m - +x -x +B m In he example given, we onside a sysem whih eains he uppe sideband as he pinipal band and he lowe sideband onibues he vesigial sideband pa. oweve, eihe sideband an be used o om he pinipal band in paie.

6 In he example given, he ile has a gain o K o all equenies > +x passband. Fo he equenies < < +x, whih eside in he pinipal band, he gain is less han K, so some loss ompaed o he passband ous. Fo he equenies -x < <, he ile s gain is no 0, heeoe some o he ohe sideband s equeny omponens ae passed by he ile and s has a vesigial sideband. The bandwidh o s = B m +x. Geneally, sine we ae ying o edue he bandwidh ompaed o DSB- SC, x is smalle han B m.

7 S S S SC DSB whee e We noe ha due o emeian symmey in he equeny esponse o eal sysems. *

8 Demodulaion o We use he same demodulao as DSB-SC s S x X LPF z = Gm Z os I we wan z = Gm, hen we need o impose a onsain on he equeny esponse o he modulao s ile.

9 X S S X Baseband

10 Z Z Z We wan Z = G, whee G is a onsan. I we ensue ha K 4 K K K Z

11 Theeoe z = K/4m. * Le us eplae by and by D and we ge * This ieia only need be ue ove he equeny ange o he signal. x x 1 K D D K *** -D +D x 1* +x = K

12 Possible USB iles Files wih linea ansiion bands Raised osine iles -x +x

13 Example The signal m = os10+3os30. We wish o ansmi his signal using wih aie = 5os500. The ile s esponse is shown below. Find s as well as is bandwidh

14 Soluion Sine uses equeny disiminaion, i is pobably bes o wok in he equeny domain. Le us ind and S DSB-SC. S DSB-SC

15 Nex we ind S = S DSB-SC. S DSB-SC S =0 490=1/4 510=3/4 530= / 5/ / 15/ s = 7.5os os os490

16 Example Show ha we an demodulae s o he pevious example o obain Gm. s os500 = 7.5os530 os os510os os490os500 = 3.75os os os os os os990. e lowpass ileing z = 3.75os os os10 = 3.75os30 +.5os10 = 1.5m.

17 Inoduion o ngle odulaion In angle modulaion, he ampliude o he modulaed signal emains ixed while he inomaion is aied by he angle o he aie. The poess ha ansoms a message signal ino an angle modulaed signal is a nonlinea one. This makes analysis o hese signals moe diiul. oweve, hei modulaion and demodulaion ae ahe simple o implemen.

18 The angle o he aie Le q i epesen he insananeous angle o he aie. We expess an angle modulaed signal by: s os q whee is he aie ampliude. i

19 Insananeous equeny One yle ous when q i hanges by adians, heeoe he aveage equeny o s on he ineval o +D is: Theeoe he insananeous equeny is ound in he limi as D ends owads 0. i i D D D q q d d i i 1 q

20 Phase modulaion Thee ae wo angle modulaion ehniques. Phase modulaion P Fequeny modulaion F In P, he phase o he aie is a linea union o he message signal, m. Theeoe s P is: s P os k m p whee k p is he phase sensiiviy and is he phase o he unmodulaed aie. To simpliy expessions, we will assume ha = 0. Theeoe he angle o a P signal is given by q i = + k p m.

21 F Fo F, he insananeous equeny is a linea union o he message: whee k is he equeny sensiiviy. m k i i i d m k d q F d m k s os

22 Insananeous equeny o a P signal / Insananeous phase o an F signal Fom s P, we ind i P k p dm d Fom s F, we ind k m d i F m d/d k p /k od F m k /k p od P 0 s P s F

23 Example Find s F and s P i m = os m. SOLUTION s P os k os p m s F os d sin m m k os sin m m m

24 The P and F o he example ae shown hee o = 5, = 1, = 1 kz, m = 100 z, k p = ads/v and k = 500 z/v. s P s F en seondes en seondes