The starting phase of each symbol is determined by past bits. This memory can improve detection by giving additional hints about past symbol values.

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Robyn Copeland
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1 8.2 Start With CPFSK CPFSK Sigals Three pictures of the same CPFSK sigal: istataeous frequecy istataeous phase trajectory Trajectory is a circle with radius A = P. CPFSK moves at a steady speed aroud the circle, with reversals. The startig phase of each symbol is determied by past bits. This memory ca improve detectio by givig additioal hits about past symbol values

2 Relatio betwee badpass sigal ad complex basebad: () 2Re 2 cos 2 () ( π +φ ) jφ() t j2πfct s t = Ae e = A fct t st () o Istataeous values: total phase: 2 π f t+φ ( t) c frequecy (rad/s): 2 π fc +φ ( t) = 2π fc ± 2πfd frequecy (Hz): fc ± f d o To shift frequecy up or shift frequecy dow: phase icreases liearly s() t rotates i positive directio, costat speed or phase decreases liearly or s() t rotates i egative directio, costat speed 8.2-2

3 PSK carries iformatio i phase, FSK i the directio of travel: PSK FSK The product fdt has a strog effect o power spectrum: o Geerally, a greater max frequecy deviatio f d produces a wider spectrum. o Multiples of 1/2 create discrete spectral compoets (toes). Cosider the value 1 ft= d cycle : 2 Oe half-cycle (half-circle) forward or backward i each bit. Start o the same phase, ed o the same phase. No phase memory

4 o Cotiuig with the fc The 1 ft= d cycle example, suppose the carrier is = 1 T (for illustratio oly, f c is ot relevat to presece of toes). 1 3 fc fd = ad fc + f d =, ad the badpass sigal is 2T 2T A tued circuit will rig at fc f c + f, hece there are toes i the sigal. d f ad aother oe will rig at d o Rigig occurs if the total phase π f t+φ ( t) (the cosie argumet) is 2 c the same at the ed of a symbol (i.e., the start of the ext symbol), o matter which data value was set: ( ) ( ) 2π f + f T = 2π f f T + 2 π, a iteger So we get toes if c d c d ft d = 2 fd ft d = = is a iteger multiple of 1 R 2 2. This exteds to multilevel (i.e., o-biary) FSK, too. s 8.2-4

5 o Why we do t like toes: They mix i oliearities to produce uexpected ad uwelcome images. They cosume power without carryig iformatio. Formalize all these observatios as expressios defiig CPFSK. o Use I = 2m 1 M, for m= 1,, M as the data at symbol time. Odd itegers from M + 1 to M 1 (e.g., -5, -3, -1, +1, +3, +5) o Alterative possibilities for the symbol (if M = 2): frequecy phase The modulatio idex h is the accumulated phase differece (i cycles) betwee the alteratives over oe symbol. It is dimesioless. Iteger h causes toes

6 8.2-6 The modulatio idex h is related to the spacig Δ f of CPFSK sigal frequecies ad the symbol rate R: o Waveforms: Istataeous frequecy is a rectagular pulse of height f I, but we d like to use h ( f = h 2T ). So ormalize, by defiig a d frequecy pulse g() t with area 1/2. d The istataeous frequecy is f () t = hi g() t = 2 f T I g() t. Larger h greater frequecy swig. d Extedig this to a sequece of pulses, f () t = h I g( t T)

7 Next, the istataeous phase φ () t = 2 π f( α) dα (radias): t T φ () t = 2 πh I q( t kt) = 4 π f T I q( t k ) k k d k k where the phase pulse qt () = g( α) dα: t The saturatio value of 1/2 meas that the phase shift caused by I has a cotiuig effect i the future; i.e., the modulatio has memory. The usual CPFSK defiig equatio writes phase i iterval as φ (, t I) = 2 πh I q( t kt) + φ, T t ( + 1) T where φ o k = k o is the phase at the begiig of time (but is zero if the receiver s phase trackig works) ad I is the symbol sequece. But all qt ( kt) up to k = 1 have already saturated at 1/2, so t T φ ( t, I) =θ + 2 πh I i T t ( + 1) T 2T θ =πh 1 k = I k 8.2-7

8 8.2-8 Examples of phase trajectories: Above: biary ( M = 2) CPFSK Right: quaterary ( M = 4 ) CPFSK Note phase states. Looks like a tree, but 8.2-8

9 8.2-9 the phases are modulo 2π, so the sigals wrap aroud a cylider i time. If h= 2 f d T is ratioal, the phase state values recur, ad we have a fiite set of phase state values. If h = m p (ad m, p are coprime), the o for eve m, we get p differet phase states πm πm πm θ 0,, 2,,( p 1) p p p o for odd m, we get 2 p differet phase states, alteratig eve ad odd πm πm πm θ 0,, 2,,(2 p 1) p p p 8.2-9

10 Fially, ote that CPFSK is ot a liear modulatio The phase is liear i the trasmitted symbols φ () t = 2 πh I q( t T ) but the actual sigal is st (, I ) = Pe φ j (, t I )

11 8.2.2 CPFSK Sigal Space, States ad Trellis A set of basis fuctios for CPFSK: o I iterval, the complex lowpass sigal is jθ t T s() t = Pe exp j2 πhi, T t < ( + 1) T 2T o Therefore, a basis set is the time traslates of ( ( ) ) g () t = exp jπh 2i M 1 t T,0 t < T, i = 1,, M i These basis fuctios are ot orthogoal, except for some choices of h, but they are liearly idepedet. For example, for M = 4 ad h = 12: o The coefficiets j P e θ with respect to this basis carry the memory of past data values, so the curret waveform sheds light o the past. Igore this memory durig detectio (e.g., by use of a differetial detector) ad you pay a price of several db i lost SNR margi

12 o Implemetatios either use a set of filters matched to the gi () t or they compute those ier products implicitly, as part the metric calculatios i tree or trellis search. The ifluece of the past o the respose of a system is summarized i its state. Here θ acts as the state: θ + 1 =θ +πhi jθ t T st () = Pe exp j2πhi 2T state trasitio equatio output equatio i iterval o If h is ratioal, the there is a fiite set of values for the phase state, ad we ca summarize the recurrece with a state diagram. For example, for h = 12ad M = 2 or 4,

13 The state trasitio diagram ca be urolled to form a trellis. Use the same example ( h = 12ad M = 2 or 4) For M = 2 (biary sigals): compacted or startig from a kow state ad for M = 4 (quaterary sigals): alteratig subsets, eve ad odd Show startig from state 0 at time 0. Note the parallel trasitios. o For M=2, the sequece of states σ uiquely determies the data sequece I. o For M = 4, the parallel trasitios mea the state sequece oly partially determies the data. Also, which of the two I values i a (e.g.) +1, -3 trasitio leaves o trace i the memory (the state), so o help from cotext (the future ad past) i decidig which it was

14 8.2.3 MSK A Special Case FSK ad QPSK are quite differet or are they? For oe selectio of parameters MSK CPFSK is equivalet to precodig applied to offset QPSK (OQPSK). MSK (miimum shift keyig): M = 2 ad h = 12. [ ] Re s( t) = P J p( t T) [ ] eve Im s( t) = P J p( t T) odd where the { 1, 1} J +, too. Looks like 2PAM o each axis idepedetly

15 It ca also be cosidered as offset QPSK, i which the imagiary compoet is delayed by half a symbol. If s() t = P Jp( t T) + j Jkp( t kt) eve k odd the s() t is like 4QAM (or QPSK) of symbol duratio 2 T P bpt ( T), b = J + jj + eve but with the imagiary part delayed by half a symbol (i.e., by T), so it s offset QPSK (OQPSK). QPSK 1 Offset QPSK If the receiver is coheret, it ca tell Re [ s( t )] from Im [ s( t )], ad it s easy to detect the J values, as the sigs of the matched filter output samples. But what is the relatio betwee these J values ad the I values of the MSK sequece, the oes we wat to detect?

16 How to get the I values from the J values? Recall that I =+ 1 is aticlockwise rotatio, I = 1 is clockwise. So: o for eve, I = JJ + 1 ; o for odd, I = JJ + 1 ; o hece, I = ( 1) JJ + 1. o It s differetial decodig with a alteratig sig flip. Easy. From this, the BER follows as Pb 2Q( 2Eb N0) 2Q( 2PT N0) = =. If we had detected I usig the sigal i iterval oly, it would have bee 2FSK with o-orthogoal sigals, with the two alterative sigals startig from the same ukow phase. See Assigmet 2 from 2005 for its BER: b ( b 0 ) P = Q E N (MSK detectio without makig use of memory)

17 The similarity with QPSK also gives us a easy calculatio of the MSK power spectrum: they are the same. See Sectio

Frequency Response of FIR Filters

Frequency Response of FIR Filters EEL335: Discrete-Time Sigals ad Systems. Itroductio I this set of otes, we itroduce the idea of the frequecy respose of LTI systems, ad focus specifically o the frequecy respose of FIR filters.. Steady-state