Digital Modulators & Line Codes

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1 Digital Modulators & Line Codes Professor A. Manikas Imperial College London EE303 - Communication Systems An Overview of Fundamental Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

2 Table of Contents 1 Introduction First Modelling of Digital Modulators Examples of Binary Modulators Second Modelling of Digital Modulators: Signal Constellation Distance Between two M-ary signals Examples of Signal Constellation Diagram 4ASK and QPSK Block Structures 2 QPSK: Comments Examples of QPSK-Rx s Constellation Diagram 3 Performance Evaluation Criteria 4 Performance of Binary Digital Modulators/Demodulators 5 Line Codes PSD(f) of "line-code" signals Main Types of Line Codes and their PSD Popular Line Codes HDB3 Line Codes Example: Autocorr. & PSD(f) of a Bipolar Line Code Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

3 Introduction Introduction General Block Structure of a Digital Communication System Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

4 Introduction Let us focus on the "discrete channel" Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

5 Introduction With reference the previous figures: at point T : s(t) waveform. The digital modulator takes γ cs -bits at a time at some uniform rate r cs = T 1 cs and transmits one of M = 2 γ cs distinct waveforms s 1 (t),..., s M (t) i.e. we have an M-ary communication{ system. 0 s1 If γ cs = 1 we have one bit at a time 1 s 2 i.e. a binary comm. system A new waveform (corresponding to a new γ cs -bit-sequence) is transmitted every T cs seconds at point T : noisy waveform r(t) = βs(t) + n(t). The transmitted waveform s(t), affected by the channel, is received at point T at point B2 : a binary sequence. based on the received signal r(t) the digital demodulator has to decide which of the M waveforms s i (t) has been transmitted in any given time interval T cs Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

6 Introduction If M = 2 Binary Digital Modulator Binary Comm. System If M > 2 M-ary Digital Modulator M-ary Comm. System B2 B2 Pr(D 1 H 1 ), Pr(D 1 H 2 ),..., Pr(D 1 H M ) Pr(D 2 H 1 ), Pr(D 2 H 2 ),..., Pr(D 2 H M ) F =.....,.. (1) Pr(D K H 1 ), Pr(D K H 2 ),..., Pr(D K H M ) Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

7 Introduction Binary Com. Systems: use M = 2 possible waveforms T cs T cs { s 0 (t), s 1 (t)}; T cs = T b (2) M-ary Com. Systems: use M possible waveforms {s 1 (t), s 2 (t),..., s M (t)}; T cs = γ cs T b (3) Tcs Tcs Tcs The M signals (or channel symbols) are characterized by their energy E i Tcs E i = si 2 (t) dt; i (4) 0 Furthermore their similarity (or dissimilarity) is characterized by their cross-correlation ρ ij = 1 Ei E j Tcs 0 s i (t) s j (t) dt (5) Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

8 Introduction B2 B2 Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

9 Introduction First Modelling of Digital Modulators First Modelling of Digital Modulators Note: Transforming a sequence of 0 s and 1 s to a sequence of ±1 s o/p = 1 2 i/p (6) Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

10 Introduction Examples of Binary Modulators First Modelling of Digital Modulators A binary digital modulator maps 0 s and 1 s { onto two analogue 0 so (t) symbol-waveforms s 0 (t) and s 1 (t), that is 1 s 1 (t) 0 t T cs The three basic binary digital modulators ASK (Amplitude Shift-Keyed) 0 s o (t) = A 0 cos(2πf c t) 1 s 1 (t) = A 1 cos(2πf c t) for 0 t T cs PSK (Phase Shift-Keyed) 0 s o (t) = A c cos(2πf c t) 1 s 1 (t) = A c cos(2πf c t ) for 0 t T cs FSK (Frequency Shift-Keyed) 0 s o (t) = A c cos(2πf 0 t) 1 s 1 (t) = A c cos(2πf 1 t) for 0 t T cs Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

11 Introduction First Modelling of Digital Modulators The above binary digital modulators using complex representation: ASK (Amplitude Shift-Keyed) PSK (Phase Shift-Keyed) =+A c {}}{ A c exp(j0 ) exp(j2πf c t) 0 s o (t) = 1 s 1 (t) = A c exp(j180 ) exp(j2πf c t) }{{} = A c for 0 t T cs FSK (Frequency Shift-Keyed) 0 s o (t) = A 0 exp(j2πf c t) 1 s 1 (t) = A 1 exp(j2πf c t) for 0 t T cs 0 s o (t) = A c exp(j2πf 0 t) 1 s 1 (t) = A c exp(j2πf 1 t) for 0 t T cs Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

12 Introduction First Modelling of Digital Modulators Communication Systems Classification: 1 Coherent (if demodulator is coherent, i.e. it uses a copy of the carrier) 2 Non-coherent (if demodulator is non-coherent, e.g. it does not use the carrier, e.g. envelope detector) Note: Optimum demodulators are coherent. Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

13 Introduction Second Modelling of Digital Modulators: Signal Constellation Second Modelling of Digital Modulators: Signal Constellation We may represent the signal s i (t) by a point (w si ) in a D-dimensional Euclidean space with D M. The set of points (vectors) specified by the columns of the matrix W = [ w s1, w s2,..., w sm ] is known as signal constellation. (7) E i = w H s i w si = w si 2 (8) ρ ij = 1 w H Ei E s i w sj = w H s w i sj j w si (9) w sj Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

14 Introduction Second Modelling of Digital Modulators: Signal Constellation Distance Between two M-ary Signals The distance between two signals s i (t) and s j (t) is the Euclidean distance between their associate vectors w si and w sj i.e. d ij = w si w sj = (w si w sj ) H (w si w sj ) = w H s i w si + w H s j w sj 2w H s i w sj = E i + E j 2ρ ij Ei E j d 2 ij = E i + E j 2ρ ij Ei E j It is clear from the above that the Euclidean distance d ij associated with two signals s i (t) and s j (t) indicates, like the cross-correlation coeffi cient, the similarity or dissimilarity of the signals. Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

15 Introduction Second Modelling of Digital Modulators: Signal Constellation Examples of Signal Constellation Diagram Consider an M-ary System having the following signals {s 1 (t), s 2 (t),..., s M (t)} with 0 t T cs M-ary ASK channel symbols: s i (t) = A i cos(2πf c t) where A i =given= 2i 1 M (say) dimensionality of signal space = D = 1 if M = 4 then Es 1 Es 2 Es 3 Es 4 Origin s1( t) s2( t) s3( t) s4( t) if M = 8 then Es 1 Es 2 Es 3 Es 4 Es 5 Es 6 Es 7 Es 8 Origin 000 s1( t) 001 s2( t) 011 s3( t) 010 s4( t) 110 s5( t) s t 6( ) 101 s7( t) 100 s8( t) Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

16 Introduction Second Modelling of Digital Modulators: Signal Constellation M-ary PSK ) channel symbols: s i (t) = A. cos (2πF c t + 2π M. (i 1) + φ for i = 1, 2,.., dimensionality of signal-space = D = 2 M = 4 & φ = 0 M = 4 & φ = 45 M = 8 & φ = 0 01 s2( t) Es 2 Es 2 01 s2( t) Es1 s1( t) 00 Es s4( t) Es s3( t) Es s2( t) Es 3 11 s3( t) Origin s1( t) Es 1 00 Origin 110 Es 5 s5( t) 45 0 Origin 000 s1( t) Es 1 s4( t) 10 Es 4 s3( t) 11 Es 3 s4( t) 10 Es 4 s6( t) 101 Es s t 6 7( ) 100 s8( t) Es 8 Es 7 Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

17 Introduction Second Modelling of Digital Modulators: Signal Constellation M-ary FSK: very diffi cult to be represented using constellation diagram with f i f j = n 1 2T cs f i + f j = m 1 2T cs where n, m = integers Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

18 Introduction Second Modelling of Digital Modulators: Signal Constellation M-ary Amplitude & Phase - M-ary QAM: 1 channel symbols: s i (t) = A i cos(2πf c t + ϕ i + φ) i = 1, 2,..., M n = 1, 2,..., M 1 or s nm (t) = A n cos(2πf c t + ϕ m + φ) m = 1, 2,..., M 2 M = M 1 M 2 2 dimensionality of signal space = D = 2 PAM-PSK QAM Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

19 Introduction Second Modelling of Digital Modulators: Signal Constellation 4ASK and QPSK Block Structures An ASK (M=4) Communication System Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

20 Introduction Second Modelling of Digital Modulators: Signal Constellation A QPSK Communication System Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

21 QPSK: Comments From this figure it is clear that the constellation diagram shows the mapping of binary digits to QPSK channel symbols (constellation points) as well as the square root of the energy E s of the channel symbols. The diagram also indicates a Gray code mapping from binary digits to channel symbols (constellation points). Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51 QPSK: Comments A QPSK modulator has four channel-symbols which are described by the following equation: s i (t) = A cos(2πf c t + 2π (i 1) + φ) for i = 1, 2, 3, 4 (10) M with M = 4 and 0 t T cs and the modulation process is described by the so called constellation diagram. The previous figure (page 20) shows the constellation for φ = 45.

22 QPSK: Comments Overall, using complex number representation, it is clear from the constellation diagram that 00 s 1 (t) = E s exp(j45 ) exp(j2πf c t) }{{} =m 1 01 s 2 (t) = E s exp(j135 ) exp(j2πf c t) }{{} =m 2 11 s 3 (t) = E s exp(j225 ) exp(j2πf c t) }{{} =m 3 10 s 4 (t) = E s exp(j315 ) exp(j2πf c t) }{{} =m 4 (11) In other words, s i (t) = m i exp(j2πf c t) for i = 1, 2, 3 and 4 (12) Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

23 QPSK: Comments It is important to point out that the four symbols m 1, m 2, m 3 and m 4 are known as baseband QPSK channel symbols and are used by the QPSK constellation symbol mapping block shown in the previous figure. For instance, if the bit-pair at the input of the QPSK modulator is 01 then the output is the baseband channel symbol m 2. With reference to the figure given in page 20, the term exp(j2πf c t) (see Equations 11 and 12) is shown at the Transmitter as the up-conversion from baseband to bandpass. In a similar fashion the down-conversion from baseband to bandpass is shown at the receiver s front-end using the complex conjugate of the transmitter s carrier, i.e. using exp( j2πf c t). Thus, overall, we have exp(j2πf c t) exp( j2πf c t) = 1. Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

24 QPSK: Comments Based on the above discussion it is clear that the presence of the carrier does not affect the analysis of the system. Therefore, it is common practice to ignore the carrier when analyzing communication systems, by working on the baseband. For the rest of this explanatory note the carrier term will be ignored from both Tx and Rx. Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

25 QPSK: Comments Summarising a QPSK modulator/demodulator is represented by its constellation diagram and the QPSK symbol mapper transforms the binary sequence to a sequence of QPSK complex channel symbols m i, forming the baseband QPSK message signal m(t) of bandwidth B i.e. m(t) = n m 1,m 2,m 3,m 4 {}}{ a[n] c(t n T cs ); nt cs t < (n + 1) T cs where c(t) = rect { t T cs } {a[n]} = sequ. of independent data symbols (m i ) B = r cs 2 = 1 2T cs Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

26 QPSK: Comments Thus with reference to the binary sequence of bits (message), by looking at the QPSK constellation diagram it is clear that we have the following mapping where Note that, for a binary system where m(t) n ±1 00 m 1 = A exp(j45 ) 10 m 4 = A exp(j315 ) 01 m 2 = A exp(j135 ) A = E s {}}{ a[n] c(t n T cs ); nt cs t < (n + 1) T cs c(t) =rect{ t T cs } {a[n]} = sequ. of independent data bits (±1s) B = r cs 2 = 1 2T cs Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

27 QPSK: Comments Examples of QPSK-Rx s Constellation Diagram Examples of Plots of QPSK- Receiver s Constellation Diagram x Im 0 Im Re "good" Re x 10 6 "bad" Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

28 Performance Evaluation Criteria Performance Evaluation Criteria In general the quality of a digital communication system is expressed in terms of the accuracy with which the binary digits delivered at the output of the detector represent the binary digits that were fed into the digital modulator. It is generally taken that it is the fraction of the binary digits that are delivered back in error that is a measure of the quality of the communication system. This fraction, or rate, is referred to as the bit error probability p e, or, Bit-Error-Rate BER. Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

29 Performance Evaluation Criteria The performance of M-ary communication systems is evaluated by means of the average probability of symbol error p e,cs, which, for M > 2, is different than the average probability of bit error (or Bit-Error-Rate BER), p e. That is { pe,cs = p e for M > 2 p e,cs = p e for M = 2 (i.e. Binary Communication Systems) However because we transmit binary data, the probability of bit error ρ e is a more natural parameter for performance evaluation than ρ e,cs. Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

30 Performance Evaluation Criteria Although, these two probabilities are related i.e p e = f {p e,cs } their relationship depends on the encoding approach which is employed by the digital modulator for mapping binary digits to M-ary signals (channel symbols) where. γ cs = log 2 M Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

31 Performance Evaluation Criteria An Important Bound involving d ij (distance between the i th and j th constellation point): The symbol error probability p e,cs is bounded as follows: p e,cs M Pr(s i ) M i=1 j=1 j =i T { } dij 2N0 (13) definition of the minimum distance : d min = min {d ij } i,j Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

32 Performance of Binary Digital Modulators/Demodulators Performance of Binary (equiprobable) Digital Modulators/Demodulators Consider a Binary Communication system { 0 s1 1 s 2 or a more popular notation: Constellation diagram: 0 1 E0 s 0 s 1 E1 { 0 s0 1 s 1 } { p e =T{ d (1 } 2N0 =... =T ρ)eue Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

33 Performance of Binary Digital Modulators/Demodulators Thus, at the output of an optimum digital demodulator the probability of error can be calculated by using the following expression: { } p e = T (1 ρ)eue (14) where EUE= E b N 0 and PSD ni (f ) = N 0 2 E b = 1 2 T cs (s 0 0 (t) 2 + s 1 (t) 2 )dt = average signal energy with ρ = 1 E b T cs s 0 0 (t)s 1 (t)dt = the time cross-correlation between signals Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

34 Performance of Binary Digital Modulators/Demodulators N.B.: (1 ρ) E b N 0 = = p e = i.e. if E b N 0 = fixed then the optimum system is that for which the correlation coeff is -1 ρ = 1 = s 0 (t) = s 1 (t) This is known as optimum, or ideal binary Communication System Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

35 Performance of Binary Digital Modulators/Demodulators Examples Baseband MODEMS: { 0 s0 (t) = A Antipodal c 0 t T 1 s 1 (t) = A cs c s(t) = A c m(t) ({a[n]} = sequ. of independent data bits (±1s)) } p e =T{ Ac σ COHERENT MODEMS: 1. Amplitude Shift-Keyed (ASK) or On-Off Keying (OOK) { 0 s0 (t) = 0 (ASK or OOK) 1 s 1 (t) = A c cos(2πf c t) s(t) = A c m(t) } cos(2πf c t) Es1 p e =T{ 2N 0 0 t T cs Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

36 Performance of Binary Digital Modulators/Demodulators 2. Biphase Shift-Keyed: { 0 s0 (t) = A general c cos(2πf c t θ) 1 s 1 (t) = A c cos(2πf c t + θ) { } p e =T 2 EUE sin 2 ( θ) Phase-Reversal { Keying (RSK) 0 s0 (t) = A (RSK) c sin(2πf c t) 1 s 1 (t) = A c sin(2πf c t) { } p e =T 2 EUE 0 t T cs 0 t T cs Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

37 Performance of Binary Digital Modulators/Demodulators N.B.: for θ = ± π 2 then general = RSK and it is called BPSK BPSK s(t) = A c sin(2πf c t + m(t) π 2 ) (15) Equation 15 can be written as follows: BPSK can be considered as BPSK s(t) = A c m(t) cos(2πf c t) (16) { PM AM The PSD(f ) s of m(t) and s(t) are shown below: Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

38 Performance of Binary Digital Modulators/Demodulators Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

39 Performance of Binary Digital Modulators/Demodulators 3. Frequency Shift-Keyed (FSK) { 0 s0 (t) = A (FSK) : c cos(2πf c t) 1 s 1 (t) = A c cos(2π(f c + f )t) 0 t T cs, f = m 2T cs coherent {}}{ { } CFSK : p e =T EUE discontinuous FSC Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

40 Performance of Binary Digital Modulators/Demodulators continuous FSC Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

41 Performance of Binary Digital Modulators/Demodulators { } 1 1 ASK p e =T 2 EUE 2 non-coherent ASK p e = 0.5 exp( E s 1 4N 0 ) + 0.5T { } 3 FSK p e =T EUE 4 FSK (non-coherent) p e = 1 2 exp { { 1 2} EUE} 5 BPSK p e =T 2EUE { } Es1 2N 0 6 BPSK (differential) p e = 1 2 exp { EUE} { } 7 QPSK p e =T 2EUE { } 8 MSK p e =T 1.7EUE { } 9 Gaussian MSK p e T 1.36EUE } 10 M-ary PSK (coherent) p e 2T{ 4EUE sin( π 2M { ) } 11 M-ary QAM p e 4(1 1 3 )T M M 1 EUE 12 DS-BPSK and DS-QPSK SSS p e =T { } 2EUE equ Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

42 Line Codes PSD(f) of "line-code" signals Line Codes PSD(f) of Line Code Signals definitions: where s(t) = a[n]c(t nt cs ); nt cs < t < (n + 1)T cs (17) n c(t) is an energy signal of duration T cs. This implies that c(t nt cs ) is defined in the interval nt cs < t < (n + 1)T cs {a[n]} is a binary sequence having an autocorrelation function R aa [k] = E {a[n]a[n + k]} = I (a[n]a[n + k]) i th -pair. Pr(ith -pair) i =1 where I denotes the total number of "pairs" Insert Figure Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

43 Line Codes PSD(f) of "line-code" signals PSD s (f ) = FT(c(t)) 2 T cs R aa [0] + + k= k =0 R aa [k] exp( j2πfkt cs ) (18) Note that if R aa [k] = { E { a 2 n } for k = 0 E {a n } E {a n+k } for k = 0 { µ 2 i.e. R aa [k] = a + σ 2 a for k = 0 where µ a = mean and σ a = std for k = 0 µ 2 a then (18) PSD s (f ) = σ 2 FT(c(t)) 2 a T }{{ cs } Continuous Spectrum + µ2 a Tcs 2 comb 1 ( FT(c(t)) 2 ) Tcs } {{ } Discrete Spectrum Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

44 Line Codes PSD(f) of "line-code" signals EXAMPLE... FOR YOU... if a n = 0, 1 with Pr {a n = 0} = Pr {a n = 1} = 1 2 of unipolar RZ line-code signal. find the spectrum Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

45 Line Codes Main Types of Line Codes and their PSD Main Types of Line Codes and their PSD Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

46 Line Codes Popular Line Codes Popular Line Codes Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

47 Line Codes HDB3 Line Codes HDB3 Line Codes Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

48 Line Codes Example: Autocorr. & PSD(f) of a Bipolar Line Code Example: Autocorr. & PSD(f) of a Bipolar Line Code 1 Show that for a bipolar line code the autocorrelation function of the code sequence {a[n]} is as follows: R aa [k] = 1/2 if k = 0 1/4 if k = 1 0 if k 2 2 If Pr(0) = Pr(±1) = 0.5 and c(t) =rect{ t T cs } (19), derive an expression for the power spectral density PSD(f ) for the bipolar line code waveform m(t) = a[n] c(t nt cs ) (20) n= Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

49 Line Codes Example: Autocorr. & PSD(f) of a Bipolar Line Code Solution: Autocorrelation R[0], R[1] R aa [0] = E { a[n] 2} = { 0 0 = (±1) (±1) = } = = 1 2 R aa [1] = E {a[n]a[n + 1]} = 1 st 2 nd 1 st 2 rd Pr 0, 0, = 0 1/4 0, ±1, = 0 1/4 ±1, 0, = 0 1/4 ±1, 1, = 1 1/4 (21) = ( 1) 1 4 = 1 4 (22) Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

50 Line Codes Example: Autocorr. & PSD(f) of a Bipolar Line Code Solution: Autocorrelation R[k],k>1 1 st 2 nd 3 rd 1 st 3 rd Pr 0, 0, 0 = 0 1/8 0, 0, ±1 = 0 1/8 R aa [k] = 0, ±1, 0 = 0 1/8 0, ±1, 1 = 0 1/8 with k>1 ±1, 0, 0 = 0 1/8 ±1, 0, 1 = 1 1/8 ±1, 1, 0 = 0 1/8 ±1, 1, ±1 = +1 1/8 = (ignoring the 2nd column and multiplying 1st with 3rd we have) = ( 1) 1 8 = 0 (23) Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

51 Line Codes Example: Autocorr. & PSD(f) of a Bipolar Line Code Solution: PSD(f) { FT(rect t T cs }) 2 PSD(f ) = {R[0] + 2R[k] cos(2πkt cs )} T cs = T cssinc 2 2 { (ft cs ) 1 T cs } 4 cos(2πt cs ) { 1 = T cs sinc 2 (ft cs ) 2 1 } 2 cos(2πt cs ) = T cs sinc 2 (ft cs ) sin 2 ( 2πTcs 2 ) (24) Prof. A. Manikas (Imperial College) EE303: Dig. Mod. and Line Codes / 51

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