S Transmission Methods in Telecommunication Systems (5 cr) Tutorial 4/2007 (Lectures 6 and 7)

Transcription

1 S Transmission Methods in Telecommunication Systems (5 cr) Tutorial 4/007 (Lectures 6 and 7) 1

2 1. Line Codes / Johtokoodit Sketch beneath each other line codes Manchester, Differential Manchester and AMI when the bit train is transmitted. A: See L6, S Remarks: At the start of the Differential Manchester or AMI pulse train can be inverse and so the pulse train is different as above sketched. In the AMI pulse train the '1' pulse can have no transition to zero during the '1' bit. It's also possible that the pulse and the rest time above are vice versa.

3 . Abbreviations and terms / Lyhenteitä ja käsitteitä Explain briefly what do the following abbreviations and terms mean: a) ISI b) Matched filter a) See lecture 6 notes, slide 9: ISI = Intersymbol interference [pulssien välinen keskinäisvaikutus], in the ideal case the previous and the following pulses doesn't affect to the amplitude value of the pulse issued at the sampling time. b) See lecture 6 notes, slide 1: Matched filter [sovitettu suodatin] is needed in the receiver when the SNR value is desired to have maximum value. In the ideal case the impulse response of matched filter is reverse in the time domain than received pulse has. Because of the convolution in the time domain maximum correlation is then achieved. 3

4 3. Assume that noise has spectrum density of N 0 = W/Hz and receiver s bandwidth is W = 00 khz and impedance level is Z 0 = 50 Ω. Calculate a) the (average) noise power P N at the receiver and b) the probability that noise voltage exceeds +00 mv. See L6, S W 5 a) Noise power PN = N0 df = 00 khz 10 = 4 10 W = 40 μw Hz W b) Noise s RMS voltage = standard deviation σ is URMS = σ = PN Z0 = 40 μw 50 Ω 44,7 mv m = mean value = 0 V, k σ = 00 mv -> k 4,5 To get the desired probability, we have to use Q-function and it s curves (on the next page). So the probability p = Q(k) 3, m λ 1 Q( k) e d π k m+ kσ λ = X σ m λ 4

5 Q-function Q(k) 5

6 4. Signal-to-noise ratio (SNR, k, κ) A DVB-S receiver has BER = 10 - without channel coding. When interleaving, Reed Solomon and Convolution codes are used BER = Modulation method is QPSK (4-PSK) which has the 1 following proberty: BER = Q( SNR ) Calculate by using Q-function diagram how much SNR seems to be improved when channel coding is used. See lecture 6 slide 38 a) Without coding: ( ) ( ) BER = Q SNR = 10 = Q k ( ) SNR = k and SNR ( db) = 10 lg SNR db b) With channel coding: 10 ( ) ( ) BER = Q SNR = 10 = Q k So the improvement is ( ) SNR = k and SNR ( db) = 10 lg SNR db ( ) ( ) SNR db SNR db db 1 6

7 Q-function Q(k) 7

8 4. Signal-to-noise ratio (SNR, k, κ) A DVB-S receiver has BER = 10 - without channel coding. When interleaving, Reed Solomon and Convolution codes are used BER = Modulation method is QPSK (4-PSK) which has the 1 following proberty: BER = Q( SNR ) Calculate by using Q-function diagram how much SNR seems to be improved when channel coding is used. See lecture 6 slide 38 a) Without coding: ( ) ( ) BER = Q SNR = 10 = Q k ( ) SNR = k,07 = 4, 8 and SNR ( db) = 10 lg SNR 6,3 db b) With channel coding: 10 ( ) ( ) BER = Q SNR = 10 = Q k So the improvement is ( ) SNR = k 6,3 = 39,7 and SNR ( db) = 10 lg SNR 16 db ( ) ( ) SNR db SNR1 db 9,7 db 8

9 5. A NRZ-signal (Non Return to Zero) / NRZ-signaali Calculate by giving formulas the power spectrum (density) of a NRZ-signal {0 V,1 V} when the bits 0 and 1 have the same probability. 1/ See L6, S15 Because of the probabilities for the bits '0' and '1' are the same, we can conclude that the random signal has DC-value = <v(t)> = 0,5 V. Now we can divide the random signal into the two parts: vi( t) = va( t) + vb( t ) = 05, V + vb( t) and vb ( t ) = { 05,, + 05, } = the random signal. Average value and autocorrelation: <v(t)> = R( ± ) = 05, R( ± ) = 05, v v ( ) When the random signal is vb t, the autocorrelation function is b τ R ( τ) = σ tria where D is the duration of a bit. And v D b R 0 = σ = P = ± 0, 5 = 0, 5 v ( ) ( ) 9

10 5. A NRZ-signal (Non Return to Zero) / NRZ-signaali Calculate by giving formulas the power spectrum (density) of a NRZ-signal {0 V,1 V} when the bits 0 and 1 have the same probability. / So the autocorrelation function for the NRZ-signal is τ R v ( τ ) = 05, tria + 05, D Thus the power spectrum density is ( ) = ( τ) = 05 ( ) + 05 δ( ) Gv f F R v, D sinc fd, f 10

11 6. Transferring a file in an AWGN channel / Tiedoston siirto kohinaisessa kanavassa 1/3 A file which size is 1,4 MB (about 11,74 Mbits transferred in an AWGN channel which has the attenuation of 16 db. The power of the transmitter is 30 dbm. The noise has 1-sided power spectrum 0 density N0 = 410 W / Hz in the receiver. The modulation system sends two bits (= 1 symbol) at the same time and the symbol error probability is E rx, where E rx is the average energy of a Ps = Q N0 9 symbol. How long it takes to transfer the file when P s 10 is needed? See L6, S0 1 6 The channel attenuation is A , S = db AS =. The transmitted power is PRX = 30 dbm = 1W. PTX So the received average power is PRX =. AS Rb Because of two bits are sent at the same time, the symbol rate is R S =, where is the bit rate. R b 11

12 6. Transferring a file in an AWGN channel / Tiedoston siirto kohinaisessa kanavassa /3 A file which size is 1,4 MB (about 11,74 Mbits transferred in an AWGN channel which has the attenuation of 16 db. The power of the transmitter is 30 dbm. The noise has 1-sided power spectrum 0 density N0 = 410 W / Hz in the receiver. The modulation system sends two bits (= 1 symbol) at the same time and the symbol error probability is E rx, where E rx is the average energy of a Ps = Q N0 9 symbol. How long it takes to transfer the file when P s 10 is needed? Because of E N RX (see Q-function figure): The average energy for a received symbol is Thus 9 ( 60) 10 Q, ERX = 6, 0 = SNR= 36 ERX = 36N N 0 0 P P 36N0 = R = 348, 9 kbit/ s AR TX TX b S b 18NA 0 S 0 E P T P PTX A R RX RX = RX S = = RS S b 1

13 6. Transferring a file in an AWGN channel / Tiedoston siirto kohinaisessa kanavassa 3/3 A file which size is 1,4 MB (about 11,74 Mbits transferred in an AWGN channel which has the attenuation of 16 db. The power of the transmitter is 30 dbm. The noise has 1-sided power spectrum 0 density N0 = 410 W / Hz in the receiver. The modulation system sends two bits (= 1 symbol) at the same time and the symbol error probability is E rx, where E rx is the average energy of a Ps = Q N0 9 symbol. How long it takes to transfer the file when P s 10 is needed? The time needed for transmission is T 6, bit = 33, 7 s 3 bit 348, 9 10 s 13

14 7. 8-PSK, 16-QAM and noise / 8-PSK, 16-QAM ja kohina 1/ a) See lecture 6 notes and slide 38. Let A = 1 and 8-PSK is used. Calculate the symbol error rate of 8-PSK when a white Gaussian noise having the voltage rms-value 97,7 dbμv is added to the signal. b) See lecture 6 notes and slide 39. Let a = 1 and BER (Bit Error Rate or Ratio) = Calculate by giving formulas the maximum allowed rms-value for white Gaussian noise and give SNR (Signalto-Noise Ratio) in db's when 16-QAM is considered. See L6, S38 a) To find out probability Q(k) = p when SNR = κ = k is known and when k > 3, k e Qk ( ) A π 1 π pε = Q sin Q sin 97,7 k π σ M = Q( 5) e 6, π 7 14

15 7. 8-PSK, 16-QAM and noise / 8-PSK, 16-QAM ja kohina / a) See lecture 6 notes and slide 38. Let A = 1 and 8-PSK is used. Calculate the symbol error rate of 8-PSK when a white Gaussian noise having the voltage rms-value 97,7 dbμv is added to the signal. b) See lecture 6 notes and slide 39. Let a = 1 and BER (Bit Error Rate or Ratio) = Calculate by giving formulas the maximum allowed rms-value for white Gaussian noise and give SNR (Signalto-Noise Ratio) in db's when 16-QAM is considered. b) The bit error rate (ratio) is 3 E 3 a 6 pε b = Q = Q = N0 4 σ So 6 ( ) κ ( ) Q SNR = 1,33 10 SNR 4,7 = SNR,1 10lg,1 13,4 db a a 1 = κ σ = = 13 mv σ κ 4,7 Please use Q-function figure here to get SNR! 15

16 8. HDB-3 and NRZ coding / HDB3- ja NRZ-koodaus a) HDB-3 coding is used for the bit train Sketch the RZ-pulse train. 1/ b) A NRZ-coded base band signal is transferred in a cable, which attenuation is 0 db/km. In a receiver the level of the white Gaussian noise is 150 fw. The error probability in the receiver is Calculate the maximum length of a cable when the transmitted power to the cable is 16 dbm. a) See L6, S7-8, HDB-3 is derived from AMI When four (4) zeros are going to send consecutively [peräkkäin], last zero is coded using a violation pulse (same direction as an earlier one pulse has). Violation pulses are added to AMI code because of the synchronization in a receiver is easier. If you think for example that there are 500 sequential zeros the receiver can lost the synchronization when AMI is used. AMI is used in USA in PSTN systems, HDB-3 in Europe. 16

17 8. HDB-3 and NRZ coding / HDB3- ja NRZ-koodaus a) HDB-3 coding is used for the bit train Sketch the RZ-pulse train. / b) A NRZ-coded base band signal is transferred in a cable, which attenuation is 0 db/km. In a receiver the level of the white Gaussian noise is 150 fw. The error probability in the receiver is Calculate the maximum length of a cable when the transmitted power to the cable is 16 dbm. See L6, S15 8 b) Let p = 10. Thus (from the Q function figure) k = SNR = κ 5,5. So the SNR = κ 30,5 and 4,54 pw PR = SR = κ NR 4,54 pw 10lg dbm 83, 4 dbm 1 mw db PS PR ,4 PS = PR + l 0 l = km km 4,97 km 5 km km 0 db 0 17

18 9. Block Codes / Lohkokoodit A block code consists of the following codes: 10011, 11101, 01110, a) How many errors can be detected/corrected by this code? b) Is this a linear code? a) See L7, S11-1 dmin = 3 l = dmin 1= t = l/ = 1 d min = the minimum number of bits that are different in code words, Hamming distance l = the number of errors that can be detected at reception (but not corrected) t = the integer part of calculation l/, the number of errors that can be corrected. b) This is a linear code, because two conditions are satisfied: - It includes the all-zero vector - The sum of any two code vectors produces another vector in the code 18