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S-72.1140 Transmission Methods in Telecommunication Systems (5 cr)

Tutorial 4/2007(Lectures 6 and 7)

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1. Line Codes / JohtokooditSketch beneath each other line codes Manchester, Differential Manchester and AMI when the bit train 1101000101 is transmitted.A: See L6, S7-8

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.

1 1 1 1 10 0 0 0 0

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2. Abbreviations and terms / Lyhenteitä ja käsitteitäExplain 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 21: 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 matchedfilter is reverse in the time domain than received pulse has. Because of the convolution in the time domain maximum correlation is then achieved.

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3. Assume that noise has spectrum density of N0 = 2·10-10 W/Hz and receiver´s bandwidth is W = 200 kHz and impedance level is Z0 = 50 Ω. Calculatea) the (average) noise power PN at the receiver andb) the probability that noise voltage exceeds +200 mV.

See L6, S11-12a) Noise power

b) Noise´s RMS voltage = standard deviation σ is

m = mean value = 0 V,k·σ = 200 mV -> k ≈ 4,5

To get the desired probability,we have to use Q-function and it´scurves (on the next page).So the probability p = Q(k) ≈

10 50 200 2 10 4 10 40N

W

WP N df kHz W WHz

μ− −= = ⋅ ⋅ = ⋅ =∫

0 40 50 44,7RMS NU P Z W mVσ μ= = ⋅ = ⋅ Ω ≈

m k+ σm

( )2

212 k

Q k e dλ

λπ

∞ −

≡ ∫

X mλ

σ

−

−−

=

63,6 10−⋅

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Q-function Q(k)

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4. Signal-to-noise ratio (SNR, k, κ)A DVB-S receiver has BER = 10-2 without channel coding. When interleaving, Reed Solomon and Convolution codes are usedBER = 10-10. Modulation method is QPSK (4-PSK) which has the

following proberty: Calculate by using Q-function diagram how much SNR seems to be improved when channel coding is used.See lecture 6 slide 38a) Without coding:

b) With channel coding:

So the “improvement” is

( )12

BER Q SNR=

( ) ( )( )

21 1 1

21 1 1 1

2 2 10

( ) 10 lg

BER Q SNR Q k

SNR k and SNR dB SNR dB

−⋅ = = ⋅ =

⇒ = ≈ = ⋅ ≈

( ) ( )( )

102 2 2

22 2 2 2

2 2 10

( ) 10 lg

BER Q SNR Q k

SNR k and SNR dB SNR dB

−⋅ = = ⋅ =

⇒ = ≈ = ⋅ ≈

( ) ( )2 1SNR dB SNR dB dB− ≈

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Q-function Q(k)

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4. Signal-to-noise ratio (SNR, k, κ)A DVB-S receiver has BER = 10-2 without channel coding. When interleaving, Reed Solomon and Convolution codes are usedBER = 10-10. Modulation method is QPSK (4-PSK) which has the

following proberty: Calculate by using Q-function diagram how much SNR seems to be improved when channel coding is used.See lecture 6 slide 38a) Without coding:

b) With channel coding:

So the “improvement” is

( )12

BER Q SNR=

( ) ( )( )

21 1 1

2 21 1 1 1

2 2 10

2,07 4,28 ( ) 10 lg 6,3

BER Q SNR Q k

SNR k and SNR dB SNR dB

−⋅ = = ⋅ =

⇒ = ≈ = = ⋅ ≈

( ) ( )( )

102 2 2

2 22 2 2 2

2 2 10

6,3 39,7 ( ) 10 lg 16

BER Q SNR Q k

SNR k and SNR dB SNR dB

−⋅ = = ⋅ =

⇒ = ≈ = = ⋅ ≈

( ) ( )2 1 9,7SNR dB SNR dB dB− ≈

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5. A NRZ-signal (Non Return to Zero) / NRZ-signaaliCalculate 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. 1/2

See L6, S15Because 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:

and =the random signal.

Average value and autocorrelation:<v(t)> =

When the random signal is , the autocorrelation function iswhere D is the duration of a bit. And

( ) ( ) ( ) ( )0 5i a b bv t v t v t , V v t= + = + ( ) 0 5 0 5bv t , , ,= − +

0 5 0 25v vR ( ) , R ( ) ,±∞ = ⇒ ±∞ =

( )bv t( ) 2

v

bR triaDττ σ ⎛ ⎞= ⋅ ⎜ ⎟

⎝ ⎠( ) ( )220 0 5 0 25

v

bR P , ,σ= = = ± =

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5. A NRZ-signal (Non Return to Zero) / NRZ-signaaliCalculate 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. 2/2

So the autocorrelation function for the NRZ-signal is

Thus the power spectrum density is

( ) 0 25 0 25vR , tria ,Dττ ⎛ ⎞= ⋅ +⎜ ⎟

⎝ ⎠

( ) ( ) ( ) ( )20 25 0 25v vG f F R , D sinc fD , fτ δ⎡ ⎤= = ⋅ ⋅ + ⋅⎣ ⎦

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6. Transferring a file in an AWGN channel / Tiedoston siirtokohinaisessa kanavassa 1/3A file which size is 1,4 MB (about 11,74 Mbits transferred in an AWGN channel which has the attenuation of 126 dB. The power of the transmitter is 30 dBm. The noise has 1-sided power spectrum density in the receiver. The modulation system sends two bits (= 1 symbol) at the same time and the symbol error probability is , where Erx is the average energy of a

symbol. How long it takes to transfer the file when is needed?

200 4 10N W / Hz

See L6, S20The channel attenuation is . The transmitted power is .So the received average power is .

Because of two bits are sent at the same time, the symbol rate is , where is the bit rate.

−= ⋅

0

rxs

EP QN

⎛ ⎞= ⎜ ⎟

⎝ ⎠910sP −≤

12 6126 1 10 ,S SA dB A= ⇒ = ⋅

30 1RXP dBm W= =TX

RXS

PPA

=

2b

SRR =

bR

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6. Transferring a file in an AWGN channel / Tiedoston siirtokohinaisessa kanavassa 2/3A file which size is 1,4 MB (about 11,74 Mbits transferred in an AWGN channel which has the attenuation of 126 dB. The power of the transmitter is 30 dBm. The noise has 1-sided power spectrum density in the receiver. The modulation system sends two bits (= 1 symbol) at the same time and the symbol error probability is , where Erx is the average energy of a

symbol. How long it takes to transfer the file when is needed?

200 4 10N W / Hz

Because of (see Q-function figure):

The average energy for a received symbol is

Thus

−= ⋅

0

rxs

EP QN

⎛ ⎞= ⎜ ⎟

⎝ ⎠910sP −≤

( ) 96 0 10Q , −≈

00 0

6 0 36 36RX RXRX

E E, SNR E NN N

= ⇒ = = ⇒ =

2RX TXRX RX S

S S b

P PE P TR A R

= ⋅ = =

00

236 348 918

TX TXb

S b S

P PN R , kbit / sA R N A

= ⇒ = ≈

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6. Transferring a file in an AWGN channel / Tiedoston siirtokohinaisessa kanavassa 3/3A file which size is 1,4 MB (about 11,74 Mbits transferred in an AWGN channel which has the attenuation of 126 dB. The power of the transmitter is 30 dBm. The noise has 1-sided power spectrum density in the receiver. The modulation system sends two bits (= 1 symbol) at the same time and the symbol error probability is , where Erx is the average energy of a

symbol. How long it takes to transfer the file when is needed?

200 4 10N W / Hz

The time needed for transmission is

−= ⋅

0

rxs

EP QN

⎛ ⎞= ⎜ ⎟

⎝ ⎠910sP −≤

6

3

11 744 10 33 7348 9 10

, bitT , sbit,s

⋅= ≈

⋅

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7. 8-PSK, 16-QAM and noise / 8-PSK, 16-QAM ja kohina 1/2a) 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) = 10-6. Calculate by giving formulas the maximum allowed rms-value for white Gaussian noise and give SNR (Signal-to-Noise Ratio) in dB's when 16-QAM is considered.

See L6, S38a) To find out probability Q(k) = p when SNR = κ = k2 is known andwhen k > 3,

2

2( )

2

k

eQ kk π

−

≈⋅

( )2

97,76 20

572

12 sin 2 sin8

10 10

12 5 2 6,35 105 2

Ap Q QM

Q e

επ π

σ

π

−

−−

⎡ ⎤⎢ ⎥

⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥= = ≈⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎛ ⎞⎝ ⎠ ⎝ ⎠⎣ ⎦ ⋅⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

≈ ⋅ ≈ ⋅

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7. 8-PSK, 16-QAM and noise / 8-PSK, 16-QAM ja kohina 2/2a) 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) = 10-6. Calculate by giving formulas the maximum allowed rms-value for white Gaussian noise and give SNR (Signal-to-Noise Ratio) in dB's when 16-QAM is considered.b) The bit error rate (ratio) is

So

Please use Q-function figure here to get SNR!

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0

3 2 3 104 10 4b

E ap Q QNε σ

−⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

( ) ( )61,33 10 4,7 22,1 10lg 22,1 13,4Q SNR SNR SNR dBκ−= ⋅ ⇒ ≈ ⇒ = ≈ ≈ ≈

1 2134,7

a a mVκ σσ κ= ⇒ = = ≈

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8. HDB-3 and NRZ coding / HDB3- ja NRZ-koodausa) HDB-3 coding is used for the bit train 011000000011. Sketch the RZ-pulse train. 1/2b) A NRZ-coded base band signal is transferred in a cable, which attenuation is 20 dB/km. In a receiver the level of the white Gaussian noise is 150 fW. The error probability in the receiver is2 ⋅ 10-8. 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.

1 010 0 0 0 0 0 0 1 1

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8. HDB-3 and NRZ coding / HDB3- ja NRZ-koodausa) HDB-3 coding is used for the bit train 011000000011. Sketch the RZ-pulse train. 2/2b) A NRZ-coded base band signal is transferred in a cable, which attenuation is 20 dB/km. In a receiver the level of the white Gaussian noise is 150 fW. The error probability in the receiver is2 ⋅ 10-8. Calculate the maximum length of a cable when the transmitted power to the cable is 16 dBm. See L6, S15b) Let . Thus (from the Q function figure) .So the and

82 10p −= ⋅ 5,5k SNR κ= = ≈30,25SNR κ= ≈

4,544,54 10lg 83,41R R R

pWP S N pW dBm dBmmW

κ⎛ ⎞

= = ⋅ ≈ ≈ ≈ −⎜ ⎟⎝ ⎠

16 83,420 4,97 520 20S R

S RP PdBP P l l km km km km

km dB− +

= + ⋅ ⇒ = ≈ ≈ ≈

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9. Block Codes / LohkokooditA block code consists of the following codes: 10011, 11101, 01110, 00000.a) How many errors can be detected/corrected by this code?b) Is this a linear code?

a) See L7, S11-12

dmin = the minimum number of bits that are different in code words, Hamming distancel = the number of errors that can be detected at reception (but not corrected)t = the integer part of calculation l/2, 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

min min3 1 2 / 2 1d l d t l= = − = = =⎢ ⎥⎣ ⎦