Baseband Data Transmission I After this lecture, you will be able to

Transcription

1 Baseband Data Transmission I After this lecture, you will be able to describe the components of a digital transmission system Information source, transmitter, channel, receiver and destination calculate the signaling rate and bit rate of a system design the matched filter of a receiver derive the condition for maximum signal-to-noise ratio at the receiver determine the error rate Error rate versus received signal energy per bit per hertz of thermal noise D.1

2 Reference Reference Chapter , S. Haykin, Communication Systems, Wiley. D.

3 Introduction Digital communication system Information source Transmitter channel Receiver Destination Transmitted message Noise Received message Information source produces a message (or a sequence of symbol) to be transmitted to the destination. Example 1 Analog signal (voice signal): sampling, quantizing and encoding are used to convert it into digital form D.3

4 Introduction (1) Sampling and quantizing 87 Quantization noise t Digits D.4

5 encoding Digits Introduction () Binary code Return-to-zero D.5

6 Introduction (3) Example digital source from a digital computer Transmitter operates on the message to produce a signal suitable for transmission over the channel. D.6

7 Introduction (4) Channel medium used to transmit the signal from transmitter to the receiver Attenuation and delay distortions Noise Receiver performs the reverse function of the transmitter determine the symbol from the received signal Example: 1 or for a binary system Destination the person or device for which the message is intended. D.7

8 Signaling Rate Digital message An ordered sequence of symbols drawn from an alphabet of finite size µ. Example Binary source: µ= for alphabet,1 where and 1 are symbols A 4 level signal has 4 symbols in its alphabet such as ±1, ±3 Signaling Rate The symbols are suitably shaped by a shaping filter into a sequence of signal-elements. Each signal-element has the same duration of T second and is transmitted immediately one after another, so that the signal-element rate (signaling rate) is 1/T elements per second (bauds). D.8

9 Bit Rate Binary digit: and 1 Symbols: 1,,3 and 4 D.9

10 Bit Rate Bit Rate The bit rate is the product of signaling rate and no of bit/symbol. Example A 4-level PAM with a signaling rate = 4 bauds/s. Bit rate (Data rate) =4 X log (4) = 48 bits/s (bps) D.1

11 Matched Filter (1) A basic problem that often arises in the study of communication systems is that of detecting a pulse transmitted over a channel that is corrupted by channel noise t t Square pulse Signal at the receiving end D.11

12 Matched Filter () A matched filter is a linear filter designed to provide the maximum signal-to-noise power ratio at its output. This is very often used at the receiver. x ( t) = g( t) + w( t) y( t) = go ( t) + n( t) Consider that the filter input x(t) consists of a pulse signal g(t) corrupted by additive noise w(t). It is assumed that the receiver has knowledge of the waveform of the pulse signal g(t). The source of uncertainty lies in the noise w(t). The function of receiver is to detect the pulse signal g(t) in an optimum manner, given the received signal x(t). D.1

13 Matched Filter (3) The purpose of the circuit is to design an impulse response h(t) of the filter such that the output signal-to-noise ratio is maximized. Signal Power Let g(f) and h(f) denoted the Fourier Transform of g(t) and h(t). ( t) H ( f ) G( f )exp( jπ g = ft) df The signal power = g t) H ( f ) G( f )exp( jπft) ( = df D.13

14 Matched Filter (4) Noise Power N Since w(t) is white with a power spectral density, the spectral density function of Noise is N S N ( f ) = f H ( ) The noise power = N E[ n ( t)] = H ( f ) df D.14

15 Matched Filter (5) S/N Ratio Thus the signal to noise ratio become η = H ( f ) G( N f )exp( jπft ) df H ( f ) df....(1) (the output is observed at t = T ) D.15

16 Matched Filter (6) Our problem is to find, for a given G(f), the particular form of the transfer function H(f) of the filter that makes η at maximum. Schwarz s inequality: If φ ( x dx < and φ ( x dx <, 1 ) 1 ( x) φ ( x) ) φ dx φ ( x dx 1 ) φ ( x) * This equality holds, if and only if, we have φ 1( x) = kφ ( x) where k is an arbitrary constant, and * denotes complex conjugation. dx D.16

17 Matched Filter (7) Applying the schwarz s inequality to the numerator of equation (1), we have H ( f ) G( f )exp( jπft ) df H ( f ) df G( f ) df () j ft (Note: e π = 1) D.17

18 Matched Filter (8) Substituting () into (1), The S/N ratio η N G( f ) df or η E N where the energy E= (3) G( f ) df is the input signal energy D.18

19 Matched Filter (9) Notice that the S/N ratio does not depend on the transfer function H(f) of the filter but only on the signal energy. The optimum value of H(f) is then obtained as * H ( f ) = kg ( f ) exp( jπft ) D.19

20 Matched Filter (1) Taking the inverse Fourier transform of H(f) we have h(t)=k * G ( f ) exp[ jπf ( T t)] df * and G ( f ) = G( f ) for real signal g(t) h(t)=k G ( f ) exp[ jπf ( T t)] df h(t)=kg(t-t)..(4) Equation (4) shown that the impulse response of the filter is the time-reversed and delayed version of the input signal g(t). Matched with the input signal D.

21 Matched Filter (11) Example: The signal is a rectangular pulse. g(t) A T t The impulse response of the matched filter has exactly the same waveform as the signal. h(t) ka T t D.1

22 Matched Filter (1) The output signal of the matched filter has a triangular waveform. g o (t) ka T T t D.

23 Matched Filter (13) In this special case, the matched filter may be implemented using a circuit known as integrate-anddump circuit. r(t) T Sample at t = T D.3

24 Realization of the Matched filter (1) Assuming the output of y(t) = r(t) h(t) t = r( ) h( t τ ) dτ Substitute (4) into (5) we have t y( t) = r( τ ) g[ T ( t τ )] dτ When t = T y(t)= T r( τ ) g( τ ) dτ τ...(5) D.4

25 Realization of the Matched filter () r(t) T g(t) Correlator D.5

26 Error Rate of Binary PAM (1) Signaling Consider a non-return-to-zero (NRZ) signaling (sometime called bipolar). Symbol 1 and are represented by positive and negative rectangular pulses of equal amplitude and equal duration. Noise The channel noise is modeled as additive white Gaussian noise of zero mean and power spectral density N o /. In the signaling interval t T b, the received signal is x( t) = + A + A + w( t) w( t) symbol1 was sent symbol was sent A is the transmitted pulse amplitude T b is the bit duration D.6

27 Error Rate of Binary PAM () Receiver x(t) T y Decision device 1if if y y > λ < λ Sample at t = T b λ It is assumed that the receiver has prior knowledge of the pulse shape, but not its polarity. Given the noisy signal x(t), the receiver is required to make a decision in each signaling interval D.7

28 Error Rate of Binary PAM (3) In actual transmission, a decision device is used to determine the received signal. There are two types of error Symbol 1 is chosen when a was actually transmitted Symbol is chosen when a 1 was actually transmitted Case I Suppose that a symbol is sent then the received signal is x(t) = -A + n(t) If the signal is input to a bandlimited low pass filter (matched filter implemented by the integrate-and-dump circuit), the output y(t) is obtained as: y(t)= t 1 x( t) dt = A + T b T b n( t) dt D.8

29 Error Rate of Binary PAM (4) As the noise n(t) is white and Gaussian, we may characterize y (t) as follows: y (t) is Gaussian distributed with a mean of A N the variance of y(t) can be shown as σ y = Tb (Proof refers to p.54, S. Haykin, Communication Systems) D.9

30 Error Rate of Binary PAM (5) The Probability density function of a Gaussian distributed signal is given as 1 ( y y) f y ( y ) = exp( ) πσ σ y y f y ( y ) = πn 1 / T ( y + A) exp( N / T b b ) where f y (y ) is the conditional probability density function of the random variable y, given that was sent D.3

31 Error Rate of Binary PAM (6) Let p 1 denote the conditional probability of error, given that symbol was sent This probability is defined by the shaded area under the curve of f y (y ) from the threshold λ to infinity, which corresponds to the range of values assumed by y for a decision in favor of symbol 1 D.31

32 Error Rate of Binary PAM (7) The probability of error, conditional on sending symbol is defined by P = P( y > Symbol was sent) 1 λ = λ f y ( y ) dy 1 = πn / T b λ exp( ( y + A) N / T b ) dy D.3

33 Error Rate of Binary PAM (8) Assuming that symbols and 1 occur with equal probability, i.e. P = P 1/ 1 = If no noise, the output at the matched filter will be A for symbol and A for symbol 1. The threshold λ is set to be. D.33

34 Error Rate of Binary PAM (9) Define a new variable z = y + A N o / T b and then dy = N dz. T b We have 1 P1 = exp( z π E b / N o ) dz where E b is the transmitted signal energy per bit, defined by E = A T b b D.34

35 Error Rate of Binary PAM (9) At this point we find it convenient to introduce the definite integration called complementary error function. = erfc( u ) exp( z ) dz π u Therefore, the conditional probability of error P 1 1 = erfc( E N ( Note: erf( u) = exp( z π u b ) ) dz and erfc(u)=1-erf(u) ) D.35

36 Error Rate of Binary PAM (1) In some literature, Q function is used instead of erfc function. 1 u Q( x) = exp( ) du π 1 x Q( x ) = erfc( ) and erfc( x ) = Q( x ) x D.36

37 Error Rate of Binary PAM (11) Case II Similary, the conditional probability density function of Y given that symbol 1 was sent, is 1 ( y A) f y ( y1) = exp( ) π N / T N / T b b P 1 1 = πn / T b λ exp( ( y N A) / T b ) dy D.37

38 Error Rate of Binary PAM (1) By setting λ = and putting y A = z N / Tb we find that P 1 = P1 The average probability of symbol error P e is obtained as P e = P P + P 1 1P1 If the probability of and 1 are equal and equal to ½ 1 Eb Pe = erfc( ) N D.38

39 Error Rate of Binary PAM (13) D.39