CS3282: Digital Communications 5.1 BMGC/ 25/08/06

University of Manchester CS3282: Digital Communications ’05-‘06

Section 5: Detection of Binary Signals in AWGN

Consider the transmission of a single bit using binary signalling, i.e. sending one particular symbol to represent "1" or a different symbol to represent "0". As would be seen at the receiver in the absence of any noise, possibly after regeneration and equalisation, let the two possible symbols be s1(t) and s0(t). If the received signal is affected by additive noise n(t), instead of s1(t) or s0(t) we would receive:

01)()()( oritntstr i =+= Assume n(t) is zero mean additive white Gaussian noise (AWGN) of ‘2-sided’ power spectral density N0/2 Watts/Hz. Note that the bandwidth of the noise n(t) need not be specified and may in fact be considered infinite. Probably the simplest example we could consider is where the possible symbols are as follows:

0 logicfor ) t all(for 0)( and 1 logicfor otherwise:0

0:)( 01 =

≤≤+

= tsTtV

ts

This is a form of 'unipolar' signalling with a 'rectangular' pulse shape. An alternative, almost as simple, is to receive

0 logicfor otherwise:0

Tt0:V)( and 1 logicfor

otherwise:00:

)( 01 ≤≤−

= ≤≤+

= tsTtV

ts

This is a form of 'bipolar' signalling with a rectangular pulse shape. A rectangular pulse-shape will prove unsuitable for many practical situations, but it provides a good starting point. Taking the example of unipolar signalling given above, with noise, the received signal r(t) will be as shown on the left below for logic '1' and as shown on the right for logic '0'. Note that we are res

t

+V+V

s0(t)+n(t)

t

For this unipolar sigto sample the receivat t = T/2, and coms1(t) with noise andnot matter whether t=T). But whereverwill cause s1(t) to bbetween zero volts of receiving s1(t) is

T

s1(t)+n(t)

tricting our analysis to a single bit to begin with, not a sequence of bits. nalling example, a viable strategy for detecting whether a 1 or 0 was sent would be ed signal r(t) at a suitable point in time, say in the middle of the pulse, i.e. pare it with a 'threshold' of say +V/2. If r(T/2) > +V/2 we are likely to be receiving if r(T/2) < +V/2 we are likely to be receiving s0(t). For rectangular pulses, it does we sample in the middle, at the beginning ( close to t=0) or at the end (close to we sample, there is a risk that the noise signal n(t) will have such a value that it e mistaken for s0(t) or vice versa. The choice of a threshold of +V/2, i.e. half way and +V, is a good choice when the probability of receiving s0(t) and the probability known to be equal, i.e. 0.5. This is commonly the case.

CS3282: Digital Communications 5.2 BMGC/ 25/08/06

Exercise 5.1: What is the best threshold for 1 volt & 0 volt rectangular binary symbols if the probabilities are 0.9 and 0.1 respectively (instead of 0.5 and 0.5) and that the noise variance is 1/16 at the threshold detector? Solution method: Either use trial & error with Q(z) curve, devise a simple MATLAB program or show that a theoretical solution is γ best = 0.5 − σ2 loge(prob(1 volt) / prob(0 volt) ) Remember that Q(z) = 0.5*erfc( z / (√2) ) & that MATLAB provides the erfc function. Note that d Q(z) / dz = − (1/√(2π)) exp(−z2 / 2) and that the theoretical formula can be derived by differentiating the bit-error probability: PB = prob(0)*Q(γ / σ) + prob(1)*Q((1-γ)/σ) with respect to γ to seek a minimum. An improved detection process for unipolar rectangular symbols s1(t) or s0(t) is illustrated below:

z(T) z(t) = ai(t)+n0(t)

r(t) = si(t)+n(t)

Decide z(T)>Vthres z(T)<Vthres

Sample at t=T Averager Denote the response of the averager to si(t) by ai(t) for i = 1,0, and the response to n(t) by n0(t). The averager (or smoother) could ideally produce an output

∫=t

drTtz0

)()1()( ττ

To generate we can use an integrator as illustrated below ∫t

dr0

)( ττ

Integrator

r(t) -z(t)

Dumper

-

+

As the rectangular pulse starts at t=0, it is a good idea to have a means of forcing the integrator output to start at exactly zero volts at t=0. This is just a switch which connects the output to ground (zero volts) to "dump" any charge on the capacitor. An integrator with this 'charge dumping' facility is an 'integrate and dump' circuit. For the unipolar signalling example with rectangular pulse shape, the response a0(t) of the 'I & D' averager to s0(t) without noise would be exactly zero and the response a1(t) to s1(t) would be as shown below. The response to a1(t) starts at zero and reaches +V at t=T. Check this from the averager equation given above. For t > T, the averager output will remain at +V until we apply the charge dumper again. a1(t)

+V

T

t

CS3282: Digital Communications 5.3 BMGC/ 25/08/06

In practice, instead of a clean version of a1(t) as shown above, the 'I & D' averager will produce an output z(t) = a1(t) + n0(t) where n0(t) is due to the noise signal n(t) entering the 'I & D' averager along with s1(t). The effect of the averager will be to reduce the level of the noise since positive and negative noise voltages within n(t) will tend to cancel each other out in the averaging process. Hence if we sample z(t) at t = T, the chances of noise causing a wrong decision are reduced. Note that it is now much better to sample at t=T rather than in the middle of the pulse since the averaging process must be allowed to finish. Once we have sampled z(t) and made the decision, the charge dumping mechanism can be applied to set the averager output back to zero, and the circuit is then able to receive another pulse representing another bit of information. 'Matched filter' method for detecting a rectangular pulse shape. Consider an alternative approach to the detection process for a single bit. (We will consider sequences of bits later). This is the 'matched filter' approach and employs a filter rather than an 'I & D' averager. Denote the filter response to r(t) as ψ(t) as shown below. Note that ai(t) and n0(t) may now be different from what they were with the averager. r(t)

= si(t)+n(t)

ψ(t) = ai(t)+n0(t)

ψ(T) Decide ψ(T)>Vthres

ψ(T)<Vthres

Sample at t=T Filter A low-pass filter can be considered as a sort of smoother or averager, but there are differences. Let the frequency-response of the filter be H((f)) which is the Fourier transform of its impulse-response h(t). Now assume that the impulse-response is:

≤≤

=otherwise

TtTth

:00:/1

)(

The reason for choosing this shape of impulse-response will be made clear later. In response to an input signal r(t), the filter's output will be the time-domain convolution between r(t) and h(t) defined as follows:

∫∞

∞−−= τττψ dtrht )()()(

In this case,

∫ −=T

dtrTt0

)()/1( )( ττψ ∫ −=∴T

dTrTT0

)()/1( )( ττψ

Substituting t = T-τ which means that dτ = -dt, we obtain:

∫∫ =−=T

TdttrTdttrTT

0

0)()/1( )()/1( )(ψ

which is identical to the value of z(T) obtained using the 'I & D' averager. So we now have another way of generating z(T) for a single pulse, and it uses a filter rather than an 'I & D' averager. Note that ψ(t) is expected to be equal to z(t) only at t=T, but this is the only time point we are interested in anyway. It is the point where we sample the output to make the decision. It is easily shown that the response of the filter to s1(t) is as follows: Note that the filter output has decayed to zero at t=2T without the need to apply a 'charge dumper'.

T

+V a1(t)

2T

t

CS3282: Digital Communications 5.4 BMGC/ 25/08/06

Estimating the bit-error probability when an I&D averager or 'rect-pulse' matched filter is employed. First note that since the decision is made on the basis of z(T) or equivalently ψ(T), the bit-error probability will be exactly the same for both approaches. Now analyze the matched filter approach. The ‘2-sided’ power spectral density of n0(t), where n0(t) is the filter's response to n(t), is: PSD0(ƒ) = PSD(f) |H((f))|2 where PSD(ƒ) is the ‘2-sided’ power spectral density of the noise signal n(t). The reason for this may be found in Section 2. We know that PSD(f) = N0/2 for all f. The power of n0(t) is therefore:

TNdffHN

dffHfdff

/5.0)(5.0

|)(|)PSD( )(PSD

02

0

20

==

=

∫∫∫

∞

∞−

∞

∞−

∞

∞−

since by Parseval’s Theorem,

TdtTdtthdffHT

/1 )/1( )()(0

222 === ∫∫∫∞

∞−

∞

∞−

When Gaussian noise is applied to an LTI filter, the output is also Gaussian noise, though not necessarily white even when the input is white. This is a consequence of the “central limit theorem”. To summarise, if a white Gaussian noise signal n(t) with 2-sided PSD N0/2 Watts/Hz were applied to a filter with the impulse-response

≤≤

=otherwise

TtTth

:00:/1

)(

the output would be a spectrally coloured Gaussian noise signal n0(t) of power 0.5N0/T Watts. A filter with this impulse-response is a type of matched filter matched to a rectangular pulse. We will call this a 'rect-pulse' matched filter. Therefore, if a signal r(t) = si(t) + n(t), with i=1 or 0, is applied to the same filter, the output will be ψ(t) = ai(t)+n0(t) where ai(t) is the filter's response to si(t) for i = 1 or 0. If we sample ψ(t) at t=T, we obtain ψ(T) = ai(T)+n0(T) and ψ(T) is exactly the same value that would be obtained by sampling the output z(t) from an I&D averager at t=T. In the case of unipolar signalling with +V and zero valued rectangular pulses as illustrated above, we know precisely that ai(T) is +V or zero, but what can we say about n0(T)? It is just a single voltage obtained by sampling a random signal which may or may not cause a bit-error depending on how large it is. We cannot say what this voltage will be exactly so we cannot say definitely that it will or will not cause an error. But we can estimate the probability of it being large enough to cause an error We can do this because we can say something about the properties of a statistical process that describes the production of this voltage ; i.e. a statistical process which is Gaussian with zero mean and variance 0.5N0/T. In general, subject to certain conditions, if the average value of a random voltage signal is zero, its power will be equal to the variance (σ2) of a statistical process that describes it. So the effect of using an 'I&D' averager or 'rect-pulse' matched filter is to produce a signal which, when sampled at t=T, gives a value ai(T) + n0(T) where n0(T) is a random variable of variance 0.5N0/T. If ai(T) is +V and a0(T) is zero, and we take a threshold at +V/2 the probability of a bit-error is: Pb = ' prob(n0(T) < −V/2) when s1(t) is received' + 'prob(n0(T) > V/2) when s0(t) is received'. Since there is likely to be no correlation between the noise and whether we transmit '1' or '0', we can say that Pb = prob(n0(T) < −V/2) * prob ( transmitting 1) + 'prob(n0(T) > V/2)* prob(transmitting 0)

CS3282: Digital Communications 5.5 BMGC/ 25/08/06

If a '1' and a '0' is equally likely, i.e. if prob(transmitting 1) = prob(transmitting 0) then both are equal to 0.5 and Pb = 0.5 [ prob(n0(T)<-V/2) + prob(n0(T)>+V/2) ]. Finally, if the statistical process underlying the generation of n0(T) is Gaussian with zero mean, then it follows that prob(n0(T)<-V/2) = prob(n0(T)>+V/2) from the symmetry of the probability density function (pdf) graph. Therefore for the unipolar rectangular signalling example, the probability of a bit-error is: Pb = prob( n0(T) > +V/2 ) If the noise n0(t) had power equal to 1 Watt, and therefore variance σ0

2 = 1 then Pb would be equal to Q(V/2) where Q(z) is the 'Q' function whose graph is presented in the appendix. However, since we know that the noise power is 0.5N0/T, then σ0

2 = 0.5N0/T and the standard deviation σ0 is the square root of this value. Therefore we can finally say that:

=

TNVQPb /5.0

2/

0

Example 5.2A: A receiver receives 1 volt and 0 volt rectangular binary symbols at 100 Baud. The transmission is distorted by additive zero mean white Gaussian noise (AWGN) with N0=0.00025 Watts/Hz. Estimate the bit-error probability that would be obtained with an 'I&D' averager or a 'rect-pulse' matched filter. Assume equal occurrence of 1’s and 0’s and an appropriate threshold. Solution: Consider the output from the I&D averager (or MF). The noise has variance:

20

20 1025.102.000025.0)T2/(N −×=×==σ

Therefore, σ0=0.112. The 1Volt and 0Volt pulses are unaffected by the averaging process, though to gain maximum benefit, we must sample at the end of the pulse rather than in the middle. The decision threshold will be 0.5Volts. The error probability is:

6

00

00

105.3)46.4(Q

5.0Q5.05.0Q5.0

))5.00()T(n(P)"1("P))5.01()T(n(P)"0("P

−×≈

=

σ

×+

σ

×=

−<×+−>×

The error-rate is one bit in 200,000. (About one character wrong in a 6-page document, or about one serious speech sample error in a G711 64kb/s transmission about every 6 seconds, assuming errors in the least significant 4 bits are not serious). It is interesting to compare this result with what would be obtained using the simple method of sampling in the centre of each rectangular symbol without a matched filter or 'I&D' averager. But we

CS3282: Digital Communications 5.6 BMGC/ 25/08/06

must note first of all that under the assumption that the noise n(t) is AWGN, its bandwidth, and therefore its power and its variance are infinite also. Therefore Pb = Q(0) = 0.5, the worst we can get. Not 1?? However, we get a more sensible result when we realise that the 'front-end' of the receiver would almost certainly have a filter to restrict the bandwidth of the received signal and remove most of the much wider bandwidth noise. Assume that the noise band-width is restricted to ±B Hz, for some value of B, by a 'front-end' filter The noise power would now be restricted to N0B Watts meaning that the variance at the decision stage of the detector would also be N0B. Now consider the following example: Example 5.2B: A receiver receives 1 volt and 0 volt rectangular binary symbols at 100 Baud. The transmission is distorted by additive white Gaussian noise (AWGN) with N0=0.00025 Watts/Hz and with zero mean. Estimate the error probability that would be obtained without a matched filter or averager but when an ideal low-pass filter H((f)) is employed at the receiver with bandwidth ±B Hz with B=500 Hz. Assume an equal occurrence of 1’s and 0’s and an appropriate threshold. Compare with the error probability obtained in the previous example. Solution: With an ideal low-pass filter, the filtered noise would have power = 0.00025 x 500 = 0.125 Watt. Its variance σ0

2 = 0.125 and the decision would be on the basis of 0 Volt and 1 Volt pulses. The threshold would be 0.5 Volts and the bit-error probability would be the probability of a noise sample (whose standard deviation is 0.35) exceeding 0.5 Volts when “0” is transmitted or being less than –0.5 Volts when a “1” is transmitted. The bit-error probability without the averager or matched filter is therefore,

2108)42.1(35.05.0 −×≈=

QQ

This is an bit-error-rate (BER) of one bit in 12.5 (About one character wrong every one or two, or about 1 in 3 serious speech sample errors at 8 x 8k b/s per second with G711.) Example 5.3A: Repeat the previous example where the rectangular pulses are changed to:

a. +3 Volts for logic “1” and +2 Volts for “0”. b. +0.5 Volts for “1” and –0.5 Volts for “0”. All other parameters are unchanged.

Solution: Although the decision thresholds change, (2.5 V for (a) and 0 V for (b) the solutions are identical. Example 5.3B: A receiver receives AMI coded data with ±1 volt and 0 volt rectangular binary symbols at 100 Baud. (AMI stands for “alternate mark inversion” and this means that instead of transmitting 0,+V,+V,+V, for 0111, we send 0, +V, -V, +V, to try to keep the average voltage close to zero). The transmission is distorted by zero mean AWGN with N0=0.00025 Watts/Hz. An integrate and dump averager or matched filter is employed. Estimate the bit-error probability assuming an equal occurrence of 1’s and 0’s and an appropriate threshold. Compare with what would be obtained if the “integrate and dump” averager or matched filter is replaced by an ideal low-pass filter band-limiting the received signal from –500Hz to +500Hz, assuming that this bandwidth is wide enough to avoid imposing significant changes on the shape of the rectangular pulses. Solution: With the “I & D” averager or matched filter, the bit-error probability is:

CS3282: Digital Communications 5.7 BMGC/ 25/08/06

[ ][ ]

changenoagain105.3)46.4(Q

5.0Q5.05.0Q5.0

)5.0)T(n(P5.0)5.0)T(n(P5.0)"1("P)5.0)T(n(P5.0)5.0)T(n(P5.0)"0("P

6

00

00

00

−×≈

=

σ

×+

σ

×=

−<×+>×+−<×+>×

Example 5.4: In solving example 5.1, the decision threshold was set half way between the two received levels. Assume this threshold remains fixed and that due to a circuit malfunction the dc level of the transmission drifts so that the rectangular pulses become 1.2volts and 0.2volts rather than 1volt and 0volts. How would the error rate be affected? Solution: With the “I & D” averager or MF, the bit-error probability is:

3

103

00

1075.11011075.1

)25.6(Q5.0)68.2(Q5.0112.07.0Q5.0

112.03.0Q5.0

)7.0)T(n(P)"1("P)3.0)T(n(P)"0("P

−

−−

×≈

×+×≈

+=

×+

×=

−<×+>×

Example 5.5: Considering yet again Example 5.1, when the probabilities of receiving “0” and “1” are equal, the choice of a decision threshold half way between 1volt and 0volts is reasonable. Now assume that we are informed that the probability of receiving a “0” will be twice that of receiving “1”. If we keep the same half way threshold, the bit-error probability will be as before. But can the bit-error probability be decreased by adjusting the decision threshold? Solution: Exercise Matched filter Consider signalling with two possible shapes, p(t) for logic ‘1’ and q(t) for logic ‘0’. The signal is received with zero mean AWGN of ‘2-sided’ PSD N0/2 Watts/Hz. The received signal, r(t), is passed through a filter H((f)) to produce z(t) which is passed on to a sampler and threshold detector. Denote the response of the filter H((f)) to p(t), q(t) & n(t) by p0(t), q0(t) & n0(t) respectively. For each symbol, he threshold detector must choose a sampling point, t=t0 say, and set a threshold normally at γ = (p0(t0) + q0(t0)/2. It is convenient to assume that each pulse starts at t=0 and ends at t=T, so that the sampling point should be taken at t=T. Assuming p0(T) > q0(T): if z(T) ≥ γ the threshold detector delivers logic ‘1’ and if z(T) < γ, logic ‘0’ is delivered. The delivered result will be correct unless n0(T) > α for q(t) or n0(T) <−α for p(t) where α is the “headroom” between γ and p0(T) or q0(T).

CS3282: Digital Communications 5.8 BMGC/ 25/08/06

α

'headroom' α

p0(T)

γ = ( p0(T)+q0(T) ) / 2

q0(T) Therefore α = p0(T) - γ = γ - q0(T) = (p0(T) - q0(T) )/2 Assuming equal probabilities for ‘1’s and ‘0’s, the bit-error probability is PB = 0.5Q(α/σ0) + 0.5Q(-α/σ0) = Q(|α/σ0|) where σ0 is the standard deviation ( σ0

2 is the variance ) of the filtered noise n0(t). H((f)) could be an

ideal low-pass filter with pass-band ±B Hz (or maybe a Butterworth approximation to this) but we have already seen that for rectangular pulse-shapes a matched filter is better. But the 'I&D' averager and the matched filter we have used so far only work for rectangular pulses. To generalise this idea to pulses p(t) and q(t) of any shape, we can generalise the concept of a matched filter. We need a matched filter which is “tuned” to the difference in shape between p(t) and q(t) when these pulses are not necessarily rectangular. The matched filter we require is a filter which maximises |α/σ0| = (p0(T) - q0(T)/(2σ0)| since this will clearly minimise the bit-error probability which equals Q(|α/σ0|). If p(t) and q(t) have Fourier transform P((f)) and Q((f)) respectively, it follows that, by the inverse FT :

∫∞

∞−−=−= dfefQfPfHTqTp fTj πα 2

00 ) ))(())(( ))(((5.0))()((5.0

If white Gaussian noise (WGN) n(t) with two-sided PSD equal to N0 /2 Watts/Hz is applied to the same filter, the power of the output n0(t) (also zero mean Gaussian noise) is:

∫∞

∞−= dffHNPowerNoise 2

0 ))((5.0

The power of n0(t) will be equal to the variance σ02 .

To minimise PB, we would like to maximize |α/σ0| or equivalently |α/σ0|2 which is equal to:

∫∫

∞

∞−

∞

∞−−

=dffHN

dfefQfPfH Tfj

20

2 2

20

))((5.0

) ))(())(( ))(((5.0 |/|

π

σα

This expression may be studied using the well-known “Schwartz inequality”. Cauchy-Schwartz’s inequality for complex valued functions x(t) and y(t) For any real or complex function x(t) of t, the following inequality applies for any real or complex function y(t): ∫∫∫ ≤ dtdtdttytx 22 2

y(t) x(t) )()(

CS3282: Digital Communications 5.9 BMGC/ 25/08/06

Equality applies if x(t) = k y*(t) i.e. complex conjugate, for any constant k To put this another way, suppose we have two 'finite energy' signals x(t) and y(t) whose energies are fixed at say Ex and Ey. Then the minimum value of

2

)()(∫∞

∞−dttytx

occurs when x(t) = k y*(t) (for all t) for some constant k and this minimum is equal to ExEy. We can also use the Couchy-Schwartz inequality in the frequency-domain, i.e.

dffYEdffXEEEdffYfX YXYX

222

))(( and ))(( where))(())(( ∫∫∫∞

∞−

∞

∞−

∞

∞−==≤

which means that the maximum value of

k((f)) kYX((f) EEdffYfX YX somefor when occurs i.e. ))(())((2

∗∞

∞−=∫

Proof for real x(t) and y(t):

( )0)()()(2)(

0)()(222

2

≥+−=

≥−

∫∫∫∫

dttxdttytxdttyQ

anyfordttytx

λλ

λλ

Q(λ) is a quadratic function of λ. Q(λ) tends to +∞ as λ tends to ±∞. Therefore Q(λ) can only have a minimum, never a maximum. The minimum cannot be negative, therefore the curve for Q(λ) against λ cannot cross the λ axis. The minimum must be positive or zero. Therefore either both roots of the quadratic equation ‘ Q(λ) = 0 ’ must be complex, or there can be at most one real root where Q(λ) = 0. (If there were 2 real roots, the curve would cross the axis & become negative for some values of λ).

The roots of ‘aλ2 + bλ + c = 0’ are a

acbb2

42 −±−=λ

It follows that b2 – 4ac ≤ 0. Substituting for a, b and c gives Schwartz’s inequality as quoted above. If x(t) = ky(t) for any constant k, then x(t) y(t) = k |y(t)|2 and |x(t)|2=k2|y(t)|2. It follows that equality applies in this case. Exercise 5.6: Repeat this proof for complex valued x(t) and y(t) remembering that x(t)x*(t) = |x(t)|2. Solution: Exercise for you. Applying the frequency-domain version of the Cauchy-Schwartz inequality to the expression for |α/σ0|2 above, taking X((f)) as H((ƒ)) and Y((f)) as Si((ƒ))e2πjƒT gives:

∫∫

∫∫∫

∞

∞−

∞

∞−

∞

∞−

∞

∞−

∞

∞−

−=−≤∴

−≤

dffQfPN

dfefQfPN

dffHN

dfefQfPdffH

jfT

jfT

2

0

22

0

20

2

0

222

20

))(())((2

1)))(())(((2

1 |/|

))((2

))(())((())(( |/|

π

π

σα

σα

CS3282: Digital Communications 5.10 BMGC/ 25/08/06

For equality, i.e. to give us the maximum |α/σ0|2 for given symbol shapes p(t) & q(t), we must have a filter whose frequency-response is

kefQfPkfH jfT constantsomefor) ))(())(( ())(( 2** π−−= Taking the inverse FT of this would give us the impulse response h(t) of the filter we require to maximise |α/σ0|2 . We call this the “matched filter” for the difference between p(t) and q(t). We note that the maximum value of |α/σ0|2 is:

dEN

dttqtpN

dffQfPN 0

2

0

2

0

2

0 21 )()(

21 ))(())((

21 max =−=−=

∫∫∞

∞−

∞

∞−σα

by Parseval's theorem (see Section 2) and denoting by Ed the 'energy of the difference signal p(t)-q(t):

∫ ∫∞

∞−

∞

∞−−=−= dttqtpdffQfPEd

22 |)()(||))(())((|

Therefore if we employ a matched filter, we achieve the minimum possible error probability, which is

)2( 0N

EdQ

If the FT of p(t) is P((ƒ)), then the FT of p(-t) is P*((ƒ)), i.e. the complex conjugate of P((ƒ)). Similarly for q(t). Taking the complex conjugate in the frequency-domain corresponds to a time-reversal in the time-domain. Also, if the FT of p(t) is P((ƒ)), the Fourier transform of p(t-T) is P((ƒ))e-2jπƒT . Therefore multiplying a FT spectrum by e-2jπƒT corresponds to a delay in the time-domain of T seconds. This means that the ideal matched filter H((ƒ)) for given symbols p(t) and q(t) that we are trying to distinguish is one whose impulse response h(t) is equal to the difference p(t)-q(t) modified in the following 3 ways:

(i) reversed in time, (ii) delayed by T seconds, and (iii) multiplied by any positive constant k.

Going back to the transmission of a rectangular symbol for p(t) and zero for q(t) this result now tells us that an ideal matched filter for such symbols would have the impulse-response:

≤≤

=otherwise

TtTth

:00:/1

)(

However, we can now design matched filters for other symbol shapes. Correlation Detector The ideal matched filter for symbol shapes p(t) and q(t) at a transmission rate 1/T Baud has impulse response:

kanyfor))tT(q)tT(p(k)t(h −−−=

CS3282: Digital Communications 5.11 BMGC/ 25/08/06

When the received signal is r(t), the output from the matched filter is the convolution between r(t) and h(t):

∫∫ −=−=∞

∞−

tdthrdthrtz

0)()()()()( ττττττ

if the filter is causal; i.e. ( )0)( tforth >=− ττ and r(t) starts at t=0. This can also be viewed as a the “cross-correlation at delay t seconds” between r(t) and h(-t).

∫∫∫

−=ττ−ττ=∴

ττ−−−τ−−τ=∴T

0

T

0

t

0

dt))t(q)t(kr(t)(p d))(q)()(pkr( z(T)

d)))t(T(q))t(T(p(k)(r)t(z

Replacing τ by t does not change the integral at all. We can now see that z(T) can be produced in a different way, using a multiplier and integrator as shown below.

This is called a “correlation detector”, and is often more practical and easier to realise than a matched filter. It is similar to a “coherent detector”. The output depends on the energy of the symbol difference relative to the noise power, which means that the actual shape of the difference p(t)-q(t) is now not of primary importance.

Sample at t=T to obtain z(T)

p(t)-q(t)

r(t)

∫t

0

When p(t) and q(t) are rectangular symbols, this becomes equivalent to an 'I&D' averager. To summarise this section so far, we consider an input signal consisting of p(t) or q(t) corrupted by AWGN n(t) of two-sided PSD N0 /2 Watts/Hz. A matched filter H((ƒ)) is a filter with the property that the ratio |α/σ0 | (or equivalently α2/σ0

2) is maximised when α=(p0(t)-q0(t)/2 and p0(t) & q0(t) are the filter’s responses to p(t) & q(t) respectively, σ0 is the standard deviation (σ0

2 is the variance) of n0(t), the filter’s response to n(t), and the symbol rate is 1/T. A correlation detector is alternative way of calculating the matched filter output required at the decision points. Considering binary transmission, when the symbols being transmitted are p(t) for logic “1” and q(t) volts for logic “0”, and defining a threshold γ = (p0(T)+q0(T))/2 (assuming an equal number of “1”s and “0”s) this allows us to calculate the error probability as Q(|α/σ0|). Remember that Q(z/σ0) is the probability of a Gaussian variable of zero mean and variance σ0 2 being greater than z. If Ed denotes the energy of the difference between pulses p(t) & q(t), the bit-error probability with a matched filter is:

=

02NEQP d

B

When p(t) is rectangular of height A and duration T, and q(t)=0, this formula gives us

=

0

2

2NTAQPB

Exercise 5.7: A binary signal with NRZ ‘+1 volt’ and ‘0 volt’ rectangular symbols p(t) and q(t) and a symbol rate of 1/T Baud is corrupted by AWGN with two-sided PSD N0 /2 = 1x10-3 Watts/ Hz. If the received signal is detected with a matched filter, what is the maximum bit-rate that can be sent with a

CS3282: Digital Communications 5.12 BMGC/ 25/08/06

bit-error rate of less than 1 bit in 1000? (We need to find T). What would be the error-rate with no matched filter if the noise bandwidth B=10kHz ? Solution:

filtermatchedwithratebitMaximumHzT

T

TQPB

−=≤

≥

≤

××

= −

2611.3250

001.0104

13

With no MF, PB = Q(1/(2σ)) with σ2 = 0.2. Exercise 5.8: Consider again Exercise 5.7 where instead of being a rectangular symbol, p(t) is a triangular symbol of height √3 T and duration T as shown below. Sketch the impulse-response of the matched filter and calculate the minimum bit-rate achievable using this matched filter.

1 .7 3 2 / T

T

t

Example 5.9: In a binary digital communication system, the signal component obtained from a correlation receiver is expected to be p0(T) = 1 volt for logic “1” and q0(T) = -1 volt for logic “0”. Assume equal probability of 1s and 0s and take a threshold γ=0 volts. If the Gaussian noise at the correlator output has unit variance, find the probability of a bit-error, and the corresponding bit-error rate. Solution: PB = Q(1/1) = Q(1) = 0.18. Error rate = 1 bit in 5.5 bits. Easy because we are told σ0

2. Exercise 5.10: How would the bit-error probability obtained in problem 5.9 be affected if the threshold γ were inappropriately chosen to be 0.2 volts instead of zero. Solution: PB = 0.5Q(1.2/1) + 0.5 Q(0.8/1) = 0.19. Bit-error rate: 1 bit in 5. Note that:

∫∫∞

∞−

∞

∞−− dttqdttp 22 |)(||)(|

i.e. the difference between the energy of p(t) and the energy of q(t) is not necessarily equal to Ed since

we cannot say that is always equal to zero. If this expression is zero, p(t) and q(t) are

said to be orthogonal. It may be argued that unipolar signalling is orthogonal since the product of p(t) and q(t) is zero for all t because one of the symbols is zero for all t. Bipolar signalling is definitely not orthogonal.

∫∞

∞−dttqtp )()(

CS3282: Digital Communications 5.13 BMGC/ 25/08/06

Average Energy per bit Clearly, the higher the power of the transmitted signal conveying a sequence of symbols, the higher will be the voltages of the symbols (e.g. rectangular symbols), the better will they be seen above the noise (which remains fixed in power) and therefore the lower will be the bit-error probability. If the average power of the signal is P Watts and the bit-rate is 1/T bits/second (bit-rate, not symbol-rate), we can say that the “average energy per bit”, Eb, is P divided by (1/T). Units are Joules/bit. This is because Watts are Joules/second which, when divided by bits/second, become Joules/bit. It is very useful to know how much energy each bit of information is going to cost us to send. It will be different for different signalling techniques as we shall see. The required power of the transmission is clearly Eb times the bit-rate. Do not confuse Eb (average energy per bit) with Ed (energy of difference signal). Unipolar signalling Let p(t) be a shaped pulse and q(t) = 0 for 0 ≤ t ≤ T. This is unipolar signalling. We have shown that when a matched filter is employed,

=

0

dB N2

EQP

with [ ] [ ] )t(pofenergydt)t(pdt)t(q)t(pET

0

22d ==−= ∫∫

∞

∞−

Assuming an equal number of ones & zeros, for unipolar signalling, the average energy per bit, Eb = Ed/2. Expressing the error probability in terms of Eb now, we obtain for unipolar signalling:

) /EQ( = 0b NPB Exercise 5.11: Let p(t) = A and q(t) = 0 for 0 ≤ t ≤ T. This is NRZ unipolar signalling with rectangular pulses. Using the formula just established, show that PB=Q( )2/( 0

2 NTA ), assuming an equal number of ones and zeros. Solution: Eb = Average of {A2T and 0} = A2T/2. Therefore PB=Q( )2/( 0

2 NTA ). This agrees with earlier results. Bipolar signalling Let p(t) = s(t) and q(t) = -s(t) for 0 ≤ t ≤ T. This is binary signalling with anti-podal signals i.e. two signals which are the negative of each other. Signal s(t) may have any shape. When a matched filter is employed with impulse response 2s(T-t); i.e. it is matched to p(t)-q(t):

=

0

dB N2

EQP

with [ ] [ ] b

T

0

22d E4)t(pofenergy4dt)t(p4dt)t(q)t(pE =×==−= ∫∫

∞

∞−

since the energy of p(t) and q(t) are equal.

CS3282: Digital Communications 5.14 BMGC/ 25/08/06

Expressing the bit-error probability in terms of Eb, we obtain for bipolar signalling: ) /E 2Q( = 0b NPB Exercise 5.12: Let p(t) = A and q(t) = -A for 0 ≤ t ≤ T. This is bipolar signalling with antipodal rectangular NRZ pulses. Calculate PB assuming equal numbers of “1”s and “0”s. Solution: The average energy per bit, Eb, equals A2T because the energy of a rectangular symbol of height ±A and duration T seconds is A2T. Therefore: ) /A 2Q( = 0

2 NTPB

To do this another way, note that E [ ] [ ] TA4dt)t(q)t(pdt)t(q)t(p 2T

0

22d =−=−= ∫∫

∞

∞−

Therefore

=

=

=

0

2

0

2

0

224

2 NTAQ

NTAQ

NE

QP dB

Comments about uni-polar and bipolar signaling: It is now interesting to consider which of the two signalling techniques analysed above is the more efficient in terms of the having the lower bit-error probability obtained for a given average energy per bit. On page 95 of “Digital Communications” by Sklar is a graph of bit-error probability PB against 10 log10 (Eb/N0) for matched filter reception of (i) unipolar and (ii) bipolar base-band signalling. The horizontal axis tells us how much greater, in dBs, the average energy / bit is than the power spectral density N0 (sometimes call this Nn) of the noise. It is a ratio of Eb in Joules/bit to N0 in Watts/Hz. Remember, if the noise bandwidth is 0 to B Hz, and its power is its variance σ2, then N0=σ2/B. You can draw this graph yourself with the aid of a graph of the co-error function Q(z) against z. We note from this graph that when Eb = N0, i.e. for a ratio of 0 dB, PB is about 0.8x10-1 for bipolar and about 0.2 (1 bit in 5 in error) for unipolar. When Eb is 10 times N0, the ratio is 10 dB, and we find that PB is 0.3x10-5 for bipolar and about 10-3 for unipolar. An error probability of 10-3 means an bit error rate of 1 bit in 1000 bits. Therefore for a given average energy / bit, bipolar is significantly better than unipolar, even though we found that for the same rectangular symbol height A, the average energy was higher for bipolar than for unipolar. This means that for bipolar, A can be reduced from that used by unipolar, to achieve the same error probability. Reading across the graph, if PB=10-4, i.e. a bit-error rate of one error in 1000 bits, the ratio must be 8.2dB for bipolar and 11.2dB for unipolar. If N0 is the same for both readings (i.e. for the same noise conditions), unipolar signalling must have energy per bit 3dB higher than bipolar to achieve this same bit-rate. This is obvious from the formulae: to have the same error probability, ) /)(E 2 0b NbipolarQ( ) /)(E Q( = 0b Nunipolar which means that Eb(unipolar) = 2Eb(bipolar). This factor of two is an energy difference of 3dB. These results will stand us in good stead when we look at modulated data transmission over a radio channel, for example. We can refer to the unipolar signalling considered above as form of “base-band orthogonal signalling” and to bipolar signalling as a form of “base-band antipodal signalling”. The results we have obtained for the matched filter (or correlation) detection of these signals will be the same for coherently detected band-pass antipodal signalling (e.g. binary phase shift keying, (PSK)) and band-pass orthogonal signalling (e.g. binary frequency shift keying (FSK)), respectively.

CS3282: Digital Communications 5.15 BMGC/ 25/08/06

Exercise 5.13: (Jan 1996 exam question) A binary digital communication system transmits equally likely symbols p(t) & q(t) at 1/T Baud where p(t) = At/T for 0 ≤ t≤ T and q(t) = 0. On the channel, white Gaussian noise with a two-sided power spectral density of N0/2 = 10-15 Watts/Hz is added. The signals are detected by a matched filter detector.

a) Show that the optimal threshold γ = A2T/6 b) Give the impulse-response of the matched filter required and hence show that the average bit-

error probability is PB = Q( )6/( 02 NTA )

c) If A = 0.2mV find the maximum binary bit-rate possible such that PB does not exceed 10-3. d) Is this orthogonal signalling?

Exercise 5.14: How would your answer to Exercise 5.13 be affected if a correlation detector rather than a matched filter were to be used. Give a block diagram of the detector. Exercise 5.15: A binary digital communication system transmits equally likely symbols p(t) & q(t) at 1/T Baud where:

≤<≤≤

≤<≤≤

T t T/2 :A +T/2t 0 :A -

= )t(q and T t T/2 :A -

T/2t 0 :A + = )t(p

a) What is the name for this form of coding and what are its advantages and disadvantages? On the channel, WGN with a 2-sided PSD of N0/2 WHz-1 is added. The signals are detected by a matched filter detector.

b) Give the impulse-response of the matched filter required and determine the optimal threshold γ. c) Derive an expression for the average bit-error probability, PB, in terms of the average energy/bit

and N0. d) Compare this expression with what was obtained for bipolar signalling using rectangular

symbols of height A and duration T.