Application of Matched Filter

Transcription

1 Application of Matched Filter Lecture No. 12 Dr. Aoife Moloney School of Electronics and Communications Dublin Institute of Technology

2 Overview This lecture will look at the following: Matched filter vs correlator Application of Matched Filter Performance of unipolar binary signalling Performance of bipolar binary signalling February 2005 Slide: 1

3 Matched Filter vs Correlator For an input signal, x(t), the output of the filter matched to s(t) is identical at time T only to the output of a correlator performing the integration of the product x(t)s(t) over the duration of the symbol, T z(t) = r(t) h(t) = t 0 r(τ)h(t τ)dτ February 2005 Slide: 2

4 z(t) = t r(τ)h(t τ)dτ = t r(τ)s [T (t τ)]dτ 0 This becomes, at t = T: z(t) = T 0 0 r(τ)s[τ]dτ which is the correlation of r(t) with s(t). February 2005 Slide: 3

5 Application of Matched Filter The variance σ of z(t) is equivalent to the noise. The mean square noise power is therefore given by σ 2 and the signal power to the average noise power is given by: ( S N ) T = a2 i n 2 = a2 i σ 2 In order to minimise the probability of error, ie., [ ] a1 a 2 P B = Q 2σ we must use in the receiver a linear filter that will max- February 2005 Slide: 4

6 imise the difference (a 1 a 2 )/σ or equivalently, that maximises: [ a1 a 2 A filter matching the difference signal (s 1 (t) s 2 (t)) will maximise the signal to noise ratio, given by: ( ) [ ] 2 S (a1 a 2 ) = N σ T σ which will then take the value 2E d /N 0, noting E d is the energy of the difference signal at the filter input. ] 2 February 2005 Slide: 5

7 Therefore if we want to maximise (a 1 a 2 )/σ we should use a filter matched to the signal (s 1 (t) s 2 (t)). We then have: (a 1 a 2 ) 2σ = (a1 a 2 ) 2 4σ 2 = And therefore [ ] (a1 a 2 ) P B = Q 2σ 1 4.2E d N 0 = = Q ( Ed 2N 0 Ed We now have the detection system shown below: ) 2N 0 February 2005 Slide: 6

8 Filter matched to r(t) s 1 (t)-s 2 (t) z(t) z(t) Decision z(t) > < (a 1 +a 2 )/2 Sample at t=t February 2005 Slide: 7

9 Performance of Unipolar Binary Signalling We consider the case where: s 1 (t) = A, 0 t T, for binary 1 s 2 (t) = 0, 0 t T, for binary 0 with A > 0 The unipolar signal is present with the addition of zero mean Gaussian noise at the input of a matched filter. The output is sampled at a time T. The filter is imple- February 2005 Slide: 8

10 mented with a correlator that multiplies and integrates the incoming signal r(t) with the difference of the reference signals (s 1 (t) s 2 (t)) = s 1 (t) After a symbol duration T, the output z(t) is compared with the threshold γ. When s 1 has been sent the mean value of the random February 2005 Slide: 9

11 variable z is: a 1 (T) = E [z(t)] T [ = E s 2 1 (t) + s 1 (t)n(t) ] dt = T 0 0 A 2 dt = A 2 T similar calculation will show: a 2 (T) = 0 February 2005 Slide: 10

12 The threshold γ is given by: γ = (a 1 + a 2 ) 2 = A2 T 2 If the correlator output is greater than A 2 T/2, the signal is declared to be s 1 (t), otherwise s 2 (t) The energy of the difference signal is E d = A 2 T The probability of error can be expressed using the co error (complementary error) function, taking into account February 2005 Slide: 11

13 that the average energy per bit, E b, can be expressed as: E b = A2 T 2 Therefore: P B = Q ( Ed 2N 0 ) = Q ( Eb N 0 ) = Q ( A 2 T 2N 0 ) So we have the following detection system: February 2005 Slide: 12

14 Filter matched to r(t) s 1 (t) z(t) z(t) Decision z(t) > < A2 T/2 Sample at t=t February 2005 Slide: 13

15 Performance of Bipolar Binary Signalling We consider the case where: s 1 (t) = +A, 0 t T for binary 1 s 2 (t) = A, 0 t T for binary 0 with A > 0 Binary waveforms that are the negative from one another are called antipodal signals The receiver for this type of waveform can be imple- February 2005 Slide: 14

16 mented using a correlator that multiplies and integrates the incoming signal r(t) with the difference of the reference signals (s 1 (t) s 2 (t)) = 2s 1 (t) After a symbol duration T, the output z(t) is compared with the threshold γ. When s 1 has been sent the mean February 2005 Slide: 15

17 value of the random variable z is: a 1 (T) = E [z(t)] = E T 2 [ s 2 1(t) + s 1 (t)n(t) ] dt = T 0 0 2A 2 dt = 2A 2 T Similar calculation will show that a 2 (T) = 2A 2 T For antipodal signals, a 1 = a 2 and the threshold γ is February 2005 Slide: 16

18 given by: γ = (a 1 + a 2 ) = 0 2 If the correlator output is greater than 0, the signal is declared to be s 1 (t), otherwise s 2 (t) The energy of the difference signal is E d = 4A 2 T The probability of error can be expressed using the co error function: ( ) ( ) ( ) Ed 4A P B = Q = Q 2 T 2A = Q 2 2N 0 2N 0 N 0 February 2005 Slide: 17

19 Conclusion This lecture has looked at the following: Matched filter vs correlator Application of Matched Filter Performance of unipolar binary signalling Performance of bipolar binary signalling February 2005 Slide: 18