1. ECEN4523 Commo Theory Lecture #42 30 November 2015 Dr. George Scheets www.okstate.edu/elec-engr/scheets/ecen4533 n Read 11.3 n Problems: 11.1-3 & 4 n Final Exam, 1000 – 1150, Friday, 11 December (Live) u Remote DL students < 18 December n Quiz 9 Results u Hi = 10, Low = 1, Ave = 6.33, σ = 3.40

2. ECEN4523 Commo Theory Lecture #43 2 December 2015 Dr. George Scheets www.okstate.edu/elec-engr/scheets/ecen4533 n Problems: 11.2-3 & 6 n Final Exam, 1000 – 1150, Friday, 11 December (Live) u Remote DL students < 18 December

3. ECEN4523 Commo Theory Lecture #44 4 December 2015 Dr. George Scheets www.okstate.edu/elec-engr/scheets/ecen4533 n Problems: 11.2-7 & 11.3-1 n Final Exam, 1000 – 1150, Friday, 11 December (Live) u Remote DL students < 18 December

4. Sum of S.I. Random Variables n Given a Uniformly Distributed noise voltage N... Time Volts 0

5. Bin Count Time Volts 0

6. ...With Uniform Voltage PDF... n volts f N (n) 0+1 1/2 Suppose you add the number 6 to this. What's the output PDF of N+6?

7. What Happens to Time Domain Signal? n Noise voltage N, centered about 0 volts n Now you add 6 volts, yielding N + 6 Time Volts 0

8. Time domain signal is now... Time Volts 6

9. PDF of Y = N + 6 y volts f Y (y) 6+7 5 1/2

10. f Y (y) = f N (n) f 6 (n) n volts f N (n) 0+1 1/2 n volts f 6 (n) +6 1 Convolution of an arbitrary function with a delta function Shifts and centers function's graph to delta function's location.

11. PDF of Y = N + 6 y volts f Y (y) 6+7 5 1/2

12. Add together two die rolls x dots f die roll #1 (x) 1/6 x dots f die roll #2 (x) 1 2 3 4 5 6 1/6 0 1 2 3 4 5 6

13. Add together two S.I. die rolls y dots f sum (y) 2 3 4 5 6 7 8 9 10 11 12 6/36 4/36 2/36 1/36 3/36 5/36

14. Ex) System with Ideal LP Filter W = 10*R b = 10 MHz n Signal Power out ≈ 625 watts u +25 v-peak square waves: Logic 1 u -25 v-peak square wave: Logic 0 n Noise Power out u Gaussian Distributed u Zero Mean (V DC = 0) u σ 2 N = E[N 2 ] = A[n(t) 2 ] = 500(10 9 )No watts

15. BPSK with LP Filter Example n Suppose Noise Power out of LP Filter = 60 watts u SNR = 625/60 = 10.42 n Set σ N 2 = 500*10 9 No = 60 u No = 120(10 -12 ) watts/Hz n P(BE) = Q(25/7.746) = Q(3.227) = 625.5(10 -6 )

16. Matched Filters nFnFnFnFilter where h(t) matches the pulse shape uhuhuhuh(t) = p(T-t) nNnNnNnNote h(t) here is the impulse response of an Integrate and Dump circuit. nMnMnMnMatched Filters maximize output SNR t p(t) p(0) p(3) 30 t p(-t) α -30 p(3) p(0) t p(3-t) = h(t) p(3) p(0) 30

17. Matched Filters n Filter where h(t) matches the pulse shape u h(t) = p(T-t) t p(t) p(0) 30 p(3) t p(-t) -30 p(0) p(3) t p(3-t) = h(t) 30 p(0) p(3)

18. BPSK with LP Filter Example n Suppose Noise Power out of LP Filter = 60 watts u SNR = 625/60 = 10.42 n Set σ N 2 = 500*10 9 No = 60 u No = 120(10 -12 ) watts/Hz n P(BE) = Q(25/7.746) = Q(3.227) = 625.5(10 -6 )

19. BPSK with Matched Filter Example n Suppose No = 120(10 -12 ) watts/Hz (same value as with LP Filter example) n P(BE) = Q[(25*10 -6 )/(54.77*10 -9 )] = Q(456.4) = 0 n Signal Power out of integrator = (25*10 -6 ) 2 = 625(10 -12 ) watts n Noise Power out of integrator = 25No/10 6 = 3(10 -15 ) watts n SNR = 208,333

20. Final Exam n Open Book & Notes n Comprehensive u Readings u Notes u HW u Power Point u Overlaps more likely!