SYSC5606-4-2-with-notes

SYSC5606-4-2-with-notes - Key Points Multipath causes the...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
SYSC 5606 - 4 Copyright © 1995-2009 by R.H.M. Hafez 1 Key Points Key Points ± Multipath causes the fading phenomenon ± In heavy clutter with no line-of-sight (LOS) the fading distribution is Rayleigh ± If LOS exists fully or partially, the effect of fading is reduced significantly ± Rayleigh fading is the most important mathematical model ± In a Rayleigh channel, "Fading" dominates the error performance ± A 2-state model (Good / Bad) is reasonable ± We have simple formulas to calculate the average fading and non-fading intervals
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is a simple model of a wireless fading channel. The transmitted signal is a single frequency sinusoidal signal. Multipath propagation changes the amplitude of the signal by a random factor, A, and the phase by a random phase shift “fi”. The received signal is the sum of the multipath corrupted signal plus additive white Gaussian noise, n(t). The transmitted power is constant, but the received power depends on the random factor A. At any instant, the received power is (A o A) 2 /2. Since A changes from location to location and from time to time, the average received power A o 2 /2 * the average value of A 2 . The noise process has an average power N o SYSC 5606 - 4 Copyright © 1995-2009 by R.H.M. Hafez 2 )] t ( t cos[ A ) t ( S 0 θ + ω = Fading Channel RX TX Noise, n(t) S(t) R(t) (A, φ ) ) t ( n ] ) t ( t cos[ A A ) t ( R 0 + φ + θ + ω = 2 0 2 0 A N 2 A SNR _ Inst = ) A ( N 2 A SNR _ Average 2 0 2 0 = A Model For Fading Channel A Model For Fading Channel
Background image of page 2
The bit error rate of an optimum receiver of binary data in an AWGN channel is simply (1/2).erfc(sqrt(signal-to-noise-ratio)) as shown on the slide. In order to obtain the average BER we must average the instantaneous BER over all possible values of the amplitude A. Since A has a Rayleigh distribution, we must average the instantaneous BER over the probability density funtion of A, P(A) as shown in the second equation. Notice that, average the signal to noise ratio first then calculating the probability or error using that average value is the wrong procedure. Rather, we should calculate the probability of error for every value of A then average the error probabilities as shown in the slide. This correct procedure results in a much worse probability of error as will be illustrated in the next slide. SYSC 5606 - 4 Copyright © 1995-2009 by R.H.M. Hafez 3 Basic Receiver Performance Basic Receiver Performance ± Instantaneous Bit Error Rate = 2 0 2 A N 2 A erfc 2 1 ) A ( BER 0 ± Average Bit Error Rate dA ) A ( P ) A ( BER BER 0 = Much worse performance than most channels
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This slide compares the BER of an AWGN channel and a Rayleigh fading channel having the same signal to noise ratios. The dotted line curve is for an ideal AWGN channel with an optimum binary
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/24/2009 for the course SCE Sysc5606 taught by Professor Roshdy during the Spring '09 term at Carleton CA.

Page1 / 23

SYSC5606-4-2-with-notes - Key Points Multipath causes the...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online