Communication Systems Engineering Episode 3 Part 6 pps
Communication Systems Engineering Episode 2 Part 6 pps
... CONGESTION CONTROL BER EFFICIENCY 0 0.1 0.2 0 .3 0.4 0.5 0 .6 0.7 0.8 0.9 1 1.00E-07 1.00E- 06 1.00E-05 1.00E-04 1.00E- 03 1,544 64 KBPS 16 KBPS 2.4 KBPS KBPS • TCP assumes dropped ... - 19 83 First widely available release – 4.3BSD Tahoe - 1988 Slow start and congestion avoidance – 4.3BSD Reno - 1990 Header compression – 4.4BSD - 19 93 Multicast support, R...
Ngày tải lên: 07/08/2014, 12:21
Communication Systems Engineering Episode 1 Part 1 pps
... Section 6. 1 Sec. 2.2, 2.4 Sec. 6. 5 Sec. 6. 2 -6 .3 Sec. 7.1 - 7 .3 Sec. 7.5 Sec. 7.5 Sec. 7 .6 Chapter 9 Sec. 9.5 - 9 .6 Sec. 7.7 Spectra of digitally modulated signals Sec. 8.1 - 8 .3 Packet ... L1 6- Feb L2 11-Feb L3 13- Feb L4 18-Feb 20-Feb L5 25-Feb L6 27-Feb L7 4-Mar L8 6- Mar L9 11-Mar L10 13- Mar L11 18-Mar L12 20-Mar L 13 25-Mar 27-Mar 1-Apr...
Ngày tải lên: 07/08/2014, 12:21
Communication Systems Engineering Episode 1 Part 2 ppsx
... ∑ P i Log ( ) ≤ ∑ P i Log(M) = Log(M) i=1 P i i=1 Eytan Modiano Slide 6 1 16 . 36 : Communication Systems Engineering Lecture 2: Entropy Eytan Modiano Eytan Modiano Slide 1 HX px ... rule: H(X 1 , , X n ) = H(X 1 ) + H(X 2 |X 1 ) + H(X 3 |X 2 ,X 1 ) + …+ H(X n |X n-1 …X 1 ) 2. H(X,Y) = H(X) + H(Y|X) = H(Y) + H(X|Y) 3. If X 1 , , X n are independent then: H(X...
Ngày tải lên: 07/08/2014, 12:21
Communication Systems Engineering Episode 1 Part 6 potx
... x y ,4 − QAM ⇒ A m , A m ∈ { + / − 1 } x y 16 − QAM ⇒ A m , A m ∈ { + / − 1, + / − 3 } Eytan Modiano Slide 16 -3 -1 1 3 -3 -1 1 3 Bandpass signals • To transmit a baseband signal ... M − 1 , K = [( [( y 3 3 M E = 2 ( M − 1 ) E s 3 g Transmitted energy = E s = ( M − 1 ) E g 2 3 E b ( QAM ) = Energy / bit = ( M − 1 ) E g 3Log 2 () M...
Ngày tải lên: 07/08/2014, 12:21
Communication Systems Engineering Episode 1 Part 7 pps
... for M-PAM M 2 − 1 M 2 − 1 E av = 3 E g => E bav = 3Log 2 () E g M E = 3Log 2 () E bav M g M 2 − 1 Log M M E bav − 2 2 0 6 1 () ( N) Pe P e = 2Q , P eb ... = 2 Q sin( π / M), P eb = es Log 2 ()M Eytan Modiano Slide 29 Noise in communication systems S(t) Channel r(t) = S(t) + n(t) r(t) n(t) • Noise is additional “unwanted”...
Ngày tải lên: 07/08/2014, 12:21
Communication Systems Engineering Episode 1 Part 10 ppsx
... multiple of G • Choice of G is a critical parameter for the performance of a CRC 16 . 36 : Communication Systems Engineering Lectures 14: Cyclic Codes and error detection Eytan Modiano Performance ... or 3 errors (d > 3) 2) All bursts of errors of r or fewer bits 3) Random large numbers of errors with prob. 1-2 -r • Standard DLC's use a CRC with r= 16 with option o...
Ngày tải lên: 07/08/2014, 12:21
Communication Systems Engineering Episode 2 Part 4 pps
... mixed voice and data Slot 1 2 3 4 5 6 frame 1 frame 2 frame 3 frame 4 frame 5 15 3 20 2 15 7 3 9 7 9 7 9 18 7 3 15 9 6 18 idle idle idle idle 2 3 3 6 Eytan Modiano Slide 17 Analysis ... Comparison of MAC protocols Load (packets/second) Delay in seconds 0 2 4 6 8 10 12 14 16 18 20 0 0.2 0.4 0 .6 0.8 ALOHA SCHEMES TDMA (10 USERS) P...
Ngày tải lên: 07/08/2014, 12:21