Class Notes_Lect13

Class Notes_Lect13 - hers-3\D\X 00 T‘V”¢S\A‘$YV\Q...

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Unformatted text preview: hers-3 \D \X 00...... T‘V”¢S\A‘$YV\Q{\. NR2 ( NO“ kw” TC) .ZQTDB. 1"]. “‘“V A . $3M‘m‘ :3) tack $39006 3: \—\O\+ 0"” \H‘L‘fW-A'k.’ S$m¥o\ Raf; = :L‘: b'fl‘ (031. :1: bebo‘ rake (-Q«-\‘M’s nxqmguy = J1: Channel Parameters N_RZ ' 1L2 Symbol ./ 111 0 ‘T4- Symbol Period = T Symbol Rate =1/T - Bit rate, R - Defines the number of b_it§ transmitted per second - Aka. data rate - Example: W - Audio system with 8—bit sampler (8 bit A/D) and signal sampling rate of 8 kg -) R= 8*8000= 64 kbits/sec - Symbol (aka. Baud) rate, 1/T - Defines the number of symbols transmitted per second - R= @aud ratex n (# of bits per symbol) - If one bit is transmitted per symbol % R= Baud rate = 1/T Spring, 09 4 Léx——-——‘§‘g’g§$ig":€ @@ Binary vs. M-ary Signaling I 9‘25” _ Binary MEL! Symbol 00 11 10 oo 01 oo :2) “:2- M) : .( v(t) Symbol . . . : . 4T} 4T} - Binary: Symbol has 2 allowed states (i.e. level), ex. 1 or O - n (# of bits persymbol)=1 - M-agy: Each symbol has m finite states - n (# of bits per symbol)= log2 m - EX. : 1 bit/symbol (binary) 9 m = 2 (Le. 0 or 1) . EX.: 3 bits/symbol 9m = 8 (Le. 0, 1, 2, 3, 4, 5, 6, or 7) R = Baud rate x® = (an) 5 mm AAAAAAAAAA Spring, 09 (E and V0 \A'HA 'fi‘US‘N-Vt O'Q S um ‘Q \(Kuhey Y\ —- Em) [\E = W p—d E = n + S‘5Wbo\ RA“? .. E- — ”—— ‘ :—'> ‘OPS/ JAZ- @D Example - The GSM standard specifies a data rate of 270.83 kbps, SNRmin ofgaclB and B of 200 kHz. ”‘3‘" - Compare the actual capacity to maximum capacity C: B toszuausw) ; zoogg \bcngvr t0 permlR 2 ”no ~23 J, qZ‘I. Q 6’39... ii?) 2 £32 HOPE - If SNR increased to 20 dB, calculate the channel capacity L— ZOOé'; K°°31K\’<‘ \00) ’3- \-23 Mb?$‘ ’— ArC/Ei Q g 207. C. ’_ N27g \vxcm Nola: 11710 dg \“Cm \“ $NQ‘" 0079’“; \KA Q. Spring, 09 3 £lettto$ciene AAAAAAAAAA D “bd‘A $fi§m s‘ . 5E? ,15 P‘CT‘QWAkkm-Q}(LQ 7:) @ QW'LV‘ EXQNO Eb é imrfib Qex YDF‘Y 2 E No 9:. Moi“. A ' m3 WMr (ha-B3 -__._ Nod-B k_\:pu( MSMI'HZ Q) Hz. . m ...
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