Logarithms
• Log Review
if x ay then y loga x
so
if x 210 then 10 log2 x
Logarithms
• For example
find log2 4096
4096 2y
ln 4096 y ln 2
ln 4096ln 2
y12
Logarithms
so
log2 x ln xln 2
if base = B then
logB x ln xln B
log10 xlog10 B
Logarithms
• Laws of Logarithms
loga (xy) loga x loga y
loga (x / y) loga x loga y
loga xn n loga x
• Intermodulation noise– results when signals at different
frequencies share the same transmission medium
• the effect is to create harmonic interface at
f1 f 2 and / or f1 f 2
f1 frequency of signal 1
f2 frequency of signal 2
• cause– transmitter, receiver of intervening
transmission system nonlinearity
• Crosstalk– an unwanted coupling between signal
paths. i.e hearing another conversation on the phone
• Cause– electrical coupling
• Impluse noise– spikes, irregular pulses
• Cause– lightning can severely alter data
Channel Capacity
• Channel Capacity– transmission data rate of a channel (bps)
• Bandwidth– bandwidth of the transmitted signal (Hz)
• Noise– average noise over the channel
• Error rate– symbol alteration rate. i.e. 1-> 0
Channel Capacity
• if channel is noise free and of bandwidth W, then maximum rate of signal transmission is 2W
• This is due to intersymbol interface
Channel Capacity
• Example
w=3100 Hz
C=capacity of the channel
c=2W=6200 bps (for binary transmission)
m = # of discrete symbols
C = 2Wlog2m
Channel Capacity
• doubling bandwidth doubles the data rate
if m=8c 2(3100 hz)log2 8 18,600 bps
Channel Capacity
• doubling the number of bits per symbol also doubles the data rate (assuming an error free channel)
(S/N):-signal to noise ratio
(S / N)dB 10logsignal powernoise power
Hartley-Shannon Law
• Due to information theory developed by C.E. Shannon (1948)
C:- max channel capacity in bits/second
C w log2 (1SN
)
w:= channel bandwidth in Hz
Hartley-Shannon Law
• Example
W=3,100 Hz for voice grade telco lines
S/N = 30 dB (typically)
30 dB = 10 logPsPn
Hartley-Shannon Law
3 logPsPn
log10
PsPn
3
103 PsPn
1000
C 3100 log2 (11000) 30,898 bps
Hartley-Shannon Law
• Represents the theoretical maximum that can be achieved
• They assume that we have AWGN on a channel
Hartley-Shannon Law
C/W = efficiency of channel utilization
bps/Hz
Let R= bit rate of transmission
1 watt = 1 J / sec
Eb=enengy per bit in a signal
Hartley-Shannon Law
S = signal power (watts)Tb the time required to send a bit
R =1
T b
Eb STb
EbN0
energy per noise power density per hertz
Hartley-Shannon Law
EbN0
S / RN0
SkTR
k=boltzman’s constantby
Eb STb
S EbTb
S / R Eb
N0 kTR
Hartley-Shannon Law
assuming R=W=bandwidth in HzIn Decibel Notation:EbN0
S 10 log R 228.6dbW 10 logT
Hartley-Shannon Law
S=signal powerR= transmission rate and -10logk=228.6So, bit rate error (BER) for digital data is a decreasing function of Eb
N0
For a given , S must increase if R increases
EbN0
Hartley-Shannon Law
• Example
For binary phase-shift keying =8.4 dB is needed for a bit error rate of
EbN0 10 4
let T= k = noise temperature = C, R=2400 bps & Pe 10 4 BER
Hartley-Shannon Law
• Find S
S EbN0
10 logR 228.6dbW 10logT
S 8.4 10 log2400 228.6dbW 10 log 290
S=-161.8 dbw
ADC’s
• typically are related at a convention rate, the number of bits (n) and an accuracy (+- flsb)
• for example– an 8 bit adc may be related to +- 1/2 lsb
• In general an n bit ADC is related to +- 1/2 lsb
ADC’s
• The SNR in (dB) is therefore
SNRdB 10 log10SN
whereS 2n
N 12
2 n 2 n 1
SNRdB 10 log10 22n1 (20n 10)log10 2
SNRdB 6n 3about
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