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