程序代写代做代考 scheme EE450 Discussion #6 – cscodehelp代写
EE450 Discussion #6
Shannon’s Theorem Modulations
Multiplexing
Shannon’s Theorem
C = B log2(1 + SNR)
Theoretical Maximum Capacity that can be
obtained on a line
Sets an Upper Bound on the capacity given the conditions
Used for Calculating the
Signal to Noise Ratio – Given the Bandwidth and capacity
of the channel
Bandwidth – Given the SNR and Channel Capacity Capacity – Given the SNR and the Bandwidth
Problem #1
What SNR is needed to put a T-1 carrier on a 50 khz line?
What do we know?
T-1 Capacity = 1.544 Mbps Bandwidth = 50 KHz
Move them around and Solve:
1,544,000 = 50,000 log 2 (1+SNR)
2^30.88 –1 = SNR
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C Blog
I
Continued
So SNR = 1976087931
SNR is typically measured in DB
Use SNR dB = 10 log 10 (SNR)
In this case
SNR dB = 10 log 10 (1976087931) SNR aprox. 92.9 dB
However you must NOT plug SNR into Shannon’s theorem in dB format
Problem #2
Calculate the maximum rate supported by a telephone line with BW of 4 KHz. When the signal is 10 volts, the noise is 5 milivolts.
SNR=Signal power/Noise Power
Power is proportional to square of the voltage
S/N = (10^2)/(0.005^2) = 4000000
B = 4000 Hz
C = B log 2 (1+S/N)
Reminder: log 2 x = ln x / ln 2
C = 4000 log 2 (1+4000000)= 87726 bps
Review on Modems
Modem Stands for
MOdulator / DEModulator
Uses Sine wave As the carrier Signal
Digital to Analog Encoding
WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998
Modulation
Need to Encode Digital Data in an Analog Signal
In modem transmission we use different techniques for modulation
Amplitude Modulation Frequency Modulation Phase Shift
Amplitude Modulation
Varies the Amplitude of the Signal
Amplitude Modulation
Same Signal Greater Amplitude
Amplitude = 1 Amplitude = 2
Amplitude Modulation
Amplitude 2 = 1 Amplitude 1 = 0
This signal Represents:
0011010
ASK
WCB/McGraw-Hill
The McGraw-Hill Companies, Inc., 1998
FSK
WCB/McGraw-Hill
The McGraw-Hill Companies, Inc., 1998
Phase-Shift Modulation
Start with our normal sine wave
The sine wave has a period of P
P may be denoted as T instead in the equations
Phase-Shift Modulation
Shift the Phase of the
Shifted diagram shows that the cycle starting at 1 vs. starting at 0
PSK
WCB/McGraw-Hill
The McGraw-Hill Companies, Inc., 1998
PSK Constellation
WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998
4-PSK
WCB/McGraw-Hill
The McGraw-Hill Companies, Inc., 1998
4-PSK Constellation
WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998
8-PSK Constellation
WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998
PSK
WCB/McGraw-Hill
The McGraw-Hill Companies, Inc., 1998
Combining Both
Modulation used in Modern Modems
Uses:
Amplitude Modulation Phase Shift Keying
QAM
Quadrature Amplitude Modulation Big Name – Simple Concept
4-QAM and 8-QAM Constellation
WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998
8-QAM Signal
WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998
16-QAM Constellation
WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998
Bit Rate and Baud Rate
WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998
Problem #3
A modem uses an 8-PSK modulation scheme supporting data rate of 4800 bps. What is the signaling rate (aka baud rate)?
8 PSK – (Phase Shift Keying)
8 different encoding levels
Each encoding has log2 8 = 3 bits 4800 / 3 = 1600 Baud Rate
Statistical TDM Parameters
I = Number of Input Sources
R = Data rate of each source (bps)
α (Alpha) = mean fraction of time each source is transmitting
M = Effective capacity of multiplexed line
K = M / (I x R) = Ratio of multiplexed line
capacity to total input rate
λ (lambda) = α x I x R = Average Arrival rate
Ts = 1 / M = Service time in seconds
ρ : Line Utilization
ρ = Fraction of total link capacity being used
Many different forms to express line utilization
ρ = λ Ts
ρ = (α x I x R) / M ρ = α/ K
ρ = λ/ M
Sample Problem #5
Ten 9600 bps lines are multiplexed using TDM. Ignoring overhead bits what is the total capacity required for Synchronous TDM?
Simple: 10 X 9600 = 96 kbps (96,000)
Sample Problem #6
Ten 9600 bps lines are multiplexed using TDM. Assuming that we limit line utilization to 0.8 and each line is busy 50 % of the time. What is the capacity required for Statistical TDM?
What do we know?
Line utilization – ρ = .8
Fraction of time transmitting – α = .5
R Data Rate of each input source = 9600 bps I number of Input Sources = 10
Continued
The Equation:
ρ = α x I x R x /M
Where M is the capacity of the multiplexed line Rearrange for M
M = α x I x R /ρ
Plug in the given parameters
M = 0.5 x 10 x 9600 / 0.8
M = 60 kbps
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