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
l
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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