cc notes lecture 04

7/29/2019 CC Notes Lecture 04

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Computer Communication &

Networks

Lecture # 04

Physical Layer: Signals & Digital Transmission

Course Instructor:

Engr. Sana Ziafat


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Physical Layer Topics to Cover

Signals

Digital Transmission

Analog TransmissionMultiplexing

Transmission Media


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Analog & Digital Data

Data can be analog or digital. The term

analog data refers to information that is

continuous; digital data refers to information

that has discrete states. Analog data take oncontinuous values. Digital data take on

discrete values.


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To be transmitted, data must betransformed to electromagnetic signals.

Note


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Signals can be analog or digital.Analog signals can have an infinitenumber of values in a range; digital

signals can have only a limited

number of values.

Note


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Analog Vs Digital


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Analog Signals


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Sine Wave


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Cont

Three parameters to describe a sine wave

1. Peak amplitude

2. Frequency and time period

3. Phase


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Bandwidth


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Digital Signals


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Digital Signals

In addition to being represented by an analog

signal, information can also be represented

by a digital signal. For example, a 1 can be

encoded as a positive voltage and a 0 aszero voltage. A digital signal can have more

than two levels. In this case, we can send

more than 1 bit for each level.


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Digital Signal


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Bit Rate & Bit Interval (contd.)


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Bit Interval and Bit Rate

ExampleA digital signal has a bit rate of 2000 bps. What is the

duration of each bit (bit interval)

Solution

The bit interval is the inverse of the bit rate.

Bit interval = 1/ 2000 s = 0.000500 s

= 0.000500 x 106ms = 500 ms


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The bit rate and the bandwidth are

proportional to each other.

Note


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Analog Vs Digital


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Analog versus digital signals


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Low Pass & Band Pass


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Data Rate Limits


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Noiseless Channel: Nyquist Bit Rate

Defines theoretical maximum bit rate for

Noiseless Channel:

Bit Rate=2 X Bandwidth X log2L


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Example 8

Consider the same noiseless channel, transmitting a signal

with four signal levels (for each level, we send two bits).

The maximum bit rate can be calculated as:

Bit Rate = 2 x 3000 x log2 4 = 12,000 bps


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Increasing the levels of a signal mayreduce the reliability of the system.

Note


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Noisy Channel: Shannon Capacity

Defines theoretical maximum bit rate for

Noisy Channel:

Capacity=Bandwidth X log2(1+SNR)


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Example

We can calculate the theoretical highest bit rate of a

regular telephone line. A telephone line normally has a

bandwidth of 4KHz. The signal-to-noise ratio is usually

3162. For this channel the capacity is calculated as

C = B log2 (1 + SNR) = 3000 log2 (1 + 3162)= 3000 log2 (3163)

C = 3000 11.62 = 34,860 bps


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Example

We have a channel with a 1 MHz bandwidth. The SNR

for this channel is 63; what is the appropriate bit rate and

signal level?

Solution

C = B log2 (1 + SNR) = 106 log2 (1 + 63) = 10

6 log2 (64) = 6 Mbps

Then we use the Nyquist formula to find thenumber of signal levels.

6 Mbps = 2 1 MHz log2L L = 8

First, we use the Shannon formula to find our upper

limit.


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The Shannon capacity gives us the

upper limit; the Nyquist formula tells ushow many signal levels we need.

Note


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Transmission Impairments


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Transmission Impairments

Signals travel through transmission media,

which are not perfect. The imperfection

causes signal impairment. This means that

the signal at the beginning of the medium isnot the same as the signal at the end of the

medium. What is sent is not what is received.

Three causes of impairment areattenuation,

distortion, andnoise.


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Example

Suppose a signal travels through

transmission medium and its power is

reduced to one half. Calculate attenuation

loss?


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Performance

Bandwidth

Throughput

Latency (Delay)

Bandwidth-Delay Product


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Throughput


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Latency

Latency = propagation time

+

transmission time

+

queuing time

+

processing time


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Propagation Time


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The bandwidth-delay product defines

the number of bits that can fill the link.

Note


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Physical Layer Topics to Cover

Signals

Digital Transmission

Analog Transmission

Multiplexing

Transmission Media


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Analog To Digital

Conversion


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Sampling

Pulse Code Modulation

Sampling Rate: Nyquist Theorem


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PCM


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According to the Nyquist theorem, thesampling rate must be

at least 2 times the highest frequencycontained in the signal.

Note


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Transmission Modes


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Line Coding & Decoding


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Signal Levels (Elements) Vs Data Levels

(Elements)


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Data Rate Versus Pulse Rate

Number of bits per second = Data Rate

Number of signal elements per second =

Pulse Rate


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Pulse Rate Vs Bit Rate

Example

A signal has two data levels with a pulse duration of 1

ms. We calculate the pulse rate and bit rate as follows:

Pulse Rate = 1/ 10-3= 1000 pulses/s

Bit Rate = Pulse Rate x log2 L = 1000 x log2 2 = 1000 bps


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DC Component

k f h


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Lack of Synchronization


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Example 3

In a digital transmission, the receiver clock is 0.1 percentfaster than the sender clock. How many extra bits per

second does the receiver receive if the data rate is 1

Kbps? How many if the data rate is 1 Mbps?

Solution

At 1 Kbps:

1000 bits sent1001 bits received1 extra bpsAt 1 Mbps:

1,000,000 bits sent1,001,000 bits received1000 extra bps

Li C di S h


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Line Coding Schemes

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