Upload
rrs1988
View
228
Download
0
Embed Size (px)
Citation preview
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
1/32
DATA RATE LIMITS
A very important consideration in data communications is
how fast we can send data, in bits per second, over a
channel.
Data rate depends on three factors:
1. The bandwidth available
2. The level of the signals we use
3. The quality of the channel (the level of noise)
1
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
2/32
Conti
Two theoretical formulas were developed to calculate thedata rate:-
1. By Nyquist for a noiseless channel
2. By Shannon for a noisy channel
2
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
3/32
Two digital signals: one with two signal levels and the other
with four signal levels
3
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
4/32
Noiseless Channel : Nyquist Bit rate
For a noiseless channel, the Nyquist bit rate formula definesthe theoretical maximum bit rate
Bitrate = 2* bandwidth * log2L
Where
Bandwidth is the bandwidth of the channel.
L is the number of signal levels used to represent data.
BitRate is the bit rate in bits per second.
4
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
5/32
Conti
According to a formula, we can have any bit rate byincreasing the number of signal levels.
It is theoretically correct but practically there is limit.
When we increase the number of signal levels, we impose
the burden on receiver.
If the number of levels in a signal is just 2 then the receiver
can easily distinguish between 0 and 1.
5
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
6/32
Conti
If the level of signal is 64 then receiver must be able todistinguish between 64 different levels.
IMP Note:-
Increasing the levels of a signal reduces the reliability of the
system.
6
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
7/32
Example1
Consider a noiseless channel with a bandwidth of 3000 Hztransmitting a signal with two signal levels. The maximum bit
rate can be calculated as
Bit Rate = 2 3000 log22 = 6000 bps
7
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
8/32
Example2
Consider the same noiseless channel, transmitting a signalwith four signal levels (for each level, we send two bits). The
maximum bit rate can be calculated as:
Bit Rate = 2 x 3000 x log24 = 12,000 bps
8
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
9/32
Noisy Channel: Shannon Capacity
In reality we do not have noiseless channel : the channel is
always noisy.
In 1944, Shannon introduces a formula called the Shannon
capacity to determine the theoretical highest data rate for a
Noisy Channel:
Capacity=Bandwidth X log2(1+SNR)
Where
Bandwidth is the bandwidth of the channel.
SNR is the signal-to-noise ratio and
Capacity is the capacity of the channel in bits per second.
9
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
10/32
Conti
In Shannon formula there is no indication of a signal level.
This means that no matter how many levels we have, we can
not achieve a data rate higher than the capacity of the
channel.
10
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
11/32
Example1
Consider an extremely noisy channel in which the value ofthe signal-to-noise ratio is almost zero. In other words, the
noise is so strong that the signal is faint. For this channel the
capacity is calculated as
C = B log2(1 + SNR)
= B log2(1 + 0)
= B log2(1)
= B 0= 0
11
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
12/32
Example2
We can calculate the theoretical highest bit rate of a regulartelephone 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
12
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
13/32
Example3
We have a channel with a 1 MHz bandwidth. The SNR for thischannel is 63; what is the appropriate bit rate and signal
level?
First, we use the Shannon formula to find our upper limit.
C = B log2(1 + SNR) = 106log2(1 + 63) = 106log2(64) = 6 Mbps
Then we use the Nyquist formula to find the
number of signal levels.
6 Mbps = 2 1 MHz log2L L = 8
13
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
14/32
Conti
Note:-
The Shannon capacity gives us the upper limit
Whereas
The Nyquist formula tells us how many signal levels
we need.
14
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
15/32
Performance
One important issue in networking is the performance of thenetworkhow good is it?
1. Bandwidth
2. Throughput3. Latency (Delay)
4. Bandwidth-Delay Product
5. Jitter
15
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
16/32
1. Bandwidth
One characteristics that measures network performance isbandwidth.
However the term can be used in two different context with
two different measuring values:
1. Bandwidth in Hertz
2. Bandwidth in bits per second
16
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
17/32
Bandwidth in Hertz
Bandwidth in Hertz is the range of frequencies contained in acomposite signal or the range of frequencies a channel can
pass.
Bandwidth of a subscriber telephone line is 4kHz.
17
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
18/32
Bandwidth in bits per second
Bandwidth can also refer to the number of bits per secondthat a channel, a link or even a network can transmit.
Eg.
Bandwidth of a fast Ethernet network is a maximum of
100mbps.
This means that the network can send 100mbps.
18
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
19/32
Relationship between B in Hz and B in bps
There is a explicit relationship in Bandwidth in Hertzand Bandwidth in bits per second.
Basically an increase in Bandwidth in Hertz means an
increase in Bandwidth in bits per second.
19
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
20/32
2. Throughput
Throughput is a measure of how fast we can actually senddata through a network.
Network throughput is measured in bits per second (bps).
Throughput and bandwidth are different.
Eg. A link may have a bandwidth of B bps, but we can only
send T bps through this link with T always less than B.
In other words, badwidth is the potential measurement of a
link. 20
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
21/32
Conti
Throughtput is the actual measurement of how fast we cansend data.
E.g.
We may have a link with a bandwidth of 1Mbps, but thedevices connected to the end of the link may handle only
200kbps.
This means that we can not send more than 200kbps throughthis link.
21
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
22/32
Conti
22
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
23/32
3. Latency[Delay]
The latency or delay defines how long it takes for an entiremessage to completely arrive at the destination from the
time the first bit is sent out from the source.
Latency is made of four components:-
A. Propagation Time
B. Transmission Time
C. Queuing Time
D. Processing Time
Latency=Propagation time + Transmission time + Queuing Time
+ Processing Time
23
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
24/32
A. Propagation Time
Propagation time measures the time required for a bit totravel from the source to the destination.
Propagation Time= Distance/Propagation Speed
The propagation speed of electromagnetic signals depends
on the medium and the frequency of the signal.
E.g.
In a vacuum , light is propagated with a speed of 3*10m/s.
It is lower in air; much lower in cable.
24
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
25/32
B. Transmission Time
In data communication we dont send just 1 bit, we send amessage.
The first bit may take a time equal to the propagation time to
reach its destination; the last bit may also take the same
amount of time.
However , there is the time between the first bit leaving the
sender and the last bit arriving the receiver.
25
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
26/32
Conti
The first bit leaves earlier and arrives earlier and the last bitleaves later and arrives later.
The time required for transmission of a message depends on
the size of the message and the bandwidth of the channel.
Transmission time=Message size/Bandwidth
26
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
27/32
C. Queuing Time
Queuing Time is the time needed for each intermediate orend device to hold the message before it can be processed.
It is not a fixed factor.
It changes with the load imposed on the network.
When there is heavy traffic on the network, the queuing time
Increases.
27
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
28/32
Conti
An intermediate device such as router, queues the arrivedmessages and processes them one by one.
If there are many messages, each message have to wait.
28
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
29/32
29
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
30/32
30
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
31/32
31
8/13/2019 Chapter 3 - The Physical Layer-Data Rate Limits
32/32
32