Data Communication
Prof. Dr. Hassan Al-Mahdi
Lecture 5
▪ Bit Rate Limits of Digital Signal
▪ Performance
From chapter 3
Signal Element and Data Element
Data element: a piece of information (bit) need to send
signal element : shortest time unit of a digital signal
represent one or more bits that we can send
Data elements: are being carried;
Signal elements are the carriers
r: a ratio define the number of data elements carried by signal
element.
(r = number of data elements / number of signal elements)
Data Rate Versus Signal Rate depend
• N:
on r value
Data rate : number of data elements (bits)/1s (bps)
called bit rate
• S: Signal rate: the number of signal elements sent in
1sec called baud rate
• baud rate = signal rate = pulse rate= modulation rate
Relationship between data rate & Baud rate depends on:
1- the value of r.
2-the data pattern.
Transmitting data of all 1s or all 0s may need baud rate
different from other data pattern.
The formula of the relationship between data rate and
signal rate S = c x N x 1/r
Data Rate Versus Signal Rate depend r value
The formula of the relationship between data rate and
baud rate S = c x N x 1/r (S = Bmin)
where :
S: the number of baud (signal) rates (minim bandwidth)
N : the date rate (bit rate) bps;
r :the number of data elements /number of signal element
(baud tare)
c: the case factor, which varies depend on 3 cases :
The best case: when we need the minimum signal rate
The worst case: when we need the maximum signal rat
The average case factor: usually used c=1/2
Note:
Decreasing signal rate decreases the required4-5bandwidth.
Relationship between data (bit) rate and signal rate
Example:
A signal is carrying data in which one data
element is encoded as one signal element. If the bit
rate is 100 kbps, what is the average value of the
baud rate if c is between 0 and l?
Solution
Assume the average value of c =1/2.
The baud rate is :
S =c x N x 1/r = 1/2 x 100,000 x 1/1 = 50,000 =50 kbaud
Relation between maximum bit rate and signal rate
• The minimum bandwidth: Bmin = cx N x 1/r
• The maximum bit rate (i.e. capacity): Nmax = 1/c x B x r
• The maximum data rate of a channel defined by the
Nyquist formula is C = 2 x B x log2L.
Does this agree with the previous formula for Nmax?
Answer
Yes . Let c=1/2
So, we have: Nmax = C = 1/c x B x r = 2 x B x log2L (prove)
If L=4 then log2L = log2 4 =2 this mean each signal
element carry 2 bits then r= 2. so equality is correct
DATA RATE LIMITS
• A very important consideration in data
communications is how fast we can send data, in
bits per second (bps), 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)
Increase the number of signal levels increases the bit rate.
but,
the probability of an occurring error also increase.
This lead to reduces the reliability of the system. Why??
Capacity of a System
◼ A symbol of signal may carry a single bit or “n” bits.
◼ The number of signal levels = 2n.
◼ As the number of levels increased the spacing between
level decreases. So, the probability of error is increased
due to the presence of transmission impairments.
Threshold
Level
Nyquist Theorem (Noiseless channel)
◼ Nyquist Theorem calculate the upper bound of the bit rate
in the transmission system from the number of bits in a
symbol (or signal levels) and the bandwidth of the system
in the case of noiseless channel
◼ The Nyquist bit rate formula defines the theoretical
maximum bit rate (C)
L=2n= number of signal levels
n=number of bits/symbol
C= capacity in bps = upper bit rate
B = bandwidth in Hz
Example
Consider a noiseless channel with a bandwidth of 3000
Hz transmitting a signal with two signal levels (one bit/
signal level). The maximum bit rate can be calculated as
Example
Consider the same noiseless channel transmitting a
signal with four signal levels (for each level, we send 2
bits). The maximum bit rate can be calculated as
Example
We need to send 265 kbps over a noiseless channel with
a bandwidth of 20 kHz. How many signal levels do we
need?
Solution
We can use the Nyquist formula as shown:
❖ Since this result is not a power of 2, we need to either increase the number of
levels or reduce the bit rate.
❖ If we have 128 levels, the bit rate is 280 kbps.
❖ If we have 64 levels, the bit rate is 240 kbps.
Shannon’s Theorem (noisy channel)
◼ Shannon’s theorem gives the capacity (upper bound of bit
rate) of a system in the presence of noise.
C = B log2(1 + SNR)
Example
The signal-to-noise ratio is often given in decibels. Assume that
SNRdB = 36 and the channel bandwidth is 2 MHz. The theoretical
channel capacity can be calculated as
Example
Consider an extremely noisy channel in which the value of
the signal-to-noise ratio is almost zero.
In other words, the noise is so strong that the signal is
faint (very weak). What is capicty of this channel
This means that the capacity of this channel is zero
regardless of the bandwidth.
In other words, we cannot receive any data through this
channel.
Example
We can calculate the theoretical highest bit rate of a
regular telephone line.
A telephone line normally has a bandwidth of 3 KHz. The
signal-to-noise ratio is usually 3162.
For this channel the capacity is calculated as
This means that the highest bit rate for a telephone line is 34.860 kbps.
If we want to send data faster than this, we can either increase the bandwidth
of the line or improve the signal-to-noise ratio.
Example
We have a channel with a 1-MHz bandwidth. The SNR for
this channel is 63. What are the appropriate bit rate and
signal level?
Solution
First, we use the Shannon formula to find the upper limit.
The Shannon formula gives us the upper limit of 6 Mbps,. Find the
number of signal levels.
For better performance we choose something lower, 4 Mbps, for
example.
Then we use the Nyquist formula to find the number of signal levels.
Example
For practical purposes, when the SNR is very high, we can
assume that SNR + 1 is almost the same as SNR. In these
cases, the theoretical channel capacity can be simplified to
calculate the theoretical capacity of channel with a 2-MHz
bandwidth and that SNRdB =36.
Note
The Shannon capacity gives us the upper limit;
the Nyquist formula tells us how many signal
levels we need.
PERFORMANCE
One important issue in networking is the performance of
the network: how good is it? We discuss quality of service
(QoS) an overall measurement of network performance.
QoS measured through bandwidth, throughput, Delay
times
Note 1- Bandwidth: In networking, we use the
term bandwidth in two contexts.
▪ The first, bandwidth in hertz: when refers to the range of frequencies
in analogue signals of the range of frequencies that a channel can
pass.
▪ The second, bandwidth in bits/second: when refers to the speed of bit
transmission in a channel or link. Often referred to as Capacity.
PERFORMANCE
Relationship
- Bandwidth in hertz and bandwidth in bits per second are related.
- Increasing bandwidth in hertz leads to an increase in bandwidth in
bits per second.
- The specific relationship between them depends on the type of
transmission, whether it's baseband or with modulation.
Example
The bandwidth of a subscriber line is 4 kHz for voice or data.
For data transmission. This line Can carry be up to 56 kbps using a
sophisticated modem to change the digital signal to analog.
Example
If the telephone company improves the quality of the line and
increases the bandwidth to 8 kHz,
we can send 112 kbps by using the same technology of previous
Example.
2- Throughput
what is amount of Data can be get out using the available
Bandwidth
A network with bandwidth of 10 Mbps can pass only an average of
12,000 frames per minute.
Each frame carrying an average of 10,000 bits.
What is the throughput of this network?
Solution
We can calculate the throughput as
The throughput is almost one-fifth of the
bandwidth in this case.
Propagation & Transmission Delay
◼ Propagation speed: speed at which a bit travels
though the medium from source to destination.
◼ Transmission speed: the speed at which all the
bits in a message arrive at the destination.
(difference in arrival time of first and last bit)
• Propagation Delay Time = Distance/Propagation speed
• Transmission Delay Time = Message size/bandwidth bps
• Latency = Propagation delay + Transmission delay +
Queueing time + Processing time
Example
What are the propagation time and the transmission time
for a 2.5-kbyte message if the bandwidth of the network is
1 Gbps? If the distance between the sender and the
receiver is 12,000 km and the bit travels at 2.4 ×108 m/s.
Solution
Propagation Delay Time = Distance/Propagation speed
Transmission Delay Time = Message size/bandwidth bps
Note:
Because the message is short and the bandwidth is high, the dominant
factor is the propagation time, not the transmission time. So, The
transmission time can be ignored.
Example
What are the propagation time and the transmission time for a 5-Mbyte message
if the bandwidth of the network is 1 Mbps? Assume that the distance between the
sender and the receiver is 12,000 km and that light travels at 2.4 × 108 m/s.
Solution
We can calculate the propagation and transmission times as shown on the next
slide.
Note:
Because the message is very long and the bandwidth is not very
high, the dominant factor is the transmission time, not the
propagation time. The propagation time can be ignored.
The bandwidth-delay product
We can think about the link between two points as a pipe. The cross
section of the pipe represents the bandwidth, and the length of the
pipe represents the delay.
We can say the volume of the pipe defines the bandwidth-delay
product.
The bandwidth-delay product defines the
number of bits that can fill the link.
❖Bandwidth-Delay Product
❖Link bandwidth: 1 bps (bits
per second). Figure: Filling the link with bits for case 1
❖Link delay: 5 seconds
❖what the bandwidth-delay
product means in this case.
❖Bandwidth-delay product: 1
bps × 5 s = 5 bits.
❖This means, the maximum
number of bits that can fill the
link no more than 5 bits.
❖Bandwidth-Delay Product
❑ Now assume we have a bandwidth of 5
bps. There can be maximum 5 × 5 = 25
Figure: Filling the link with bits for case 1
bits on the line.
❑ At each second, there are 5 bits on the
line; the duration of each bit is 0.20 s.
❖ Jitter
❑ Jitter is a problem if different packets of data encounter
different delays and the application using the data at the
receiver site is time-sensitive (audio and video data, for
example).
❑ If the delay for the first packet is 20 ms, for the second is
45 ms, and for the third is 40 ms, then the real-time
application that uses the packets endures jitter.
❖Jitter
❑Jitter is the variation in the arrival time of data packets or signals.
❑It can cause inconsistent delays in the delivery of data.
❑Jitter is constitute a problem for real-time applications like VoIP
calls, video streaming, and online gaming.
To reduce or solve jitter in data communication:
- Implement Quality of Service (QoS) for prioritizing time-sensitive traffic.
- Use buffering to smooth out variations in packet arrival times.
- Consider error correction techniques.
- Increase available bandwidth.
- Ensure time synchronization.
Thanks
