MD03 Data and Signals

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CNT 3004

Module 3

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Chapter 3

Data and Signals

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Chapter 3

IntroductionTo

PhysicalLayer

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To be transmitted, data must be transformed to electromagnetic

signals.

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

signals can have only a limited number of values.

Analog and Digital

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Comparison of analog and digital signals

The curve representing the analog signal is passing through an infinite number of points.

The vertical lines of the digital signal demonstrates the sudden jump the signal makes from value to value.

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Periodic and Nonperiodic Signals

A periodic signal completes a pattern within ameasurable time frame, called a period, andrepeats that pattern over subsequent identicalperiods. The completion of one full pattern iscalled a cycle.

A nonperiodic (or aperiodic) signal changes withoutexhibiting a pattern or cycle that repeats overtime.

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Periodic Analog Signals

Periodic analog signals can be classified assimple or composite. A simple periodic analogsignal, a sine wave, cannot be decomposed intosimpler signals. A composite periodic analog signalis composed of multiple sine waves.

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A sine wave

A signal is periodic if it consists of a continuously repeating pattern. A sine wave is an example of a periodic and continuous signal.

A signal is called aperiodic (or nonperiodic) if it changes without exhibiting a pattern or a cycle that repeats over time.

Example:Value= 5 Sin (6 t )

Value= A Sin (2 f t + ) where t is time

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The peak amplitude of a signal represents the absolute value of its highest intensity, proportional to the energy it carries.

Amplitude

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Period and frequency

Each sine wave can be characterized by its amplitude A, frequency f, and phase .

Frequency and period are inverses of each other; as frequency increases, the period decreases (and vice versa)

A periodic signal with a frequency of 6 Hz

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Period and frequency

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Frequency f and period T are inverses of each other.

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Units of periods and frequencies

Period is formally expressed in seconds

Frequency is formally expressed in hertz (Hz)

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In data communication, we commonly use periodic analog signals and

aperiodic digital signals.

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The power we use at home has a frequency of 60 Hz. Theperiod of this sine wave can be determined as follows:

This means that the period for our lights at home is 0.0166 second or 16.6 milliseconds. Our eyes are not sensitive enough to distinguish these rapid changes in amplitude.

Example

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Example Express a period of 100 ms in microseconds, and express the corresponding frequency in kilohertz.

SolutionThe equivalent of 1 ms is 10-3 s. We make the following substitutions:100 ms = 100 10-3 s = 100 10-3 10 s = 105 s

Now we use the inverse relationship to find the frequency, changing hertz to kilohertz100 ms = 100 10-3 s = 10-1 s f = 1/10-1 Hz = 10 10-3 KHz = 10-2 KHz

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Frequency is the rate of change with respect to time. Change in a short span of

time means high frequency. Change over a long span of time means low frequency.

If a signal does not change at all, its frequency is zero. So a constant value can be considered a sine wave whose frequency is zero. For example, The

voltage of an AA battery is constant which is considered a sine wave with a peak value of 1.5 V

and a frequency of zero.

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Phase

The term phase, or phase shift, describes theposition of the waveform relative to time 0. If wethink of the wave as something that can be shiftedbackward or forward along the time axis, phasedescribes the amount of that shift. It indicates thestatus of the first cycle.

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Relationships between different phases

Phase is measured in degrees or radians. The unit of radian is also known as rad. We have the following relationship

2 rad = 360 degrees

1 rad = 360/(2) = 57.27 degrees

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Three sine waves with different phases

= /2 rad

= rad

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Example

A sine wave is offset one-sixth of a cycle with respect to time zero. What is its phase in degrees and radians?

Solution

We know that one complete cycle is 360 degrees.

Therefore, 1/6 cycle is

(1/6) 360 = 60 degrees = 60 x 2 /360 rad = 1.046 rad

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Sine wave example 1

S(t) = A Sin (2 f t + )

S(t) = 5 Sin (25.1428 t )

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Sine wave example 2

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Horizontal Shifting: 0

S(t) = A Sin (2 f t + )

Sine wave example 3

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We have been showing a sine wave by using what is called a time domain plot. The time-

domain plot shows changes in signal amplitude with respect to time (it is an

amplitude-versus-time plot).

An analog signal is best represented in the frequency domain.

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Time and frequency domains

In a frequency-domain plot, the horizontal axis measures the frequency.

In a time-domain plot, the horizontal axis is a measure of time.

When f= 0 S(t) = A Sin (2 f t+/2 ) = A sin (/2) = A

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Time and frequency domains (continued)

A time-domain graph plots amplitude as a function of time.

A frequency-domain graph plots each sine wave’s peak amplitude against its frequency.

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Time and frequency domains (continued)

Frequency can be easily determined from a frequency-domain graph of a signal.

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The time-domain and frequency-domain plots of a sine wave

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Example

The frequency domain is more compact and useful when we are dealing with more than one sine wave.

For example, the figure on the next slide shows three sine waves, each with different amplitude and frequency. All can be represented by three spikes in the frequency domain.

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The time domain and frequency domain of three sine waves

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A single-frequency sine wave is not useful in data communications; we need to change one or more

of its characteristics to make it useful.

When we change one or more characteristics of a single-frequency signal, it becomes a composite

signal made of many frequencies.

According to Fourier analysis, any composite signal can be represented as a combination of simple sine waves with different frequencies, phases, and amplitudes. A composite signal is composed of more than one simple sine wave.

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If the composite signal is periodic, the decomposition gives a series of simple sine waves with discrete frequencies; if the composite signal is nonperiodic, the decomposition gives a combination of sine waves with continuous frequencies.

When a periodic signal is decomposed, the resulting sine waves are called harmonics. The sine wave with the smallest frequency is called the first harmonic and its frequency is called the fundamental frequency f. The frequencies of the other sine waves are integral multiples of the fundamental frequency. For example, the third harmonic has frequency 3fand the ninth harmonic has frequency 9f. When decomposing a periodic signal, there are no frequencies with fractional values such as 3.2f.

Decomposition of Composite Signals

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Example: Adding three harmonics

A Composite Periodic Signal

The above figure shows a periodic composite signal with fundamental frequency f. It is the combination of three simple sine waves with frequencies f, 3f and 9f as shown on the next slide. Thus this signal is decomposed into the first, third and ninth harmonics.

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Decomposition of a composite periodic signal in the time and frequency domains

Frequency

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Square wave

According to Fourier Analysis, this periodic square signal can be decomposed into a series of discrete sine harmonics.

The three odd harmonics f, 3f, 9f gave a signal that has the general shape of a square wave. We shall see later that to get a square wave, we need to use all odd harmonics: f, 3f, 5f, 7f, 9f,11f, …

signal with f, 3f, 9f harmonics

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The time and frequency domains of a nonperiodic signal

The above figure shows a nonperiodic composite signal. It can be the signal created by a microphone or a telephone set when a word or two is pronounced. Voice composite signals are usually nonperiodic, because we do not usually keep repeating the same word or words with exactly the same tone.

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Bandwidth

The bandwidth is a property of a medium. The bandwidth of a medium is the difference between the highest and the lowest frequencies that the medium can satisfactorily pass (i.e., these frequencies are permitted to pass through this medium).

The bandwidth is also a property of a signal. The bandwidth of a signal is the difference between the highest and the lowest frequencies of the simple sine waves that compose this signal. Thus a simple sine wave has a bandwidth of zero. For example, if a composite signal contains frequencies between 1000 and 5000, its bandwidth is 5000 − 1000, or 4000 Hz.

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Bandwidth

The bandwidth of a signal is the range of frequencies the signal occupies. Bandwidth is determined by finding the difference between the highest and lowest frequency components.

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The bandwidth of periodic and nonperiodic composite signals

A periodic signal is decomposed into a finite number of sine waves with discrete frequencies.

A nonperiodic signal is decomposed into an infinite number of sine waves with continuous frequencies.

The periodic signal in part (a) has a bandwidth of 4000 Hz and is decomposed into a finite number of harmonics. The nonperiodic signal in part (b) also has a bandwidth of 4000 Hz but it is decomposed into an infinite number of sine waves.

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If a periodic signal is decomposed into five sine waves with frequenciesof 100, 300, 500, 700, and 900 Hz, what is its bandwidth? Draw thespectrum diagram (frequency domain plot), assuming all componentshave a maximum amplitude of 10 V.SolutionLet fh be the highest frequency, fl the lowest frequency, and B thebandwidth. Then

Example

The spectrum has only five spikes, at 100, 300, 500, 700, and 900 Hz.

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ExampleA periodic signal has a bandwidth of 20 Hz. The highest frequency is 60 Hz. What is the lowest frequency? Draw the spectrum if the fundamental frequency is 1 Hz and the signal contains all integral frequencies of the same amplitude. What is the number of harmonics of this composite signal?

Solution B = fh fl 20 = 60 fl

fl = 60 20 = 40 Hz

# of harmonics = 60 - 40 + 1 = 21

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Example

A signal has a spectrum with frequencies between 1000 and 2000 Hz (bandwidth of 1000 Hz). A medium can pass frequencies from 3000 to 4000 Hz (a bandwidth of 1000 Hz). Can this signal faithfully pass through this medium?

SolutionThe answer is definitely no. Although the signal and the medium have the same bandwidth (1000 Hz), the range does not overlap. The medium can only pass frequencies between 3000 and 4000 Hz; the signal is totally lost.

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ExampleA nonperiodic composite signal has a bandwidth of 200 kHz, with a middle frequency of 140 kHz and peak amplitude of 20 V. The amplitude changes linearly and the two extreme frequencies have an amplitude of 0. Draw the frequency domain of the signal.

SolutionThe lowest frequency must be at 40 kHz and the highest at 240 kHz. The figure below shows the frequency domain and the bandwidth.

20 V

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ExampleAn example of a nonperiodic composite signal is the signal propagated by an AM radio station. In the United States, each AM radio station has a bandwidth of 10 kHz bandwidth.

The AM spectrum assigned by FCC ranges from 530 KHz to 1700 kHz.

ExampleAnother example of a nonperiodic composite signal is the signal propagated by an FM radio station. In the United States, each FM radio station has a bandwidth of 200 kHz = 0.2 MHz.

The FM spectrum assigned by FCC ranges from 88 MHz to 108 MHz for a total FM bandwidth of 20 MHz covering 101 radio channels.

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A digital signal with two levels; each level represents one bit

Digital Signals

In addition to being represented by an analog signal, information can alsobe represented by a digital signal. For example, a 1 can be encoded as apositive voltage and a 0 as zero voltage. A digital signal can have more thantwo levels. In this case, we can send more than 1 bit for each level.

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Bit rate and bit interval

The bit rate is the number of bits transmitted per second. The bit interval is the duration needed to transmit one bit.

A digital signal is described by its bit rate (instead of frequency) and its bit interval.

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Two digital signals: one with two signal levels and the other with four signal levels

A digital signal with 2 levels; each level represents one bit

bit rate = 1 x 8 = 8 bps

A digital signal with 4 levels; each level represents two bits

bit rate = 2 x 8 = 16 bps

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Example A digital signal has eight levels. How many bits can berepresented by one level?Solution:

We calculate the number of bits from the formula

Each signal level represents 3 bits. This means thatwhen the receiver receives a signal with certain level, ithas received the value of three consecutive bits.

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The analog bandwidth of a medium is expressed in hertz (recall that bandwidth is the difference between the highest and the

lowest frequencies that the medium can pass)

The digital bandwidth of a medium is its bit rate and is expressed in bits per second. The bit rate depends on the

encoding scheme used to transmit data.

ExampleAssume we need to download text documents at the rate of 100 pages per second. A page is an average of 24 lines with 80 characters in each line and each character requires 8 bits. What is the required bit rate of the channel?Solution

The bit rate is

Nyquist Sampling Theorem

Consider a composite analog signal whose highest frequency is fh =W. In the above figure, W= 4000 Hz. Rather than transmitting the signal in analog form, we can send the signal in a digitized form by sampling the amplitude of the signal at a series of points spaced at equal intervals.

To reconstruct the digitized signal successfully, the sampling rate must be at least two times the highest frequency contained in the signal. For the signal shown in the above figure, the sampling rate should t be equal to 2W = 24000 = 8000 samples/second.

Example:

For an intuitive example of the Nyquist theorem, letus sample a simple sine wave with frequency f atthree sampling rates:

• fs = 2f = Nyquist sampling rate • fs = 4f = two times the Nyquist sampling rate • fs = f = one-half the Nyquist sampling rate

The figure on the next page shows the sampling andthe subsequent recovery of the signal.

Recovery of a sampled sine wave for three different sampling rates

•fs = f = one-half the Nyquist sampling rate

• fs = 2f = Nyquist sampling rate

•fs = 4f = two times the Nyquist sampling rate

1. It can be seen that sampling at the Nyquist rate can create a good approximation of the original sine wave (part a).

2. Oversampling at double the Nyquist rate (part b) can create a better approximation or the same approximation, but it requires more storage and is often redundant and unnecessary.

3. Sampling below the Nyquist rate (part c) does not produce a signal that looks like the original sine wave.

Conclusions: part a

part b

part c

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ExampleA digitized voice channel is made by digitizing a 4-kHz bandwidth analog voice signal. We need to sample the signal at twice the highest frequency (two samples per hertz). We assume that each sample requires 8 bits. What is the required bit rate?SolutionThe bit rate can be calculated as

ExampleHDTV uses digital signals to broadcast high quality video signals. The HDTV screen is normally a ratio of 16 : 9. This ratio is achieved by having 1920 by 1080 pixels per screen, and the screen is renewed 30 times per second. Twenty-four bits represents one color pixel. What is the bit rate for high-definition TV (HDTV)?SolutionThe bit rate is

The TV stations reduce this rate to 20 - 40 Mbps through compression.

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The time and frequency domains of periodic and nonperiodic digital signals

A digital signal is a composite signal with an infinite bandwidth.

A periodic signal is decomposed into a finite number of sine waves with discrete frequencies.

A nonperiodic signal is decomposed into an infinite number of sine waves with continuous frequencies.

This square signal is decomposed into odd harmonics.

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Baseband transmission

Baseband transmission means sending a digital signal directly over a channel without changing it to an analog signal.

According to Fourier analysis, a digital signal is a composite analog signal with an infinite bandwidth (frequencies between 0 to ).

Digital signals used in data communications are mostly nonperiodic. We can transmit a digital signal in two ways: baseband transmission or broadband transmission.

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Requirement of Baseband Transmission

To preserve the shape of the digital signal, baseband transmission requires the following low-pass channel, i.e., a channel that passes frequencies between 0

and f. The lowest frequency of a low-pass channel is zero. the low-pass channel must be a wide bandwidth channel (i.e., f is

infinite or is very large).

A narrow-bandwidth low-pass channel is not suitable for digital transmission

A wide-bandwidth low-pass channel is suitable for digital transmission

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Baseband transmission using a dedicated medium

Consider a dedicated medium (e.g., coaxial cable or fiber optic) with a very wide bandwidth (the minimum frequency f1 is close to zero and the maximum frequency f2 is very high). Baseband transmission using this medium can be used to deliver the digital signal with good accuracy. The digital data can be correctly deduced from the received signal even though it is not an exact replica of the original signal.

In baseband transmission, the required bandwidth is proportional to the bit rate; if we need to send bits faster, we need more bandwidth.

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Low-pass and band-pass ChannelsA channel or link is either low-pass: if the lowest frequency f1 is zero (or practically

very close to zero)band-pass: if the lowest frequency is larger than zero

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Broadband Transmission (Using Modulation)

If the available channel is a band-pass channel, we cannot send the digital signal directly to the channel; we need to convert the digital signal to an analog signal before transmission. This process is called modulation.

Band-pass channels cannot be used to transmit digital signals with good quality. Digital

transmission needs a low-pass channel with wide bandwidth.

Analog transmission, on the other hand, can easily use a band-pass channel.

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Broadband Transmission (Using Modulation)

The figure on the next slide shows the modulation of a digital signal. The digital signal is first converted to a composite analog signal. In the shown scheme, a single-frequency analog signal (called the carrier) is used. The amplitude of the carrier has been changed to look like the digital signal. The result is a composite analog signal that is capable to propagate smoothly through the medium. At the receiver, the received analog signal is converted to digital signal and the result is a replica of what has been sent. This type of modulation is called Amplitude Modulation.

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Modulation of a digital signal for transmission on a band-pass channel

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Examples of Amplitude Modulation

Data

Modulated Signal 1 is high amplitude0 is zero amplitude

1 1 1

0 0 0

Modulated Signal 1 is high amplitude0 is low amplitude

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Transmission Impairment

Attenuation

Distortion

Noise

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Signal corruption

No transmission media is perfect; each medium passes some frequencies, weakens other, and blocks still others.

Causes of impairment

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Attenuation

Attenuation is the loss of a signal’s energy (strength) due to the resistance of the transmission medium.

To show that a signal has lost or gained strength, we use the unit of decibel (dB) which measures the relative strength of two signals or one signal at two different points. If P1 and P2 are the powers of a signal at Points 1 and 2, then

attenuation in dB = 10 log10 (P2/P1)

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ExampleImagine a signal travels through a transmission medium and its power is reduced to half. This means that P2 = 0.5 P1. In this case, the attenuation (loss of power) can be calculated as

10 log10 (P2/P1) = 10 log10 (0.5P1/P1) = 10 log10 (0.5) = 10(–0.3) = –3 dB

ExampleImagine a signal travels through an amplifier and its power is increased ten times. This means that P2 = 10 P1. In this case, the amplification (gain of power) can be calculated as

10 log10 (P2/P1) = 10 log10 (10P1/P1)

= 10 log10 (10) = 10 (1) = 10 dB

a loss of 3 dB implies losing one half the power

Remark: when the attuation is too large, the signal at the reciever is totally lost, i.e., P20.10 log10 (P2/P1) = 10 log10 (0) = - dB

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ExampleDecibel numbers can be added (or subtracted) when we are dealing with several points instead of just two. In the figure below, a signal travels a long distance from point 1 to point 4. The signal is attenuated by the time it reaches point 2. Between points 2 and 3, the signal is amplified. Again, between points 3 and 4, the signal is attenuated. We can find the resultant decibel for the total attenuation of the signal just by adding the decibel measurements between each set of points.

dB = –3 + 9 – 3 = +3

9

Note: +3 dB implies doubling the power of the signal while -3 dB implies losing one half the power

3 dB

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Distortion means that the signal changes its form or shape. Distortion can only occur in a composite signal made of different frequencies. The different frequencies have different propagation speeds and therefore different delays in arriving to the destination.

If the differences in delay are not exact multiple of the period, phase differences are created. This leads to a change in the shape of the received composite signal.

Distortion (Propagation Distortion)

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Noise

Noise is the external energy that corrupts a signal. Thermal noise is the extra signal created by the random motion of

electrons in the wire. Induced noise is produced by external sources such as motors and

appliances. Crosstalk noise is the effect of one wire on the other. One wire

acts as a sending antenna that corrupts the signal in the other wire. Impulse noise is a spike that comes from power lines or lightning.

signal + noisesignal

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Signal-to-Noise Ratio (SNR)The metric used to measure noise is called the signal-to-noise ratio (SNR).

SNR is defined as

SNR = (average signal power)/(average noise power)

The larger the value of SNR the better the quality of the signal. Note that SNR is the ratio between two power (energy) values. If however the signal and noise are described in terms of voltage values (rather than power values), then the correct value of SNR should be computed as the ratio between the square of the voltage values.

Because SNR is the ratio between two powers, it is often described in decibels.

SNR dB = 10 log10 SNR

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Two cases of SNR: a high SNR and a low SNR

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The power of a signal is 10 mW and the power of the noise is 1 μW;what are the values of SNR and SNRdB ?SolutionThe values of SNR and SNRdB can be calculated as follows:

Example

The values of SNR and SNRdB for a noiseless channel areExample

We can never achieve this ratio in real life; it is an ideal value.

μ

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Data Rate Limit

Noiseless Channel: Nyquist Bit Rate

Noisy Channel: Shannon Capacity

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Nyquist Bit Rate for Noiseless Channels The Nyquist formula determines the theoretical data

rate for a noiseless channel.

BitRate = 2 x BW x log2 L

where BW is the bandwidth of the channel in Hz, L is the number of signal levels used to represent the data, and BitRate is the bit rate in bits per second (bps).

Increasing the number of signal levels L would help increase the bit rate but would also increase the complexity of the receiver and may reduce the reliability of the system.

Note: a noiseless channel has SNR =

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Shannon Capacity for Noisy Channels

The Shannon formula determines the theoretical maximum data rate for a noisy channel.

Capacity= C = BW x log2 (1 + SNR)

where BW is the bandwidth of the channel in Hz, SNR is the signal-to-noise ratio, and C is the capacity (maximum transmission speed) of the channel in bps.

The number of signal levels L does not appear in the Shannon formula and does not influence the theoretical maximum data rate of the channel.

When the signal is equal to the noise (SNR=1), the capacity of the noisy channel is equal to its bandwidth.

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ExampleConsider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. The maximum bit rate can be calculated asNyquist formula:

Bit Rate = 2 BW log2 L

= 2 3000 log2 2 = 6000 bps

ExampleConsider a noiseless channel with bandwidth of 3000 Hz, 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 3000 log2 4 = 12,000 bps

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ExampleConsider 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. For this channel the capacity C is calculated asShannon Formula

C = BW log2 (1+SNR)

= BW log2 (1+0)

= BW log2 (1)

= BW 0

= 0 This means that the capacity of this channel is zeroregardless of the bandwidth. In other words, we cannotreceive any data through this channel.

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ExampleWe can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000 Hz (300 Hz to 3300 Hz). The signal-to-noise ratio is usually 3162. For this channel the capacity is calculated as

Shannon FormulaC = BW log2 (1 + SNR)

= 3000 log2 (1 + 3162) = 3000 log2 (3163)= 3000 11.62

= 34,860 bps

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PERFORMANCE

One important issue in networking is the performance ofthe network—how good is it and how can we measure it?

BandwidthThroughputLatency (Delay)Bandwidth-Delay Product

Topics discussed in this section:

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In networking, we use the term bandwidth in two contexts.

❏ The first, bandwidth in hertz, refers tothe range of frequencies in acomposite signal or the range offrequencies that a channel can pass.

❏ The second, bandwidth in bits persecond, refers to the speed of bittransmission in a channel or link.

Bandwidth

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Throughput

Throughput is a measure of how fast we can send data through a network.

Throughput has units of bits per second (bps). Example: we may have a link of digital bandwidth BW=1

Mbps but the device connected on the other side of the link can only handle 200 kbps. Then the maximum throughput of this link is 0.2 Mbps which is one-fifth of the bandwidth.

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A network with bandwidth of 10 Mbps can pass only anaverage of 12,000 frames per minute with each framecarrying an average of 10,000 bits. What is thethroughput of this network?

SolutionWe can calculate the throughput as

Example

The throughput is almost one-fifth of the bandwidth inthis case.

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Propagation time

Propagation time is directly proportional to distance and inversely proportional to propagation speed.

The propagation time has units of seconds. When propagation speed is multiplied by propagation time,

we get the distance a signal or a bit has traveled. The propagation speed has units of meters/second or

kilometers/second.

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The bandwidth-delay product of a link defines the number of bits that can fill the link.

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 one-way delay. We can say the volume of the pipe defines the bandwidth-delay product.

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Bandwidth-Delay Product: Filling the link with bits

Case 1: the link has a bandwidth of 1 bps, the one-way delay of the link is 5 seconds. Note that the duration of the bit is 1 second.The product 15 is the maximum number of bits that can fill the link, i.e., there can be no more than 5 bits at any time on the link.

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Bandwidth-Delay Product: Filling the link with bits

Case 2: the link has a bandwidth of 5 bps, the one-way delay of the link is 5 seconds. Note that the duration of the bit is 0.2 second.The product 55 = 25 is the maximum number of bits that can fill the link, i.e., there can be no more than 25 bits at any time on the link.

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For a single link, the delay is mostly the one-way propagation delay, i.e., the time it takes for a bit of data to travel from the source to the destination. For multiple links, there are other types of delays such as queuing time and processing delays.

Latency = propagation time + transmission time + queuing time + processing delay

The queuing time is the time needed for each intermediate or end device to hold the message before it can be processed and transmitted. The value of the queuing time depends on the load on the network; when there is heavy traffic, the queuing time increases.

The end-to-end delay in a network, or latency, is the sum of the different types of delays.

Packet Latency

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Propagation time of a packet in one link =

(physical length of the link)/(speed of signal propagation)

The propagation time of a packet in a link does not depend on the size of the packet. The speed of signal propagation is the same as the speed of travel of the signal in the link.

Transmission time of a packet =

(size of packet)/(transmission speed of sender)

The transmission speed is the same as the transmission bandwidth (in bps).

Propagation Time and Transmission Time

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What is the total delay (latency) for a packet of size 50,000 bits that is being sent on a multiple-link path with 10 routers each having a queuing time of 2 ms and a processing time of 1 ms. The total length of all links is 2000 Km. The speed of light inside the links is 2 x 108 m/s. All links have a transmission bandwidth of 0.5 Mbps.

Example

SolutionWe have Latency = processing time + queuing time + transmission time + propagation timeProcessing time = 10 1 ms = 10 ms = 0.010 sQueuing time = 10 2 ms = 20 ms = 0.020 sTransmission time = 50,000 / (500,000 bps) = 0.1 sPropagation time = (2000 Km) / (2 108) = 0.01 sLatency = 0.010 + 0.020 + 0.1 + 0.01 = 0.140 second

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Jitter

Another performance issue that is related to delayis jitter. We can roughly say that jitter is a problemif different packets of data encounter differentdelays and the application using the data at thereceiver site is time-sensitive (audio and videodata, for example). If the delay for the first packetis 20 ms, for the second is 45 ms, and for the thirdis 40 ms, then the real-time application that usesthe packets endures jitter.