McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 UNIT - 2 | Website for students | VTU NOTES

McGraw-Hill ©The McGraw-Hill Companies, Inc., 2004

UNIT - 2

www.bookspar.com | Website for students | VTU NOTES

Unit 2Title: Data, Signals & Digital Transmission

Syllabus:

Analog & Digital Signals, Transmission impairments, Data rate limits, Performance, Digital to analog conversion, Analog to digital conversion, Transmission modes.


Analog & Digital data

Data can be in analog or digital.

i) Analog Data – refers to information that is continuous and takes continuous values.

Example: Human voice.

ii) Digital data – refers to information that has discrete states and take discrete values.

Example: Data stored in a computer memory.



Analog & Digital signals

Signals can be Analog or Digital.

Before transmitting the data over a medium, the data must be converted in to electromagnetic signals.



Analog signalAn analog signal is a continuous signal. It has an infinite number of values in a range.

Example - Human Voice.



Digital signal

A digital signal is a discrete signal. It has a limited number of values.


A sine wave

Mathematically, a Sine wave can be represented by

S(t) = A sin(2Π f t + Φ)

Where, S – instantaneous amplitude, A – peak amplitude,

f – frequency and Φ - phasewww.bookspar.com | Website for students |

VTU NOTES

AmplitudeHighest intensity of a signal represent the peak amplitude of the signal. The intensity is proportional to the energy it carries.Amplitude is measured in volts.


Frequency and period are inverses of each other.

Period & Frequency

Period refers to the time taken by a signal to complete one cycle & expressed in seconds. It is denoted by ‘ T ‘ .

Frequency refers to numbers of signals produced in one second & and expressed in hertz ( Hz ). It is denoted by ‘ f ‘.

The relation between period and frequency is given by

T x f = 1.


Period and frequency


Units of periods and Units of periods and frequenciesfrequencies

Unit Equivalent Unit Equivalent

Seconds (s) 1 s hertz (Hz) 1 Hz

Milliseconds (ms) 10–3 s kilohertz (KHz) 103 Hz

Microseconds (ms) 10–6 s megahertz (MHz) 106 Hz

Nanoseconds (ns) 10–9 s gigahertz (GHz) 109 Hz

Picoseconds (ps) 10–12 s terahertz (THz) 1012 Hz


Example 1Example 1Express a period of 100 ms in microseconds, and express the corresponding frequency in kilohertz.

SolutionSolution

We Know that, 1 ms = 10 3 µs. 100 ms = 100 103 µs = 105 s

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


• Frequency is the rate of change of the

signal with respect to time.

• Change in a short span of time means

high frequency.

• Change over a long span of time means

low frequency.

Note:Note:


• If a signal does not change at all, its frequency is zero.

• If a signal changes instantaneously (it jumps from one level to another in no time), its frequency is infinite, because its period is zero.

Note:Note:


Phase

Phase describes the position of the waveform relative to time zero. It is measured in

degrees or radians.


Signals with different phases


Example Example

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

SolutionSolution

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


Sine wave examples


Sine wave examples (continued)


Sine wave examples (continued)


Time and frequency domains Time-domain plot shows changes in signal amplitude with respect to time.

Frequency-domain plot show a relations between amplitude and frequency .

A signal with peak amplitude= 5 volts, and frequency =0 www.bookspar.com | Website for students |

VTU NOTES

Time and frequency domains (continued)

A signal with peak amplitude= 5 volts, and frequency = 8


Time and frequency domains (continued)

A signal with peak amplitude= 5 volts,

and frequency = 16www.bookspar.com | Website for students |

VTU NOTES

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.

Note:Note:


When we change one or more When we change one or more characteristics of a single-characteristics of a single-frequency signal, it becomes a frequency signal, it becomes a composite signalcomposite signal made of many made of many frequencies.frequencies.

Note:Note:


According to Fourier analysis, any composite signal can be represented as a combination of simple sine waves with different frequencies, phases, and amplitudes.

Note:Note:

Any composite signal is a sum of set sine waves of different frequencies, phases and amplitudes. Mathematically it is represented by

S(t) = A 1 sin(2Π f1 t + Φ1) + A2 sin(2Π f2 t + Φ2) + A3 sin(2Π f 3 t + Φ3 ) +…..


Three harmonics

A graph with three harmonic waves


Adding first three harmonics

Fig. Adding first three harmonics

To create a complete square wave sum up all the odd harmonics up to infinity.


Signal corruption & Bandwidth

A signal has to pass through a medium. One of the characteristics of the medium is frequency. The medium needs to pass every frequency and also preserve the amplitude and phase.

No medium is perfect. A medium passes some frequencies and blocks some others.


The bandwidth is a property of a The bandwidth is a property of a medium. Bandwidth is the difference between the Bandwidth is the difference between the highest and the lowest frequencies that the highest and the lowest frequencies that the medium can satisfactorily pass.medium can satisfactorily pass.

Bandwidth

Bandwidth of a composite signal is the difference between the highest and the lowest frequencies contained in that signal.

For example, Voice normally has a spectrum of 300 – 3300 Hz. Thus, requires a bandwidth of 3000 Hz.


Bandwidth (continued)

The medium can pass some frequencies above 5000 and below 1000, but the amplitude of those frequencies are less than those in the middle.


Example: If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is the bandwidth? Draw the spectrum, assuming all components have a maximum amplitude of 10 V.

Solution B = fh fl = 900 100 = 800 Hz The spectrum has only five spikes, at 100, 300, 500, 700, and 900


Example

A signal has a bandwidth of 20 Hz. The highest frequency is 60 Hz. What is the lowest frequency? Draw the spectrum if the signal contains all integral frequencies of the same amplitude.

SolutionSolution

B = fB = fhh f fl l →→ 20 = 60 20 = 60 ffl l →→ ffll = 60 = 60 20 = 40 Hz20 = 40 Hz


Example 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?

SolutionSolution

The answer is definitely no. Although the signal The answer is definitely no. Although the signal can have the same bandwidth (1000 Hz), the range can have the same bandwidth (1000 Hz), the range does not overlap. The medium can only pass the does not overlap. The medium can only pass the frequencies between 3000 and 4000 Hz. The signal frequencies between 3000 and 4000 Hz. The signal with frequency 1000 & 2000 Hz is totally lost.with frequency 1000 & 2000 Hz is totally lost.www.bookspar.com | Website for students |

VTU NOTES

Digital Signals

Data can be represented by a digital signal.

Bit 1 can be encoded by positive voltage and

bit 0 can be encoded by zero voltage.


Bit rate & Bit interval Bit interval - is the time required to send one bit.

Bit rate - is the number of bits sent in 1 second. It is expressed in bps.


Example Example

A digital signal has a bit rate of 2000 bps. What is the duration of each bit (bit interval)

SolutionSolution

The bit interval is the inverse of the bit rate.

Bit interval = 1/ 2000 s = 0.000500 s = 0.000500 x 106 s = 500 s



Example

Assume we need to download text documents at the rate of 100 pages per minute. What is the required bit rate of the channel?.

Answer

A page is an average of 24 lines with 80 characters in each line. If we assume that one character require 8 bits.

The bit rate of the channel =

100 x 24 x 80 x 8 = 1,636,000 bps = 1.636 Mbps



Bit Length

It is the distance one bit occupies on the transmission medium.

Bit Length = Propagation speed x Bit interval.

Transmission of Digital Signals

A digital signal can transmit by using either baseband transmission or Broadband transmission.

Baseband transmission -means sending a digital signal with out changing it to an analog signal.

Baseband transmission requires a low-pass channel.

Digital transmissionDigital transmission use a low-pass channel. use a low-pass channel.



Broadband transmission (modulation)- means sending a digital signal after changing it to an analog signal.

Broadband transmission requires a bandpass channel.

Analog transmission use a band-pass channel.


Low-pass & Band-passA channel is either low pass or band pass.

A low – pass channel has a B/W With frequencies between 0 & f.

A band pass has B/W with frequencies between f1 & f2.



Data Rate Limit Data Rate Limit The data rate over a channel depends on 3 factors. i) The Band width available. ii) The levels of signals that can use for data transmission.iii) The quality of the channel.Two theoretical formulas were developed to calculate the data

rate.

1) Nyquist Bit Rate - Noiseless Channel

2) Shannon Capacity - Noisy Channel

Nyquist Bit RateMaximum Bit rate = 2 x Bandwidth x log 2L where L is number of

levels used to represent data.


Example Example

Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. Calculate the bit rate.

Answer:Answer:

BitBit Rate = 2 Rate = 2 3000 3000 log log22 2 = 6000 2 = 6000

bpsbps


Example Example

Consider the same noiseless channel, transmitting a signal with four signal levels (for each level, we send two bits). Calculate the bit rate.

Answer: Answer:

Bit Rate = 2 x 3000 x logBit Rate = 2 x 3000 x log22 4 = 12,000 bps 4 = 12,000 bps


Shannon capacity of Bit rate

Maximum data rate of a noisy channel,

C=Band width x log 2 (1+SNR)

Where SNR is the Signal to Noise Ratio.



Signal – to – Noise Ratio (SNR)

SNR is defined as

SNR = Average signal power / Average noise power.

It is described in decibel units, SNR dB.

SNR dB = 10 log 10 SNR


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. Calculate the channel capacity.

Answer:Answer:

C = B logC = B log22 (1 + SNR) = B log (1 + SNR) = B log22 (1 + 0) (1 + 0)

= B log= B log22 (1) = B (1) = B 0 = 0 0 = 0


Example:

A telephone line has a bandwidth of 3000 Hz (300 Hz to 3300 Hz). The signal-to-noise ratio is 3162. Find the channel capacity.

Answer:Answer:

C = B logC = B log22 (1 + SNR) = 3000 log (1 + SNR) = 3000 log22 (1 + 3162) (1 + 3162)

= 3000 log= 3000 log22 (3163) (3163)

= 3000 = 3000 11.62 = 34,860 bps 11.62 = 34,860 bps


ExampleExample

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?

SolutionSolution

C = B logC = B log22 (1 + SNR) = 10 (1 + SNR) = 1066 log log22 (1 + 63) = 10 (1 + 63) = 1066 log log22 (64) = 6 Mbps (64) = 6 Mbps

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

6 6 Mbps = 2 Mbps = 2 1 MHz 1 MHz log log22 LL L = 8 L = 8

First, we use the Shannon formula to find our upper First, we use the Shannon formula to find our upper limit.limit.


Transmission Impairments


Attenuation

Attenuation means loss of energy. When a signal (simple /composite) travels through a medium, it loses some of its energy.

To compensate the loss, amplifiers are used to amplify the signal. It is measured in decibels (dB)



Decibel (dB)

To show that a signal has lost or gained strength, we use the unit called decibel. The decibel measures the relative strengths of two signals or one signal at two different points. The decibel is negative if a signal is attenuated and positive if a signal is amplified.

db = 10 log 10 p2 / p1 where p1 & p2 are the powers of a signal at point 1 & 2 respectively.


ExampleExampleImagine a signal travels through a transmission medium and its power is reduced to half. This means that P2 = 1/2 P1. Find the attenuation (loss of power).

SolutionSolution

Attenuation Loss =10x logAttenuation Loss =10x log1010 (P2/P1) (P2/P1)

= 10x log= 10x log1010 (0.5P1/P1) (0.5P1/P1)

= 10x log= 10x log1010 (0.5) = 10x(–0.3) = –3 dB (0.5) = 10x(–0.3) = –3 dB


Example Example

Imagine a signal travels through an amplifier and its power is increased 10 times. This means that P2 = 10 x P1. find the amplification (gain of power).

Answer:Answer:

Amplification= 10 logAmplification= 10 log1010 (P2/P1) (P2/P1)

= 10 log= 10 log1010 (10P1/P1) (10P1/P1)

= 10 log= 10 log1010 (10) = 10 (1) = 10 dB (10) = 10 (1) = 10 dBwww.bookspar.com | Website for students |

VTU NOTES


Example

The loss in a cable is measured in decibels per kilometer

( dB/km). If the signal at the beginning of a cable with -0.3 db/km has a power of 2 mW. What is the power of the signal at

5 km?.

Solution

The loss of energy in the cable in decibels is = 5 x (-0.3) = -1.5 dB.

Power of the signal at 5 kms= dB = 10 x log p2/p1 = - 1.5

or p2/p1 = 10 -0.15 = 0.71

P2 = 0.71 P1 = 0.7 x 2 = 1.4 mW


DistortionDistortion means the signal changes its shape. Distortion occurs in a composite signal because a it is made up of signals of different frequencies. Each signal has its own propagation speed. So, its own delay in arriving at the destination.


NoiseSeveral types of noise such as thermal noise, induced noise, crass talk and impales noise may corrupt the signals.



Example

The power of a signal is 10 mW and the power of the noise is 1 µW. What are the values of SNR & SNR dB?.

Solution

SNR = Average signal power / Average noise power.

= 10 x 10 -6 W / 10 -9 W = 10,000

SNR dB = 10 x log 10 SNR = 10 x log 10 10000 = 40


Example

What is the channel capacity for a teleprinter channel with a 300 Hz bandwidth and a signal to noise ratio of 3 dB.

Solution

Using Shannon's equation: C = B log 2 (1 + SNR)

We have B = 300 Hz, (SNR) dB = 3

C = 300 log 2 (1 +3) = 300 log 2 (4) = 600 bps


Example

A digital signaling system is required to operate at 9600 bps.

a) If a signal element encodes a 4 bit word, what is the minimum required bandwidth of the channel?

b) Repeat part (a) for the case of 8 – bit words.

Solution By Nyquist's Theorem: C = 2 x B x log2M

We have C = 9600 bps

a) log 2M = 4, because a signal element encodes a 4-bit word

Therefore, C = 2 x B x 4 9600 = 2xBx4 Hence, B = 1200 Hz

b) 9600 = 2 x B x 8 and B = 600 Hzwww.bookspar.com | Website for students |

VTU NOTES

Example

Given a channel with an intended capacity of 20 Mbps, the

bandwidth of the channel is 3 MHz. Assuming thermal noise, what

signal to noise ratio is required to achieve this capacity?

Solution

C = B log2 (1 + SNR)

20 x 106 = 3 x 106 x log 2 (1 + SNR)

log 2 (1 + SNR) = 6.67

1 + SNR = 2 6.67

SNR = 2 6.67 - 1www.bookspar.com | Website for students | VTU NOTES


Performance

Bandwidth is a potential measure of a link.

Bandwidth in Hz – The range of frequencies that a channel can pass.

Bandwidth in bps – Number of bits per second that a channel can transmit.


Throughput

Throughput is an actual measurement of data received at the receiver.

Example

A network with bandwidth 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network?

Solution

Throughput = ( 12000 x 10000) /60 = 2 Mbps



Latency ( Delay )The latency the time taken for an entire message to arrive completely at the destination taken from the time the first bit is sent out from the source.

Latency = propagation time + transmission time +

queuing time + Processing delay.


Propagation timePropagation time is the time required for a bit to travel from source to the destination.

Propagation time = Distance / Propagation speed.


Wavelength

Wave length is the distance a simple signal can travel in one period.


Examples

Question paper: January / February 2005

Using Shannon’s theorem, compute the maximum bit

rate for a channel having a bandwidth of 3100 Hz and

signal to noise ratio is 20 dB.


DigitalTransmission


Line coding

Line Coding is the process of converting binary data into digital data.

Characteristics of line coding techniques are

1) Signal level Vs Data level

2) DC ( Direct Current ) – component & self synchronization

3) Pulse rate Vs Bit rate


Signal level versus data level

The number of values allowed in a signal are called Signal Levels.

The number of values used to represent data are called Data Levels.


DC component

Some line coding schemes leave have residual dc component. It is undesirable because,

i) it causes distortion and may create errors in the output. ii) This component is the extra energy residing on the line and useless. a) + voltage is not cancelled by – voltage b) + voltage are cancelled by - voltage.


Pulse Rate & Bit Rate

A pulse is the minimum amount of time required to transmit a symbol.

The number of pulses per second is called pulse rate.

Bit Rate- The number of bits transmitted per second is called bit rate.

If a pulse carries a bit, then the pulse rate = bit rate. If pulse carries more than one bit, then bit rate > pulse rate.

Bit Rate = Pulse Rate x log 2 L


Example

A signal has two data levels with a pulse duration of 1 ms. Calculate the pulse rate and bit rate.

Answer:Answer:

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

Bit Rate = Pulse Rate x logBit Rate = Pulse Rate x log22 L = 1000 x log L = 1000 x log22 2 = 1000 bps 2 = 1000 bps


Example Example

A signal has four data levels with a pulse duration of 1 ms. Calculate the pulse rate and bit rate.Answer:Answer:Pulse Rate = = 1000 pulses/sPulse Rate = = 1000 pulses/sBit Rate = PulseRate x log2 L = 1000 x log2 4 = 2000 bpsBit Rate = PulseRate x log2 L = 1000 x log2 4 = 2000 bps

Example Can the bit rate be less than pulse rate.

Solution

The bit rate is always greater than or equal to the pulse rate because a pulse contains one or more bits.


SynchronizationAt the receiver, to interpret the signal correctly , the receiver bit interval must match exactly to the senders bit interval. If the receiver clock is faster or slower, the bit intervals are not matched. So the output will be wrong.


Line coding schemes

There are three classifications of line coding

1) Unipolar 2) Polar & 3) Bipolar


Unipolar encoding

The signal levels are on one side of the time axis either above or below.

The 1’s are encoded as positive voltage value and 0’s are encoded as zero voltage value.

Disadvantages

The average amplitude of a unipolar encoded signal is nonzero. This method lacks synchronization.


Polar encoding uses two voltage levels Polar encoding uses two voltage levels (positive and negative).(positive and negative).

Note:Note:

Polar Encoding

The average voltage level on the line is reduced.

The D.C component problem is alleviated.


polar encoding

There are 4 types of polar encoding schemes

i) NonReturn to Zero ( NRZ )

ii) Returned to Zero ( RZ )

iii) Manchester

iv) Differential Manchester


A positive voltage represent the bit 0,

a negative voltage represent the bit 1.

Disadvantage:

When the data contains a long stream of 0’s or 1’s cases the problem of synchronization with the sender clock.

NRZ – L ( NonReturn Zero – Level ) Encoding


NRZ – I ( NonReturn Zero – Invert ) Encoding

The inversion is the transition between a positive and a negative voltage.

A 1 bit is represented when there is a transition.

A 0 bit is represented when no change.


NRZ-L and NRZ-I encoding


Example

How does NRZ – L differs from NRZ – I ?

Solution

In NRZ-L the signal depends on the state of the bit: a positive voltage is a 0, and the negative a 1.

In NRZ-I the signal is inverted when a 1 is encountered.


RZ ( Return to Zero ) encoding

RZ encoding uses three values- positive, negative & zero. Signal changes for each bit. At the mid of each bit interval, the signal returns to zero. A transition from positive to zero represents bit 1. A transition from negative to zero represents bit 0.


Manchester encoding Manchester encoding uses two level of amplitude. Signal changes for each bit. Inversion takes place at the middle of each bit interval. A transition from negative to positive represents bit 1. A transition from positive to negative represents bit 0.


Differential Manchester encoding The differential Manchester encoding uses two level of amplitude. Signal changes for each bit. Inversion takes place at the middle of each bit interval is used for synchronization. The additional transition at the beginning of the interval represents bit 0. The absence of additional transition at the beginning of the interval represents bit 1.


In bipolar encoding, we use three In bipolar encoding, we use three levels: positive, zero, levels: positive, zero,

and negative.and negative.

Note:Note:

Bipolar EncodingBipolar Encoding


Bipolar AMI (Alternate Mark Inversion) encoding

The AMI encoding uses three voltage levels, positive, negative & zero. The zero voltage level represents bit 0. The alternate positive and negative voltage levels represent bit 1.


2B1Q ( two binary, one quaternary)

2BIQ encoding uses four voltage levels. Each pulse represent 2 bits.

-

-


MLT-3(Multi Line Transmission three level) signal

MLT-3 encoding uses three level signals, +1,0 & -1. The signal transitions from one level to the next at the beginning of a bit interval represent bit 1 , there is no transition at the beginning of a bit 0


Example

Assume a data stream is made of ten 0’s. Encode this stream, using the following encoding schemes. How many changes (vertical line) can you find for each scheme?

a) Unipolar b) NRZ- L c) NRZ-I d) RZ e) Manchester f) Differential Manchester g) AMI


ExampleAssume a data stream is made of ten alternate 0’s & 1’s. Encode this stream, using the

following encoding schemes. How many changes (vertical line) can you find for each scheme?

a) Unipolar b) NRZ- L c) NRZ-I d) RZ e) Manchester f) Differential Manchester g) AMISolution


EXAMPLE Jan/Feb 2005

Sketch the signal waveforms when 00110101 is transmitted in the following signal codes.

i) NRZ – L ii) Manchester Code


Block Coding


Block coding Block coding is used to improve the performance of line coding. The method has three steps- Division, Substitution & Line coding.

Division – The sequence of bits is divided into group of m bits.


Substitution in block coding

Substitute an m-bit code for an n-bit group. For example in the figure, 4B/5B ,a 4-bit group is substituted to a 5-bit code. While substituting choose a policy such that it helps in synchronization and error detection.


Table : 4B/5B encodingTable : 4B/5B encoding

Data Code Data Code

0000 1111011110 1000 1001010010

0001 0100101001 1001 1001110011

0010 1010010100 1010 1011010110

0011 1010110101 1011 1011110111

0100 0101001010 1100 1101011010

0101 0101101011 1101 1101111011

0110 0111001110 1110 1110011100

0111 0111101111 1111 1110111101www.bookspar.com | Website for students |

VTU NOTES

Analog to Digital conversionAnalog to Digital conversion

SAMPLING

•Pulse Amplitude Modulation•Pulse Code Modulation•Sampling Rate: Nyquist Theorem


PCM Technique


PAM ( Pulse Amplitude Modulation) PAM is a method used to convert analog signal level in to discrete digital signal value at constant interval of time called sample. PAM uses a method called sample and hold. At a given interval, the signal level is read, then held briefly.


Quantized PAM signal

Quantization is a method of assigning integral values in a specific range to the sampled instances.


Binary Encoding

The quantized samples are assigned by sign and magnitude. Each value is translated into 8 bit representation. In eight bits first bit is used to indicate the sign and the other seven bits to represent the quantized value.


PCM

The binary digits are transformed to a digital signal by using a line coding method called Pulse Coding Method.


According to the Nyquist theorem, the According to the Nyquist theorem, the sampling rate must be at least 2 times sampling rate must be at least 2 times

the highest frequency.the highest frequency.

Note:Note:

Sampling Rate : Nyquist Theorem

Example:We want to sample telephone voice with a maximum frequency of 3000 Hz, we need a sampling rate of 6000 samples.


Nyquist theorem

For voice-over –phone –lines, take a sample for every 1/6000 s.


Example Example

What sampling rate is needed for a signal with a bandwidth of 10,000 Hz (1000 to 11,000 Hz)?

SolutionSolution

The sampling rate must be twice the highest frequency in the signal:

Sampling rate = 2 x (11,000) = 22,000 samples/sSampling rate = 2 x (11,000) = 22,000 samples/s


Example Example

We want to digitize the human voice. What is the bit rate, assuming 8 bits per sample?

SolutionSolution

The human voice normally contains frequencies from 0 to 4000 Hz. Sampling rate = 4000 x 2 = 8000 samples/sSampling rate = 4000 x 2 = 8000 samples/s

Bit rate = sampling rate x number of bits per sample Bit rate = sampling rate x number of bits per sample = 8000 x 8 = 64,000 bps = 64 Kbps= 8000 x 8 = 64,000 bps = 64 Kbps

Bit Rate –

Bit Rate = Sampling rate x Number of bits per sample


Transmission ModeTransmission Mode

Parallel Transmission

Serial Transmission


Data transmission

The transmission of binary data across a medium can be made in either parallel or serial mode.


Parallel transmission

In parallel mode, a group of n bits are sent with each clock tick. It needs n lines

Advantage - Speed. Disadvantage – Cost.

So, parallel transmission restricted to short distance.


Serial transmission

In serial mode, 1 bit is sent with each clock tick. There are two classes of serial transmission – i) Synchronous & ii) Asynchronous

Advantage – Less cost.


Asynchronous transmission

The start bits are 0’s and the stop bits are 1’s.The gap is represented by an idle line.


Synchronous transmission

In Synchronous transmission, the bit stream is a combination of many bytes. Each byte is introduced with out a gap. The receiver reconstruct the information.

The advantage of synchronous transmission is speed. Synchronous transmission is more faster than Asynchronous transmission. So, Synchronous transmission is useful for high-speed applications.


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McGraw-Hill©The McGraw-Hill Companies, Inc., 2004 UNIT - 2 | Website for students | VTU NOTES