SatyugDarshan Technical Campus,Faridabad

B.Tech(CSE 3rdSem) 1st Sessional Examination Sep 2015

Subject: Digital Electronics

Sub Code: EE - 204-F

Time: 1 Hr. 30 Min MM: 30Note: Question number 1 is compulsory and attempt one question each from section B and Section C. Marks for each question are shown against each.

SECTION ‘A’

1. Write short note on the following:a. Reflected Gray Code

Answer: Gray code is also called as reflected gray code as the (N+1) bit code can be formed by reflecting an N bit code and appending ‘0’ with old code and ‘1’ with newly formed code. The code thus formed differs in one bit value only from one number to next or previous number. An example of forming gray code is shown below:

b. BCD Addition process, and perform the 68 - 24using 9’s complement.Answer: For +ve numbers write the 4-bit values for digits

For –ve numbers, take 9’s complement, and write the 4-bit values for digitsAdd the numbersIf addition of each nibble > 9 or Carry generated, Add 6 to result

+68 = 068-24 = 9’s complement of 024 = 999 – 024 = 975

c. What do you understand by Don’t Care condition, how are they helpful in solving the equation using K-MapAnswer: Don’t care conditions are those terms whose presence or absence does not matter much, however if they are used properly, they may help in simplifying the equation to great extent. They may be taken as ‘1’ for SOP and ‘0’ for POS if they help in forming the octet, quad or pair.

d. Signed number representation, and write the advantages of 2’s complement representation over 1’s complement system.Answer:Signed representation is used to represent both the +ve and the –ve numbers. There are following three type of signed number representation in binary system. These are:a. Signed magnitude Number: In this representation one additional bit is needed at

MSB position to indicate whether the number is +ve or –ve. Thus if there are N bits then N-1 bits are used for magnitude and Nth bit indicate sign.The range of representable numbers are +/- 0 to +/- 2N – 1.

b. 1st complement representation: In this the magnitude is complemented to get the –ve number. The sign bit is also made equal to ‘1’ to indicate –ve number. For example to represent – 35 using 1’s complement we will do as:

+35 = 0100011

-35 in 1’s complement = 1011100 ; obtained by making ‘0’ as ‘1’ and vice versa

c. 2’s Complement representation: The –ve number is representing by 1st taking the 1’s complement of the magnitude and appending 1 at MSB and adding 1 to it. As an example -35 in 2’s complement is written as:+35 = 0100011-35 in 1’s complement = 1011100Adding ‘1’ +1We get -35 in 2’s complement= 1011101

Advantage of 2’s complement system: i. There is only one zero as against two zero in other representationj. Addition and subtraction is performed only by addition processk. End around carry is ignored

SECTION ‘B’

2. (a) Why is it necessary to use Hamming code? A receiver received the following Hamming code 0011100101101 with odd parity. Find the error in the received code and write the correct code that was transmitted by the receiver.Answer:13 12 11 10 9 P8 7 6 5 P4 3 P2 P1 0 0 1 1 1 0 0 1 0 1 1 0 1

We find P1= check ‘1’ leave ‘1’: check 1,3,5,7,9,11,13 = Even (Four ‘1’s) ParityP2= Check 2, leave 2 I.e check 2,3,6,7,10,11 = Even (Four ‘1’s) ParityP4= Check 4, leave 4 I.e check 4,5,6,7,12,13 = Even (Two ‘1’s) Even parityP8= Check 8, leave 8 I.e check 8,9,10,11,12,13 = Odd

Parity P8 P4 P2 P10 1 1 1 = 7; therefore 7th bit is in error

So the correct code transmitted is: 0011101101101

(b) Simplify F(ABCD) = ∑m (0,1,5,9,13,14,15) +d(3,4,7,10,11 using K-Map.Answer:1st of all name the boxes as per the minterms. These are shown in bottom right corner of each box.

Enter the ‘1’ against the minterms and ‘X’ against the don’t care terms.

If we consider don’t care terms 3,7,11 as ‘1’s then we can form an octet with terms (1,3,5,7,13,15,9,11)Similarly, Quad at the top is formed with terms (0,1,4,5) and Quad at bottom (10,11,14,15)The solution obtained is :F(ABCD) = A’C’ + AC + D

3. (a) Explain the Mc-Clusky method of solutions of Boolean equation. Solve the following equation using the Mc-Clusky method.F(ABCD) = ∑m (20,28,37,39,48,56)

Answer:

Minterms m20 m28 m37 m39 m48 m56Binary 010100 011100 100101 100111 110000 111000No. of 1’s 2 3 3 4 2 3

Group Minterms BinaryABCDEF

Cheque if used

pair Minterms

2 m20 010100 √ m20-m28 01-100 A’BDE’F’m48 110000 √ m48-m56 11-000 ABD’E’F’

3 m28 011100 √ m37-m39 1001-1 AB’C’DFm37 100101 √m56 111000 √

4 m39 100111 √

(b) Minimize the following standard POS expression using K-Map and implement the circuit using NOR gates.F = ΠM (0,2,3,5,7)Answer: i. First of all we enter ‘0’ at the given Maxterms. j. Now we find that three pair can be formed with ‘0’s

The solution obtained is:F(ABC) = (A’ + C’) + (A + B’) + (A + C)

SECTION ‘C’

4. (a) Design a Half Adder.

A half adder is a combinational circuit that adds two bits as shown in the following truth table.

A B Sum Carry0 0 0 00 1 1 01 0 1 01 1 0 1

Now from the Truth Table we form the following SOP equations for Sum and Carry

Sum = A’B + AB’ = A xor B

Carry = AB

(b) Show that a full Adder can be constructed with two half Adders and an OR-gateAnswer

A B C Sum Carry0 0 0 0 00 0 1 1 00 1 0 1 00 1 1 0 11 0 0 1 01 0 1 0 11 1 0 0 11 1 1 1 1

The SOP equation from the truth table are written below:Sum = A’B’C + A’BC’ + AB’C’ + ABCCarry = A’BC + AB’C + ABC’ + ABCSimplifying for Sum

The logic circuit diagram using two half adder is shown below

5. (a) What is a magnitude comparator? Design a 2-bit magnitude comparatorA magnitude comparator is a combinational circuit that compares two numbers and results in equal, less than or greater than depending on the magnitude of the two numbers. This is shown in the truth table below where A and B are two 2-bit numbers.

Functional table for two bit comparisonInput A Input B Outputs

A1 A0 B1 B0 A>B  A<B A=B0 0 0 0 0 0 10 0 0 1 0 1 00 0 1 0 0 1 00 0 1 1 0 1 00 1 0 0 1 0 00 1 0 1 0 0 10 1 1 0 0 1 00 1 1 1 0 1 01 0 0 0 1 0 01 0 0 1 1 0 01 0 1 0 0 0 11 0 1 1 0 1 01 1 0 0 1 0 01 1 0 1 1 0 0

1 1 1 0 1 0 01 1 1 1 0 0 1A>B = A1'A0B1'B0' + A1A0'B1'B0' + A1A0'B1'B0 + A1A0B1'B0' +A1A0B1'B0 + A1A0B1B0'A<B = A1'A0'B1'B0 + A1'A0'B1B0' + A1'A0'B1B0 +A1'A0B1B0' + A1'A0B1B0 + A1A0'B1B0A=B=A1'A0'B1'B0' + A1'A0B1'B0 + A1A0'B1B0' + A1A0B1B0The Simplifications for the above Boolean equations:

(a).       A>B G=A1B1' + A0B1'B0' + A1A0B0'(b)        A<BL=A<B = A1'A0'B0 +A1'B1 + A0'B1B0 (c )       A==BE=A==B         =          A1'A0'B1'B0' + A1'A0B1'B0 + A1A0'B1B0' + A1A0B1B0=          A’1A’0B’1B’0 + A’1A0B’1B0 + A1A’0B1B’0  + A1A0B1B0 =          A'0B'0( A'1B'1 + A1B1) + A0B0( A'1B'1+A1B1)=          ( A'1B'1 + A1B1)( A'0B'0 + A0B0)=          (A1 ⊕ B1)' ( A0 ⊕ B0)'

(b) Write short note on Multiplexer. A multiplexer (or mux) is a device that selects one of several analog or digital

input signals and forwards the selected input into a single line. A multiplexer of 2ninputs has n select lines, which are used to select which input line to send to the output. In general, a MUX (multiplexer) has 2^N inputs, N control lines and a single output. This is shown in the following figure.

B.Tech(CSE, EE,ECE 5thSem) 1st Sessional Examination Sep 2015

Subject: Microprocessor and Interfacing

Sub Code: EE - 309-F

Time: 1 Hr. 30 Min MM: 30Note: Question number 1 is compulsory and attempt one question each from section B and Section C. Marks for each question are shown against each.

SECTION ‘A’

1. Write short note on the followings

a. Bus system of 8085 MicroprocessorAnswer :8085 is an 8-bit processor which has 16-bit address line, 8-bit data data bus. The data bus is multiplexed with the lower byte of the address bus, and require external hardware to demultiplex the address and data bus for external interface.

b. Interrupts in 8085 microprocessorAnswer: An interrupt is a request from an external device for attention of the CPU for carrying out some task for the device. Interrupt can be hardware or software interrupts 8086 supports 5 hardware interrupts and 8 software interrupts. These interrupts are further categorized as maskable or non-maskable which means that they may be blocked (maskable) and which cannot be blocked (Non-maskable), another way of classigying these interrupts are vectored or non-vectored interrupts. A vectored interrupt is one whose address of the ISR is known and the call to specific location is made to service the interrupt. Non-vectored interrupt address has to be supplied by the interrupting device.The interrupts of 8085 system are:TRAP: Non MaskableRST 7.5: Vectored but maskableRST 6.5: Vectored but maskableRST 5.5: Vectored but maskableINTR:

c. Flag register of 8086 MicroprocessorAnswer: Flag register indicate the status of the current operation performed by the CPU(to be more specific inside the ALU). 8086 has a 16-bit flag register of which only 9-bits are used as Flag. 5 of the flags are same as in 8085 microprocessor and four new flags are added. The flag register of 8086 microprocessor is shown below:

15 14 13 12 11 10 9 8 7 6 5 4 3 2 1- - - OF DF IF TF S Z - AC - P - Cy

These flags are control Flags:i. DF: data direction flag used to point to the beginning or end of array

ii. IF: Interrupt Flag, by setting or clearing this flag the interrupt can be disabled or enabled.

iii. TF: Trap Flag is used for single step for debugging operations.Data Flags: i. Carryi. Parity

ii. Auxiliary Flagiii. Zero Flagiv. Sign Flagv. Overflow Flag

d. Calculation of Physical address in 8086 Microprocessor . 2Answer: The physical address in 8086 system is calculated using the following formulaPhysical Address = Segment Register Value x 10 + OffsetFor example, if Value in Segment register=1234h, offset is 2345h, then :Segment address x 10 = 1234h x 10h = 12340

offset = + 2345Physical Address = 14385h This is a 20-bit address.

SECTION ‘B’

2. (a) Explain the function of ALE and IO/M’ signals of the 8085 microprocessor. Also explain the need to demultiplex the AD7-AD0 lines. 5Answer : A high ALE signal is generated during the first T state of a machine cycle by the CPU to enable the latch. During this cycle this signal will activate the latch so that the low order address bus can be separated from the data bus.IO/M’ signal is a control signal which along with S1,S0, RD’ and WR’ is used for selecting memory or I/O device for read write operation.Need for Multiplexing: The low order address and the data bus are internally multiplexed in 8086 microprocessor. This multiplexing is done to reduce the number of pins on the IC package. However to interface with the external environment, these buses are required to be separated for proper timing synchronization and operation. This separation of the low order address bus and data bus is called de-multiplexing the buses. To de-multiplexing the address and data bus the CPU generates a high ALE signal to enable the latch to latch the address during the first T state of the machine cycle. This de-multiplexing is shown in the following figure.

(b) (c) WAP to add 10 numbers stored in memory, and store the sum at address 2050h and carry if any at address 2051h. 5

Answer: Assuming that the 10 numbers are stored in memory starting at address 2020h. As we add the carry if any will be incremented in register ‘B’ and accumulator will contain the sum of the numbers. Finally the Sum will be stored at 2050h and carries will be stored at 2051h. The following chart shows the logic for the operation.

The complete program is given beside the flowchart.

3. (a) Write an assembly language program to count the number of 1’s and zeros in a given data. 5

(b) What do you understand by the word addressing mode? Explain the different addressing mode of 8085 microprocessor. 5

Answer:

Addressing Modes:

As we are aware that a data item may reside in internal registers, RAM, may come from input device or can be assigned as immediate as part of an instruction. For accessing each type there is a well defined way or method. Therefore, we can define addressing modes as the methods of

accessing the data byte(operands). Addressing mode are of the following type in 8085 microprocessor.

1. Immediate addressing mode

In this addressing mode the data byte is defined as part of the instruction. 8085 microprocessor supports an immediate to register or immediate to memory assignment, and use the instruction MVI for this purpose.

The general Syntax is : MVI Reg/Mem 8-bit Data

Example:

MVI A, 55h ; Transfers an immediate number 55h to Accumulator

MVI B, 55h ; Transfers an immediate number 55h to Register B

MVI M, 55h ; Transfers an immediate number 55h to Memory pointed by HL pair

2. Implicit addressing mode:

This addressing mode does not specify any additional data, but the data is implied with the instruction, the data is referred by the name of the instruction. As an example CLA stands for clear the accumulator. Here no additional reference is made to locate the data. Instruction of this type are 1-byte instructions.

Examples:

CMA,STC, CLC, RLC, RRC,RAL, RAR

3. Register addressing mode:

This addressing mode is used t transfer the data item from source register to a destination register. The source register must be loaded with data before using this instruction. This addressing mode uses MOV instruction for transferring data among registers.

Syntax:

MOV Rd, Rs

Here Rs is source register from the general purpose registers and Rd is the destination register from the general purpose registers.

Examples:

MOV A, B ; Transfers 8-bit data from register B to accumulator

MOV D, E ; Transfers 8-bit data from register E to register D

4. Direct addressing Mode

The direct addressing mode refers to the data from / to the memory for transfer to or from the registers. Use of this addressing mode require setting the memory pointer. 8085 uses MOV instruction for this purpose.

Syntax:

MOV Rd, M; where Rd is the destination register and M is the memory location pointed by HL

MOV M, Rs; where M is the memory location pointed by HL, and Rs is the source register.

Usage example:

LXI H, 2050 ; Set the memory pointer

MOV B, M ; Transers data from memory location 2050h into register B. This instruction also expects the data byte at location 2050.

5. Indirect memory addressing mode

In this addressing mode the data is accessed indirectly from the memory. The address from where the data is to be fetched is to be found at other place. 8085 uses instructions such as LDAX, STAX, LHLD, SHLD for indirect addressing. These are explained below:

Example

LDAX [Rp] ; Will load accumulator with data from address given in register pair Rp(BC, DE, HL)

STAX [Rp] ; Will store data given in accumulator in memory at address given in register pair Rp.

LHLD [Rp] ; Loads Register L from address given in Rp, then increment Rp and load H from there.

LHLD [Rp] ; Stores Register L from at address given in Rp, then increment Rp and stores H there.

SECTION ‘C’

4. (a) Explain the internal architecture of 8086 Microprocessor 5Answer:The system architecture defines how the processor and components relates to each other. The different components of 8086 are grouped and partitioned logically into two processing units. They are the Bus Interface Unit (BIU) and the Execution Unit (EU) as shown in figure below:

Execution Unit (EU):The EU consists of a set of general purpose registers (AX, BX, CX and DX) the base and index pointers, the ALU, control unit and the flag register. The EU is where actual processing take place. The instructions from the instruction queue and the data that have been fetched by the BIU are accepted by EU and operate on the data as per the instructions. After an operation the processed data (result) can be taken to memory or other peripheral device for later use. As stated above it has 4 components: Control circuitry, ALU, Flag registers and general purpose registers. The basic functions of these different components are: : 

1. Control Circuits: it directs all the internal operations.2. ALU: It is where all logic and arithmetic operations are performed.

3. General purpose registers: They are used to store data during execution.4. Flag registers: It has a 16bit flag register containing 9 flags that are set for certain conditions

during any operation.

Bus interface unit (BIU):The basic purpose of this unit is to generate the address of the instructions and the data. It provides a full 16 bit bidirectional data bus and 20 bit address bus (i.e. Address/Data bus is de-multiplexed here). The BIU is responsible for the fetching of the data and instructions from the memory. It contains the following components.

1. Segment registers: There are four segment register namely code segment register(CS), data segment register, extra segment register, stack segment register.These register contain the 16 bit segment begining address. The full 20 bit address is however obtained by multiplying segment register value by (10)hex and adding the offset/(instruction pointer value)

2. Multiplier/adder circuit: The BIU contain a multiplier/ adder unit for calculation of the instruction/data address.

3. Instruction Pointer: The instruction pointer points to the address of the instruction. It is a 16 bit address and thus can point to the 64KB address within a segment.

4. Queue : 8086 has a 6 byte queue, use of the queue allows prefetching of instruction from the memory while the execution of previous instruction is going on in the execution unit.

(b) Explain the Programming Model of 8085 Microprocessor 5

The programming model of 8085 consist of an 8-bit accumulator, one flag register,  6 general 8-bit purpose registers, and two 16-bit special purpose registers. These registers are critically required when programming a 8085 processor. The different register in the programming model are explained below:Accumulator: Accumulator is an 8-bit register. For processing the numbers, 8085 assumes that one of the numbers is in accumulator. After processing, the result is accumulated in accumulator.General purpose registers:8085 has six general purpose registers these are Register B,C,D,E,H,L.All these registers are 8-bits each. These registers can also be used in pair as BC, DE and HL for use as 16-bit data operations.

Program Counter (PC)This 16-bit register deals with sequencing the execution of instructions. This register is a memory pointer. Memory locations have 16-bit addresses, and that is why this is a 16-bit register.The microprocessor uses this register to sequence the execution of the instructions. The function of the program counter is to point to the memory address from which the next byte is to be fetched. When a byte (machine code) is being fetched, the program counter is incremented by one to point to the next memory location Stack Pointer (SP)The stack pointer is also a 16-bit register used as a memory pointer. It points to a memory location in R/W memory, called the stack. The beginning of the stack is defined by loading 16-bit address in the stack pointer. The 8085 uses PUSH and POP instructions for stack operations.  Flag Register:ALU of 8085 contains a special register called a “Flag register” and is an 8-bit register. At the time of design of 8085 only five flags of this register were used and the rest three were reserved. These flags are set or reset after an operation according to data conditions of the result in the accumulator and other registers. They are called Zero(Z), Carry (CY), Sign (S), Parity (P), and Auxiliary Carry (AC) flags; their bit positions in the flag register are shown in the Figure below. The most commonly used flags are Zero, Carry, and Sign. The microprocessor uses these flags to test data conditions

5. Explain the flag register of 8086 microprocessor. Register B has 65H and the accumulator has 97H. Subtract the content of register B from the content of the accumulator. Also give the flag status in 8086 Microprocessor system 10

Answer:Register B = +65 = 01000001Register C = +97 = 01100001C-B using 2’s complement-65 10111110 + 1 = 10111111

Adding +97 = 01100001- 65 = 10111111

97-65 = 1 00100000 Neglecting the carry, we get the answer as 00100000, here MSB is ‘0’ indicating result is +ve, and the magnitude of the result is 32The status as indicated by the flag register is :

15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0OF DF IF TF S Z AC P Cr0 0 0 0 0 0 1 0 1

B.Tech(ECE-7thsem) 1st Sessional Examination Sep 2015

Subject: Mobile CommunicationSub Code: ECE-419-F

Time: 1 Hr. 30 Min MM: 30Note: Question number 1 is compulsory and attempt one question each from section B and Section C. Marks for each question are shown against each.

SECTION ‘A’

1. Write short note on the followingsa) The basic propagation mechanisms of mobile radio propagation. 3

Answer:Reflection, Differeaction and Scattering are the three basic mechanism which impact the propagation in mobile communication system. These mechanisms are defined below:i. Reflection: It occurs when a propagating electromagnetic wave impinges upon an

object which has very low dimensions when compared to the wavelength of the propagating wave. Reflection occur from the surface of the earth and from buildings and walls.

ii. Differaction: It occurs when the radio path between the transmitter and receiver is obstructed by a surface that has sharp irregularities(edges). The secondary waves resulting from the obstructing surface are present throughout the space and even behind the obstacle, giving rise to a bending of waves around the obstacle, even when a LoS path does not exist between Tx and Rx. At high frequencies differaction like reflection, depends on the geometry of the object, as well as the amplitude, phase, and polarization of the incident waves at the point of differaction.

iii. Scattering: Scattering occurs when the medium through which the wave travels consist of objects with dimensions that are small compared to the wavelength, and where the number of obstacles per unit volume is large. Scattered waves are produced by rough surfaces, small objects, or by other irregularities in the channel.

In practice, foliage, street signs, lamp posts induce scattering in mobile communication system.

b) Doppler Spread and Coherence Time. 3AnswerDoppler spread BD is a measure of the spectral broadening caused by the time rate of change of mobile radio channel and is defined as the range of frequencies over which the received Doppler spectrum is essentially non-zero. When a pure sinusoidal tone of frequency fc is transmitted, the received signal spectrum, called Doppler spectrum, will have components in the range of fc – fd to fc + fd, where fd is the Doppler shift. The amount of spectral broadening depends on fd which is a function of relative velocity of the mobile, and the angle θ betweeen the direction of motion of the mobile and direction of arrival of scatter waves. If the baseband signal bandwidth is much greater than BD, the effect of Doppler spread are negligible at the receiver. In this case this is a slow fading channel.Coherence Time: Coherence time Tc is the time dual of the Doppler Spread and is used to characterize the time varying nature of the frequency disperssiveness of the channel in time domain. The Doppler spread and Coherence time are inversely proportional to one another. That is :

Tc = 1/fm --------(1)Coherence time is actually the a statistical measure of the time duration over which the channel impulse response is essentially invariant, and qualifies the similarity of the channel response at different times.If the coherence time is defined as the time over which the time correlation function is above 0.5, then coherence time is approx:

Tc= 916 Πfm ----------(2)

where fm is Doppler shift given by fm = v/λA popular rule of thumb for modern digital communication is to define the coherence time as the geometric mean eqn (1) and (2). That is:

Tc=√ 916Π fm2 =¿ 0.423

fm¿

This means two signals arriving with a time separation greater than Tc are affected differently by the channel

c) Reference model of a mobile communication 2Answer:The reference model is used to structure communication system. Figure below shows the simplified reference model of the wireless communication system. In the figure a personal digital assistant (PDA) communicates with the base station shown in the middle. The base station consists of a radio transceiver and an internetworking unit connecting the wireless link with the fixed link. The communication partner of the PDA, a conventional computer, is

shown on the right of picture. Also shown below each unit (PDA, base station and computer) are the protocol stack implemented in the system according to the reference model, such as the PDA and computer need full protocol stack. Applications on the end systems communicate with each other using the lower services. Intermediate systems, such as internetworking units, do not necessarily need all layers. During communication, as only the entities at the same level communicate with each other, end system applications do not notice the intermediate systems directly in this scenario.

d) Time dispersion parameters. 2Answer:In order to compare different multipath channels and to develop some general design guidelines for wireless systems, parameters which quantify the multipath channel are used. The mean excess delay, rms delay spread, and excessive delay spread(X dB) are multipath channel parameters that can be determined from a power delay profile. The time dispersive properties of wide band multipath channels are most commonly quantified by their mean excess delay(τ ) and rms delay spread(σ τ)i. The mean excess delay is defined to be:

τ =∑kak

2 τ k

ak2 =

∑kp (τ ¿¿ k)τ k

∑kp(τk )

¿

ii. The rms delay spread is the square root of the second central moment of power delay profile and is defined to be:

σ τ=√τ� 2−¿¿

Where τ� 2=¿∑kak

2 τk2

ak2 =

∑k

p(τ¿¿k )τ k2

∑k

p (τ k )¿

iii. The maximum excess delay (X dB) of the power delay profile is defined to be the time delay during which the multipath energy falls to X dB below the maximum. In other words, the maximum excess delay is defined as (τ¿¿x−τ0) ,whereτ 0¿ is the first arriving signal and τ x is themaximumdelay at which a multipath component is within X dB of the strongest arriving multipath signal

SECTION ‘B’

2. a). Level Crossing in mobile communication system. 5Answer:Rice computed joint statistics which provided simple expression for computing the average number of level crossings and the duration of the fades. The level crossing rate (LCR) and average fading duration are the two statistics which are useful for designing error control codes and diversity scheme to be used in mobile communication system, since it becomes easy to relate the time rate of change of received signal to the signal level and velocity of mobile.Level crossing rate(LCR) is defined as the expected rate at which the Rayleigh fading envelop, normalized to the local rms signal level, crosses a specified level in a positive going direction. The number of level-crossings per seconds is given by:

NR = ∫0

∞

r p (R , r )dr=√2π fmρe−ρ2

where

r is the time derivative of r(t) i.e. the slopep (R ,r ) is the joint density function of r and r at r = RFm is the maximum Doppler frequencyρ = R / Rrms is the value of the specific level R, normalized to local rms

amplitude of the fading envelopThe level crossing rate is the function of the mobile speed. There are certain level crossings at both high

and low levels, with a maximum rate occurring at ρ=1√2

b). For a Rayleigh Fading signal, find (a) number of zero level crossings and (b) the average fade duration for the threshold levels, ρ=0.1 and ρ=1, when the Doppler frequency is given as 20 Hz. 5

Answer:

a) for ρ=1, the number of zero level crossing, NR= √2π fmρ e−ρ2

=√2π X 20 (1 ) e−12

¿18.44 crossings /secAverage fade duration,

T=[e12

−1 ] /(1)(20)¿

= 1.71/50.1= 34.1 ms

b) For ρ=0.1, the number of zero level crossing,

NR= √2π fmρ e−ρ2

=√2π X 20 (1 ) e−0.12

¿45.25crossings /secAverage fade duration,

T=[e0.12

−1 ] /(0.1)(20)¿

= 0.002ms = 2 microseconds3. (a) Explain the factors influencing the small scale fading. 5Answer:

There are many physical factors in the radio propagation channel that influence the small-scale fading. These are explained below:

a. Multipath Propagation: The presence of reflecting objects and scatterers in the channel creates a constantly changing environment that dissipate the signal energy in amplitude, phase and frequency. These effects t results in radio signals reaching

the receiving antenna by two or more paths. The effects of multipath include constructive and destructive interference, and phase shifting of the signal.

b. Speed of the Mobile: The relative motion between the base station and the mobile results in random frequency modulation due to different Doppler shifts on each of the ,ultipath components. Doppler shift will be positive or negative depending on whether the mobile receiver is moving towards or away from the base station.

c. Speed of the surrounding objects: If the objects in the radio channel are in motion, they induce a time varying Doppler shift on multipath components. If the surrounding objects move at a greater rate than the mobile, then this effect dominates the small-scale fading. Otherwise, the motion of the surrounding objects may be ignored, and only the speed of the mobile need be considered.

d. Transmission Bandwidth of the signal: If the transmitted radio signal bandwidth is greater than the bandwidth of the multipath channel, the received signal will be distorted, but the received signal strength will not fade much over a local area. If the transmitted signal has a narrow bandwidth compared to the channel, the amplitude of the signal will change rapidly, but the signal will not be distorted in time.

(b) What do you understand by small-scale fading? Explain the types of the small scale fading in mobile radio communication. 5

Answer:Small-Scale Fading: Small-scale fading, or simply fading is used to describe the rapid fluctuations of the amplitude, phase or multipath delays of radio signal over a short period of time or travel distance, so that large-scale path loss effects may be ignored. Fading is caused by interference between two or more versions of the transmitted signal which arrive at the receiver at slightly different times. As a mobile moves over very small distances, the instantaneous received signal strength may fluctuate rapidly giving rise to small-scale fading.Type of Small-Scale Fading: Fadings are classified depending on multipath time delay spread or based on Doppler Spread and are given below:

SECTION ‘C’

4. (a) What do you understand by free space path loss. 3Answer:In free space radio waves travel as light does, it follows a LoS communication between the sender and receiver. Even if no matter exists between the communicating stations, the signal still experiences a free space path loss. Thus in telecommunication, free space path loss is the loss in signal strength that would result if all absorbing, differacting, obstructing, refracting, scattering, and reflecting influences were sufficiently removed having no effectnon its propagation. It does not take into account any path gain of antenna. The free space loss (FSL) is proportional to square of distance and is given as:

FSL = (4 Πd)2

λ = (4 Πdf )2

c

The free space power received by a receiving antenna which is separated from the radiating antenna by a distance d, is given by Friis free space equations as:

Pr(d) =PtGtGr λ2

(4 πd)2 L where

Pt is transmitted power, Gt, Gr are gain of radiating and receiving antenna, L is system loss other than environment, λis wavelength, d is separation

(b) Find the Fraunhofer distance for an antenna with maximum dimension of 1 meter and operating frequency of 900 MHz. If the antenna have a unity gain, calculate the path loss. 7Answer:Since the operating frequency f = 900 Mhz,

the wavelength λ = 3×108

900 X 106 = 0.33m .

Thus, with the largest dimension of the antenna, D=1m, the far field distance is df = 2D2 / λ = 2(1)2 / 0.33 = 6m

Path Loss PL(dB) = -10log (λ2

(4 Π ¿¿¿2d2¿)¿

= -10log(0.332

(4 x3.14 ¿¿¿2 X62¿)¿= 47 dB

5. (a) What is the purpose of the propagation model. Explain the Okumara’s Propagation Model and also write the advantages of this model. 10Answer:Propagation model have focused on predicting the average received signal strength at a given distance from the transmitter, as well as the variability of the signal strength in close spatial proximity to a particular location. Propagation model that predict the mean signal strength for an arbitrary T-R separation distance are useful in estimating the radio coverage area of a transmitter and are called large-scale propagation model. On the other hand, propagation model that characterize the rapid fluctuations of the received signal strength over very short travel distances are called small-scale propagation model.Okumra’s Model: Okumara model for urban areas is a radio propagation model that was built using the data collected in city of Tokyo, japan. The model is ideal for using in cities with many tall blocking buildings. The model is applicable for frequencies in the range of 150 MHz to 1950 MHz and distance of 1Km to 100 Km. it can be used for base station antenna heights ranging from 30m to 1000mtr.Okumara developed a set of curves giving the median attenuations relative to free space (Amu), in the urban area over a quassi smooth terrain with base station affective antenna height (hte) of 200 m and mobile antenna height (hre) of3m. These curves were developed from experimental data using vertical omnidirectional antenna at both sender and receiver end and are plotted as a function of frequencies in the range of 100Mhz to 1920MHz and as a function of distance from the base station in the range of 1 Km to 100 Km.To determine the path loss using Okumara model, the free space path loss is first determined between points of interest, and then the values of Amu(f,d) is added to it along with correction factor to account for the type of terrain. The model can be expressed as:

L50(dB)= LF + Amu(f,d) – G(hte) – G(hre) - GAREA

Where L50 is the 50 percentile(i.e. median) value of propagation path loss.

LF is the free space path loss

Amu is the median attenuation relative to free space

G(hte), G(hre) are the base station and mobile antenna heightGAREA is the gain due to type of environment.Furthermore Okumara found that G(hte) varies at the rate of 20dB/decade and G(hre) varies at a rate of 10dB/decade for heights less than 3 m.

G(hte) = 20 log (hte200

¿ 1000m > hte > 30m

G(hre) = 10 log (hre3

¿ hre <= 3m

G(hre) = 20 log (hre3

¿ 10 m > hre > 3m

Advantages: Okumara’s model is wholly based on measured data and does not provide any any analytical explanations.This model is considered to be the simplest modelThis model is best in terms of accuracy in path loss prediction for mature cellular and land mobile radio system in clustered environmentThis model is very practical and has become a standard for systems planning in modern land mobile radio system in JapanThis model is very good for urban area.Disadvantages:Slow Response to rapid changes in terrainIt is not very good for rural area

ORWhat is the difference between large-scale and small-scale propagation model. Explain the Hata propagation Model. Also write the advantages and disadvantages of this model. 10Answer:Difference between large-scale and small-scale propagation model

large-scale Propogation Model small-scale Propogation ModelPropagation model that predict the mean signal strength for an arbitrary T-R separation distance are useful in estimating the radio

Propagation model that characterize the rapid fluctuations of the received signal strength over very short travel distances are called

coverage area of a transmitter and are called large-scale propagation model.

small-scale propagation model.

This model provide path loss over very large T-R separation

This model provide path loss for very small T-R distance variations

Hata Model for Path Loss calculations:This model is an empirical formulation of graphical path loss provided by Okumara and is valid for following data ranges:frequencies of 150 MHz to 1500 MHzBase station antenna heights (hte) 30m to 200mMobile antenna height (hre) from 1m to 10mThis model is most widely used in radio frequency propagation for predicting the behavior of cellular transmission in built up areas. This omodel also has two more varieties for transmission in suburban and open areas. Hata model predicts the path loss along a link of terrestrial microwave or other type of cellular communications.This model is applicable for transmission inside the cities and is suited for point-to-point and broadcast transmission.Hata Model for urban areas is formulated as:L50 (urban)= 69.55 + 26.16 log (f) – 13.82 log (hte) - a (hre) + (44.9 – 6.55 log (hte) ) log d ----(1)For small and medium cities, the mobile antenna correction factor is given as:a(hre) = (1.1 log fc – 0.7) hre – (1.56 log fc – 0.8) dB ----(2)For large citiesa(hre) = 8.29 (log 1.54 hre)2 - 1.1 dB for fc <= 300 MHz ----(3)a(hre) = 3.2 (log 11.75 hre)2 - 4.97 dB for fc > 300 MHz ----(4)

To obtain path loss in suburban areas, the standard Hata model in above eqn-1 is modified as:L50dB = L50(Urban) – 4.78 (log fc)2 + 18.33 log fc - 40.94 Advantages: Okumara model does not have any of the path-specific correction which are available in Okumara model.Hata Model though modified version of Okumara model but compares very closely wit the original Okumara Model.This moel is well suited to large cell mobile systems.Disadvantages:Equations are hard to remember as compared to easy readable graphical model given by OkumaraHata model is not suited for personal communication system which have cell on the orders of 1Km radius.