1 S.NO SUBJECT CODE SUBJECT NAME PAGE NO 1 EC8353 Electron Devices and Circuits 2 2 EE8301 Electrical Machines-I 19 3 EE8391 Electromagnetic Theory 46 4 ME8792 Power Plant Engineering 65 5 EE8351 Digital Logic Circuits 86 6 MA8353 Transforms and Partial Differential Equations 112 CLASS II YEAR/ III SEMESTER 2 MARK AND 16 MARK QUESTION BANK DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING V.S.B. ENGINEERING COLLEGE, KARUR
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S.NO SUBJECT CODE SUBJECT NAME PAGE NO
1 EC8353 Electron Devices and Circuits 2
2 EE8301 Electrical Machines-I 19
3 EE8391 Electromagnetic Theory 46
4 ME8792 Power Plant Engineering 65
5 EE8351 Digital Logic Circuits 86
6 MA8353 Transforms and Partial Differential
Equations 112
CLASS II YEAR/ III SEMESTER
2 MARK AND 16 MARK QUESTION BANK
DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
V.S.B. ENGINEERING COLLEGE, KARUR
2
V.S.B ENGINEERING COLLEGE, KARUR
Department of Electrical and Electronics Engineering
ELECTRONIC DEVICES AND CIRCUITS
Two marks Question and answers
UNIT -I
1. What are conductors? Give examples?
Conductors are materials in which the valence and conduction band overlap each
other so there is a swift movement of electrons which leads to conduction. Ex. Copper, silver.
2. What are insulators? Give examples?
Insulators are materials in which the valence and conduction band are far away from
each other. So no movement of free electrons and thus no conduction. Ex glass, plastic.
3. What are Semiconductors? Give examples?
The materials whose electrical property lies between those of conductors and
insulators are known as Semiconductors. Ex germanium, silicon.
4. What are the types of Semiconductor?
1. Intrinsic semiconductor
2. Extrinsic semiconductor.
5. What is Intrinsic Semiconductor?
Pure form of semiconductors are said to be intrinsic semiconductor.
Ex: germanium, silicon
6. What is Extrinsic Semiconductor?
If certain amount of impurity atom is added to intrinsic semiconductor the resulting
semiconductor is Extrinsic or impure Semiconductor.
7. What are the types of Extrinsic Semiconductor?
1. P-type Semiconductor 2. N- Type Semiconductor.
8. What is P-type Semiconductor?
The Semiconductor which are obtained by introducing pentavalent impurity atom
(phosphorous, antimony) are known as P-type Semiconductor.
9. What is N-type Semiconductor?
The Semiconductor which is obtained by introducing trivalent impurity atom
(gallium, indium) are known as N-type Semiconductor.
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10. What is doping?
Process of adding impurity to a intrinsic semiconductor atom is doping. The impurity
is called dopant.
11. Define drift current.
When an electric field is applied across the semiconductor, the holes move towards
the negative terminal of the battery and electron move towards the positive terminal of the
battery. This drift movement of charge carriers will result in a current termed as drift current.
12. Give the expression for drift current density due to electron.
Jn = q n μnE
Where,
Jn - drift current density due to electron
q- Charge of electron
μn - Mobility of electron
E - applied electric field
13. Define the term diffusion current.
A concentration gradient exists, if the number of either electrons or holes is greater in
one region of a semiconductor as compared to the rest of the region. The holes and electron
tend to move from region of higher concentration to the region of lower concentration. This
process in called diffusion and the current produced due this movement is diffusion current.
14. What is recovery time? Give its types.
When a diode has its state changed from one type of bias to other a transient
accompanies the diode response, i.e., the diode reaches steady state only after an interval of
time “tr” called as recovery time. The recovery time can be divided in to two types such as
(i) forward recovery time (ii) reverse recovery time
15. Define storage time.
The interval time for the stored minority charge to become zero is called st2orage
time. It is represented as t s.
16. Define transition time.
The time when the diode has normally recovered and the diode reverse current
reaches reverse saturation current I0 is called as transition time. It is represented as t t.
17. What is zener breakdown?
When a small value of reverse bias voltage is applied , a very strong electric field is
set up across the thin depletion layer. This electric field is enough to break the covalent
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bonds. Now extremely large number of free charge carriers are produced which constitute the
zener current. This process is known as zener break down.
18. What is avalanche break down?
When bias is applied, thermally generated carriers which are already present in the
diode acquire sufficient energy from the applied potential to produce new carriers by
removing valence electron from their bonds. These newly generated additional carriers
acquire more energy from the potential and they strike the lattice and create more number of
free electrons and holes. This process goes on as long as bias is increased and the number of
free carriers gets multiplied. This process is termed as avalanche multiplication. Thus the
break down which occurs in the junction resulting in heavy flow of current is termed as
avalanche break down.
19. What is rectifier? What are its types?
Rectifier is an electronic device which converts an alternating (ac) voltage or current
into a unidirectional (dc) voltage or current.
Types of rectifier:
1. Half wave rectifier
2. Full wave rectifier
i. Full wave rectifier with center tapped transformer
ii. Full wave bridge rectifier
20. Define rectifying efficiency.
Rectifying efficiency is defined as the ratio of DC output power into AC input power
of a rectifier.
21. What is the function of filters?
Filter is used to reduce the ripple contents in the output of a rectifier to obtain a pure
dc.
22. List the advantages of Zener regulator.
1. Simple circuits
2. Only 2 or 3 components are required to be used
3. Low cost.
23. What is PN junction diode?
A PN junction diode is a two terminal device consisting of a PN junction formed
either in germanium or silicon crystal. A PN junction is formed from a piece of
semiconductor by diffusing P-type material to one half sides and N type material to other half
side.
24. What is depletion region in a PN junction diode?
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The region around the junction from which the charge carriers are depleted is called
as depletion region. When a PN junction is forward biased, the depletion region width
decreases When a PN junction is reversed biased the depletion region width increase.
25. Define the term transition capacitance CT of a diode.
When a PN junction is reverse biased the depletion layer acts like a dielectric material
while P and N type region on either side have low resistance acts as the plates. In reverse
biased PN junction may be regarded as parallel plate capacitor. This junction capacitance is
called transition capacitance. It is denoted by CT and is also called as space charge
capacitance or depletion layer capacitance.
26. List the application of PN junction diode.
Used as rectifier diodes in dc power supplies
Used as signal diodes in communication circuits for modulation and demodulation
Used in clipped and clamper circuits
Used as a switch in logic circuits used in computers
UNIT-II
1. What is bipolar junction transistor?
A bipolar junction transistor (BJT) is a three terminal semiconductor device in which
the operation depends on the interaction of both majority and minority carriers and hence the
name bipolar.
2. What are the different configurations of BJT?
Common emitter configuration
Common collector configuration
Common base configuration
3. What is thermal runaway?
The continuous increase in collector current due to poor biasing causes the
temperature at collector terminal to increase. If no stabilization is done, the collector leakage
current also increases. This further increases the temperature. This action becomes
cumulative and ultimately the transistor burns out. The self destruction of an un stabilized
transistor is known as thermal runaway.
4. Define the different operating region of transistor.
Active region: The collector junction is reverse biased and emitter junction is forward biased.
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Cut off region: The collector and emitter junction are both reverse biased.
Saturation region: The collector and emitter junction are forward biased.
5. List the uses of emitter follower (common collector configuration) circuit.
It is widely used in electronic instruments because of low output impedance and high
input impedance.
It is used of impedance matching.
6. Define alpha and beta of the transistor.
The ratio of change in collector current IC to the change in emitter current IE at
constant collector base voltage VCB ( α = IC/ IE)
Base current amplification factor (β)
The ratio of change in collector current IC to the change in base current IB ( β = IC/
IB)
7. What is meant by early effect?
When the collector base voltage is made to increase, it increase the depletion region
across the collector base junction, with the result that the effective width of base terminal
decreases. This variation of effective base width by collector base voltage is known as base
width modulation or early effect.
8. Explain the significance of early effect or base width modulation.
It reduces the charges recombination of electron with holes in ht base region, hence
the current gain increase with increase in collector base voltage. The charge gradient is
increased within the base; hence the current due to minority carriers across emitter junction
increases.
9. Which configuration provides better current gain?
CB configuration
10. What is the significance of VBE and ICO?
VBE and ICO are significant because any changes in VBE and ICO cause a drastic
change in temperature and collector current IC. It leads to thermal runaway problem.
11. List out the different types of biasing.
Voltage divider bias.
Base bias
Emitter feedback bias
Collector feedback bias.
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12. Why is the transistor called a current controlled device?
The output characteristics of the transistor depend on the input current. So transistor
is called a current controlled device.
13. Define current amplification factor.
It is defined as the ratio of change in output current to the change in input current at
constant other side voltage.
14. What are the requirements for biasing circuits?
The q point must be taken at the Centre of the active region of the output
characteristics.
Stabilize the collector current against the temperature variations.
Make the q point independent of the transistor parameters.
When the transistor is replaced, it must be of same type.
15. When does a transistor act as a switch?
The transistor acts as a switch when it is operated at either cutoff region or saturation
region.
16. What is biasing?
To use the transistor in any application it is necessary to provide sufficient voltage
and current to operate the transistor. This is called biasing.
17. What is operating point?
For the proper operation of the transistor a fixed level of current and voltages are
required. This values of currents and voltages defined at a point at which the transistor
operate is called operating point.
18. What is stability factor?
Stability factor is defined as the rate of change of collector current with respect to the
rate of change of reverse saturation current.
19. What is d.c load line?
The d.c load line is defined as a line on the output characteristics of the transistor
which gives the value of Ic & Vce corresponding to zero signal condition.
20. What are the advantages of fixed bias circuit?
This is simple circuit which uses a few components. The operating point can be fixed
anywhere on the Centre of the active region.
21. Explain about the various regions in a transistor.
The three regions are active region, saturation region and cutoff region.
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22. Explain about the characteristics of a transistor.
Input characteristics: it is drawn between input voltage & input current while keeping
output voltage as constant.
Output characteristics: It is drawn between the output voltage &output current while
keeping input current as constant.
23. What is the necessary of the coupling capacitor?
It is used to block the c signal to the transistor amplifier. It allows ac &blocks the d c.
24. What is reverse saturation current?
The current due to the minority carriers is called the reverse saturation current.
25. What is a FET?
A field effect (FET) is a three terminal semiconductor device in which current
conduction takes place by one type of carriers (either holes or electron) and is controlled by
an electric field.
26. Why FET is called an unipolar device?
The operation of FET depends upon the flow of majority carriers only (either holes or
electrons) the FET is said to be unipolar device.
27. Why the input impedance of FET is more than that of a BJT?
The input impedance of FET is more than that of a BJT because the input circuit of
FET is reverse biased whereas the input circuit of BJT is forward biased.
28. What is meant by gate source threshold voltage of a FET?
The voltage at which the channel is completely cur off and the drain current becomes
zero is called as gate source threshold voltage.
29. Why N channel FET’s are preferred over P channel FET’s?
In N channel FET the charge carriers are the electrons which have a mobility of about
1300 cm2/ VS, whereas in P channel FET’s the charge carriers are the holes which have a
mobility of about 500 cm2 /VS. the current in a semiconductor is directly proportional to
mobility. Therefore the current in N channel FET is more than that of P channel FET.
30. What is JFET? And What are the terminals and types in JFET?
JFET- Junction Field Effect Transistor. And the terminals are Gate, Drain and Source
and the types are N- Channel JFET and P- Channel JFET.
31. What are all the types of MOSFET?
i) Enhancement type ii) Depletion type
32. Differentiate Enhancement and Depletion MOSFET.
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Enhancement MOSFET Depletion MOSFET
Positive voltage at the gate Negative voltage at the gate
Inversion layer is made Depletion of majority carriers happens
Negative charges are formed Positive charges are formed
UNIT-III
1. What is an amplifier?
An amplifier is a device which produces a large electrical output of similar
characteristics to that of the input parameters.
2. How are amplifiers classified according to the input?
1. Small– signal amplifier
2. Large – signal amplifier.
3. How are amplifiers classified according to the transistor configuration?
1. Common emitter amplifier
2. Common base amplifier
3. Common collector amplifier.
3. What is the different analysis available to analyze a transistor?
1. AC analysis. 2. DC analysis.
4. How can a DC equivalent circuit of an amplifier be obtained?
By open circuiting the capacitor.
5. How can a AC equivalent circuit of a amplifier be obtained?
By replacing dc supply by a ground and short- circuiting capacitors.
6. What is an amplifier?
An amplifier is a device which produces a large electrical output of similar
characteristics to that of the input parameters.
7. How are amplifiers classified according to the input?
1. Small – signal amplifier 2. Large – signal amplifier
8. How are amplifiers classified according to the transistor configuration?
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1. Common emitter amplifier. 2. Common base amplifier. 3. Common collector amplifier.
9. List out the biasing schemes available to achieve the required bias in a FET.
Voltage divider bias, Base bias, Emitter feedback bias, Collector feedback bias,
Emitter bias.
10. Mention the parameters of JFET.
A.C. drain resistance
Transconductance
Amplification factor
11. What is transconductance in JFET?
It is the ratio of small change in drain current to the corresponding change in drain to
source voltage.
12. What is amplification factor in JFET?
It is the ratio of small change in drain to source voltage to the corresponding change
in Gate to source voltage.
13. Why do we choose q point at the center of the load line?
The operating point of a transistor is kept fixed usually at the center of the active
region in order that the input signal is well amplified. If the point is fixed in the saturation
region or the cut off region the positive and negative half cycle gets clipped off respectively.
14. Define MOSFET and what are all the terminals.
Metal oxide semiconductor field effect transistor. The terminals are gate, Drain and
source.
15.Why bypass and coupling capacitor are used in amplifier circuits?
Bypass capacitor CE:
The capacitor connected in parallel with the emitter resistor RE is called as the emitter bypass
capacitor.
This capacitor offers a low reactance to the amplified ac signal. Therefore the emitter resistor
RE gets bypassed through CE.
16. How does the MOSFET has high input impedance?
The input impedance of a MOSFET is higher than that of FET since the gate is
insulated from the channel by thin layer of silicon di oxide.
17. Define stability factor of an amplifier? What is its ideal value.
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It is rate of change of collector current with respect to the reverse saturation current
ICO or β or V be. This is the factor which is used to monitor the thermal stability of the
amplifier circuit. The ideal value of stability factor 1.
18. What is the advantage of using emitter resistance in the context of biasing?
It is used to increase the stability by providing negative feedback.
19. What is bandwidth of an amplifier?
The range of frequencies between the upper cutoff frequency and lower cutoff
frequency is known as bandwidth.
20. What are the features of cascode amplifier?
It is another type of wide band amplifier where the first stage is a CE amplifier and
the second stage is the CB amplifier stage. This arrangement is designed to provide high
input impedance with lower voltage gain to ensure that the miller capacitance is at a
minimum with the CB stage providing good high frequency operation.
UNIT-IV
1. What is a differential amplifier?
An amplifier that has two inputs and produces on output signal that is a function of
the difference between the two given output.
2. What are the applications of difference amplifier?
Medical electronic field
Input stage in the measuring instruments
Analog computation
Linear integrated circuit
3. What are the advantages of differential amplifier?
It uses no frequency dependent coupling or bypassing capacitors.
It can compare any tow signals and detect the difference.
It gives higher gain than two cascaded stages of ordinary direct coupling.
4. What is operational amplifier?
An op amp to perform mathematical operation like summation, multiplication,
differentiation and integration etc. in analog computers. It is very high directly couple
negative fee back amplifier, which can amplify signals having frequency ranging from 0Hz to
1 MHz.
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5. What are the specifications for an ideal operational amplifier?
Open loop gain = ∞, Input impedance = ∞
Output impedance = 0, Band width = ∞, CMRR = ∞
6. What is common mode voltage swing?
The common mode voltage swing is defined as the maximum peak input voltage
which may be applied to either input terminal without causing abnormal operation or
damage.
7. Define slew rate.
It measure of an operational amplifier’s switching speed defined as the maximum
time rate of change of the output voltage when subjected to a square wave input signal when
the closed loop gain is unity. Unit is V/msec.
8. Define input off set voltage.
The algebraic difference between the currents into the (-) input and (+) input is
referred to as input offset current.
9. Is the practical op-amp on ideal op-amp?
A practical op-amp is not ideal and has finite value of input offset voltage input offset
current and input bias current. These produce a dc offset voltage at the output.
10. Can op-amp be used to amplify AC as well as DC output?
Op amp can be used to amplify AC and DC for amplifying AC .we use a capacitance
coupled amplifier.
11. What is phase shift distortion?
If the phase shift introduced by the amplifier for different input frequencies are not
proportional to frequency then phase distortion will take place. The phase distortions are not
detectable by the human ears as they are insensitive to the phase changes.
Therefore, phase shift distortion takes place due to unequal phase shifts of the input
signal at different frequencies.
12. What is difference between voltage amplifier and power amplifier?
Small signal amplifiers are also known as “Voltage amplifiers”. This is because
these amplifiers are used primarily for voltage amplification but they are not capable of
supplying a large power to the loads such as loud speakers. The large signal amplifier (power
amplifier) will increase the current sourcing and sinking capability. So at its output we get a
high voltage, high current signal that means a high power signal. Thus the power amplifier is
basically a current amplifier.
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13. What are the types of bias method?
1. Fixed bias circuit (single base resistor biasing)
2. Collector to base bias circuit
3. Voltage divider bias (self-bias) circuit.
14. Define pinch off voltage.
The drain source voltage(VDS) at which the drain current (ID )reaches to its constant
saturation level is called as “pinch off voltage, VP”
VP = (q ND a2)/2є
15. Why thermal runaway not present in FET?
Thermal runaway does not exist in JFET, because drain resistance rd increases with
the temperature, which reduces ID. Thus with the reduction of ID the temperature of the
device is reduced.
16. What is meant by monostable, bistable, astable multivibrator?
Bistable multivibrator-It has two stable states. The multivibrator can exist indefinitely
ineither of the two stable states .It requires an external triggering pulse to change from one
state to another.
Monostable Multivibrator: It has one stable state and one quasi state. The
multivibratorremains in a stable state and when external triggering is applied, then
multivibrator goes to quasi state .After some time interval, the circuit automatically returns to
normal state.
Astable Multivibrator-The astable multivibrator has both the states as the quasi stablestates.
None of the state is stable. Due to this, the multivibrator automatically makes the successive
transition from one quasi stable state to other, without any triggering pulse
17. Mention few applications of UJT.
1. Phase control 2.Saw – tooth generators 3.Non-sinusoidal oscillators 4.Triggering
device for SCR and DIAC.
18. List the various square wave generator circuits.
Astable multivibrator.
Monostable multivibrator.
Bistable multivibrator.
Schmitt trigger
19. List the various saw tooth generator circuit.
Exponential charging
Miller circuit.
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Bootstrap circuit.
Phantastron circuit.
Inductor circuit
20. How the frequency of oscillation varied in an astable multivibrator?
1/T = 1/ 1.38RC, so by varying the value of R or C, the frequency of oscillation can
be varied.
UNIT-V
1. Define positive feedback.
If the feedback signal is in phase with input signal, then the net effect of the feedback
will increase the input signal given to the amplifier. This type of feedback is said to be
positive or regenerative feedback.
2. Define negative feedback.
If the feedback signal is out of phase with the input signal then the inputvoltage
applied to the basic amplifier is decreased and correspondingly the output isdecreased. This
type of feedback is known as negative or degenerative feedback.
3. Define sensitivity.
Sensitivity is defined as the ratio of percentage change in voltage gain with feedback
to the percentage change in voltage gain without feedback.
4. What are the types of feedback?
i. Voltage-series feedback
ii. Voltage-shunt feedback
iii. Current-series feedback
iv. Current-shunt feedback
5. Define feedback.
A portion of the output signal is taken from the output of the amplifier and is
combined with the normal input signal. This is known as feedback.
6. Give an example for voltage-series feedback.
The Common collector or Emitter follower amplifier is an example for voltage series
feedback.
7. Give the effect of negative feedback on amplifier characteristics.
i. Negative feedback reduces the gain
ii. Distortion is very much reduce
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8. What is Oscillator circuit?
A circuit with an active device is used to produce an alternating current is called an
oscillator circuit.
9. What are the classifications of Oscillators?
Based on wave generated:
(i) Sinusoidal Oscillator (ii)Non-sinusoidal Oscillator or Relaxation Oscillator
Ex: Square wave, Triangular wave, Rectangular wave etc.
10. Give the properties of negative feedback.
i. Negative feedback reduces the gain
ii. Distortion is very much reduced.
11. What are the types of feedback oscillators?
i. RC-Phase shift Oscillator
ii. LC-Oscillators
a. Tuned collector Oscillator
b. Tuned emitter Oscillator
c. Tuned collector base Oscillator
d. Hartley Oscillator
e. Colpits Oscillator
f. Clap Oscillator.
12. What are the conditions for oscillation?
The total phase shift of an oscillator should be 360o. For feedback oscillator i should
satisfies Barhausen criterion.
13. What is Miller crystal oscillator? Explain its operation.
It is a Hartley oscillator its feedback Network is replaced by a crystal. Crystal
normally generate higher frequency reactance due to the miller capacitance are in effect
between the transistor terminal.
14. Define Oscillator.
A circuit with an active device is used to produce an alternating current is called an
oscillator circuit.
15. What is feed back?
It is the process of injecting some energy from the output and then returns it back to
the input.
16. What is the disadvantage of negative feedback?
16
Reduces amplifier gain
17. Define Blocking Oscillator.
A special type of wave generator which is used to produce a single narrow pulse or
train of pulses.
18. What are the two important elements of Blocking Oscillator?
Transistor and pulse transformer.
19. What are the applications of blocking Oscillator?
It is used in frequency dividers, counter circuits and for switching the other circuits.
21. Define Hartley oscillator.
A LC oscillator which uses two inductive reactance and one capacitive reactance in
its feedback network is called Hartley Oscillator.
22. Define Colpitts oscillator.
A LC oscillator which uses two capacitive reactance and one inductive reactance in
its feedback network is called Hartley Oscillator.
23. What are the main advantages of crystal oscillator?
The main advantages of crystal oscillator are frequency accuracy, stability and low
power consumption.
24. What do you mean by Multivibrators and mention its types?
The Multivibrators are used to produce the non – sinusoidal input signal. Types:
magnetic reluctance.Itisanalogoustoelectricresistance.
5. Drawthetypicalnormalmagnetizationcurve of ferromagneticmaterial.
B(T) Saturationzone
Linear Zone (const μ)
Initialnonlinearzone H(A/m) 6. Whatisfringing? In the air gap the magnetic flux fringes out into neighboring air paths due to thereluctanceofair gap which causesa non uniform flux densityin the air gap of amachine.Thiseffectiscalledfringing effect.
1. Draw a general lay out of a thermal power plant and explain the working of different
circuits. 2. What factors are considered for selecting a site for a big thermal power plant?
3. How much coal, cooling water and combustion air are required for a thermal power
station of 500 MW capacity per hour. 4. How much ash and SO2 are produced per day from a plant of Koradi size if Indian low
grade coal is used. 5. What is the importance of thermal power plant in the national power grid?
6. What is meant by overfeed and underfeed principles of coal firing? Which is preferred
for high volatile coal and why. 7. What are the advantages of burning the fuels in pulverized form? 8. Why ash and dust handling problem is more difficult than coal handling problems.
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9. What are different ash handling systems? Discuss the relative merits and demerits. 10. How the ash produced carries the importance in the selection of thermal power plant
site 11 Draw a general lay out of a thermal power plant and explain the working of different
circuits. 12 What factors are considered for selecting a site for a big thermal power plant?
13 How much coal, cooling water and combustion air are required for a thermal power
station of 500 MW capacity per hour. 14 How much ash and SO2 are produced per day from a plant of Koradi size if Indian low
grade coal is used. 15 What is the importance of thermal power plant in the national power grid?
16 What is meant by overfeed and underfeed principles of coal firing? Which is preferred
for high volatile coal and why. 17 What are the advantages of burning the fuels in pulverized form? 18 Why ash and dust handling problem is more difficult than coal handling problems. 19 What are different ash handling systems? Discuss the relative merits and demerits. 20. How the ash produced carries the importance in the selection of thermal power plant
site.
UNIT – 2: HYDROELECTRIC POWER PLANTS
1. What are the different factors to be considered while selecting the site for hydroelectric
power plant? 2. How the hydroelectric power plants are classified. 3. How the most economical capacity of hydroelectric power plant is decided. 4. What do you understand by run-off river power plant and how its performance is
increased by introducing a pondage in the plant? 5. Explain in detail about pump storage plant. 6. Draw a neat diagram of storage type hydroelectric power plant and describe the function
of each component used in the plant. 7. Mention the advantages and disadvantages of hydroelectric power plants compared with
thermal power plants.
8. Why the combined operation of hydro and thermal plants is more economical than
individual operation of the plant. 9. What do you understand by pump storage plant and what are the advantages and
limitations of this plant. 10. What are the specific advantages of storage reservoir type power plant? How they differ
from other types of hydro power plant?
UNIT – 3: NUCLEAR POWER PLANTS
1. Why uranium oxide is preferred over uranium as fuel. 2. Why cladding is necessary. What are the requirements of a good cladding material? 3. What properties are required for a good coolant? Which gases are used as coolant?
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4. What are the desirable properties of a good moderator? Compare H2O, D2O and C as
moderators. 5. What are the desirable properties of control rod materials? Compare the merits and
demerits of different control rod materials. 6. Why shielding of a reactor is necessary. What do you understand by thermal shielding? 7. Compare the properties of stainless steel and zirconium for use as reactor fuel element
cladding. 8. How induced radioactivity affects the cost of shielding. 9. Considering the problem of induced radioactivity which coolant among water and
sodium is more desirable and why. 10. Discuss the advantages and disadvantages of Lithium, Bismuth and sodium as coolants
for nuclear reactors.
UNIT – 4: GAS AND DIESEL POWER PLANTS
1. What are the main advantages of a combined cycle system in the present power picture
of the world? 2. Draw the line diagrams of repowering system using steam turbine only and boiler only.
Discuss the merits and demerits also. 3. What is the gasification of coal and explain in detail. 4. What are the merits and demerits of using air or O2 in a gasification plant when the
gasification plant is integrated with closed cycle? 5. What do you understand by PFBC, Explain in detail? 6. Draw the line diagrams of two different PFBC systems which are commonly used and
discuss their merits and demerits. 7. What are the main difficulties faced in developing the combined cycles with PFBC. 8. Why and when organic fluid is preferred over water in the bottoming cycle. What are its
advantages? 9. Discuss the part behavior of combined cycle plant and compare with conventional gas
turbine plant of the same capacity. 10. What future developments are expected in combined cycle plants?
UNIT – 5: NON-CONVENTIONAL POWER GENERATION 1. What are the non-conventional sources of energy and why are they seriously thought
throughout the world. 2. What are the different sources of geothermal energy? 3. Discuss the different systems used for generating the power using geo-thermal energy. 4. What are the specific environmental effects if the geothermal source of energy is used for
power generation? 5. What factors are considered for selecting a suitable site for tidal power plants? 6. Differentiate with neat sketches the difference between single basin and double basin
systems.
86
7. List out the advantages of tidal power plants over the conventional hydel power plants. 8. What are the basic requirements for locating a wind power plant? What factors affect
them? 9. What methods are used to overcome the fluctuating power generation of a wind mill? 10. Explain the working of a fuel cell and list out its advantages over other non-
conventional systems of power generation.
V.S.B ENGINEERING COLLEGE, KARUR
DEPARTMENT OFELECTRICALAND ELECTRONICS ENGINEERING
II YEAR / III SEMESTER
DIGITAL LOGICCIRCUITS TWO MARK QUESTION AND ANSWER
UNIT I
NUMBER SYSTEMS AND DIGITAL LOGIC FAMILIES
1. What is meant by parity bit?
A parity bit is an extra bit included with a message to make the total number of 1’s
either even or odd. Consider the following two characters and their even and odd parity:
with even parity with odd parity.
ASCII A = 1000001 01000001 11000001
ASCII T = 1010100 11010100 01010100
In each case we add an extra bit in the left most position of the code to produce an
even number of 1’s in the character for even parity or an odd number of 1’s in the character
for odd parity. The parity bit is helpful in detecting errors during the transmission of
information from one location to another.
2. What are registers?
Register is a group of binary cells. A register with n cells can store any discrete
quantity of information that contains n bits. The state of a register is an n-tuple number of
1’s and 0’s, with each bit designating the state of one cell in the register.
3. Define binary logic?
Binary logic consists of binary variables and logical operations. The variables are
designated by the alphabets such as A, B, C, x, y, z, etc., with each variable having only
two distinct values: 1 and 0. There are three basic logic operations: AND, OR, and NOT.
4. Convert (4021.2)5 to its equivalent decimal.
(4021.2)5 = 4 x 53 + 0 x 52 + 2 x 51 + 1 x 50 + 2 x 5-1
= (511.4)10
87
5. Represent binary number 1101 - 101 in power of 2 and find its decimal equivalent.
N = 1 x 23 + 1 x 22 + 0 x 21 + 1 x 20 + 1 x 2-1 + 0 x 2-2 + 1 x 2-3
= 13.625 10
6. Convert (634)8 to binary.
6 3 4
110 011 100
Ans = (110 011 100)2
7.Convert (9 B 2 - 1A) H to its decimal equivalent.
N = 9 x 16 2 + B x 16 1 + 2 x 16 0 + 1 x 16 -1 + A (10) x 16 -2
= 2304 + 176 + 2 + 0.0625 + 0.039
= 2482.1 10
8.What are the different classifications of binary codes?
Weighted codes
Non - weighted codes
Reflective codes
Sequential codes
Alphanumeric codes
Error Detecting and correcting codes.
9. Convert 0.640625 decimal number to its octal equivalent.
0.640625 x 8 = 5.125
0.125 x 8 = 1.0
Ans. = 0.640 625 10 = 0.51
10. Convert 0.1289062 decimal number to its hex equivalent.
0.1289062 x 16 = 2.0625
0.0625 x 16 = 1.0
Ans. = 0.21 16
11. Convert decimal number 22.64 to hexadecimal number. 22/ 16=6
1/ 16 =1
0.64 x 16 = 10.24
0.24 x 16 = 3.84
0.84 x 16 = 13.44
0.44 x 16 = 7.04
Ans. = (16.A3D7)16.3
12. What are the two steps in Gray to binary conversion?
The MSB of the binary number is the same as the MSB of the gray code number.
So write it down. To obtain the next binary digit, perform an exclusive OR operation
between the bit just written down and the next gray code bit. Write down the result.
13. Convert gray code 101011 into its binary equivalent.
88
Gray Code: 1 0 1 0 1 1
Binary Code:1 1 0 0 1 0
14. Convert 10111011 is binary into its equivalent gray code.
Binary Code: 1 0 1 1 1 0 1 0 1 1
Gray code: 1 1 1 0 0 1 1 0
15. Find 2’S complement of (1 0 1 0 0 0 1 1) 2
0 1 0 1 1 1 0 0 1 1’s Complement
+ 0 1
0 1 0 1 1 1 0 1 0 2’s Complement.
16. What are the advantages of 1’s complement subtraction?
The 1’s complement subtraction can be accomplished with an binary adder.
Therefore, this method is useful in arithmetic logic circuits.
The is complement of a number is easily obtained by inverting each bit in the
number
17.Classify thelogicfamily by operation?
TheBipolarlogicfamilyis classified into
Saturated logic
Unsaturated logic.
TheRTL,DTL, TTL,I2
L, HTLlogiccomes underthesaturated logicfamily.
TheSchottkyTTL,andECLlogic comes undertheunsaturated logicfamily.
18.State the classifications ofFET devices. FET is classifiedas
PART A 1. Form a PDE by eliminating the arbitrary constants ‘a’ and ‘b’ from z = ax2+by2. (AU-A/M- 2017) Given z = ax2+by2
p = 2ax 2
p = ax----------------------------- (1)
q = 2by2
q = by------------------------------- (2)
From (1) and (2) , eliminate a and b
z =22
qypx
2z = px+qy 2. Form the partial differential equation from (x-a)2+(y-b)2+z2 = 1,by eliminating a and b. Partial differentiation w.r.to x and y gives (AU-M/J-2013)-2 2(x-a)+2zp=0 ; (x-a)= - pz 2(y-b)+2zq=0 ; (y-b)= - qz Using these in the given equation we get , p2z2+q2z2+z2=1 3. Form the p.d.e from z=ax3+by3(AU-M/J-2014)-2 z=ax3+by3
………(1)
; ie) p= 3ax2 q= 3ay2
ie)
(1) z= ie) 3z=px+qy
4. Form a PDE by eliminating the arbitrary function f from z = eayf(x+by) (AU-
A/M-2017)
PDE required theis
)(
..)( q
.)(.)().(
1).(
)(
'
'
apbq
e
zbyxf
ae
pbq
aebyxfpb
aebyxfbe
peaebyxfbbyxfe
y
zq
byxfex
zp
byxfez
ayay
ay
ay
ay
ayayay
ay
ay
5. Find the PDE of all spheres whose centre lie on x=y=z. (AU-N/D-2016)-3
23axx
z
23ayy
z
ax
p
23b
y
q
23
b
qypx
3
113
General equation of the sphere is
Here centre is (a,b,c) and radius r. Centre lies on x =y = z . i.e a = b = c.
(2) x – a+zp –ap = 0 ; x + zp = a+ap ; x+zp = a(1+p) ------------- (4) (3) y –a+ zq - aq = 0 ; y + zq = a + aq; y + zq = a(1+q) ------------(5) (4) / (5) x+xq+zp+zpq = y+yp+zq+zpq x+xq+zp-y-yp-zq = 0 ; x-y+(x-z)q+(z-y)p = 0 x-y = (z-x)q+(y-z)p ; (z-x)q+(y-z)p = x-y. 6. Form the partial differential equation by eliminating the arbitrary constants a and b from log(az-1) = x+ay+b (AU-A/M-2015) Given log(az-1) = x+ay+b -------------------------------(1)
Differentiating w.r.t x : 11
1
ap
az --------------------(2)
Differentiating w.r.t y : aaqaz
1
1 ---------------------(3)
From (2) : pz
a
1
--------------------------------------(4)
From (3) : q= az-1 -----------------------------------------(5) Solving (4) and (5) and eliminate ‘a’. p(q+1) = zq This is the required PDE. 7. Form the p.d.e by eliminating the arbitrary function f from z = f(y/x).(AU- N/D- 2012)-2
Given:
x
yfz . -------------- (1)
Diff. (1) p.w.r.to x we get
2
'
x
y
x
yfp
x
z -------- (2)
Diff. (1) p.w.r.to x we get
xx
yfq
y
z 1' -------- (3)
x
y
xx
y
q
p 12
; xp + yq = 0 is the required p.d.e.
8. Form the p.d.e by eliminating the function f from z = f(x2-y2) (AU-N/D- 2017)
Given z =f(x2-y2) Differentiation w.r.to x : p = f’(x2-y2).2x----------------(1) Differentiation w.r.to y : q = f’(x2-y2).-2y ----------------(2) Eliminating f from (1) and (2)
2222 )()()( rczbyax
2222 )()()( rczbyax
x
z
y
z
114
y
x
q
p
2
1
Px – qy = 0. This is the required p.d.e. 9. Form the p.d.e by eliminating the arbitrary function f(x2-y2, z) = 0. (AU -N/D-2014) Given x2-y2 = f(z) Partially differentiating w.r.to x , 2x = f’(z)p------(1) Partially differentiating w.r.to y , -2y = f’(z)q------(2) (1)/(2)implies that p/q = -x/y py = -qx qx+py = 0.This is the required p.d.e. 10. Form the p.d.e by eliminating the arbitrary function from f(x2+y2, z-xy) = 0. (AU-M/J-2016) Given x2+y2 = f(z-xy) Partially differentiating w.r.to x, 2x = f’(z-xy)(p-y)-------(1) Partially differentiating w.r.to y, 2y = f’(z-xy)(q-x)-------(2) (1)/(2)implies that x/y = (p-y)/(q-x) qx-x2 = py-y2 This implies that x2+y2 = qx- py. This is the required p.d.e. 11.Find the complete integral of p+q = 1.(AU- N/D- 2014) Given p+q = 1--------------------------(1) Let z = ax+by+c-----------------------(2)
apx
z
and bq
y
z
-----(3)
Substitute equation (3) in equation (1) , we get a+b = 1 That is b = 1- a--------------------(4) Substitute equation (4) in equation (2),we get Z = ax + (1-a) y+ c is the complete integral. 12. Find the complete solution of the partial differential equation p3-q3 = 0. (AU-A/M-2016) This equation is of the form F(p,q) = 0 Hence the trial solution is z = ax+by+c p= a and q = b Therefore a3 – b3 = 0
13. Find the complete integral of 1 qp (AU-N/D-
2017)
1 qp --- (1)
This is of the type F(p,q)=0
The trial solution is z=ax+by+c
Sub. p=a and q=b in (1)
Therefore (1) implies 21 ab
115
Then, cyaaxz 2
1 which is Complete Integral
14. Find the complete integral of pqp
y
q
x
pq
z
(AU-N/D-2016) This is of the form z = px + qy + f(p,q) Given z = px+qy+(pq)3/2
Hence the complete integral is z = ax+by+(ab)3/2
15. Find the complete integral of p+q=x+y (AU-N/D-2013)-2 Let p+q=x+y=k p-x=k, y-q=k p=k+x, q=y-k
z=
=
= ckyyx
kx 22
22
16. Find the complete solution of q = 2px. (AU-A/M-2015) q=2px=a (say) q=a; p=a/2x dz = (a/2x) dx + ady Integrating, Z = (a/2) logx + ay +b This is the complete solution. 17.Find the general solution of the Lagrange linear equation given by pyz+qzx = xy.
(AU-N/D-2013) This is of the form : Pp+Qq= R
Auxiliary equation is : R
dz
Q
dy
P
dx
xy
dz
zx
dy
yz
dx
Group 1: zx
dy
yz
dx
xdx = ydy Integrating , x2/2 = y2/2+c1
2/2 x2-y2 = u
Group 2: xy
dz
yz
dx
xdx = zdz Integrating, x2-z2 = v Therefore the solution is φ(u,v) = 0 Φ(x2-y2, x2-z2) =0 18. What is the C.F of (D2-DD’)z=x+y
The A.E is m2-m=0 implies m=2(twice)
qdypdx
dykydxxk )()(
116
Therefore C.F = f1(y+2x)+xf2(y+2x)
19. Solve (D4-D’4)z = 0. (AU- M/J -2014) A.E is m4-14 = 0 (m2)2-(12)2 = 0 implies that (m2+1)(m2-1) = 0 m= 1,-1 and m = i,-i z = f1(y+x)+f2(y-x)+f3(y+ix)+f4(y-ix). 20. Solve (D3-3DD’2+2D’3)z = 0.(AU A/M 2018)-3 A.E is m3-3m+2 = 0 m= 1,1,-2 The solution is z = f1(y-2x)+f2(y+x)+xf3(y+x) 21. Solve (D3-D2D’-8DD’2+12D’3)z=0
The A.E is m3-m2-8m+12=0
m=2,2,-3
The Solution is , z= )3()2()2( 321 xyfxyxfxyf
22. Find the particular integral of (D2-D’2+DD’)z = cos(x+y).(AU-N/D- 2012)
P.I =
yxDDDD
cos1
2''2[Replace D2,DD’,D’2 by -1,-1,-1]
= yx cos0
1= yx
DD
cos
2
1'
= yxDD
DD
cos
4
22'2
'
= yxDD
cos
3
2 '
=3
sinsin2
yx
23. Solve x
z
yx
z
x
z
2
2
2
= 0
(AU-N/D- 2013) (D2-DD’+D)=0 D(D-D’+1) = 0 that implies (D-m1D’-0)(D-D’+1) = 0 m1 = 0 , c1 = 0, m2 = 1 ,c2 = -1 The solution is z = e0xf1(y)+e-xf2(y+x)
PART – B 1. a. Form the PDE by eliminating the arbitrary function φ from the relation Φ(x2+y2+z2, xyz) = 0. (AU-M/J-2016)-2-(8) b. Find the partial differential equations of all planes which are at a constant distance ‘k’ units from the origin. (AU-A/M-2016)(8) 2. a. Form the PDE by eliminating the arbitrary function ‘f’ and ‘g’ from z = x2f(y) +y2g(x) (AU-N/D-2013)(8) b. Form the PDE by eliminating the arbitrary functions ‘f’ and ‘ф’ from the relation z = x f(y/x) +y ф(x). (AU-A/M-2016)(8) 3. a. Form the partial differential equation by eliminating arbitrary functions from z = y2+2f(1/x+log y)(AU-M/J-2014)(8) b.Find the complete solution of 9 (p2z + q2) = 4. (AU-N/D-2014)-2(8)
4. a. Find the singular solution of the p.d.e. z = px+qy+ 221 qp
(AU-N/D-2015)-4(8) b. Solve: z = px + qy +p2+p-q2. (AU-M/J-2014)-2(8) 5. a. Find the general solution of z=px+qy+p2+pq+q2 (AU A/M 2018)-3(8) b. Find the singular solution of z = px+qy+p2-q2(AU A/M 2017)-4 (8) 6. a. Solve p2x2+q2y2 = z2 . (AU-N/D-2014)-2-(8) b. Solve z = px+qy+p2q2 and obtain its singular solution.(AU-A/M-2015)- (8) 7. a. Find the complete solution of z2 (p2+q2) = x2 +y2. (AU-A/M-2015)- (8) b. Obtain the complete solution of p2+x2y2q2 = x2z2(AU –M/J-2015)-(8) 8. a. Find the general solution of (z2-2yz-y2)p+(xy+zx)q=xy-zx (AU A/M 2017)(8) b. Find the general solution of (z2-y2-2yz) p+(xy+zx)q = (xy-zx)(AU-A/M-2015)- (8) 9. a. Solve (x2-yz)p + (y2-zx)q = (z2-xy) (AU-A/M-2016)-3(8) b. Solve (x-2z)p + (2z - y)q = y-x (AU-A/M-2017)(8)
10. a. Solve x(y-z)p+y(z-x)q = z(x-y)(AU-M/J-2014)-2(8) b. Solve x(z2-y2)p + y(x2-z2)q =z(y2-x2) (AU-M/J-2016)-3(8) 11. a. Solve (D3-2D2D’)z = 2e2x+3x2y.(AU-A/M-2016)(8) b. Solve (D2-5DD’+6D’2)z = ysinx (AU N/D 2017)(8) 12. a. Solve (D2+DD’-6D’2)z = ycosx (OR) (r+s-6t)=y cosx. (AU-A/M-2018)-3(8) b.Find the general solution of (D2+2DD’+D’2)z=xy+ex-y(AU N/D 2017)(8) 13. a. Solve (D3-7DD’2-6D’3)z = sin(2x+y) (AU-M/J- 2013)(8) b. Find the general solution of (D2+2DD’+D’2)z=x2y+ex-y (AU-A/M-2017)(8) 14. a.Solve: (D2-3DD’+2D’2)z = (2+4x)ex+2y (AU-N/D-2015)-2(8) b.Find the general solution of (D2-3DD’+2D’2+2D-2D’)z=sin(2x+y)
(AU A/M 2017)(8) 15. a. Solve (D2+2DD’+D’2-2D-2D’)z = sin(x+2y)(AU-N/D-2015)(8) b. Solve : (D2+4DD’-5D’2)z=sin(x-2y)+e2x-y (AU-N/D-2017)(8)
UNIT - II FOURIER SERIES
PART A
118
1. State Dirichlet’s conditions for a given function to expand in Fourier series. (AU -N/D- 2017)-8
Let f(x) be defined in the interval c<x<c+2 π with period 2 π and satisfy the following conditions: (1) f(x) is single valued (2) It has a finite number of discontinuities in a period of 2 π. (3) It has a finite number of maxima and minima in a given period.
(4) is convergent
These conditions are Dirichlet’s conditions. 2. State the sufficient conditions for existence of Fourier series. (AU A/M 2017)-2 The sufficient conditions for existence of Fourier series is given by (i) f(x) is defined and single valued except possibly at a finite number of points in (-π, π). (ii) f(x) is periodic with period 2π. (iii) f(x) and f ’(x) are piecewise continuous in (-π, π), then the Fourier series of f(x) converges to
(a) f(x) if x is a point of continuity
(b)
2
00 xfxf if x is a point of discontinuity.
3. Find the value of the Fourier series of f(x) = 0 in (-c,0) = 1 in (0,c) , at the point of discontinuity x = 0. (AU-M/J-2016)
The value of the Fourier series is f(x) =
2
00 ff =
2
1
2
10
4. If f(x) is discontinuous at a point x=a, then what does its Fourier series represent at that point. (AU-N/D-2017) If f(x) is discontinuous at a point x=a, then at that point f(x) cannot be expanded as Fourier series. 5. Find the constant term in the Fourier series corresponding to f(x) = cos2x expressed in the interval (-π, π).(AU-M/J-2012)
Given f(x) = cos2x = 2
2cos1 x
We know that f(x) = nxbnxaa
n
n
n
n sincos2 11
0
a 0 = xdxxdx
0
22 cos2
cos1
= dxxdxx
00
2cos11
2
2cos12 = 1
Therefore the Constant term = 2
1
2
0 a
6.Write down the form of the Fourier series of an odd function in (-l,l) and associated Euler’s formulas for Fourier coefficients.(AU-N/D-2013)
f(x) = l
xnb
n
n
sin
1
dxxf
c
c
)(
2
119
bn = dxl
xnxf
l
l
l
sin
1
7. Find the co-efficient bn of the Fourier series for the function f(x)=xsinx in (-2,2). (AU-N/D-2012) f(x) = xsinx f(-x) = -xsin(-x) = xsinx = f(x) Therefore f(x) is an even function. Therefore bn = 0.
8. Find a0 in the expansion of f(x)=ex as a Fourier series in ; 0<x<2π
(AU-N/D-2013)
1111)(
1 2
2
0
2
0
2
0
0
eedxedxxfa xx
9. Determine the Fourier series for the function f(x) = x in x . (AU-N/D-2015)
f(x) = x f(-x) = -x = -f(x) Therefore f(x) is an odd function .Therefore a0 = an = 0
bn =
nxdxxnxdxxf sin1
sin1
u = x v = sin nx u’ = 1 v1 = -cosnx/n u’’ = 0 v2 = -sinnx/n2
bn =
nnn
nx
n
nxxnn
12121sincos12
f(x) =
nxnn
n
sin12
1
10. Find the value of bn in the Fourier series expansion of f(x) = x+ π in (-π, 0) = -x + π in (0, π) (AU-M/J-2016)
Let f(x)= ф1(x) , (-π, 0) = ф2(x) , (0, π) ф1(x) = x+ π , ф2(x) =-x + π ф1(-x) = - x+ π = ф2(x) f(x) is an even function. Therefore bn=0. 11. Find bn in the expansion of f(x)=x2 as a Fourier series in ; -π<x<π
(AU-N/D-2017)
f(-x)=(-x)2=x2=f(x)
Therefore f(x) is an even function.
0 nb
12. Find the sum of the Fourier series for f(x) = x+x2 in – π<x< π at x = π. (AU-A/M-2017)
120
x= π is an end point. Sum of Fourier series =
2
ff =
2222
2
2
2
13. If
1
2
22 cos
43 n n
nxx
in 0<x<2π, then deduce that the value of
12
1
n n
(AU-N/D-2014)
Put x = 0,
1
2
2
12
2
12
22 1
6
1
12
214
3 nnn nnn
14. Expand f(x) = 1 as a half range sine series in the interval (0, π) (or) Find the sine series of function f(x) = 1 , 0 ≤ x ≤ π (AU-A/M-2015)-3
The half range sine series formula is f(x) = nxbn
n sin1
Where
nnn
nxnxdxnxdxxfb
n
n
112cos2sin
2sin
2
000
= oddisnif
nn
n
411
2
, 𝑓(𝑥) = ∑4
𝑛𝜋𝑠𝑖𝑛𝑛𝑥𝑛isodd
15. Expand f(x)=x in (0,1) as a Fourier sine series
1
sin)(n
n xnbxf
0 0
sin2
sin)(2
nxdxnxdxxfbn
odd isn ,n
4
even isn ,0
112cos2
0
when
when
b
nn
nxb
n
n
n
f(x)=
1
sin44
n
nxn
16. If f(x) is expanded as a half range cosine series, express dxxf
l 2
0
)( in terms of
a0 and an.(AU-N/D-2011)
dxxf
l 2
0
)( =
1
2
2
0
24 n
nalla
17. Write the complex form of the Fourier series of f(x).(AU N/D 2017)-3 The series for f(x) defined in the interval (C,C+2π) satisfying the Dirichlet’s Conditions can be put in the complex term as
121
f(x) = where
18.Find the complex form of Fourier series for f(x) = ex ; -π<x<π and f(x+2π) = f(x)
(AU-N/D-2017)
We know that f(x) = inx
n
neC
Where Cn =
sinh1
11
12
1
2
1
2
122
11
n
in
in
edxedxee
nxinxininxx
19. Write the complex form of the Fourier series of f(x) defined in –l<x<l
(AU-N/D-2017)-3
The series for f(x) defined in the interval (-l,l) satisfying the Dirichlet’s Conditions can be put in the complex term as
n
l
xin
necxf
)( where dxexfl
c
l
l
l
xin
n
)(2
1
20. State Parseval’s Theorem on Fourier series.(AU-A/M- 2017)-3 If f(x) is expressed as a Fourier series in the interval (a,b) then
Where a0 , an , bn are the Fourier constants and is the
R.M.S. value. 21. Define root mean square value of a function f(x) in a<x<b. (AU A/M 2018)-4 Let f(x) be a function defined in an interval (a,b) then ,
R.M.S = is called the root mean square.
22. Find the root mean square value of f(x) = x2 in the interval (0, π). (AU-A/M-2017)
RMS value 5
20
4
dxx
x
23. Find the root mean square value of the function f(x) = x in the interval (0,l) (AU –N/D-2017)-3
R.M.S = in the interval (a,b) = Here a = 0 ; b = implies that
24. Find the R.M.S value of f(x) = x(l-x) in 0≤x≤l. (AU-N/D-2015)
RMS value in (0,l) is
inx
neC
dxexfC
C
C
inx
n
2
)(2
1
1
222
02
2
1
4nn ba
ay y
ab
dxxf
b
a
2
)(
ab
dxxf
b
a
2
)(
0
0
2
l
dxx
l
33
2 ll
122
dxxfl
yl 2
0
2 2 Here l = 1
30
17
453
2
4
2
53
22 45551
0
453221
0
22 llll
l
lxxxl
ldxxlx
ly
25 . What do you mean by Harmonic Analysis?(AU –M/J-2013)-2 When a function f(x) is given by its numerical values at q equally spaced points, the Process of determining the co-efficient of Fourier series representing f(x) using numerical integration is known as Harmonic Analysis.
PART-B
1. a. Determine the Fourier series for the function f(x) = xcosx in (0,2π) (AU-A/M 2017)(8) b. Find a Fourier series with period 3 to represent f(x) = 2x-x2 in (0,3). (AU-N/D-2014)(8) 2. a. Find the Fourier series expansion of f(x) = 1 for 0<x< π = 2 for π<x<2π (AU-N/D-2013)(8) b.Find the Fourier series expansion the following periodic function of period 4 f(x) = 2+x , -2≤x≤0
= 2-x , 0<x≤2. Hence deduce that 222 5
1
3
1
1
1 ---- =
8
2
(AU-A/M-2015)(8) 3. a. Find the Fourier series of f(x) = x in - π <x<π. (AU-N/D-2016)(8)
b. Find the Fourier series for the function f(x) = xcos in - π <x<π. (AU-A/M-
2016)(8) 4. a. Determine the Fourier series for the function f(x) = x2 of period 2π in - π <x<π. Hence
deduce the value of
12
1
n n
(AU A/M 2018)-2-(8) b. Expand f(x) = x2 as a Fourier series in the interval (-π,π) and hence deduce that
90..............
4
1
3
1
2
11
4
444
(AU-N/D-2015)(8)5. a. Find the Fourier series
expansion of the periodic function f(x) of the period 2 defined by f(x) = l-x , 0<x≤l = 0 , l<x≤2l . in (o.2l) (AU- N/D 2017)(8)
b. Find the Fourier series for the function f(x) = xsin over the interval (-π,π).
(AU-A/M-2015)(8)
123
6. a. Obtain the Fourier series for the function given by f(x) =
lxinl
x
xlinl
x
02
1
02
1
Hence deduce that 8
................5
1
3
1
1
1 2
222
(AU-N/D-2014)-2(8)
b. Expand f(x) = x+x2 as a Fourier series in (-π,π) and hence deduce the value of (AU- N/D 2017)(8) 7. a. Find the half range sine series of f(x) = x cosπx in (0,1). (AU-N/D-2016)(8) b. Find the half range sine series of f(x) = 4x-x2 in the interval (0,4).Hence deduce the value of
the series ...........7
1
5
1
3
1
1
13333
(AU-N/D-2014)(8)
8. a. Find the half range sine series of f(x) = x , 0<x< π/2 = π –x , π /2 < x < π.
Hence deduce the sum of the series
12
12
1
n n
(AU-A/M-2017)(8) b.Find the half range cosine series for f(x)=x( -x) in (0, )
(AU-A/M-2018)(8)
9. a. Find the half range cosine series of f(x) = x in 0<x< . Hence deduce the value of
...............5
1
3
1
1
1222
to (AU- N/D 2017)-2-(8)
b. Find the half range cosine series expansion of (x-1)2 in 0<x< l. (AU-N/D-2014)(8)
10. a. Find the Half range cosine series of f(x) = sinx in (0,π) (AU-N/D-2015)(8)
b. Expand f(x) = x, 0< x <1 = 2-x, 1 < x <2 as a series of cosine in the interval (0, 2).
(AU A/M 2017)(8) 11. a. Find the complex form of the Fourier series f(x)=e-ax in the interval;–π<x<π.(AU A/M 2017)-2(8) b. Find the complex form of Fourier series of the function f(x) = ex in -π<x<π
(AU-M/J-2016)-(8) 12. a. Find the complex form of Fourier series of the function f(x) =sinx in -π<x<π
(AU-A/M-2014)-2-(8)
b. Find the complex form of Fourier series of the function f(x) =e-x in (-l,l) (AU-M/J-2016)-2-(8) 13. a. Find the complex form of the Fourier series f(x)=eax in the interval – π<x<π
where ‘a’ is a eeal constant. Hence deduce that
aanan
n
sinh
122
(AU-N/D-2015)(8) b. Calculate the first three harmonic of Fourierseries from the following data
124
x 0
2
y 1.0 1.4 1.9 1.7 1.5 1.2 1.0 (AU A/M 2018)-7(8) 14. a. Obtain the constant term and the coefficient of the first sine and cosine terms in the Fourier expansion of y as given in the following table: (AU A/M 2017)-5(8)
b.Compute first two harmonic of the Fourier series for f(x) from the table below:
(AU-A/M-2010)(8)
x : 0 600 1200 1800 2400 3000 y: 1.98 1.30 1.05 1.30 -0.88 -0.25 15. a. Determine the first two harmonics of Fourier series for the following data (AU-A/M-2015)(8)
x 0
3
3
2
3
4
3
5
2
y 1.98 1.30 1.05 1.30 -0.88 -0.25 1.98 b. Find the Fourier cosine series up to third harmonic to represent the function given by the following table (AU-N/D-2015)-2(8)
UNIT – III APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS
PART-A
1. Classify the differential equation 026432
22
2
2
u
y
u
y
u
yx
u
x
u
(AU-N/D- 2013) A = 3, B = 4, C = 6 B2-4AC = 16-72 <0 Therefore, elliptic equation. 2. Classify the partial differential equation uxx+uyy = f(x,y) (AU-M/J- 2016) A = 1,B = 0,C = 1 B2-4AC = -4 < 0 Elliptic equation 3. Classify the PDE of uxy=uxuy+xy (AU-N/D-2017)
Here B=1,A=0,C=0
B2-4AC=1>0
Given PDE is Hyperbolic equation.
4. Classify the partial differential equation (1-x2) zxx – 2xy zxy + (1-y2) zyy + x zx
3
3
2
3
4
3
5
x 0 1 2 3 4 5 y 9 18 24 28 26 20
x 0 1 2 3 4 5 y 4 8 15 7 6 2
125
+3x2y zy -2z = 0 (AU-A/M-2015)-2
A = (1-x2) ,B = -2xy , C = (1-y2) B2-4AC = 4x2y2 – 4(1-x2)(1-y2) = 4x2y2-4+4x2+4y2-4x2y2 = 4x2+4y2-4 x = y = 0, B2-4AC = -4 < 0 , Elliptic equation x = y = positive, B2-4AC = 4 > 0, Hyperbolic equation x = y = negative, B2-4AC = 4 > 0, Hyperbolic equation 5. Find the nature of the p.d.e 4uxx+4uxy+uyy+2ux-uy = 0 A = 4, B = 4, C = 1 B2-4AC = 0 Therefore, Parabolic equation
6. Use method of separation of variables, Solve ut
u
x
u
2 , where u(x,0)=6e-3x
(AU A/M 2017)
)23(
3
2
1
6
3,6
6)0,(
)0,(
tx
x
kx
lkkx
eu
kab
exu
abexu
eabeu
7. Write down the three mathematically possible solutions of one dimensional wave equation.(AU-A/M-2015)-5 y(x,t) = (c1epx+c2e-px)(c3epat+c4e-pat) y(x,t) = (c1coskpx+c2sinpx)(c3cospat+c4sinpat) y(x,t) = (c1x+c2)(c3t+c4)
8. In the wave equation 2
22
2
2
x
yc
t
y
, what does c2 stand for?
stringtheoflengthunitmass
Tension
m
Tc
/
2
9. What is the basic difference between the solutions of one dimensional wave equation and one dimensional heat equation with respect to the time?(AU- N/D- 2017)-2 Solution of the one dimensional wave equation is of periodic in nature. But solution of the One dimensional heat equation is not of periodic in nature. 10. State the assumptions in deriving one-dimensional wave equation.
(AU- N/D- 2017)-3 (i) The motion takes place entirely in one place i.e., xy plane. (ii) Only transverse vibrations are considered. The horizontal displacement of the particles of the string is negligible. (iii) The tension T is constant at all times and at all points of the deflected string. (iv) T is considered to be so large compared with the weight of the string and hence the force of gravity is negligible.
126
(v) The effect of friction is negligible. (vi) The string is perfectly flexible,i.e., it can transmit tension but not bending or sheering forces. (vii) The slope of the deflection curve at all points and at all instants is so small that sin α can be replaced by α, where α is the inclination of the tangent to the deflection curve. 11. Write down the diffusion problem in one-dimensional as a boundary value problem
in two different forms. (AU-M/J-2013)
one dimensional heat flow
Where s
Ka 2 is known as diffusivity of the material of the bar.
In the steady state 02
2
dx
ud
12. Write down one-dimensional heat equation and all possible solution for the same. (AU-A/M-2018)-10 ut=α2uxx
.
13. How many conditions are needed to solve the one dimensional heat equation? (AU-M/J- 2009)
Totally three conditions needed.
14. State the suitable solution of the one dimensionalheat equation
(AU-A/M-2017)
The suitable solution of the given equation is u(x,t) = (Acospx+Bsinpx) 15. State the governing equation for one dimensional heat equation and necessary conditions to solve the problem.
The one dimensional heat equation is where u(x,t) is the temperature
at time t at a point distant x from the left end of the rod. The boundary conditions are
a)u(0,t) = k10C for all t≥0
b)u(l,t) = k20C for all t≥0
c)u(x,0) = f(x) , 0<x<l 16. A rod of 30cm long has its ends A and B kept at 200C and 800C respectively until steady state conditions prevail. Find the steady state temperature in the rod.
(AU-A/M-2015)-2
2
22
x
ua
t
u
txx eCeBeAtxu22
111,
teCxBxAtxu22
222 sincos,
333, CBxAtxu
2
22
x
ua
t
u
tpce22
2
22
x
ua
t
u
127
The steady state equation of the one dimensional heat flow is =0 ..........(1)
The general Solution of (1) is u(x) =ax+b ......(2) The boundary conditions are u(0) = 30, and u(30)=80
Put x=0 in (2) , u(0)=b=20 Put x=30in (2) , u(30)=30a+b =80 30a+b=80 that implies 30a=60 implies that a=2 and equation (2) implies
u(x)=2x+20 17. A rod of length 20cm whose one end is kept at 300C and the other end is kept at 700C is maintained so until steady state prevails. Find the steady state temperature. (AU-N/D-2014)-2
The steady state equation of one dimensional heat flow is =0 ......(1)
The general Solution of (1) is u(x) =ax+b ......(2) The boundary conditions are u(0) = 30, and u(l)=70 Put x=0 in (2) , u(0)=b=30 Put x=l in (2) , u(l)=al+b =70 al=40 that implies a=40/l
(2) implies u(x)= Here l=20.Therefore u(x)=2x+30
18. An insulated rod of length l cm has its ends A and B maintained at 00C and 800C
respectively. Find the steady state solution of the rod. (AU-N/D-2013)
The steady state equation of the one dimensional heat flow is =0 .......(1)
The general Solution of (1) is u(x) =ax+b ......(2) The boundary conditions are u(0) = 0, and u(l)=80
Put x=0 in (2) , u(0)=b=0 Put x=l in (2) , u(l)=la+b =80 la+b=80 that implies la=80 implies that a=80/l and equation (2) implies
u(x)=(80/l)x 19. A bar of length 50cm has its ends kept at 200C and 1000C until steady state conditions prevail. Find the temperature at any point of the bar. (AU-A/M-2014)
The steady state equation of one dimensional heat flow is =0 .......(1)
The general solution of (1) is u(x)=ax+b .........(2) Put x=0 in (2) , u(0) = b = 20 Put x=l in (2), u(l) = al+b=100 al=80 a=80/l
a=80/50 that is a=8/5. Equation (2) implies u(x) =
20. A rod of 60cm long has its ends A and B kept at 200C and 800C respectively until
steady state conditions prevail. Find the steady state temperature in the rod.
(AU-N/D-2012)
2
2
dx
ud
2
2
dx
ud
3040
xl
2
2
dx
ud
2
2
dx
ud
205
8x
128
The steady state equation of the one dimensional heat flow is =0 ..........(1)
The general Solution of (1) is u(x) =ax+b ......(2) The boundary conditions are u(0) = 20, and u(60)=80
Put x=0 in (2) , u(0)=b=20 Put x=60in (2) , u(60)=20a+b =80 20a+20=80 that implies 20a=60 implies that a=3 and equation (2) implies
u(x)=3x+20
21. In 2D heat equation or Laplace equation, what is the basic assumption. When the heat flow is along the curves instead of straight lines, the curves lying in parallel planes the flow is called two dimensional.
22. Write all possible solutions of two dimensional heat equations 02
The two-dimensional Laplace equation is given by =0 i.e.,
24. An infinitely long rectangular plate with insulated surface is 10 cm wide. The two long edges and one short edge are kept at zero temperature while the other short edge x=0 is kept at temperature given by u = 20y , 0≤y≤5 = 20(10-y), 5≤y≤10. Give the boundary conditions.
The equation to be solved is =0.The boundary conditions are
(i) u(x,0) = 0 for all x (ii) u(x,10) = 0 for all x (iii) u(∞,y) = 0 (ie) when x→∞, u→0 (iv) u(0,y) = 20y , 0≤y≤5
= 20(10-y), 5≤y≤10 25. A square plate has its faces and the edge y=0 insulated. It’s edges x=0 and x=π are kept at zero temperature and its fourth edge y =π is kept at temperature πx-x2
.Write the boundary conditions alone.
The equation to be solved is =0. The boundary conditions are
a) u(0,y) = 0 0≤ y≤π b)u(π,y) = 0 0≤ y≤π
c) 0≤ x≤π
d)u(x, π) = πx-x2 0<x< π PART-B
2
2
dx
ud
2
2
2
2
y
u
x
u
,02 u
2
2
2
2
y
u
x
u
2
2
2
2
y
u
x
u
0)( 0
y
y
u
129
1. A string is stretched and fastened to 2 points x=0 and x=l . Motion is started by displacing the string into the form y=kx(l-x) or y=l(lx-x2) from which it is released at time t=0. Find the displacement at any point on the string at a distance of x from one end at time t. (AU-A/M-2018)-7(16) 2. A string of length 2l is fastened at both ends. The midpoint of the string is taken to a height b and then released from rest in that position. Find the displacement. (AU-A/M- 2017)-3(16) 3. A string is stretched and fastened to points at a distance l apart. Motion is started by displacing the string in the form y=a sin(πx/l),0<x<l, from which it is released at time t=0. Find the displacement at any time t.(AU-M/J- 2014)(16) 4. A tightly stretched string of length ‘l’ with fixed end points is initially at rest in its equilibrium position.If it is set vibrating by giving each point a velocity
l
x
l
xvxy t
cos
3sin0, 0 ,where 0<x<l.Find the displacement of the string at a point,
at a distance x from one end at any instant‘t’.(AU –N/D-2012)(16) 5. A tightly stretched string with fixed end points x=0 and x=l is initially at rest in its equilibrium position. if it is set vibrating giving each point a velocity λx(l-x), show that
l
atn
l
xn
na
ltxy
n
12sin
12sin
12
18,
144
3
(AU –N/D-2014)(16) 6. Find the displacement of a string stretched between two fixed points at a distance of 2l apart when the string is initially at rest in equilibrium position and points of the
string are given initial velocities v = l
xin (0, l )=
l
xl 2 in (l, 2l), x being the distance
measured from one end. (AU –M/J-2016)16) 7. If a string of length l is initially at rest in its equilibrium position and each of its points is given a velocity v such that V=2 kx/l for 0<x<l/2 =2k(l - x)/l for l/2 <x <l (AU –N/D-2015)(16)
8. A bar, 10cm long with insulated sides, has its ends A and B kept at 200C and 400C
respectively until steady state conditions prevail. The temperature at A is then suddenly raised
to 500C and at the same instant that at B is lowered to 100C. Find the subsequent temperature
at any point of the bar at any time. (AU-A/M-2018)(16) 9.A rod of length l has its end A and B kept at 00C and 1000C respectively until steady state conditions prevail. If the temperature at B is reduced suddenly to 750C and at the same time the temperature at A raised to 250C find the temperature u(x,t) at a distance x from A and at time t. (AU-N/D 2017) –2(16) 10. An insulated rod of length l its ends A and B are maintained at 00C and 1000C respectively untiL steady state conditions prevail. If B is suddenly reduced to 00C and maintained so, find the temperature at a distance x from A at time t. (AU N/D 2017) – 2-(16)
11. A square plate is bounded by the lines x=0,y=0,x=20,y=20. Its faces are insulated. The
130
temperature along the upper horizontal edge is given u(x,20)=x(20-x) while the other three edges are kept at 00C. Find the steady state temperature in the plate.
(AU-N/D- 2014)-2(16) 12. A square plate is bounded by the lines x=0,y=0,x=l and y=l, it faces are insulated. The temperature along the upper horizontal edge is given by u(x,l)=x(l-x) when 0<x<l while the other three edges are kept at 00C. Find the steady state temperature in the plate. (AU-N/D- 2013)(16)
13. A long rectangular plate with insulated surface is l cm wide.If the temperature along one short edge is u(x,0) = k(lx-x2) for 0 < x < l,while the other two long edges x = 0 and x= l as well as the other short edge are kept at 00 C, find the steady state temperature function u(x,y). (AU-N/D- 2016)(16) 14.A rectangular plate with insulated surface is 20cm wide and so long compared to its width that it may be considered infinite in the length without introducing an appreciable error. If the temperature along one short edge x=0 is given by u=
2010),20(10
5100,10
yy
yy and the two long edges as well as the other short edge are kept
at 00C, find the steady state temperature distribution u(x,y) in the plate. (AU-A/M -2017) (16) 15. An infinitely long rectangular plate with insulated surface is 10 cm wide. The two long edges and on short edge are kept at zero temperature, while the other short edge
y=0 is kept at temperature given by u=
105),10(20
50,20
Xx
Xx Find the steady state
temperature distribution in the plate. (AU –M/J-2014)-2-(16)
UNIT – IV FOURIER TRANSFORMS
PART-A 1. State Fourier integral theorem
(AU-A/M-2015)-6 If f(x) is piece-wise continuously differentiable and absolutely integrable in (-∞, ∞), then
f(x) = f(t)eis(x-t)dtds
or equivalently
This is known as Fourier integral theorem or Fourier integral formula. 2. Show that f(x)=1, 0<x< cannot be represented by a Fourier integral.
00
.1)( xdxdxxf
Therefore the given function cannot be represented as Fourier integral 3. Write the Fourier transforms pair.
2
1
dtdxttfxf )(cos)(1
)(0
131
(AU–M/J-2011)-2 If f(x) is a given function then F[f(x)] and F-1[f(x)] are called Fourier transform pair
F(s)=F[f(x)]=
dxexf isx)(
2
1
; The inverse f(x) =
dsesF isx)(
2
1
4.Prove that )()( sFxfF
(AU-N/D-2012)
dxexfsF
dxexfsF
isx
isx
)(2
1)(
)(2
1)(
Taking complex conjugate on both sides we get
)()(2
1)( xfFdxexfsF isx
5. Find the Fourier transform of f(x) if AU-N/D- 2014)-3
1 ; |x|<1
f(x) = 0 ; |x|>1>0
We know that F[f(x)] = = dxe isx
1
1
=
ssi
ee
sis
e
is
e
is
e isisisisisx
sin2
2
21
1
6. Find the (complex) Fourier transform of f(x)=
bxax
bxaeikx
,0
,
(AU-A/M- 2010)
F(s)=F[f(x)]= dxexf isx
)(2
1
= dxee isxikx
2
1= dxe xski
)(
2
1
=
b
a
xski
ski
e
)(2
1 )(
= ][
)(2
)()( askibski eesk
i
= ][
)(2
)()( bskiaski eesk
i
7. State Parseval’s identity in Fourier transforms. (AU-M/J- 2011)-2 Parseval’s identity in Fourier transforms is given by
If F(s) is the Fourier transform of f(x), then
8. State and prove modulation theorem on Fourier transform. (AU-A/M-2014)-2
Statement: If F(s) is the Fourier transform of f(x), then F[f(x)cosax] = 2
1[F(s+a)+F(s-a)].
dxexf isx)(
dssFdxxf
22
)()(
132
Proof:
dxeee
xfdxaxexfaxxfF
dxexfxfF
isxiaxiax
isx
isx
22
1cos
2
1cos
2
1
= dxeexf xasixasi
2
1
2
1=
2
1[F(s+a)+F(s-a)].
9. If F(s) is the Fourier transform of f(x) , then show that the Fourier transform of eiaxf(x) is F(s+a)(AU-A/M-2015)-4
F[eiaxf(x)] = = = F(s+a)
10. State and prove first shifting theorem. (AU -A/M -2017)-3 First shifting theoremis given by F[f(x-a)] = eiasF(s)
Proof : F[f(x-a)] =
Put x-a = y when x = - , y =- dx = dy when x = , y =
= = = = eias
F(s) 11. State change of scale property of Fourier transforms. (AU-N/D-2017)-3 Change of scale property of Fourier transforms is given by
If Ff(x) = F(s) then Ff(ax)= where a≠0
Proof :
dxaxfeaxfF isx )(2
1)((
Put t=ax , dt=adx =>dx=1/a dt
dttfea
axfF asit )(2
1)(( )/(
By definition,
dttfe
sFist )(
2
1)(
0)/(1
0)/(1
)((
aforasFa
aforasFa
axfF
Ff(ax)= where a ≠ 0
12. Find Fourier sine transform of (AU A/M 2017)-5
We know that
Fs[f(x)] = f(x)sin sx dx = sinsx dx = =
dxexfe isxiax )(2
1
dxxfe xasi )(2
1 )(
2
1
dxeaxf isx)(
2
1dyeyf ayis .).( )(
dyeyfe isy
ias
).(2
dxexfe isx
ias
).(2
a
sF
ac
1
a
sF
ac
1
x
1
2
0
2
0x
1
2
2
2
133
= , a>0
13. Find the Fourier sine transform of e-ax
We know that Fs[f(x)] = f(x)sinsxdx
Fs[e-ax] = e-axsinsxdx = 22 sa
s
14. If Fs(s) is the Fourier sine transform of f(x), show that (AU- N/D-2017)-3
Fs(f(s)sinax)=
Fs(f(x)sinax) = )()(2
1asFasF CC
=
0
)()(2
1)(
2dxxasCosxasCosxf
=
0 0
)()()()(2
2
1xdxasCosxfxdxasCosxf
= )()(2
1asFasF CC
15. Define Fourier cosine transform and its inversion formula The infinite Fourier cosine transform of f(x) is defined as
Fc[f(x)] = Fc(s) = f(x)cos sx dx
The inversion formula is f(x) = Fc(s) cos sx ds
16. Find the Fourier cosine transform of(AU-N/D-2011)-3 Cosx ;if 0<x<a
f(x) = 0 ; if x ≥a
Fc(s) = cos x cos sx dx = cos x cos sx dx
= =
= = provided
s≠1;s≠ -1
17. Find Fourier cosine transform of ,a>0. (AU-N/D- 2015)-3
We know that
dxx
axsin
0
2
2
0
2
0
2
)]()([2
1asFasF ss
2
0
2
0
2
0
2a
0
2a
sxdxx0
coscos2
1
2
a
dxxsxs0
])1cos()1[cos(2
1
a
s
xs
s
xs
01
)1sin(
1
)1sin(
2
1
1
)1sin(
1
)1sin(
2
1
s
as
s
as
axe
0
cos)(2
)]([ sxdxxfxfFc
134
= since
18. Find Fourier cosine transform of xe ,a>0. (AU-N/D- 2015)-3
We know that,
1
12cos
2cos)(
2))((
2
00s
sxdxesxdxxfxfF x
c
19. Prove that Fc[f(x)cosax ] = where Fc denotes the Fourier
cosine transform f(x). (AU-M/J- 2011)-3
Fc[f(x)cosax] = f(x) cosax cos(sx) dx = f(x)
= + =
20. If Fc(s) is the Fourier cosine transform of f(x). Prove that the Fourier cosine
transform of f(ax) is (AU-N/D- 2015)
To prove: Fc (f (ax)) is
We know that Fc[f(ax)] = f(at) cosst dt
Put at=u when , adt = du
= = =
21. Find the Fourier sine transform of e-3x (AU –M/J-2013)
We know that Fs[f(x)] = f(x)sinsxdx = e-3xsinsxdx = (s/s2+9)
22.Given that of 22xe is self reciprocal under Fourier cosine transform, find
Fourier sine transform of (AU-A/M- 2015)
WKT FC [ 22xe ]= 2
2s
e
, Fs [ ]=-d/ds Fc [ [ ] =-d/ds[ 2
2s
e
]=- 2
2s
e
(-
s)=s 2
2s
e
23.Define the Convolution of two function:
dttxgtfxgf )()(2
1)(*
24. State the convolution theorem for Fourier
transforms. (AU-A/M-2018)-3
0
cos2
][ sxdxeeF axax
c
22
2
as
a
0
22cos
ba
abxdxe ax
)()(2
1asFasF cc
2
0
2
0
dxxasxas
2
)cos()cos(
0
)cos()(2
2
1xdxasxf
0
)cos()(2
2
1xdxasxf
)]()([2
1asFasF cc
a
sF
ac
1
a
sF
ac
1
2
0
utut ,00
2
0
cos)(a
du
a
suuf
0
cos)(21
tdta
stf
a
a
sF
ac
1
2
0
2
0
2
2/2xxe
2/2xxe 2/2xxe
135
If F(s) and G(s) are the Fourier transform of f(x) and g(x) respectively, then the Fourier transform of the convolution of f(x) and g(x) is the product of their Fourier transform. F[f(x)g(x] = F(S)G(S) = F[f(x)]F[g(x)] and F-1[F(S)G(S)]= F-1[F(S)] * F-1[G(S)]
25. State Parseval’s identity of Fourier transform
Let F(s) be the Fourier transform of f(x) , then
dssFdxxf22
)()(
PART-B
1. a. Express the function
1,0
1,1)(
x
xxf as a Fourier Integral. Hence evaluate
d
x
0
cossin &
d
0
sin (AU-N/D-2015)-2- (8)
b. Solve for f(x) from the integral equation (AU-A/M-2014)-(8)
2. a. Solve for f(x) from the integral equation (AU-N/D-2015)(8)
2,0
21,2
10,1sin0
s
s
ssxdxxf
b. Find the Fourier transform of 1, f(x) = 1 for |x| < 1 0 otherwise. Hence prove that
0 0
2
2
2
sinsin dx
x
xdx
x
x(AU-M/J-2016 (8)
3. a. Find the Fourier transform of e-a|x|. Hence Deduce that
(i) )(
22][
22 as
asixeF
xa
(ii)
xae
adt
ta
xt
2
cos
0
22
(AU-N/D-2014)-4(8)
b. Show that the Fourier transform of 2
2x
e
is 2
2s
e
AU-A/M-2018)-3(8) 4. a. Find the Fourier transform of f(x) given by f(x) = 1-x2 for |x| ≤1
0 for |x| ≥1
Hence evaluate (AU A/M 2018)-6 (8)
b. Find the Fourier transform of f(x) given by f(x) = a2-x2 for |x| <a 0 for |x| >a>0
P.T =3𝜋
16 (AU-N/D-2015)-5(8)
5. a. Find the Fourier transform of f(x) if (AU N/D 2017)-3(8)
0
cos)( exdxxf
dxx
x
xxx
2cos
cossin
0
3
dxx
x
xxx
2cos
cossin
0
3
136
1-|x| for |x|<1 f(x) = 0 for |x|>1 . Hence deduce that
3
sin
2
sin4
0
2
0
dtt
tanddt
t
t
b. Find the Fourier transform of e-a|x| if a>0 .Deduce that
0,4)(
13
0
222
aifa
dxax
(8)
6. a. Find Fourier transform of ,a>0 and hence find is self reciprocal under
the Fourier transform. (AU-A/M-2016-3)(8)
b. Solve the integral equation 0cos)(0
whereexdxxf (AU-A/M-2016)(8)
7. a. Find the Fourier transform of ),()( 2
2
inexf
x
(AU-A/M-2018)-2-(8)
b. Find the Fourier Sine transform of (8) f(x) = sinx, 0<x<π = 0 π<x<∞ 8. a. Find the Fourier sine transform of e-ax, a>0 and hence deduce the inversion formula. (8)
b. Find the Fourier sine transform of (AU-A/M-2016)-3(8)
9. a. Find the Fourier Sine transform of the function f(x)=x
e ax
hence deduce the
infinite Fourier Sine transform of 1/x (AU-N/D-2016)-2(8) b. Find the Fourier sine transform of e-|x|. Hence show that
0,2)1(
sin
0
3
medxx
xx a(AU-A/M-2015)(8)
10. a. Find the Fourier cosine transform of f(x)= Hence show that
(AU-N/D-2015)(8)
b. Find the Fourier sine and cosine transform of ,0<n<1.Hence Show that is
self reciprocal under both the transformation (AU-A/M-2015)-2(8)
11. a. Solve the integral equation
0
cos)( exdxxf and also show that
0
2 21
cos edx
x
x (AU-M/J-2015)(8)
22 xae 2
2x
e
2/2xxe
otherwise
xx
,0
10,1
2
sinsin
0
2
2
0
dxx
xdx
x
x
1nxx
1
137
b. Prove that is self reciprocal under Fourier Cosine transform. (AU –A/M-2014)(8)
12. a. Find Fourier Cosine transform of and hence find Fourier sine transform of x
(AU-A/M- 2018)-5(8)
b.Use transform method to evaluate
0
22 41 xx
dx(AU-A/M- 2017)-2(8)
13. Find the Fourier cosine transform of f(x) = e-ax for x > 0 . a>0 Hence deduce that
∫𝑐𝑜𝑠𝑠𝑥
𝑎2+𝑠2∞
0𝑑𝑠 and
0
22
sinds
sa
sxs (AU-M/J-2016)(16)
14. Find the Fourier Sine transform and Cosine transforms of a function f(x)=e-x.
Using Parseval’s identity , evaluate (AU-A/M- 2017)-2(8)
(i)
0
22 1x
dx and (ii)
0
22
2
1x
dxx
15. a. Find the Fourier sine transform of e-|x|. Hence S.T 0,2)1(
sin
0
3
medxx
xx a
(AU-N/D-2014)(8)
b. Using Parseval’s identity evaluate the following integrals
(i)
0
222 xa
dx (ii)
0
222
2
xa
dxx, where a>0. (AU-N/D- 2017)-2(8)
16. a. Verify the convolution theorem for Fourier transform if f(x)=g(x)=2xe
(AU-M/J-2015)(8)
b. Evaluate
0
2222 )( bxax
dx using transforms. (AU-M/J- 2015)(8)
UNIT – V
Z- TRANSFORM PART-A
1.Define Z- Transforms Let x(n) be a sequence defined for all integers then its Z-transform is defined to be
Zx(n)=X(Z)= where z is an arbitrary complex number.
= = zF(z)-zf(0) 19. Find the value of z[f(n)] when f(n) = nan (AU-N/D-2014)
z[nan] = -z [z(an)] = -z [ ] =
20.Prove that Z = -z (AU A/M 2018)-2
Given: F(Z)=Z[f(n)]
F(Z)=
=
= -Z[nf(n)]
3
)2( n
3
)5( n
3
)2( n
3
)5( n])5()2[(
3
1 nn
))((
21
bzaz
zZ
))(()(
2
bzaz
zZX
12
1
))(()(
nn z
bzaz
zzZX
))((
1
bzaz
z n
))(()()(Re
11
bzaz
zazLtzzXs
n
az
n
az
)(
1
bz
zLt
n
az
ba
a n
1
))(()()(Re
11
bzaz
zbzLtzzXs
n
bz
n
bz
)(
1
az
zLt
n
bz
ab
b n
1
ba
a n
1
ab
b n
1
][1 11
nn baba
0
)1(n
nznf
0
)1()1(n
nznfz
0
)(n
mzmfz
dz
d
dz
d
az
z
2az
az
)(nnfdz
zdF )(
0n
nn za
0
1)()()]([n
nznfnZFdz
d
0
)(n
n
z
znnf
0
)()]([n
nznnfZFdz
dz
) 5 )( 2 ( ) 5 ( ) ( lim Res
5
1
5
z z
z z z z X
n
z
n
z
142
Z[nf(n)] = -z
21.State and prove initial value theorem in Z-transforms. (AU-A/M-2017)-2 Initial value theorem in Z-transforms is given by
If Z[f(t)]=F(z),then f(0)=
Proof: F[z]=Z[f(t)]=
= =
= = f(0)
22. State final value theorem on Z-transform (AU-A/M-2017)
If Z[f(t)]-F(z) , then
)()1(lim)(lim1
zFztfzt
23. State convolution theorem on Z-transform. (AU-N/D- 2016) - 3 The convolution theorem on Z-transform is given by If )().(y(n)*x(n) then Z)(y(n) Zand )( ZYZXzYzXnxZ
24. Form the difference equation by eliminating arbitrary constants from
(AU-N/D-2017)-2
Given
..........(1)
........(2)
[Using (1)]
25. Solve yn+1-2yn=0, given that y(0)=2 (AU-N/D-2012)
Z[yn+1]-2Z[yn]=0
zY(z)-zy(0)-2Y(z)=0
Y(z)(z-2)-2z=0
Y(z)=2
2
z
z
y(n)=2(2)n=2n+1
PART-B
1. a. Find 2-t3 e Zand tnZ . (AU-N/D-2016)-(8)
b. Find Z-transform of
)2)(1(
32
nn
nby using method of partial fraction.
(AU-N/D-2017)-2-(8)
dz
zdF )(
)(lim zFz
0
)(n
nznTf
.....).2().1(
).0(2
z
Tf
z
TfTf .....
)2()()0(
2
z
Tf
z
Tff
)(lim zFz
....])2()(
)0([lim2
z
Tf
z
Tff
z
12 n
nU
12. n
n aU
2
1 2.
n
n aU
2.2. 1
1
n
n aU
nn UU 21
143
2. a. Find Z and Z . (AU-N/D-2015)-2-(8)
b. Find the Z transform of1)n(n
1 and
2cos
n(AU-A/M-2016)(8)
3. a. State and Prove initial value and Final value theorems. (AU-A/M-2010) (8)
b. Find Z
)1(
1
n and Z n (AU-A/M-2018)(8)
4. a. If U(Z)= 3
3
)1(
Z
ZZ, find the value of u0,u1 and u2. (8)
b. Find the Z transform of 1
1
n and .
2cos2
nn (8)
5.a. Find Z )sinh( Tt (8)
b. Find Z by using residue method. (AU-A/M-2015)(8)
6. a. Find Z)1()12(
42
3
zz
z by using method of partial fraction. (AU-A/M-2017)(8)
b. Find Z-1 by residue method. (AU-A/M-2018)(8)
7. a. Find Z-1 when 2<|z|<3 (AU-A/M-2015)(8)
b. Using the inversion integral method ,find the inverse Z-transform of
U(Z)= )4)(2( 2
2
zz
z (AU-A/M-2015)(8)
8. aFind Z)2()1( 2
3
zz
z by using method of partial fraction. (AU-A/M-2017)(8)
b. Find the inverse z-transform of 222 zz
zby residue method.
(AU-A/M-2015)(8) 9. a. Derive the difference equation from yn=(A+Bn)(-3)n
b. Using Convolution theorem evaluate Z 1
2
2
)( az
z(AU-M/J-2016)(8)
10. a. Find Z
4
1
2
1
2
ZZ
Z by using Convolution theorem. (AU-N/D-2017)(8)
b. Using Convolution theorem evaluate Z 1
)4)(3(
2
zz
z(AU-A/M-2015)(8)
11. a. Find Z by using Convolution theorem. (AU-M/J-2014)-3(8)
nra nn cos nra nn sin
1
)2()13(
92
3
zz
z
1
)5)(2(
32
zz
zz
)2()3(
131022
2
zz
zz
1
1
1
))((
2
bzaz
z
144
b. Using Convolution theorem evaluate Z (AU-A/M-2018)-3-(8)
12. a. Solve yn+2-4yn+1 +4 yn =0 given y0=1 and y1 =0. (AU- A/M-2018)-3(8) b. Using Z- Transform solve the equation u +3 u +2u = 0 given u(0) = 1 and
u(1) (AU- A/M-2015)-(8)
13. a. Using Z- Transform solve the equation u -5 u +6u = 4 given u(0) = 0 and
u(1) =1. (AU-M/J-2014)-2(8) b. Using Z- Transform solve the equation y +4 y -5y = 24n -8 given y(0) = 3
and y(1) = -5. (AU-M/J2010)(8) 14. a. Solve yn+2+yn=2 given y0=0 and y1=0 by using Z-transforms. (AU-M/J- 2016)(8) b. Solve using z-transform ,yn+2 -7yn+1+12yn=2n given y0= 1and y1=0 (AU-N/D-2017)-3-(8) 15.a. Solve yn+2+6yn+1 +9yn =2n, given y0= y1 =0. (AU-M/J- 2016)(8) b. Solve using z-transform, yn+2-3yn+1-10yn = 0 given y0=1 and y1=0.