Given 4 words labelled as (P), (Q), (R) and (S). One of the pairs given below has words which have similar meaning or have opposite meaning. Identify the correct option.
(P) instigate (Q) enquire
(R) construe (S) interpret
Thus, option C is the correct answer.
A job can be completed by 12 men in 12 days. How many extra days will be needed to complete the remaining job, if 6 men leave after working for 6 days?
=1 / 12×12 Work done by 12 men in 6 days = 1/2 Remaining work = 1−(1/2) = 1/2 After 6 men leave the work, time taken to complete the remaining work = Remaining Work/Remaining Men
∴ Time taken
= 12 × 12 / 6 × 2 =12days
So,
so the extra time taken =12  6 = 6 days
Which of the following phrases should be placed in the blank spaces so as to make a grammatically correct and meaningful sentence.
Even though the institute had lot of space.______
In the question, there are 4 statements followed by 4 conclusions numbered as (a), (b), (c) and (d), assume the given statements to be true even if they are at variance with commonly known facts, identify the conclusion which following from the given 3 statements
S1: Some guitars are posters.
S2: All posters are doors.
S3: Some doors are tablets.
S4: All tablets are books
Conclusions:
C1: Some doors are guitars.
C2: Some books are posters.
C3: Some tablets are guitars.
Choose one word out of the given option to replace the phrase ‘person who insists on adherence to formal rules of the literal meaning’.
Thus, option A is the correct answer.
A shopkeeper gives a discount on the marked price based on the quantity bought by a customer. Raja bought 10 pieces of a lunch box and was given a 10% discount by the shopkeeper. When Radhika bought 20 pieces of the same lunch box, she was given a 15% discount by the shopkeeper. In both the cases, the net amount of profit for the shopkeeper was identical. Which of the following is the ratio of marked price to the cost price for the lunch box?
Let 'M' and 'C' be the marked price and cost price per lunch box. We can form an equation based on the given information as
i.e.
i.e. C = 8M / 10
or M:C = 10:8 or 5: 4
It would take one machine 4 hours to complete a production order and another machine 2 hours to complete the same order. If both machines work simultaneously at their respective constant rates, the time taken to complete the same order is __________ hours.
Let t be the time taken by the machines when they work simultaneously.
∴ 1 / t = 1 / 4 + ½
∴ 1 / t = 3 / 4
∴ t = 4 / 3
Which of the options given below should replace the part of the sentence written in BOLD so as to get a meaningful and grammatically correct sentence?
We had not only helped them with money but also with new machinery and raw material.
Answer C is corrected as not only will come after the sentence.
A fair die is thrown three times and the sum of three numbers is found to be 16. Find the probability that 5 appears on the third throw.
For sum = 16, we have the following cases:
(i) 6, 6, 4
(ii) 6, 4, 6
(iii) 4, 6, 6
(iv) 5, 5, 6
(v) 5, 6, 5
(vi) 6, 5, 5
Total cases of sum (16) = 6
Favourable cases for 5 appears in 3^{rd} throw = 2
p = 2 / 6 = 1 / 3
A work of classics is split up in 3 volumeseach volume having equal number of pages. It is also known that the page numbers are running across the 3 volumes. If the sum of the first page number of the 3 volumes is 1473, the number printed on the first page of volume 3 is ________.
Let the number of pages in each volume be ′x′. The first page of the 3 volumes will be 1,1+x and 1+2x respectively. 1473 = 1 + (1 + x) + (1 + 2x) = 3 + 3xgiving
x = 1470/3 = 490
First page of volume 3 will be having pages starting from 1 + 980 = 981
Find the value of the resistance and the inductance of the branch CD if the balance is obtained under these conditions. if the arms of an ac. maxwell bridge are given as: AB is a noninductive resistance of 1000 Ω in parallel with a capacitor of capacitance 0.5 μF, BC is a non inductive resistance of 600Ω and CD is an inductive impedance which is unknown and DA is a noninductive resistance of 400 Ω?
R_{1}R_{3 }= R_{2}R_{4}
R_{3} = R_{2}R_{4}/R_{1}
R_{3} = 600 × 400/1000 = 240Ω
L_{3} = CR2R4
L_{3} = 0.5 × 10^{−6} × 400 × 600
L_{3} = 12 × 10^{−2}
L_{3} = 0.122H
A single phase semi converter is operated from 120V, 50Hz ac supply. The load current with an average value IDC is continuous and ripple free firing angle α=π/6. Determine the harmonic factor of input current?
= 0.91I_{dc}
Now the rms value of supply fundamental component of input current
Harmonic factor (HF) of input current
=
=
= 0.30
A 200V dc series motor develops its rated output at 1500 rpm while taking 20 A . Armature and series field resistances are 0.6Ω and 0.4Ω respectively. To obtain rated torque at 1000 rpm, external resistance must be added is
⊤ ∝ la^{2} ⇒ T = kla^{2}
For const torque armature I must be const.
E^{b} = 200 − 20 × (0.6 + 0.4) = 180V
l_{a} = 20A for 1000rpm. E_{a} = kϕw
180E_{a} = 1500 / 1000
E_{a }= 120V = 200 − 20(0.4 + 0.6 + R_{ext})
⇒ R_{ext} = 200 − 120 / 20 −1 = 3Ω
The value of impedance of short transmission line having supply of 6.5 kV with the rating of 10 kVA is (1+j5) p.u. What is the impedance at the new base of 13 kV and 30kVA?
So
= 0.75 + j3.75
A Signal x(t) has Fourier transform x(ω) phase and magnitude of x(ω) are shown below
Determine x(t) at t = 5_{sec}
=
x(5) = 0.5
A 3ɸ bridge converter is provided with lineline voltage of 400 V. A load which is resistive in nature of value 100 ohm takes 400 W of power from the converter, the input power factor will be
Load current
In A 3ϕ fully enthralled bridge inverter input rms current I or the
correct in each supply phase exist for 120∘ in every 180∘
Therefore rms value of input current
Input apparent power
= √3 × V_{S}I_{S}
= √3 × 400 × 1.15
= 796.72vA
∴ Input apparent power = Output power
796.72cosθ = 400
cosθ = 0.5 lagging
The zparameters of the network shown in figure are given by
z11 = v1I1∣∣t2 = 0 (Open circuit output terminal)
z11 = v1I1∣∣t2 = 0 (Open circuit output terminal)
V1 = I1R1
z11 = V1I1 = R1
z11 = v2I1∣∣I2 = 0
V2 =−αV1
V2 = −α(I1R1)
z21 = V2I1 = −αR1
z12 = v1I2∣∣I1 = 0 (Open circuit input terminal)
Since V1 = I1R1 = 0
50,Z12 = 0
z22=V2I2∣∣I1 = 0
since V1 = I1R1 = 0
Thus, α∨1 = 0
V2 = I2R2
z22 = V2I2 = R2
Alternative Method:
We have the equations for voltage V1 and V2 as
∨1 = 1R1
V2 − I2R2 + αV1=O
V2 = I2R2−αV1
From eq. (1) V2 = I2R2 − α(1R1) = (−αR1)I1 + R2I2
Comparing equation (1) and (2) to the general equation of z parameter,
The admittance parameter matrix [y] is
(Node equation at the top left node)
(Node equation at the top right node)
Comparing equation (1) and (2) with general equations, we get z
parameters as
[y] =
Geneticists say that they are very close to confirming the genetic roots of psychiatric illnesses such as depression and schizophrenia, and consequently, that doctors will be able to eradicate these diseases through early identification and gene therapy. On which of the following assumptions does the statement above rely?
A says that strategies are now available for eliminating psychiatric illnesses but it is mentioned in the very first line that the geneticists are very close but the strategy is not available till now. So, A is incorrect.
The given data says that the geneticists are working on the genetic roots of psychiatric illnesses such as depression and schizophrenia, which implies that these two diseases have genetic basis. Thus, B is the correct option.
C says that all human diseases can be traced back to genes and how they are expressed, but the data given in the question talks about psychiatric illnesses such as depression and schizophrenia only. So, C is also incorrect.
D says that in future, genetics will be the only relevant field for identifying psychiatric illnesses, which cannot be inferred from the given data. So, D is also incorrect.
So, B is the correct option
Determine the maximum drain current in mA for the JFET in the given network, if V_{GS }= −3V and V_{p }= −8V
Considering input loop, since, I_{G} = 0A and I_{S} = I_{D}
V_{GS} = −I_{D}R_{S}
⇒
Since,
Substituting values,
⇒ 2.5mA =
⇒ 2.5mA=IDSS(2564)
⇒ I_{DSS} = 64/25 × 2.5mA
⇒ I_{DSS} = 6.4mA
An ac voltmeter uses half wave rectifier and the basic meter with full scale deflection current of 1 mA and the meter resistance of 200 Ω. Calculate the multiplier resistance for a 10 V r.m.s range on the voltmeter.
the meter uses half wave rectifier and input is 10 V r.m.s
E_{av} = 1/2 (Eav over a cycle of input)
Now Ep = √2 Erms = 14.14V
E_{av} = 0.6 E_{p} = 8.99 ≈ 9V
Therefore, Eav (output) = 1/29 = 4.5V
E_{dc} = 0.45Erms
R_{s} = E_{dc}/l_{dc} − R_{m} = 0.45E_{rms}/ldc − R_{m} = 0.45×10/1×10^{−3} − 200
R_{s} = 4.3KΩ
Let f(x) = x(x  1)(x  2) be defined in [0,0.5]. Then the value of c of the mean value theorem is
f(x) = x^{3} − 3x^{2} + 2x
f′(x) = 3x^{2} − 6x + 2
f′(c) = 3c^{2 }− 6c + 2
by mean value through, we have
∴ 3c^{2} − 6c + 2 = 3/4
12c^{2} − 24c + 5=0
c = 6 ± √21/6
C = 6 − √21/6 = 0.24↔(0, 1/2)
For the flip flop configuration shown below, find the operation carried out by this configuration.
Stages
It is a gray counter.
If Laplace transform of a function is given by
Find the value of f(0^{+}) and f(∞) respectively.
Initial value = 2
Final value = 6/8 = ¾
A 70 MW steam station uses coal of calorific value 5000 Kcal/kg. Thermal efficiency of the station is 40% and electrical efficiency 70 % . calculate the coal consumption per hour when the station is delivering its fuel rated output is ___x 10^{3} kg?
Given data
Overall efficiency of the power station is n overall =η_{thermal} ×n_{electrical }
= 0.4 × 0.7 = 0.28
Units generated /hr = (70×10^{3})×1
= 70kwh
Heat produced / hr
H= electrical output in heat units /n overall
= 70 × 10^{3} × 860/0.28 = 215 × 10^{6}Kcal(1kwh = 860Kcal)
Coal consumption 1hr=H/ calorific value
= 215 × 10^{6}/5000
= 43 × 10^{3}kg
The value of is ____
=
Putting
= 0.94
Consider two real valued signals, x(t) bandlimited to [−500 Hz, 500Hz] and y(t) bandlimited to [−1kHz, 1kHz]. For z (t) = x(t). y(t), the Nyquist sampling frequency (in kHz) is __________
Multiplication in the time domain results in convolution in the frequency domain.
The range of convolution in frequency domain is [−1500Hz, 1500Hz]
So, the maximum frequency present in z(t) is 1500Hz Nyquist rate is 3000Hz or 3 kHz.
Keeping the kW demand constant if the load power factor increases the kVA demand –
kW = kVA cosϕ
K_{va} ∝ 1/cosϕ
If the power factor increases the kVA demand will decrease.
A delayed unit step function as
Its Laplace Transform is
=
=
Find the inverse Laplace transform of the function F(s) =
let G(s) = 2/(s + c)
⇒ g(t) = L^{−1}{G(s)} = 2e^{−ct}
f(t) = L^{−1}{G(s) e^{−bs}}
=2e^{− k(t − b)}u(t − b)
A discrete time signal with input as x[n] has impulse response h[n]=kδ[n] where k is constant. The output y[n] of the system is given as:
For a discrete time signal with input as x[n] has impulse response h[n] =kδ[n], the output y[n] of the system is given as:
y[n] = x[n] × h[n]
y[n] = x[n] ×kδ[n]
y[n] = k(x × δ)[n]
y[n] = kx[n]
Taking Fourier transform on both sides
Y(ω) = X(ω) x H(ω)
Therefore H(ω) = Y(ω)/X(ω)
This is known as the transfer function of the system.
h[n] = F^{1}[H(ω)].
The Potential (scalar) distribution in free space is given as V=10y^{4}+20x^{3}. if ε_{0} is permittivity of free space then the charge density ρ at the point (2,0) is
Here, we have to use Poisson's Equation,
, because
20 × 3 × 2x + 10 × 4 × 3y^{2} = −ρ/ε
At point (2,0)f
ρ = −240ε^{0}
A 4pole, 50Hz, 100MVA turbo generator has a moment of inertia 9.5 x 10^{3} kgm^{2}. What is the kinetic energy stored in the machine?
N = = 1500rpm
I = 9.5 x 10^{3} kgm^{2}
K.E. = = 117.2MJ
Determine the reciprocity and symmetry of the given twoport network.
The loop equations are:
V_{1} = 3I_{1} + 5(I_{1} + I_{2}) = 8I_{1} + 5I_{2}
V_{2} = 4I_{2} + 5(I_{1} + I_{2}) = 5I_{1} + 9I_{2}
When I_{2} = 0,
Z_{11} = V_{1}I_{1} = 8
Z_{21} = V_{2}I_{1} = 5 When I_{1} = 0,
Z_{12} = V_{1}I_{2}=5
Z_{22} = V_{2}/I_{2} = 9
Since Z11 ≠ Z_{22}, the network is not symmetrical. But, Z_{12} = Z_{21}, so the network is reciprocal.
If ω is the frequency of the 3phase source supplying to a 3phase fullwave bridge rectifier, what is the fundamental frequency of the output voltage?
Thus, the fundamental frequency of the output voltage is 6ω.
Find the complete solution of the state equation given below
Given the initial condition as
Consider the state equation be, X˙= AX + BU Where,
We know that,
Finding
=
=
State transition matrix
=
Now,
And
=
=
=
Thus,
=
In a 3phase line operating at 50 Hz, the conductors, each of 0.96 cm diameter, are arranged as given below.
Determine the inductance of the line.
Radius of the conductor = 0.96 / 2 x 100 = 0.0048m
Mutual GMD (geometric mean distance) of the conductor = = 2.296m
⇒
= 1.234 x 10^{6}H/m
= 1.234 mH/km
The divide by N counter is shown below. If initially Q_{0} = 0, Q_{1} = 1, Q_{2} = 0 the value of N is _______.
State is repeating after 5 clock pulses. N = 5
Find the difference between rank of A and columns of null space of A?
A =
We need to find the REF form of the matrix to find null space of the matrix
REF =
Clearly it has 3 pivot columns
Thus we have
Rank = 3
Column in null space = 2 = free variable in matrix.
Difference = 3  2 = 1
If an electric field is given as:
And,
Determine the work (in nJ) that is to be done in moving a 3μC charge along this path if the path is located at P(0.3, 4, 0.6).
Here, length and work is differential in nature, and hence, no integration is required. (Differential because charge is moved through ΔL ).
Therefore,
=
Now it is given that Q = 3 × 10^{−6}
So, we get:
Now, we have point P = (−0.3, −4,0.6)
x = −0.3, y = −4 and z = 0.6
Therefore,
= 4.89 x 10^{10}J
= 0.489 nJ
DC shunt motor is coupled to the identical DC shunt generator. The field of the generator is also connected to the same supply source as the motor. The armature resistance is 0.02 pu and the mechanical losses are 0.05 pu. Armature reactions can be neglected. The armature of the generator is connected to a load resistance R_{L}. With rated voltage across the motor, the load resistance across the generator is adjusted to obtain rated armature current in both motor and generator. The pu value of R_{L} is
Motor is connected across rated voltage V = 1 pu
Rated armature current flaws in bath the motor and generator Im=Ig=1pu
Back emf in motor
E_{b} = V − I_{m}R_{a} = 1 − 1 × 0.02 = 0.98pu
Mechanical output power of motor
= E_{b}I_{m} mechanicallasses
= 0.98 × 1 − 0.05 = 0.93 pu
This power is given to the generator.
Output power of generator = Output power of motor  mechanical Iasses
E_{g}I_{g }= 0.93 − 0.05 = 0.88pu ⇒ E_{g }= 0.88puTerminal voltage of generator
= E_{g} − l_{g}R_{a} = 0.88 − 0.02 = 0.86pu
Laad resistance
R_{L}= V_{9}/I_{9} = 0.86/1 = 0.86pu
Consider a 3bit number A and 2 bit number B are given to a multiplier. The output of the multiplier is realized using AND gate and one bit full adders. If minimum number of AND gates required are X and one bit full adders required are Y, then X + Y =
Number of AND gates required X = 6
Number of one bit full adders required Y = 3
X + Y = 6 + 3 = 9
X1, = X2
If z = sinh ucosv + icoshusinv then for what values of z, the function w = f(z) = u + iv is not analytic for
z=sinh ucosv+icoshusinv
By using hyperbolic properties cosh iv = cosv; sinh iv = isinv
z = sinh u cosh iv+cosh u sinh iv Now using sinh(u+iv)=sinh u cosh iv+cosh u sinh iv we get
z = sinh(u + iv)u + iv=sinh_{1} zw = sinh^{1} z
∴ z is not analytic for z^{2} = 1 ⇒ z = ±i
In the below parallel adder, A_{4} A_{3} A_{2} A_{1} is a BCD number and B_{4} B_{3} B_{2} B_{1 }= 0011, then the circuit acts as
This binary adder simply adds 0011 i.e., 3 (in decimal) to a BCD number.
When 3 is added to the BCD then it becomes excess3 code.
So, it is a BCD to excess3 Code converter.
Let = αβγ = 1,α,β,γ ∈ R and X = Then MX=0 has infinitely many solutions if trace (M) is ……………………..
MX = 0 has infinitely many solutions if M = 0 i.e., det(A) = 0
α(βγ  1)  1(γ  1) + (1  β) = 0
αβγ  α  γ + 1 + 1  β = 0
αβγ  (α + β + γ) + 2 = 0
1  (α + β + γ) + 2 = 0
(α + β + γ)=3
tr(M) = α + β + γ = 3
Compute the voltage gain for the following circuit with input frequency 2.5 KHz. [Mention the nearest Integer value in dB]
= 3.9 ≈ 4db
In the half bridge rectifier as shown below, the peak current( in A) passing through the diode is ________ (If the supply input is given as Vs = 200Sin(100πt))
We know that maximum value of A sincut + Bcosωt = √A^{2}+B^{2}
So, maximum value of diode current is
(t_{D})_{max} = 2.09A
For a given connected network and for the fixed tree, fundamental loop matrix is given by
The cut set matrix corresponding to the same tree is
Relationship between tieset matrix and cutset matrix
[B] = tieset matrix = [I:B_{T} ]
[Q] = tieset matrix = [Q_l:I]
Where [Q_{l }] =  [B_{T} ]^{T}
Relationship between tieset matrix and cutset
matrix [B] = tieset matrix =[I:B_{T}]
[Q]= tieset matrix =[Q_{1}:I]
Where [Q_{l }] =  [B_{T} ]^{T}
=
[Q] =
The three impedances Z_{1 }= 20∠30⁰Ω, Z_{2} = 40∠60⁰Ω, Z_{3} = 10∠90⁰Ω are deltaconnected to a 400V, 3 – Ø system. Determine the phase current IR.
Taking V_{RY} = V∠0⁰ as a reference phasor, and assuming RYB phase sequence,
we have
V_{RY} = 400∠0⁰V
Z_{1} = 20∠30⁰Ω = (17.32+j10)Ω I_{R} = (400∠0o)/(20∠30o)=(17.32j10) A.
Given . Find the value of ∮ on the closed path shown below.
From Stokes theorem
=
=
=
=
=
The bus admittance matrix for a power system network is given by,
A new transmission line which is represented below is connected between bus 1 and 3.
The modified bus admittance matrix is
The new transmission line is connected between bus ( 1 ) and (3), so element has to be changed, are Y_{11},Y_{13}, Y_{31}, Y_{33}
Similarly,
Modified bus admittance matrix.
A sixpole, 3−ϕ,50 Hz induction motor has a rotor resistance and rotor reactance of 0.02Ω and 0.13Ω respectively at standstill. The external resistance to be added in the rotor circuit to get 75% of maximum torque at starting is
s_{m} =
s_{m} =
s_{m} = 0.453
s_{m} = R/X = 0.453
R = 0.453 x 0.13
R = 0.058Ω
So, the external resistance to be connected to the rotor circuit would be
R_{ext} = 0.058 x 0.02 = 0.038 Ω
Consider a discrete signal,
The value of will be
Ztransform of signal will be
Putting z = 1
Z transform of signal x(n) will be:
Putting z = 1
= 2.75
The integral , where ′D′ denotes the disc x^{2} + y^{2} ≤ 4, evaluate to
Let x = rcosθ & y = rsinθ
dxdy = rdrdθ
=
=
=
= 20
Consider the discrete time signals h[n] and its shifted form h[n−k].
The value of &
h[k] = (1/2)^{k−1}(u[k+3]−u[k−10])
So, h[k] is non zero only in range − 3 ≤ k ≤ 9
So, h[−k] is non zero only in range − 9 ≤ k ≤ 3
Shifting h[k] by n, so
h[n − k] is non zero only in range, (n − 9) ≤ k ≤ (n + 3)
S_{o},A = n − 9 and B = n + 3
S_{0},A − B= (n − 9) − (n + 3)
= −12
A unity negative feedback system is described by the following state model
The steadystate error of the system due to step input of strength ' 12′ is
The transfer function is given by =
Steadystate output is
e_{88} = input  output
e_{39} = 12  12 = 0
Consider the circuit shown in figure below:
Find the phase angle of current ‘I’ with respect to voltage V_{2}.
Equivalent voltage across terminal AB is:
V = V_{1} + V_{2}
V = 150(0.5 + j0.866) + 180(1 − j)
V = 255 − j50 = 259.85 < />
Impedance across the terminal AB is
Current
Angle of voltage source V_{2} = 180√2 ∠ − 45^{∘}
So, the angle current ' " and voltage V_{2} is =11.09^{∘}.
Find V_{x} for which maximum power is transferred to load.
Apply Thevenin’s theorem in load side
To supply maximum power,
V_{AB} = 10/2 = 5V
By KCL at node A
V_{x} = 9.5 Volts
There are 3 fair coins and 1 false coin with tails on both sides. A coin is chosen at random and tossed 4 times. If ‘tails’ occurs all 4 times, then the probability that the false coin has been chosen for tossing is ___.
Required Probability = Favorable Outcomes / Total possible outcomes
Favourable outcomes = A false coin is chosen and flipped every time Probability of selecting a false coin =1/4 Probability of getting a tail on every flip of false coin =1. Favourable outcome
= 1/4 × 1 = 1/4
Total possible outcomes = Favourable outcomes + Unfavorable outcomes Unfavorable outcomes = A fair coin is chosen and flipped every time to get tail Probability of selecting a fair coin = 3/4 Probability of flipping a fair coin 4 times and getting tails every time
= (1/2)^{4} = 116 ∴ Unfavourable outcomes
= 3/4 × 1/16 = 3/64
Total possible outcomes
= 1/4 + 3/64 = 19/64
∴ Required probability
=
= 16/19
= 0.84
Consider an asymptotic Bode plot of a minimum phase linear system as shown in the figure below. The transfer function for the above system will be
The initial line has slope of −40dB/decadel i.e., the transfer function of initial line is K/s^{2}
At ω = ω1, the slope has been changed. From ω = ω_{1} to ω = 25, slope is −20dB/ decade
So,
−20 =
log25 − logω_{1}=0.8
logω_{1} = 0.597
ω_{1} = 3.95
According to the initial line.
M(dB) = 20logK − 40logω
At ω = ω_{1}M(dB) = 22
22 = 20logK = 40×0.597
logK = 2.294
K = 196.78
So, the transfer function is
The reading of the voltmeter (rms) in volts, for the circuit shown in the figure is _______
It is a Wheatstone bridge. For the bridge to be balanced, the required condition is
Z_{1}Z_{4} = Z_{2}Z_{3}
From the given circuit, we have z_{1 }= z_{4} = j1 − Ω
z_{2 }= z_{3} = (1/j) − Ω
Now substituting these value for bridge balance equation
j×j = (1/j) × (1/j)
j^{2} = 1/j^{2}
− 1 = −1
Hence, it is a balanced, Wheatstone bridge, and reading of voltmeter (rms) is
V_{rms} = Vm/(√2)
V_{rms }= 100/(1.414)
= 70.72v
Find the value of I_{x} in the given circuit.
Let the source current is I, (current direction is outward from the source)
I_{x} = 31_{0}
Apply KVL in the loop
= 2.5√2∠105°A
Consider the signal . The timeperiod of the signal will be:
⇒ = 12
⇒ = 18
Condition for periodic function:
N_{1}/N_{2} = 12/18 is rational number
Overall timeperiod, N = LCM of (12,18)
N = 36
If the maximum phase provided by the compensator is 30° and this is achieved at √6 rad/sec. The transfer function of the compensator is __________.
Given, ϕm = 30^{∘}
1/2 = 1−α/1+α
α = 13
Let assume the compensators transfer function is
Than maximum phase is provided at
So, the required compensators transfer function
Find the value of A and B for signal, g(t) = Ay(Bt), such that y(t) = x(t) × h(t) and g(t) = x(3t) × h(3t)
We know that,
So,
=
Given form is,
=
So, we can write it as,
Comparing both equations, we get,
Taking Inverse Fourier Transform.
Comparing with given signal, g(t) = Ay(Bt)
A = 1/3 and B = 3
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