Two-Port Networks
By applying KCL at node V x. V 12 V 6 VV 8 xxx c ++ =0 2 V x + 4V x + 3(V x - Vc) = 0 9V x = 3 V c V x = 1/3 V c ⇒ V x = 1/3
Chapter Highlights
List of important topics that are covered in chapter.
Exercises
CHAPTER HIGHLIGHTS
Power dissipated in the resistor (R) is
☞ Classification of Networks
☞ Network Configuration
☞ Open Circuit or Impedance (Z) Parameters
Let us consider the following example. The circuit shown in figure. Let
15
☞ Y Parameters or Short-circuit Admittance Parameters
☞ Hybrid Parameters
☞ G Parameters or Inverse Hybrid Parameters
Example 3
Introduction
Solution
We know for source-free
☞ Transmission or ABCD Parameters
☞ Inverse Transmission Parameters
☞ Interconnection of Networks
☞ ABCD Parameters in Terms of Z Parameters and Y – Parameters
☞ Network Graphs
The switch in the circuit in figure has been closed for a long time, and it is opened at t = 0. t = 0 –V 20 V + 20 mF + 1
CLassification Of Networks
Linear Circuits
It is the circuit whose parameters remain constant with change in applied voltage or current (V a I ohm’s Law). For example,resistance, inductance, and capacitance
Unilateral Circuits and Bilateral Circuit
Solution (i)
Solved Examples
Practice problems for students to master the concepts studied in chapter. Exercises consist of two levels of problems
‘Practice Problems I’ and ‘Practice Problems II’ based on increasing difficulty level.
Practice Problems 1
Non-linear Circuits
30. The incidence matrix of a graph is as follows:
When the direction of current is changed, the characteristics or properties of the circuit may change. This circuit is called unilateral circuits. For example, diode, transistor, UJT, etc. Otherwise, it is called bilateral circuit. For example, R, L, C circuits.
Active and Passive Elements
Solved problems are given topic-wise to learn and to apply the concepts learned in a particular section as per examination pattern.
For t < 0: The switch is closed, and the capacitor is open circuit to DC.
If a circuit element has the capability of enhancing the energy level of a signal passing through it, it is called an active element. For example, transistors, op-amp, vacuum tubes, etc. Otherwise, it is called passive elements. For example, resistors, inductors, thermistors capacitors, etc., are passive elements.
Lumped and Distributed Network
Circuit is in S and S. \ Vc(0 ) = 20 × 9 93 150 + =< Vfor t
Since the voltage across capacitor does not change instantaneously, Vc(0 ) = Vc(0+) = 15 V
For t ≥ 0:
The switch is opened. The circuit is shown in the figure.
Physically separable network elements such as R, L, and C are known as lumped elements. A transmission line on a cable in the other hand is an example of distributed parameter network. They are not physically separable. If the network is fabricated with its elements in lumped form, it is called a lumped network and if it is in distributed form, it is called distributed network.
Recurrent and Non-recurrent Networks
Exercises
Direction for questions 1 to 28: Select the correct alternative from the given choices.
When a large circuit consists of similar networks connected one after another, the network is called as recurrent network or cascaded network. It is also called as ladder network. Otherwise, a single network is called non-recurrent network.
Symmetrical and Asymmetrical Network
It is a circuit whose parameters changed with voltage or current. For example, diodes, transistor, etc. Non-linear circuits does not obey Ohm’s Law.
1. The maximum power transferred to the load in the circuit is given as 0.5 W. Get the values of R and RL
(A) 15 W, 10 W (B) 12.5 W, 12.5 W (C) 10 W, 15 W (D) 10 kW, 10 kW
2. Find the efficiency of the circuit given for RL = 50 W
Practice Problems 2
(A) 99% (B) 91% (C) 80% (D) 87%
Direction for questions 1 to 23: Select the correct alternative from the given choices.
3. Current in the circuit is given by the equation i(t) = 10cos(20p t + 50) and the impedance of the load is given as ZL = 5 + j3. Find the average power delivered to the load. (A) 353.5 W (B) 291.5 W (C) 250 W (D) 176.7 W
1. Which parameters are used in the analysis of transistors? (A) Z parameters (B) Y parameters (C) h parameters (D) transmission parameters
2. If a transmission line is represented by a two-port network whose parameters are A, B, C, D, then the sending- and voltage-end current are given by_____.
If the network looks the same from both the ports, then it is said to be symmetrical. Otherwise, it is called asymmetrical network. The following figures show the symmetrical and asymmetrical networks:
4. In the following circuit, the Norton equivalent current (in A) across AB is
5. Find the Thevenin’s equivalent voltage external to the load RL
The number of possible trees are
(C) It has no Y parameter. (D) It has no transmission parameter.
7. Two two-port networks have Z parameters [Z]x = ZZ ZZ xx xx 1112
(A) 25 V (B) 50
3. A two-port network is reciprocal if and only if (A) Z11 = Z22 (B) Y12 = Y21. (C) BCAD =
Then, the open-circuit transfer impedance of the cascaded network
CHapter 1 linear algebra R
In the following circuit, the adjustable resistor R is set such that the power in the 5 W resistor is 20 W. The value of R is (A) 6 W (B) 25 W (C) 4 W (D) 16 W
Previous Years’ Questions
1. The maximum power that can be transferred to the load resistor RL from the voltage source in figure is [2005]
(A) 1 W (B) 10 W (C) 0.25 W (D) 0.5 W
2. For the following circuit, Thevenin’s voltage and Thevenin’s equivalent resistance at terminals ab is [2005]
4. For the following circuit, the Thevenin’s voltage and resistance looking into XY are [2007]
5. In the following AC network, the phasor voltage VAB (in Volts) is [2007]
(A) 5 V and 2 W (B) 7.5 V and 2.5 W (C) 4 V and 2 W (D) 3
3. An independent voltage source in series with an impedance ZS = RS + jXS delivers a maximum average power to a load impedance ZL when [2007]
Hints/solutions
Practice Problems 1
Solutions for questions 1 to 65:
1. We know trace of A = sum of diagonal elements of A
⇒ 55 = 1 + 2 + 3 + + n = nn() + 1 2
On solving, we get n = 10 or 11
But n being the order of A, cannot be negative
⇒ n = 10
Hence, the correct option is (A).
Previous Years’ Questions
Contains previous 10 years’ GATE Questions at the end of every chapter that help students to get an idea about the type of problems asked in GATE and prepare accordingly.
and columns = A + iB Þ A is purely real Hence, the correct option is (C).
5. In the given matrix 2R3 = R1 We know that in any square matrix, if two rows are equal or one is the multiple of the other, then its determinant is zero. Hence, the correct option is (C).
Hints/Solutions
M01_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 51 4/11/2017 11:15:05 AM
This section gives complete solutions of all the unsolved questions given in the chapter. The Hints/ Solutions are included in the CD accompanying the book.
2. ~ nn nn nn nn nn nn n !! ! ! ! + () + () + () + () + () + () + () + () + () + () 12 1 12 13 21
2 n nn nn n + () + () + () + () + () 32 43 2
~ n! (n + 1)!(n + 2)! 11 21 12 32 13 43 nn n nn n nn n ++() + () ++() + () ++() + ()
R2 → R2 R1 and R3 → R3 R2
~ n! (n + 1)! (n + 2)! 11 21 01 22 01 32 nn n n n ++() + () + () + ()
= n! (n + 1)! (n + 2)! {2(n + 3 n 2)}
Practice Tests
= 2 n! (n + 1)! (n + 2)!
Hence, the correct option is (B).
Time-bound test provided at the end of each unit for assessment of topics leaned in the unit.
3. xx x xx x xx x CC C CC C CC C 01 1 1 12 1 2 23 1 3 22 2 66 6 + + () + () 11 21 1 31 12 11 xx xx xx x xxxx xx xx +() + () ()()()() + ()()
C3 → C3 (C1 + C2)
Þ 10 21 0 31 11 0 0 x xx x xx xx x()()()() = For any value of x = f(x) = 0. Hence, the correct option is (C).
4. A + iB = 23 1 30 1 11 1 ++
6. Given: A is a square matrix of order K We know, det (KA) = Kr det (A), where r is order of ‘A’. ∴ KK 27 ⇒ 33 27 ∴ k = 3 Hence, the correct option is (D).
7. |4AB| = 44 |A| |B| = 256 × −2 × 5 = 2560. Hence, the correct option is (C).
8. Clearly the product of the given matrix with the matrix in 2nd option results in a unit matrix.
3.340 | Part III • Unit 3 • Control Systems Test
∴ The inverse of the given matrix is option B Hence, the correct option is (B). 9. 12
Control Systems
Direction for questions 1 to 30: Select the correct alternative from the given choices.
(First C3: C3 C2, and, then C2: C2 C1) 0. A 1 does not exist. Now, if |A| ≠ 0, i.e. A is non singular and if AT A, then (AA 1)TI T
1. The transfer function gain between C(S) and R(S) in the following figure is 2 3 2 R(s) C(s) –4 (A) 5 (B) 2.5 (C) 10 (D) 2
(A 1)T = A 1 A 1 is symmetric. Hence, the correct option is (D). 10. If |A| is a 3rd order determinant, then |Adj
Time: 60 Minutes
|2
2. The Laplace transform of a transportation lag of 8 s is (A) exp (8s) (B) exp (–8s) (C) 1 38 (D) exp (–s/8)
(A) Given that |Adj A|2 = 28561 ⇒ |A|4 = 28561 (using (A))
3. The transfer function of ZOH (zero-order hold) is (A) 1 – e -Ts (B) 1 – eTs (C) 1 e s Ts (D) 1 e s Ts
Hence, the correct option is (B).
5. The transfer function of linear control system is defined as the (A) Fourier transform of impulse response (A) -6. e -3t + e -t (B) e -t + 2/3 e -3t (C) 6. e3t - e -t (D) None of these
10. The impulse response of an initially relaxed system is e -4tu(t). To produce a response of te-4t.u(t), the input must be equal to (A) e +4tu(-t) (B) e -4tu(t) (C) te-4t (D) e -4tu(t)
11. The closed loop gain of the system shown in the following figure is × 4 –2 1/2 R(s)
11. The product of a square matrix and its transpose is always symmetric.
4. The main drawback of a feedback system is (A) inaccuracy (B) inefficiency (C) unstability (D) insensitivity
GATE 2017 Solved Paper
ECE: Electronics and Communication Engineering
Set – I
Number of Questions: 65
Total Marks: 100.0
WronganswerforMCQwillresultinnegativemarks,( -1/3)for1-markquestionsand( -2/3)for2-markquestions.
General Aptitude
Number of Questions: 10
Section Marks: 15.0
Question 1 to Question 5 carry 1 mark each and Question 6 to Question 10 carry 2 marks each.
Question Number: 1
Question Type: NAT
She has a sharp tongue, and it can occasionally turn________ (A) hurtful (B) left (C) methodical (D) vital
Solution: The phrase ‘sharp’ in the given context means to be harsh or rude to someone. Hence, ‘hurtful’ is apt. The word ‘methodical’ means to be very slow. Hence, the correct option is (A).
Question Number: 2
Question Type: NAT
I __________ made arrangements had I _________ informed earlier.
(A) could have, been (B) would have, being (C) had, have (D) had been, been
Solution: The given sentence suggests that the author could have made arrangements had he been informed earlier. The word ‘could’ means a possibility, and the word ‘would’ means to have an inclination for something. The context is clearly referring to a possibility of making arrangements if the information had been passed earlier. Hence, the correct option is (A).
Question Number: 3
Question Type: MCQ
In the summer, water consumption is known to decrease overall by 25%. A water board official states that, in the summer household consumption decreases by 20%, while other consumption increases by 70%
Which of the following statements is correct?
(A) The ratio of household to other consumption is 8/17.
(B) The ratio of household to other consumption is 1/17.
(C) The ratio of household to other consumption is 17/8.
(D) There are errors in the official’s statement.
Solution: The data are tabulated below. HH is household consumption, OT is other consumption and OA is overall consumption.
OA HH OT
–25% –20% 70%
According to the official’s statement, the OA consumption lies outside the range from –20% to +70%. There have to be errors in this statement.
Hence, the correct option is (D).
Question Number: 4
Question Type: MCQ
40% of deaths on city roads may be attributed to drunken driving. The number of degrees needed to represent this as a slice of a pie chart is
(A) 120 (B) 144 (C) 160 (D) 212
Solution: 10% is represented by 36° on the pie chart.
∴ 40% is represented by 144°. Hence, the correct option is (B).
Question Number: 5
Question Type: MCQ
Some tables are shelves. Some shelves are chairs. All chairs are benches. Which of the following conclusions can be deduced from the preceding sentences?
i. At least one bench is a table
ii. At least one shelf is a bench
iii. At least one chair is a table
iv. All benches are chairs
(A) Only i (B) Only ii
(C) Only ii and iii (D) Only iv
Solution: Two possible Venn diagrams are shown below.
We see that only ii is true. At least one shelf is a bench. (Some shelves are chairs and all chairs are benches). Hence, the correct option is (B).
Question Number: 6
Question Type: MCQ
‘If you are looking for a history of India, or for an account of the rise and fall of the British Raj, or for the reason of the cleaving of the subcontinent into two mutually antagonistic parts and the effects this mutilation will have in the respective sections, and ultimately on Asia, you will not find it in these pages; for though I have spent a lifetime in the country. I lived too near the seat of events, and was too intimately associated with the actors. to get the perspective needed for the impartial recording of these matters’.
Here, the word ‘antagonistic’ is closest in meaning to (A) impartial (B) argumentative (C) separated (D) hostile
Solution: The context of the paragraph suggests that the author was present during the freedom struggle and has seen it through; however, he is unable to give an impartial perspective because he was very much involved in the freedom struggle, and has become patriotic (inferred from the paragraph). The ‘actors’ refers to all those people who made it possible to obtain independence and those who were responsible for the partition. The paragraph clearly implies that an intimate association of a person with someone will not allow him/her to be impartial. That person over a period of time becomes biased. Hence, option A is true. Options B and C negate the idea of the paragraph. Option D talks about ‘actors’ which, in the passage, is used figuratively to highlight the leaders of that time in history. Hence, option D can be eliminated.
Hence, the correct option is (A).
Question Number: 7
Question Type: MCQ
S, T, U, V, W, X, Y and Z are seated around a circular table. T’s neighbours are Y and V, Z is seated third to the left of T and second to the right of S. U’s neighbours are S and Y; and T and W are not seated opposite each other. Who is third to the left of V?
(A) X (B) W
(C) U (D) T
Solution: T’s neighbours are Y and V. or Y T V V T Y
Z is 3rd to the left of T and 2nd to the right of S. S T
U’s neighbours are S and Y. (This tells us that Y, V are to the right and left, respectively, of T, i.e. 1a is correct and not 1b.) or Y U T
T and W are not seated opposite each other. Therefore, the person 3rd to the left of V is X. Hence, the correct option is (A).
Question Number: 8
Question Type: MCQ
Trucks (10 m long) and cars (5 m long) go on a single lane bridge. There must be a gap of at least 20 m after each truck and a gap of at least 15 m after each car. Trucks and cars travel at a speed of 36 km/h. If cars and trucks go alternately, what is the maximum number of vehicles that can use the bridge in 1 hr?
(A) 1440 (B) 1200
(C) 720 (D) 600
Solution: The maximum number of vehicles corresponds to the closest spacing between the vehicles, because the speed of the traffic is constant (36 km/hr). The spacing is shown as follows.
C T
1510 520
In 1 hr, a vehicle would cover 36,000 m. Over this distance, we can have 36 000 50 , , viz 720 stretches of 50 m. Each such stretch would have 2 vehicles. Therefore, 720 stretches would have 1440 vehicles. Hence, the correct option is (A)
Question Number: 9
Question Type: MCQ
There are 3 Indians and 3 Chinese in a group of 6 people. How many subgroups of this group can we choose so that every subgroup has at least one Indian?
(A) 56
(B) 52
(C) 48
(D) 44
Solution: We can denote the 3 Indians as I1, I2, I3 and the three Chinese as C1, C2, C3. Of the 8 ways of selecting the 3 Indians, 7 are allowed by the given conditions. (The only selection not allowed is omitting all three of them.) There are 8 ways of selection the 3 Chinese. Therefore, there are 56 ways of forming the subgroup containing at least one Indian. Hence, the correct option is (A).
Question Number: 10
Question Type: MCQ
A contour line joins locations having the same height above the mean sea level. The following is a contour plot of a geographical region. Contour lines are shown at 25-m intervals in this plot.
The path from P to Q is described by
(A) Up-down-up-down
(B) Down-up-down-up
(C) Down-up-down
(D) Up-down-up
Solution: From P, the path goes down, through 575 m, 550 m, 525 m, 500 m then another 500 m, then up through 525 m, 550 m, 575 m, a bit further up and finally down through 575 m, 550 m, and a bit further down.
The path goes down, up, down. Hence, the correct option is (C).
Electronics and Communication Engineering
Number of Questions: 55
Question Number: 11
Consider the 5 × 5 matrix
Question Type: MCQ
The characteristic equation of A is
It is given that A has only one real eigenvalue. Then, the real eigenvalue of A is
(A) -2.5 (B) 0 (C) 15 (D) 25
Solution: Given matrix is
⇒ 15- l = 0 ⇒ l = 15
∴ The only real eigenvalue of A is l = 15
Hence, the correct option is (C).
Question Number: 12
Question Type: MCQ
The rank of the matrix is M = 51010 10 2 36 6
(A) 0 (B) 1 (C) 2 (D) 3
Solution: Given matrix is M =
det (M) = 0 (∵C3 = 2C1)
Therefore, Rank of M < 3
And the determinant of a submatrix 5 1 10 0 of M is 10 0
Therefore, Rank of M = 2 Hence, the correct option is (C).
Question Number: 13
Question Type: MCQ
Consider the following statements about the linear dependence of the real-valued functions y1 =1, y2 = x and 03 = x2 over the field of real numbers.
I. y1, y2 and y3 are linearly independent on -1 ≤ x ≤ 0.
II. y1, y2 and y3 are linearly dependent on 0 x 1.
III. y1, y2 and y3 are linearly independent on 0 x 1
IV. y1, y2 and y3 are linearly dependent on 1 x 0
Which one among the following is correct ?
(A) Both I and II are true
(B) Both I and III are true
(C) Both II and IV are true
(D) Both III and IV are true
Solution: Given y1 = 1, y2 = x and y3 = x2
For 1 x 0 or 0 x 1 , the linear combination of y1, y2 and y3, ay1 + by2 + cy3 = 0 only when a = b = c = 0
∴ y1, y2 and y3 are linearly independent on 1 x 0 as well as on 0 x 1
∴ Both I and III are true. Hence, the correct option is (B).
Question Number: 14
Question Type: NAT
Three fair cubical dice are thrown simultaneously. The probability that all three dice have the same number of dots on the faces showing up is (up to third decimal place)
________.
Solution: Total number of possible outcomes when three fair cubical dice are thrown = 6 × 6 × 6 = 216.
Number of possibilities in which all three dice have the same number of dots on the faces shown up = 6 (all dice showing up the face with 1 dot or 2 dots or 3 dots or 4 dots or 5 dots or 6 dots).
Required probability = 6 216 = 1 36 = 0.02778
Hence, the correct answer is 0.027 to 0.028.
Question Number: 15
Question Type: MCQ
Consider the following statements for continuous-time linear time invariant (LTI) systems.
I. There is no bounded input-bounded output (BIBO) stable system with a pole in the right half of the complex plane.
II. There is no causal and BIBO stable system with a pole in the right half of the complex plane.
Which one among the following is correct?
(A) Both I and II are true
(B) Both I and II are not true
(C) Only I is true
(D) Only II is true
Solution: Only I is true; ∵ a causal system may be stable may not be stable [e.g. et u(t) is causal but unstable due to RHP.]
Hence, the correct option is (D).
Question Number: 16
Question Type: MCQ
Consider a single-input single-output discrete-time system with x[n] as input and y[n] as output, where the two are related as: y[n] = nxn xnxn [] [] [] ⎧ ⎨ ⎪ ⎩ ⎪ , , 1 for 0 ≤ n ≤ 10 otherwise.
Which one of the following statements is true about the system?
(A) It is causal and stable
(B) It is casual but not stable
(C) It is not casual but stable
(D) It is neither casual nor stable
Solution: yn nxn xnxn () {}; {} {}; = ≤≤ ⎧ ⎨ ⎩ for 0n 10 otherwise 1
The output in both the case is depending on either the present or past values of input; so, it is causal, and in both the cases, bounded input will lead to bounded output, Hence, stable. Hence, the correct option is (A).
100cos vt 1 Ω 1 Ω 1H V2 V1
Question Number: 17
Question Type: NAT
In the circuit shown, the positive angular frequency w (in radians per second) at which the magnitude of the phase difference between the voltages V1 and V2 equals p/4 radians, is ____________.
Solution: From the given data
Let Z1 = 1 W = 1 ∠ 00W
Z2 = 1 + jw = 1 2 + w ∠ q2 W
Where qw 2 1 = - tan
VZit VZit 11 22 = ( ) = ( ) . .
Let i(t) = Im ∠ q Amp.
VIm 1 = ∠ q × 1
V2 = I m ∠q × 1 2 + w ∠ q2
VIm 2 = 1 2 + w ∠ q + q2
Given qq qp +- = 2 5 /
∴=qp 2 4 / tan - ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 1 2 1 w q
∴= w 1 rad/sec
Hence, the correct answer is 0.9 to 1.1.
Question Number: 18
Question Type: MCQ
A periodic signal x(t) has a trigonometric Fourier series expansion
x t ( ) = a o + a n cos n o t + b n sin n o t ( ) n=1
If x(t) = -x(-t) = -x(t - p/w0), we can conclude that
(A) a n are zero for all n and b n are zero for n even
(B) a n are zero for all n and b n are zero for n odd
(C) a n are zero for n even and b n are zero for n odd
(D) a n are zero for all n odd and b n are zero for n even
Solution: Given x(t) = -x(-t) means the signal is odd x(t) =- xt⎛ ⎝ ⎜ ⎞ ⎠ ⎟ p w0 , which says that the signal is half-wave symmetric.
Condition for half-wave symmetry is xtxt T () =- ± ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2
So, that signal is odd and half wave symmetric. In the resulting Fourier series expansion, a0 and a n will exist only for odd harmonics (due to half-wave symmetry). Hence, the correct option is (A).
Question Number: 19
Question Type: MCQ
A bar of gallium arsenide (GaAs) is doped with silicon such that the silicon atoms occupy gallium and arsenic sites in the GaAs crystal. Which one of the following statements is true?
(A) Silicon atoms act as p-type dopants in arsenic sites and n-type dopants in gallium sites
(B) Silicon atoms act as n-type dopants in arsenic sites and p-type dopants in gallium sites
(C) Silicon atoms act as p-type dopants in arsenic as well as gallium sites
(D) Silicon atoms act as n-type dopants in arsenic as well as gallium sites
Solution: Si is a IVth group element; so, it acts like p-type dopant in the Vth group sites (p, As, etc.) and it acts like a n-type dopant like in the IIIrd group sites (B, Al, Ga, etc.).
Hence, the correct option is (A).
Question Number: 20
Question Type: MCQ
An n+ -n silicon device is fabricated with uniform and nondegenerate donor doping concentrations of ND1 = 1 × 1018 cm -3 and ND2 = 1 × 1015cm-3 corresponding to the n+ and n regions, respectively. At the operational temperature T, assume complete impurity ionization, kT /q = 25 mV, and intrinsic carrier concentration to be ni = 1 × 1010 cm -3. What is the magnitude of the built-in potential of this device?
(A) 0.748 V (B) 0.460 V
(C) 0.288 V (D) 0.173 V
Solution: Form the given data
ND1 = 1 × 1018 atoms/cm3
ND2 = 1 × 1015 atoms/cm3
VT = 25 mv
V0 = ?
NA = n N i D 2 1
atoms/cm3
Hence, the correct option is (D).
Question Number: 21
Question Type: MCQ
For a narrow base PNP BJT, the excess minority carrier concentrations ( nE for emitter, pB for base, and nC for collector) normalized to equilibrium minority carrier
concentrations ( nE0 for emitter, nE 0 for base, and nCo for collector) in the quasi-neutral emitter, base, and collector regions are shown below. Which one of the following biasing modes is the transistor operating in? Base(N) X and Y axesarenottoscale
(A) Forward active (B) Saturation (C) Inverse active (D) Cutoff
Solution: From the given diagram
Emitter–base junction is in reverse biase and collector–base junction is in forward bias; so, it is operating in reverse active region.
Hence, the correct option is (C).
Question Number: 22
Question Type: MCQ
For the operational amplifier circuit shown, the output saturation voltage are ±15V. The upper and lower threshold voltages for the circuit are, respectively,
∴ Upper threshold voltage = +7V
Lower threshold Voltage = -3V. Hence, the correct option is (B).
Question Number: 23
Question Type: MCQ
A good transconductance amplifier should have (A) high input resistance and low output resistance (B) low input resistance and high output resistance
(C) high input and output resistances
(D) low input and output resistances
Solution: For a transconductance amplifier, input and output resistances are high. The transconductance amplifier can also be called voltage-controlled current source, i.e. VCCS. An amplifier is VC when input resistance is high, and an amplifier is CS when output resistance is high. Hence, the correct option is (C).
Question Number: 24
Question Type: MCQ
The Miller effect in the context of a common emitter amplifier explains
(A) +5 V and – 5 V (B) + 7 V and – 3V (C) +3 V and -7V
(D) + 3 V and – 3 V
Solution: By KCL
(A) increase in the low-frequency cutoff frequency
(B) an increase in the high-frequency cutoff frequency
(C) a decrease in the low-frequency cutoff frequency
(D) a decrease in the high-frequency cutoff frequency
Solution: From the high-frequency response analysis of CE amplifier we can say that fH = 1 2 pCRinin
CCCgRc inm =+ + [] p m 1
1 + gRmc is known as miller multipler; due to Miller effect in CE amplifier, input capacitance increases, and, hence, there is a decrease in high-frequency cutoff frequency. Hence, the correct option is (D).
Question Number: 25
Question Type: MCQ
In the latch circuit show, the NAND gates have non-zero, but unequal propagation delays. The present input condition is: P = Q = ‘0’. If the input condition is changed simultaneously to P = Q = ‘1’, the outputs X and Y are
(A) X = ‘1’ , Y = ‘1’
(B) either X = ‘1’, Y = ‘0’ or X = ‘0’, Y= ‘1’ (C) either X = ‘1’ , Y = ‘1’ or X = ‘0’, Y = ‘0’
(D) X = ‘0’, Y = ‘0’
Solution: When P = 0, Q = 0 ⇒ X = 1, Y =1 when P = 1, Q = 1 ⇒ X = 1, Y = 0 (or) when P = 1, Q = 1 ⇒ X = 0, Y = 1 x =1/0 p =0/1 Q =0/1 y =1/0/1
Hence, the correct option is (B).
Solution: T TTT ON
∴ Output duty cycle = T
∴ Output duty cycle in % = 3 10
Question Number: 26 Question Type: MCQ
The clock frequency of an 8085 microprocessor is 5 MHz. If the time required to execute an instruction is 1.4 ms, then the number of T-states needed for executing the instruction is
(A) 1 (B) 6
(C) 7 (D) 8
Solution: fmp = 5 MHz
T = 1 p f = 0.2 ms
1 T → 0.2 ms to complete an instruction, it took 1.4 ms; therefore, by linear relation
IT → 0.2 ms
x? → 1.4 ms
x × 0.2 = 1T X1.4
x = 7 T
Therefore, to complete instruction, we need 7 T states. Hence, the correct option is (C).
Question Number: 27 Question Type: NAT
Consider the D-Latch shown in the figure, which is transparent when its clock input CK is high and has zero propagation delay. In the figure, the clock signal CLK1 has a 50% duty cycle and CLK2 is one-fifth period delayed version of CLK1. The duty cycle at the output of the latch in percentage is _______________.
Question Number: 28
The open loop transfer function
Question Type: NAT
G s ( ) = s + 1 ( ) s p s + 2 ( ) s + 3 ( )
where p is an integer, is connected in unity feedback configuration as shown in the figure. Given that the steady state error is zero for unit step input and is 6 for unit ramp input, the value of the parameter p is __________.
G(s) –+
Solution: e k KGSHs ss p pss = + ( ) =+ ( ) ( ) =∞ 1 1 1 for unit step input for e to be ’o’ ( ) → s 0 k Lt so S ssS P pp = → + ( ) + ( ) + ( ) =∞ ≥ ( ) 1 23 ’’ must be 1 essk v == ( ) 1 6for unit reamp input
k Lt s vSGSHS = → ( ) ( ) = 0 1 6
k v = Lt s 0 S S + 1 ( ) S P S + 2 ( ) S + 3 ( ) = 1 6 for P = 1
∴= P 1
Hence, the correct answer is 0.99 to 1.01
Question Number: 29
Question Type: MCQ
Consider a stable system with transfer function
G(s) = ++ + ++ +sbsb sasa pp p qq q 1 1 1 1
where b1, … b p and a1, …, a q are real-valued constants. The slope of the Bode log magnitude curve of G(s) converges to – 60 dB/decade as w → ∞. A possible pair of values for p and q is
(A) p = 0 and q = 3
(B) p = 1 and q = 7
(C) p = 2 and q = 3
(D) p = 3 and q = 5
Solution: Slope of Polo magnitude plot at w = ∞ is ( ) ( ) =20 20 60 qp qp dB/dec qp-= 3
Hence, the correct option is (A).
Question Number: 30
Question Type: MCQ
Which of the following can be the pole-zero configuration of a phase-lag controller (lag compensator)?
(A)
s-plane jv s Pole Zero
(a)
(B) s-plane jv s Pole Zero
(b)
(C) s-plane jv s Pole Zero (c)
(D) s-plane (d) jv s Pole Zero
Solution: It is evident that, from pole zero configuration, choice (A) satisfies a phase-lag compensator. Hence, the correct option is (A).
Question Number: 31
Question Type: NAT
Let (X1, X2) be independent random variables. X1 has mean 0 and variance 1, while X2 has mean 1 and variance 4. The mutual information I (X1; X2) Between X1 and X2 in bits is
Solution: For two independent random variables,
I (x : y) = H(x) = H (x/y)
H (x/y) = H (x) for independent X and y
⇒ I(x : y) = 0
Hence, the correct answer is 0.
Question Number: 32
Question Type: MCQ
Which one of the following statements about differential pulse code modulation (DPCM) is true?
(A) The sum of message signal sample with its prediction is quantized
(B) The message signal sample is directly quantized, and its prediction is not used
(C) The difference of message signal sample and a random signal is quantized
(D) The difference of a message signal sample with its prediction is quantized
Solution: Unlike in PCM, to conserve the channel bandwidth, the difference of a message signal sample with its prediction is quantized in DPCM. Hence, the correct option is (D).
Question Number: 33
Question Type: MCQ
In a digital communication system the overall pulse shape p(t) at the receiver before the sampler has the Fourier transform P( f ). If the symbols are transmitted at the rate of 2000 symbols per second, for which of the following cases is the inter-symbol interference zero?
Solution: g =+ 01 40 11 j m
ga b =+ - j m 1
a ⎯→ attenuation constant (Np/m or dB/m)
b ⎯→ Phase constant (rad/m Or degree/M)
∴ a = 01.Np/m
1Np = 8.686 dB
∴ a = 0.8686 dB/m
Hence, the correct answer is 0.85 to 0.88.
Question Number: 35
Question Type: MCQ
Consider a wireless communication link between a transmitter and a receiver located in free space, with finite and strictly positive capacity. If the effective areas of the transmitter and the receiver antennas, and the distance between them are all doubled, and everything else remains unchanged, the maximum capacity of the wireless link
(A) increases by a factor of 2 (B) decreases by a factor of 2
(C) remains unchanged
(D) decreases by a factor of 2
Solution: P P R T =
PR → received power
PT → transmitted power
Dt → transmitting antena directivity
D r → receiving antenna directivity
l → wavelength
d K Ae = 4 2l , where A e → effective area
∴= P PR RAA T eter 1 22l
Solution: For a pulse that is free from inter-symbol interference (ISI), if P(t) is having spectrum P( f ).
Then PfkRs K () -
= constant
R s = 2 K spa
This condition is met by the pulse given in option ‘B’. Hence, the correct option is (B).
Question Number: 34
Question Type: NAT
The voltage of an electromagnetic wave propagating in a coaxial cable with a uniform characteristic impedence is V() = e-g+jwt volts, where is the distance along the length of the cable in metres, g = (0.1 + j40) m-1 is the complex propagation constant, and = 2 109 rad/s is the angular frequency. The absolute value of the attenuation in the cable in dB/metre is ______________.
Given that Aet, A er and R each is doubled.
Then P P R T remains constant
Hence, the correct option is (C).
Question Number: 36
Question Type: MCQ
Let f ( x ) = e x+x 2 for real x. From among the following, choose the Taylor series approximation of f(x) around x = 0, which includes all powers of x less than or equal to 3.
(A) 1 + x + x 2 + x 3
(B) 1 + x + 3 2 x 2 + x 3
(C) 1 + x + 3 2 x 2 + 7 6 x 3
(D) 1 + x + 3x 2 + 7 x 3
| GATE 2017 Solved Paper ECE: Set – I
Solution: Given f (x) = exx + 2 = 1 + () () ! ) ! xx xxxx + + + + +∞ 2 22 23 23
( ∵ eyyyy =+ ++ +∞ 1 23 23 !! ) = 1 + x + x2 + xxxxxxx 23 43 45 6 23 2 3 6 ++ + ++ + +∞ =+ ++ +∞ 1 3 2 7 6 23xxx .
Hence, the correct option is (C).
Question Number: 37
Question Type: NAT
A three-dimensional region R of finite volume is described by x 2 + y 2 z 3 ; 0 z 1,
where x, y, z are real. The volume of R (up to two decimal places) is _______.
Solution: The region R described by x2 + y2 ≤ z3, 0 ≤ z ≤ 1 is a solid of revolution obtained by revolving the curve Z3; where x2 + y2 = z3, 0 ≤ z ≤ 1 about z-axis. 0 x y z B A(0,0,z)
Solution:
We have I = 22 2 zdxydyxdz c ++ ( ) ∫ (1)
Where C is the straight line segment from A : (0, 2, 1) to B : (4, 1, −1)
Equation of the line segment AB is xyztsay=== ( ) 0 40 2 12 1 11 ,
⇒=== xyzzt 41 1 2
⇒ x = 4t; y = 2 − t and z = 1 − 2t
⇒ dx = 4dt, dy = −dt and dx = −2dt and t varies from t = 0 to t = 1 I = 22 2 zdxydyxdz c ++ [] ∫ = = ∫ t 0 1 [2(1 − 2t)4dt + 2(2 − t)(−dt) + 2(4t)(−2dt)]
= ∫ t 0 1 [8 − 16t − 4 + 2t − 16t]dt = = ∫ t 0 1 [4 − 30t]dt
= 4t − 30 t 2 0 1 2 ⎛ ⎝ ⎜
= 4 − 15
∴ I = −11
Hence, the correct answer is 11.1 to −10.9.
Question Number: 39 Question Type: MCQ
∴ Volume of the region R = p () ; ABdz z 2 0 1 = ∫ where AB = xy22 + = p xydz z 22 2 0 1 + () = ∫ = = ∫ p
Hence, the correct answer is 0.7 to 0.85.
Question Number: 38
Question Type: NAT
Let I = c 2 z dx + 2 y dy + 2 x dz ( ) where x, y, z are real, and let C be the straight line segment from point A : (0, 2, 1) to point B : (4, 1, –1). The value of I is __________.
Which one of the following is the general solution of the first-order differential equation dy dx = x + y 1 ( )2 where x, y are real?
(A) y = 1 + x + tan-1 (x + c), where c is a constant (B) y = 1 + x + tan (x + c),where c is a constant.
(C) y = 1 - x + tan-1 (x + c), where c is a constant
(D) y = 1 - x + tan (x + c), where c is a constant
Solution: Given differential equation is dy dxxy =+ - ( )1 2 (1)
Put x + y – 1 = u
⇒+ = 1 dy dx du dx
⇒=dy dx du dx 1
(1) becomes, du dx − 1 = u2
⇒= + du dx u 1 2 ⇒ + = 1 1 2 u dudx which is in variables separable form.
∴ Integrating on both sides, we have 1 1 2 + =+ ∫ ∫ u dudxc
⇒ tan-1 (u) = x + c
⇒ u = tan(x + c)
⇒ x + y − 1 = tan(x + c)
⇒ y = 1 − x + tan(x + c)
∴ The general solution of (1) is y = 1 − x + tan(x + c).
Hence, the correct option is (D).
Question Number: 40
Question Type: NAT
Starting with x = 1, the solution of the equation x3 + x = 1, after two iterations of Newton–Raphson’s method (up to two decimal places) is _______.
Solution: Let f (x) = x3 + x − 1 = 0
x0 = 1
f 1(x) = 3x2 + 1
∴ f (xo) = f (1) = 1 and f 1(x0) = f 1(1) = 4
By Newton–Raphson’s method, we have
x1 = x0 − fx fx 0 1 0 ( ) ( ) = 1 − 1 4 = 0.75
And the approximate root after second iteration is x2 = x1 − fx fx 1 1 1 ( ) ( ) = 0.75 − f f 075 075 1 . ( ) ( ) = 0.75 − 0 171875 2 6875
∴ x2 = 0.68604.
Hence, the correct answer is 0.65 to 0.72.
Question Number: 41
Question Type: MCQ
Let x(t) be a continuous time periodic signal with fundamental period T = 1 seconds. Let {aK} be the complex Fourier series coefficients of x(t), where k is an integer value. Consider the following statements about x(3t):
I. The complex Fourier series coefficients of x(3t) are {aK} where k is integer valued
II. The complex Fourier series coefficients of x(3t) are {3aK} where k is integer valued
III. The fundamental angular frequency of x(3t) is 6p rad/s
For the three statements above which one of the following is correct?
(A) Only II and III are true
(B) Only I and III are true
(C) Only III is true
(D) Only I is true
Solution: Fundamental period T = 1s. w0 = 2 p Timescaling property of complex Fourier series says that xta xta k k ( ) ↔ ( ) ↔ a
But T T →→ ( ) a wa w and 00 xta xta k k ( ) ←→ ( ) ←→ 3 wp p 0 32 6 ←→ ( ) =
I and III are true
Hence, the correct option is (B).
Question Number: 42
Question Type: NAT
Two discrete-time signals x[n] and h[n] are both non-zero only for n = 0, 1, 2 and are zero otherwise. It is given that x[0] = 1, x[1] = 2, x[2] = 1, h[0] = 1.
Let y[n] be the linear convolution of x[n] and h[n]. Given that y[1] = 3 and y[2] = 4, the value of the expression (10y[3] + y[4]) is ________.
Solution:
1
Given x (n)={1,2,1}
h (n)={1, b, c} usingcrosstablemethod 222b 2c
xxviii | GATE 2017 Solved Paper ECE: Set – I
y(0) =1; given y (1) = 3
⇒ 2 + b = 3 ⇒ b = 1
given y (2) = 4
⇒ 2b + c + 1 = 4
⇒ c = 4 – 2 – 1
⇒ c = 1
y (3) = b + 2c = 3
y(4) = c = 1
10y (3) + y (4) = 10 × 3 + 1 = 31.
Hence, the correct answer is 31.
Question Number: 43
Question Type: NAT
Let h[n] be the impulse response of discrete-time linear time invariant (LTI) filter. The impulse response is given by h[0] = 1 3 ; h[1] = 1 3 ; h[2] = 1 3 ; and h[ n] = 0
for n < 0 and n >2.
Let H(w) be the discrete-time fourier transform (DTFT) of h[n], where w is normalized angular frequency in radians. Given that H(w0) = 0 and 0 < w0 < p, the value of w0 (in radians) is equal to _________.
Solution: Given
if w p = 2 3 ; H (w) = 0
So, w = 2.0943 rad/s
Hence, the correct answer is 2.05 to 2.15.
Question Number: 44
Question Type: NAT
The figure shows an RLC circuit exited by the sinusoidal voltage 100 cos (3t) volts, where t is in seconds. The ratio amplitude of V2 amplitude of V1 is ________.
Hence, the correct answer is 2.55 to 2.65.
Question Number: 45 Question Type: NAT
In the circuit shown, the voltage Vin(t) is described by: V( for volts for in t t ) , , = < ≤ ⎧ ⎨ ⎩ 00 15 0 where t is in seconds. The time (in seconds) at which the current I in the circuit will reach the value 2 amperes is
Solution: From the given data, w =
Solution: From the given data
For t < 0
Vt in V so amp ( ) = () = - 00 0 , I
For t > 0:
At t = 0+, I (0+) = I 0() = 0A As t ⎯→⎯∞ iA ∞ ( ) = 15 L eq = (2 ||1) = 2 3 H R L R itiiie ite
=
( ) -∞ {} ( ) =-1 2 3 0 15 1 15 W t t sec () () () / t ⎡ ⎣ ⎤ ⎦ but It ist eAtmg () = × ( ) + =⎡ ⎣ ⎤ ⎦1 12 51 15 given i(t) = 2 A Then t = ? 2 = 5 [I – e -1.5 t] e - 1.5 t = 0.6 t = 0.34 s.
Hence, the correct answer is 0.3 to 0.4.
Question Number: 46
Question Type: NAT
The dependence of drift velocity of electrons on electric field in a semiconductor is shown as follows. The semiconductor has a uniform electron concentration of n = 1 × 1016 cm -3 and electronic charge q = 1.6 × 10-19C. If a bias of 5 V is applied across a 1-mm region of this semiconductor, the resulting current density in this region, in kA/cm2, is________.
Driftvelocity(cm/s)
Solution: Form the given data
V = 5 V, L = 1 mm E V L == ××5105 10 64 V/m V/cm
We know VE
JnqV d d = = m from the given data m
Hence, the correct answer is 1.5 to 1.7.
Question Number: 47
Question Type: NAT
As shown, a uniformly doped silicon (Si) bar for length L = 0.1 mm with a donor concentration ND = 1016 cm-3, is illuminated at x = 0 such that electron and hole pair are generated at the rate of
GL = GLO 1 x L , 0 x L ,
where GLO = 1017 cm 1 s 1 . Hole lifetime is 10-4 s, electronic charge q = 1.6 × 10-19 C, hole diffusion coefficient D p = 100 cm2/s and low-level injection condition prevails. Assuming a linearly decaying steady state excess hole concentration that goes to 0 at x = L, the magnitude of the diffusion current density at x = L/2, in A/cm2, is________.
Light Si (ND =1016cm–3)
Solution: From the given data N L GG x L xL G D LLO LO = == × =-
⎝ ⎜ ⎞ ⎠ ⎟ ≤≤ =10 01 110 10 10 16 3 5 17 cm mcm c . , m m ms31
t p = 10-4 s
q = 1.6 × 10-19c and D p = 100 cm2/sec
J p = ?
x =0 L =0.1 mm
PPP
G P PG L p Lp =+ = = 0 Δ Δ Δ t t.
PPG x noL LoP =+⎛ ⎝ ⎜ ⎞ ⎠
J p pp p . . diff diff t 0 19 417 16 10 100 10 10 01 10 -4
Jp . diff = ×16 10 10 3 4 = 16 A/cm2
Hence, the correct answer is 15.9 to 16.1.
Question Number: 48
Question Type: NAT
As shown, two silicon (Si) abrupt p-n junction diodes are fabricated with uniform donor doping concentrations of N D1 = 1014cm 3 and N D2 = 1016 cm 3 in the n-regions of the diodes, and uniform acceptor doping concentrations of N A1 = 1014 cm 3 and N A2 = 1016cm 3 in the p-regions of the diodes, respectively. Assuming that the reverse bias voltage is >> built-in potentials of the diodes, the ratio C2/C1 of their reverse bias capacitances for the same applied reverse bias, is ____________.
