InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 401. |
The equivalent circuit of this coupled network is shown in figure. Find the values of Z, Z and Z. |
| Answer» Use mesh analysis for the coupled network Apply KVL at loop 1 V1 = 5jI1 + 6jI2 Apply KVL to loop 2 V2 = 8jI2 + 6jI1 Consider given T network Apply KVL to mesh 1 V1 = (ZA + ZC)I1 + ZCI2 Apply KVL to mesh 2 V2 = (ZB + ZC)I2 + ZCI1 Compare equation (i) and (iii) ∴ ZA + ZC = 5j and ZA = 6j ∴ ZA = - j Compare equation (ii) and (iv) ZB + ZC = 8j ZB + j = 8j ZB = 2j ∴ ZA = - j ZB = 2j ZC = 6j. | |
| 402. |
The system of linear equations 4x + 2y = 1 , 2x + y = 6 has |
| Answer» 4x + 1y = 7 → (1) 2x + y = 6 → (2) (1) can be written as 2x + y = 3.5 → (3) Now LHS of equations (2) and (3) are same but RHS is different which is not possible. Hence no solution. | |
| 403. |
A control system is as shown in given figure The maximum value of gain K for which the system is stable is |
| Answer» Characteristic Equation is 1 + GH = 0 1 + or s3 + 3s2 + 2s + (1 + K) = 0 Hurwitz criterion can be applied s3 1 2 and (1 + K) both should be + ve s2 3 1 + K s1 0 ∴ K should be more than - 1 and less than 5 s0 1 + K. | |
| 404. |
The effective modulation index is: |
| Answer» μ1 = 0.2 x 3 = 0.6 μ2 = 0.2 x 4 = 0.8 μ = (0.36) + (0.64) = 1. | |
| 405. |
Consider Continuous time system = - . For the system to be linear response + should be + with relation between constant '' and '' as __________ . |
| Answer» Let ax1(t) + bx2(t) = m(t) where m(t) is new input which causes the output n(t) n(t) = m(t) - n(t) = {ax1(t) + bx2(t)} - add and subtract n(t) = x1(t) -a + x2(t) -b +a +b - n(t) = +{a + b - 1} n(t) = ay1(t) + by2(t) +{a + b - 1} For linearity, n(t) = ay1(t) + by2(t) ∴ {a + b - 1} = 0 ∴ a + b = 1. | |
| 406. |
The skin-depth of copper at a frequency of 3 GHz is 1 micron. At 12 GHz, for a non-magnetic conductor whose conductivity is times of copper, the skin depth would be : |
| Answer» . | |
| 407. |
For a medium σ = 2.1 x 10 mho/m and ε = 80 e at a frequency of 5 GHz its loss tangent is |
| Answer» = 9.45 X 10-6. | |
| 408. |
For the output F to be 1 in the logic circuit shown, the input combination should be |
| Answer» B = A F = A⊙B⊙C = A⊙A⊙C so, F = 1 when C = 1. | |
| 409. |
The ROC of Z-transform of the discrete time sequence () is |
| Answer» . | |
| 410. |
Assuming the OP-AMP to be ideal, the voltage gain of the amplifier shown in figure is |
| Answer» By rearranging | |
| 411. |
In 8085, Whenever a signal is received at TRAP terminal its program execution is transferred to a subroutine on address __________ . |
| Answer» TRAP is considered to be of the highest priority and the address of the subroutine is given by 0024 H. | |
| 412. |
If (225) = (165) find value of Y |
| Answer» (225)y => 2Y2 + 2Y + 5 = (165)8 => (117)10 ∴ 2Y2 + 2Y + 5 = 117 Y2 + Y - 56 = 0 ∴ Y = 7 or Y = - 8 ∴ Y should be 7. | |
| 413. |
Consider a parallel plate capacitor being charged. Identify the incorrect statements |
| Answer» In a parallel plate capacitor, current passes through the connectors. is the Poynting vector which points radially into the volume. | |
| 414. |
Find equivalent resistance. Each resistance is l Ω |
| Answer» Cut the cuboid at XY - AB plane. This will make two parallel and symmetrical loops. Resistance between XY and AB = 1 Ω Splitting it in two equal values resistances Solving one of the loop Use Δ - Y transformation ⇒ RP'Q' = + (1.6 || [1.2 + 2]) = 1.667 Ω ∴ Req = RP'Q' || RP'Q' = . | |
| 415. |
Find Z, V current through 10 Ω resistance |
| Answer» Apply KCL at node A VB = 5 x 1 5 V ∴ VAB = VA - VB = 1.33 - 5 = - 3.67 V | |
| 416. |
The signal flow graph of a system is shown in figure. The transfer function of the system is : |
| Answer» P1 = 1 [Path R(s to C(s)] Δ = 1 - (L1 - L2 + L3) + L1L2 P1 = 1; Δ1 = 1 [Non touching loop gain to forward path] . | |
| 417. |
The Laplace transform of the given waveform is |
| Answer» f(t) = f1(t) + f2(t) + f3(t) = 2tu(t) - 2 x 2(t - 1) + 2(t - 2)u(t - 2) f(s) = - e-s + e-2s = (1 - 2e-s + e-2s). | |
| 418. |
The region between a pin of parallel perfectly conducting planes of infinite extent is and directions is partially filled with a dielectric as shown below.A 30 GHz TE wave is incident on the air-dielectric interface. The value of reflection coefficient at the interface is |
| Answer» | |
| 419. |
Consider a Binary Symmetric Channel (BSC) with probability of error being . To transmit a bit, say 1, we transmit a sequence of three 1 s. The receiver will interpret the received sequence to represent 1 if at least two bits are 1. The probability that the transmitted bit will be received in error is |
| Answer» Probability of error = P Thus probability of no error = (1 - P) Now probability that transmitted bit, received in error = all bits are with error + one bit is with error = P3 + 3C1P2 (1 - P) = P3 + 3P2 (1 - P). | |
| 420. |
Consider the sequence whose fourier transform X() is depicted for - ≤ ω ≤ as shown in figure |
| Answer» Accordingly, we note first that periodicity in the time domain implies that the Fourier transform is zero, except possibly for impulses located at various integer multiples of the fundamental frequency. This is not true for X(ejω). We conclude, then, that x[n] is not periodic. Next, from the symmetry properties for Fourier Transforms, we know that a real valued sequence must have a Fourier transform of even magnitude and a phase function that is odd. This is true for the given |X(ejω)| and ∠X(ejω). We thus conclude that x[n] is real. Third, if x[n] is an even function, then, by the symmetry properties for real signals, X(ejω) must be real and even. However, since X(ejω) = |X(ejω)| e-j2ω, X(ejω) is not a real-valued function. Consequently, x[n] is not even. Finally, to test for the finite-energy property, we may use Parseval's relation, |X(ejω)|2 dω It is clear that integrating |X(ejω)|2 from - p to p will yield a finite quantity. We conclude that x[n] has finite energy. | |
| 421. |
The number of distinct Boolean expressions of 3 variables is |
| Answer» 22n = 223 ⇒ 28 = 256. | |
| 422. |
For the two-port network shown below, the short circuit admittance parameter matrix is |
| Answer» | |
| 423. |
The circuit given below: If R = R = 20 kΩ and C = 0.2 μF and C = 0.005 μF. Then which of the given statements given below is true? |
| Answer» = 39.7875 Hz. = 15.92 Hz. | |
| 424. |
The Nyquist sampling rate for the signal () = is given by |
| Answer» fs = 2fm | |
| 425. |
Time domain equivalent circuit of the figure is |
| Answer» In the left hand loop (- i1) enters the dot end while for the right handed loop i2 enters the dot end. This will make both the M terms at first of + ve sign but as i1 = - ve hence ultimately the M term of right hand loop involving i1 = - ve. V1 = - L1 + M V2 = L2 - M . | |
| 426. |
The region between a pin of parallel perfectly conducting planes of infinite extent is and directions is partially filled with a dielectric as shown below.A 30 GHz TE wave is incident on the air-dielectric interface. The VSWR at the interface is |
| Answer» . | |
| 427. |
A transmitted using AM has in modulated carrier output power of 10 KW and can be modulated to a maximum depth of 90% by a sinusoidal modulating voltage without causing overloading. Find the value to which an modulated carrier power can be increased without resulting in overloading if the maximum permitted modulating index is 40% |
| Answer» Now PT is fixed, μ = 0.4 Then PC is PC = | |
| 428. |
A signal with ± 10 V ranged and 1 KHz band width is being digitized using a sample and hold circuit and a 10 bit quantizer. The quantization step is : |
| Answer» Quantization step = . | |
| 429. |
The circuit I in figure is |
| Answer» Current in the circuit is [2 v at '+' terminal and '-' terminal will be same]. | |
| 430. |
The equivalent form of the logical expression (B + C + ABC + AB + A) is: |
| Answer» K-map for the expression: | |
| 431. |
For a binary symmetric channel, the error probability P(0) = P(1) = and errors are statistically independent. The channel capacity for a signalling speed 2 bits/second. |
| Answer» C = r . H r = 2 bits/second C = 2 bits/second. | |
| 432. |
In a transmission line terminated with a load equal to the characteristic impedance, the reflection coefficient is |
| Answer» . | |
| 433. |
Consider a control system shown in given figure. For slight variation in G, the ratio of open loop sensitivity to closed loop sensitivity will be given by |
| Answer» Open loop sensitivity = 1 Closed loop sensitivity = Hence ratio = 1 : (1 + GH) - 1. | |
| 434. |
Consider the signal S() shown below :The slope of the matched filter output during the interval |
| Answer» Match filter will be The output of this will be convolution of S(t) with match filter. . | |
| 435. |
The logic function (A B, C) = Σ(0, 2, 4, 5, 6) represented by |
| Answer» | |
| 436. |
An electrical system and its signal-flow graph representation are shown in figure (a) and (b) respectivelyThe values of G and H, respectively are |
| Answer» Vi(s) = Z1I1 + Z3(I1 - I2) ...(i) V0(s) = Z4I2 I2(Z3 + Z2 + Z4) = Z3I1 ...(ii) From signal flow graph I2 = I1G2 ⇒ Hence, ...(iii) From SFG ...(iv) From (ii), (iii), (iv) we get H. | |
| 437. |
Consider the circuit shown below of 2 : 1 MUX is given by the function = + Then is |
| Answer» g = ac + bc min c = 0, g = b and c = 1, g = a So f = y + ya ⇒ y x + yg ...(i) g = x b + x.a = x E + X.o ⇒ x E from (i) f = y x + y x E. | |
| 438. |
Given (224) = The value of the radix '' is : |
| Answer» (224)r = = (13)r x (13)r ∴ 2r2 + 2r + 4 = (r + 3) (r + 3) = r2 + 6r + 9 ∴ r2 - 4r - 5 = 0 ∴ (r - 5) (r + 1) = 0 r = - 1 or r = 5 Negative base is not possible. Therefore base of the equation is 5. | |
| 439. |
Which factor determines whether the medium is free space, lossless dielectric, lossy dielectric or good conductor? |
| Answer» Loss tangent (i) for lossless dielectric, tan θ ≤ 1 (ii) for good conductor, tan θ >> 1 (iii) for lossy dielectric, tan θ = 1. | |
| 440. |
Find R |
| Answer» Reduced Diagram : ∴ RAB = 3 || 6 = 2 | |
| 441. |
A 1000 H carrier wave modulated 40% at 4000 H is applied to a resonant circuit tuned to a carrier frequency and having Q = 140. What is the degree of modulation after transmission through this circuit? |
| Answer» Resulting depth of modulation | |
| 442. |
convert into spherical co-ordinates. |
| Answer» Spherical : Ar = A. a r =.sinθ.sinφ = Aθ = A. a θ =.cosφsinφ = 2 sin2φ Aφ = Aa φ =cosφ = 2 sinφ ∴ A = a r + 2sin2φa θ + 2sinφaφ A = 2sinφ[sinθ tanφ a r + sinφa θ + aφ]. | |
| 443. |
The reduced form of the functionY(A, B, C, D, E, F) = ∑(0, 1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 28, 30, 31) is given by: |
| Answer» Y = (A + D + E + F)(A + C + D + E). | |
| 444. |
For the circuit, let us assume that R = 1 Ω, L = 10 H, and V(t) = 10 cos (1000 + 30°)V and the differential equation for I() is given by |
| Answer» Replacing current and voltage by their phasors I and 10ejp/b, resp. and by jω = j1000, we obtain the phasor equation I + 10-3(j1000I) = 10ejp/6 or I(1 + j1) = 10ejp/6 Having determined the value of I, we now find the required solution to be I(t) = Rc [I ejωt] = Rc[7.07e -jp/2 ej1000t] = 7.07 cos (1000t - 15°)A. | |
| 445. |
Find equivalent capacitance if each capacitance is 1 F. |
| Answer» Use similar method. But for capacitor take inverse of value of capacitance and then apply procedure same as in resistance case. After getting the final answer take inverse of result to get Ceq ∴ In Cuboid case, For resistance For inductance For capacitance. | |
| 446. |
The inverse of the given transform is : |
| Answer» | |
| 447. |
The RMS value of the signal given below is: |
| Answer» RMS value . | |
| 448. |
Use block diagram reduction methods to obtain the equivalent T.F from R to C. |
| Answer» | |
| 449. |
For a series resonant circuit at low frequency circuit impedance is __________ and at high frequency circuit impedance is __________ Fill in the blanks respectively |
| Answer» From this curve; we can select option (a) as correct answer. | |
| 450. |
The network is |
| Answer» System T.F. = Characteristic equation: So system is second order system of the two poles are . | |