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.
| 451. |
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 minimum sampling rate is : |
| Answer» Maximum sampling rate = 2pfmax V. | |
| 452. |
The circuit below represents function X(A, B, C, D) as: |
| Answer» X = A B C(0. D + 0.D) + A B C(0.D + 0.D) + A B C(0.D + 1.D) + A B C(0.D + 0.D) + A B C(0.D + 0.D) + A B C(0.D + 1.D) + ABC(1.D + 0.D) = ∑(5, 9, 13, 14). | |
| 453. |
The truth table corresponding to the given input digital gate : |
| Answer» The circuit behaves like NAND gate. | |
| 454. |
Find E at P(1, 1, 1) caused by four identical 3 C point charges located at P(1, 1, 0), P(- 1, 1, 0), P(- 1, - 1, 0) P(1, - 1, 0) |
| Answer» E = 6.82ax + 6.82ay + 32.89az V/m r = ax + ay + az r1 = ax + ay r2 = - ax + ay r3 = - ax - ay and r4 = ax - ay ∴ r - r1 = az; r - r2 = 2ax + az; r - r3 = 2ax + 2ay + az r - r4 = 2ay + az = 6.8ax + 6.8ay + 32.9az . | |
| 455. |
The causal system given below is __________ |
| Answer» It has only zeros at origin. Hence it is minimum phase system. | |
| 456. |
Find R for maximum power transfer. |
| Answer» Use Y - Δ transformation RAB= (9 || 18) || (9 || 18) = 3 Ω. | |
| 457. |
The signal flow graph of a system is shown below:The state variable representation of the system can be |
| Answer» Let us take two state variable as shown x1 and x2 ⇒ | |
| 458. |
Consider the resonant circuit as shown belowThe circuit is scaled modified so that Z 10 and ω = 10 Radians. Find the required element values |
| Answer» Impedance scaling factor kx As = 10 Radians ∴ New value for elements are R' = kzR = 1000 x 10 = 10 kΩ . | |
| 459. |
The z-transform of a signal converges if and only if |
| Answer» For u(n), a right handled sequence, |z| > , |z| = ∴ |3z| > 1; |2z| > 1 ∴ < 1; < 1. | |
| 460. |
If the characteristic equation of a closed loop system is + 2 + 2, then the system is : |
| Answer» 1 + G(s) H(s) = 0 s2 + 2s + 2 = 0 ξ < 1, under damped. | |
| 461. |
Consider circuit with 4 : 16 Demux below: Now: |
| Answer» f1 = y2 + y6 + y11 + y12 and f2 = y1 + yl0 + yl3 f1(m, n, o, p) = (3, 4, 9, 13) f2(m, n, o, p) = (2, 5, 14) ∴ = Π(2, 3, 4, 5, 9, 13, 14). | |
| 462. |
The value of the integral is |
| Answer» Compare this with where s = 3. | |
| 463. |
The probability density function (PDF) of a random variable X is as shown belowThe corresponding cumulative distribution function (CDF) has the form |
| Answer» Cumulative distribution function (Pdf) dx Then when we integrate the line from (-1, 0) to (0, 1) we get a parabolic curve. The maximum value of Pdf can be 1. | |
| 464. |
The system with given pole-zero diagram is __________ |
| Answer» Z transform for the pole zero plot will be Degree of numerator > degree of denominator. Also ROC is not defined. Hence response of the system cannot be determined. | |
| 465. |
For the waveform shown below, the slew rate and full power bandwidth with respect to op-amp will be respectively |
| Answer» Full power bandwidth | |
| 466. |
The impulse response () of a linear invariant continuous time system is given by h() = exp (- 2) (), where () denotes the unit step function. The output of this system to the sinusoidal input () = 2cos(2) for all time , is |
| Answer» When input is sinusoidal then output is also sinusoidal with same frequency but amplitude and phase changes. Thus amplitude is Phase is tan-1 = 0.25 p Thus y(t) = 2-0.5 cos(2t - 0.25p). | |
| 467. |
Consider the z-transform X() = 5 + 4 + 3; 0 < || < ∞ The inverse transform [] is |
| Answer» X(z) = 5z2 + 4z-1 + 3 = . | |
| 468. |
Find equivalent capacitance if each capacitance is 2 F. |
| Answer» 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 ∴ Ceq = F In Cuboids case, For resistance Req = R Ω with each R = 1 For inductance Leq = L H with each L = 1 H For capacitance Ceq =CF with each C = 1 F. | |
| 469. |
In a JK flip-flop, J is connected to and K is connected to Q outputs. The JK FF converts into a |
| Answer» Both the input is opposit to each other hence it is D FF. | |
| 470. |
An infinite ladder is constructed with 1 Ω and 2 Ω resistors as shown below. Find |
| Answer» As ladder is infinite ∴ 2R + R2 = 2 + 3R R2 - R - 2 = 0 ∴ R = 2 or R = -1 Negative value of R is not possible ∴ R = 2 ∴ i = 9 A. | |
| 471. |
Consider the matrix The value of is |
| Answer» find L-1[sI -A]-1 find inverse Laplace transform of the expression. | |
| 472. |
Find current flowing through 1 H inductor at = 0, at which switch 'S' is opened. |
| Answer» For t = 0+ current in L is circuit at t = 0+ ∴ I is Io = 3 A. | |
| 473. |
A two port network is described by the relation V = 2I + 3V I = - I + 2V Then Z-parameter of such network is |
| Answer» | |
| 474. |
Find Y in the circuit below :(E, F) = ∑(0, 1, 2, 3)(A, B, E, F) = ∑(0, 1, 2, 3, 5, 6, 9)(A, B, E, F) = ∏(4, 7, 8, 10, 11, 12, 13, 14, 15) |
| Answer» | |
| 475. |
Calculate the stability factor and change in I from 25°C to 100°C for, β = 50, R/R = 250, ΔI = 19.9 A for emitter bias configuration |
| Answer» | |
| 476. |
The star equivalent C C, C of the delta network is respectively |
| Answer» ∴ . | |
| 477. |
The logic function (A, B, C) = (0, 2, 4, 5, 6) can be represented by : |
| Answer» Now the Variable Entered Map (VEM) is: | |
| 478. |
In amplitude modulation, carrier signals A cos ω has its amplitude A modulated in proportion with message bearing (low frequency) signal . The magnitude of is chosen to be __________ . |
| Answer» For proper recovery of signal |m(t)| ≤ 1. | |
| 479. |
For real values of , the minimum value of the function () exp () + exp (- ) is |
| Answer» f(x) = ex + e-x f(x) = ex - ex f(x) = 0 ⇒ ex - e-x = 0 f'(x) = ex + e-x ⇒ +ve for x = 0 Thus minimum. Minimum f(x) = e0 + e0 = 2. | |
| 480. |
For a second-order system with the closed-loop transfer function The settling time for 2-percent band, in seconds, is : |
| Answer» . | |
| 481. |
An open loop transfer function is given by has |
| Answer» The root locus is given below. From root locus, we can see that there are 3 zeros at ∞ . | |
| 482. |
The current () through a 10 Ω resistor in series with an inductance is given by () = 3 + 4 sin (100 + 45°) + 4 sin (300 + 60°) Amperes. The RMS value of the current and the power dissipated in the circuit are |
| Answer» Power = I2R = 25 X 10 = 250 Watts. | |
| 483. |
The cut off voltage for JFET is 5 V. The pinch off voltage is __________ . |
| Answer» For JFET ID = IDSS (1 - VGS/Vp)2 0 = (1 - VGS/Vp)2 ∴ VGS = VP ∴ Vp = 5 V. | |
| 484. |
The logic realized by the circuit shown in figure below is |
| Answer» F = A BC + ABC + A B C + ABC = B(AC + AC) + B(AC + AC) B(A ⊕ C) + B(A ⊕ C) (A ⊕ C) (B ⊕ B) A ⊕ C. | |
| 485. |
An image uses 512 x 512 picture elements. Each of the picture elements can take any of the 8 distinguishable intensity levels. The maximum entropy in the above image will be |
| Answer» Each picture element can be represented by 3 bits. Thus total entropy = 512 x 512 x 3 = 786432. | |
| 486. |
A tachometer has a sensitivity of 4 V/1000 . Express the gain constant of the tachometer in the units volts/(rad/sec). |
| Answer» 4 V/100 rpm = V/rad/sec = 0.0381 V/rad/sec. | |
| 487. |
Consider the signal = 1 + sin + 3 cos + cos . Then Fourier series coefficients are __________ . |
| Answer» The signal x(n) = 1 + sinn 3 cos n + cos n + cos is periodic with period N, and we can expand x[n] directly in terms of complex exponentials to obtain Collecting terms, we find that Thus for Fourier series coefficients for this example are a0 = 1, a1 = + = -j, a-1 = - = +j, a2 = j, a-2 = -j. | |
| 488. |
Find R across the terminals A and B |
| Answer» There are not independent source in the circuit. ∴ The Thevenin and Norton equivalent will have 0 A and 0 V sources. To find RTH, a 1 A source is connected as Writing a nodal equations at n, 0.01Vx + 0.003Vx + 0.0067Vx = 1 0.017Vx = 1 ∴ Vx = 58.82 V | |
| 489. |
If a signal has energy , the energy of the signal (2) is equal to __________ . |
| Answer» Energy content of a signal x(t), E = |f(t)|2 dt Now, E' = |f(2t)|2 dz for signal f(2t) Putting 2t = z, we get E' = |f(t)|2 dz = . | |
| 490. |
For a npn BJT transistor is 1.64 x 10 H. C = 10 F; C = 4 x 10 F and DC current gain is 90. Find and ( = cut off frequency, C = capacitance, C = parasitic capacitance, = transconductance, = gain BW product) |
| Answer» ∴ fT = 90 x 1.64 x 108 = 1.47 x 1010 Hz gm = 2pfT(Cμ + Cp) = 2 x p x 1.47 x 1010 = (10-14 + 4 x 10-13) gm= 38 mμ. | |
| 491. |
The impulse response () of a linear time-invariant continuous time system is described by () = exp () () + exp (β) (- ), where () denotes the unit step function, and and β are constants. This system is stable if |
| Answer» h(t) = e+atu(t) + eβtu(- t) For h(t) to be stableh(t) dt < ∞ It will happen when a is negative and β is positive. | |
| 492. |
The probability density function of a random variable is as shown. The value of A is: |
| Answer» = 1 = 1 Solve this to get value of A = 1/5. | |
| 493. |
ω = |
| Answer» R1 = (10 + 10) = 20 kΩ . | |
| 494. |
A material has conductivity of 10 mho/m and permeability of 4 x 10 H/m. The skin depth at 9 GHz is |
| Answer» | |
| 495. |
A fair coin is tossed independently four times. The probability of the event "the number of time heads shown up is more than the number of times tails shown up" is |
| Answer» 4 heads and 0 tail 3 head and 1 tail . | |
| 496. |
The steady state error of a stable type 0 unity feedback system for a unit step function is : |
| Answer» . | |
| 497. |
An 8 level encoding scheme is used in a PCM system of 10 H channel BW. The channel capacity is |
| Answer» C = 2B log2 M; M = 8 = 23, log2 M = 3 bits = 2 x 10 x 103 x 3 = 60 kbps. | |
| 498. |
For the power amplifier circuit shown below, the maximum power dissipated by both output transistor is |
| Answer» The maximum power dissipated by both output transistor is maximum | |
| 499. |
An antisymmetric filter having odd number of coefficient may have a performance as |
| Answer» Consider an anti-symmetric filter response with order 5. h(n) = [h(2), h(1), h(0), - h(1), - h(2)] We know, H(0) = addition of all component values. Hence H(0) = h(0) Similarly H(p) = addition of all component values with alternate negative sign. = h(2) - h(1) + h(0) + h(1) - h(2) = h(0) Hence we get all pass filter. | |
| 500. |
A periodic rectangular signal, () has the waveform shown in the given figure. Frequency of the fifth harmonic of its spectrum is __________ |
| Answer» The periodic time = 4 ms = 4 x 10-3 s ∴ The fundamental frequency ∴ Frequency of the 5th harmonic = 250 x 5 = 1250 Hz. | |