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.
| 351. |
If = then the sequence is |
| Answer» Consider cos= cos= cos Number of samples per period = 16p This is not a rational number. Hence signal is non periodic. | |
| 352. |
If V = 4 in the figure, the value of I is given by, |
| Answer» Using Nodal analysis At node V1 At node V2 ∴ 2V2 - V1 = 0 ∴ V1 = 2V2 Put V2 = V = 4 V1 = 8 V Put in equation (i), IS = x 8 - 4 = 10 - 4 = 6 A. | |
| 353. |
Choose correct option for a stable system : |
| Answer» From Statement 1: Roots of the characteristics equation of the system are real and negative i.e. the poles are on the left half plane. Hence the system is stable. From Statement 2: Area within the impulse response is finite i.e. it is finite duration signal. It produces a bounded output. Hence system is stable. So both the statements are true. | |
| 354. |
Locus of a point satisfying the equation arg is: |
| Answer» Solving this further we will get x2 + y2 - 3x + 3y = 0 which is the equation of circle. | |
| 355. |
A silicon PN junction with no bias have a doping concentration of N = 2 x 10 cm and N = 10 cm. The space charge width is: |
| Answer» w = 3.345 x 10-5 cm. | |
| 356. |
For a given transfer function: Damping ratio is: |
| Answer» 1 + G(s)H(s) = s2 + 2s + 2 ωn = 2 , 2εeωn = 2, εe = = 0.707. | |
| 357. |
The energy of constant-amplitude complex valued exponential sequence is __________ (where is the constant amplitude) |
| Answer» Let the constant amplitude complex valued exponential sequence. x(n) = Ae-jωn Energy = |x(n)|2 = |Ae-jωn|2 A2 |(e-jωn)|2 = A2(∞) = ∞ Hence energy in infinite. | |
| 358. |
Find the electric charge required on the earth and moon to balance their gravitational attraction if the charge on the earth is 10 times that on the moon. |
| Answer» Using law of gravitations, the force between the two point masses is F = M1 = mass of moon = 6.7 x 1022 Kg M2 = mass of earth = 6 x 1024 Kg r = distance between masses = 380 mm G = universal gravitational constant = 6.7 x 10-11 Nm2/Kg2 Let Q1 and Q2 be the charge on moon and earth respectively. Now the gravitational force must be balanced by the force of repulsion Q1Q2 = 4pε0M1M2G but Q2 = 10Q1 = 10 x 4pω0M1M2G ∴ Q2 = 173 x 1012 Q1 = 17.3 x 1012 C Q1 = 17.3 TC. | |
| 359. |
An audio signal of FM radio station is sampled at a rate of 15% above the Nyquist rate. Signal is sampled in 512 levels and bandwidth of the signal is 4 KHz. The minimum bandwidth required to transmit this signal will be |
| Answer» Nyquist rate = 2 x 4 kHz = 8 kHz Sampling rate = 8 kHz + 1.2 kHz = 9.2 kHz Number of bits required = log2512 = 9 Thus bandwidth = 9 x 9.2 = 82.8 k bits/sec. | |
| 360. |
In the given figure.The damped frequency is __________ . |
| Answer» Damped frequency ωd = 3/2. | |
| 361. |
Consider the program segment written for 8085 based system:LXI S, FFFFHMVI A, 00HMVI B, 0FHSUB BCNC GOLOCMVI A, FFH where in the program subroutine labelled ''GOLOC'' introduces delay of 28 T-states only. How many T-states are required to execute above program statement? |
| Answer» Since carry (CY) flag will not be set at the execution of "SUB B" instruction; check for CY flag fails but this requires 9(= 6 OP - CODE FETCH + 3T - states) T-states ∴ Total T-states = 10 + 7 + 7 + 4 + 9 + 7 = 44. | |
| 362. |
A 16 bit type ADC uses 1 MHz clock its maximum conversion time is __________ . |
| Answer» Conversion time = ⇒ ⇒ 6.5 m sec. | |
| 363. |
A 300 MHz plane EM wave is propagating in free space. The wave is incident normally on an infinite copper slap. The antenuation constant for the wave is (σ = 5.8 x 10 mho/ m) |
| Answer» Attenuation constant : . | |
| 364. |
Transmitted voltage level is: |
| Answer» = 61 + 0.5 = 61.5 = 7.348 V. | |
| 365. |
The signal () = (- ) is __________ . |
| Answer» For signal x(- t) to be causal, the original signal x(t) has to be anticausal. Similarly for signal x(- t) to be non causal, the original signal x(t) has to be causal. Hence response of x(t) depends solely on x(t) only. | |
| 366. |
The autocorrelation function of an ergodic random process is given by __________ . |
| Answer» The expectations or ensemble average of a random process x(t) are averages "across the process". The DC value of x(t) is defined by the time average The other time average of particular interest is the autocorrelation function Rx(t, T) defined in terms of the sample function x(t) observed over the interval - T ≤ t ≤ T. Following equation, we may formally define the time-averaged autocorrelation function of a sample function x(t) as follows : This second time-average should also be viewed as a random variable with a mean and variance of its own. In a manner similar to ergodicity of the mean, we say that the process x(t) is ergodic in the autocorrelation function if the following two limiting conditions are satisfied : . | |
| 367. |
The ratio of condition current density to displacement current density for a good conductor and a good dielectric is respectively __________ . |
| Answer» = 1 can be considered as the dividing line between the conductor and dielectrics for a good conductor >> 1 and for good dielectrics << 1 at any frequency ω. | |
| 368. |
A '' bit sequence generator employing SR flip-flop has maximum length of the sequence as : |
| Answer» Since for SR flip-flop all zero state is lock-in state; from which SR can not get out, the maximum length of sequence with 'n' bits will be 2n - 1. | |
| 369. |
A Hartley oscillator uses FET with of 2 ms and = 2 KΩ. The total coil inductance is 10 μH with turns ratio of input to output side of 1 : 10. It is tuned with a 50 pf capacitor. Find frequency of oscillations, amplifier gain margin in dB. |
| Answer» = 7.11 MHz μ = gm x rd | |
| 370. |
The recursion relation to solve = using Newton Raphson method is |
| Answer» f(xn) = xne-xn f(xn) = 1 + e-xn f'(xn) = 1 + e-xn | |
| 371. |
In the circuit shown in the given figure the diode states at the extremely large negative values of the input voltage are : |
| Answer» Input V1 is extremely large negative value, so D1 is ON. Diode D2 is shunted by R3 and the junction of R2R3 is more negative than the lower end. Diode D2's cathode is at negative potential. But the anode of D2 is less negative than E as the input voltage is extremely negative and D2 is OFF. | |
| 372. |
Consider a baseband binary PAM receiver shown below. The additive channel noise () is whit with power spectral density S() = N/2 = 10 W/Hz. The low-pass filter is ideal with unity gain and cutoff frequency 1 MHz. Let Y represent the random variable ().Y = nN if transmitted bit b = 0Y = + N if transmitted bit b = 1Where represents the noise sample value. The noise sample has a probability density function, P() = 0.5 (This has mean zero and variance 2/). Assume transmitted bits to be equiprobable and threshold is set to /2 = 10 V.The value of the parameter (in V) is |
| Answer» → exist in fourier transform Variance = which is equal to power since mean is zero ⇒ a = 107. | |
| 373. |
The inconsistency of Ampere's law can be removed by adding __________ to it. |
| Answer» The ampere's law is given as ∇.H = J Taking divergence on both sides ∇.(∇xH) = ∇.J i.e., ∇.(∇xH) = ∇.J = 0 ∴ divergence of curl is always zero But ∇.J = -rv ... continuity equation. Hence the inconsistency. It can be removed as follows → Let ∇ x H = J + G where G is some unknown quality. Then ∇.(∇ x H) = ∇.(J x G) = 0 i.e. ∇.J + ∇.G = 0 i.e. ∇.J = -∇.G or ∇.G = (-rv) i.e. ∇.G = rv ...... Gauss's law in point form Differentiating on both sides. ∇.D = rv Comparing (i) and (ii), G = D Thus ∇ x H = J + D Where J is conduction current density and D is displacement current density which is added to remove the inconsistency of Ampere's law. | |
| 374. |
If a signal (/2) has energy is E/ then energy of the signal is equal to __________ ( is constant) |
| Answer» ∴ dt = 2 dV ∴ ∴ | |
| 375. |
The maximum effective aperture of an antenna which is operating at wavelength of 2 m and has a directivity of 100 is __________ . |
| Answer» The maximum effective aperture is given by: D = 100, λ = 2 m . | |
| 376. |
A speech signal, band limited to 4kHz and peak voltage varying between + 5V and - 5V is sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits. Assuming the signal to be uniformly distributed between its peak values, the signal to noise ratio at the quantizer output is |
| Answer» SNR = 6n + 1.7 = 6 x 8 + 1.7 = 48 + 1.7 . | |
| 377. |
The drain current of a MOSFET in saturation is given by = K(V - V) where K is a constant. The magnitude of the transconductance is |
| Answer» ID = k(VGS - VT)2, . | |
| 378. |
Two infinitely long wires carrying current are as shown in the figure below. One wire is in the y-z plane and parallel to the y-axis. The other wise is in the x-y plane and parallel to the x-axis. Which components of the resulting magnetic field are non-zero at the origin? |
| Answer» Apply right hand thumb rule. | |
| 379. |
A uniform plane wave is described by the equation H = A/m. If the velocity of the wave is 2 x 10 m/s and ε = 1.8, then. The electric field intensity is |
| Answer» Using , we have, ∴ = 903.5e-j0.1pzv/m. | |
| 380. |
The surface charge density on a conductor is 1.55 x 10C/m. The conductor is immersed in water with ε = 78ε.The electric field will be __________ . |
| Answer» D = εE = 78ε0E ∴ E = 2.25 V/m. | |
| 381. |
In the given figure.The damping coefficient for the given circuit is __________ . |
| Answer» Damping coefficient is given by, Substituting R = 1 Ω, L = 1 H ∴ . | |
| 382. |
Consider the circuit The line 1 is at stuck at '0' then output Z is equal to: |
| Answer» Output of AND gate is 0 Thus Z = = (C + D). | |
| 383. |
In a infinite ladder circuit as shown above each resistance of Ω then find R |
| Answer» Let RAB = R Since ladder is infinite we can assume RCD = R ∴ R = r + (r || R) = ∴ ∴ Rr + R2 = r2 + 2rR R = 1.61r or - 0.615r R cannot be negative ∴ R = 1.61r. | |
| 384. |
In the circuit shown below current through the resistor is shown. Voltage across 4 Ω resistor is: |
| Answer» Voltage across 5 Ω = 10 V Thus voltage across 7 Ω = 10 V (5 Ω || 7 Ω) | |
| 385. |
A speech signal, band limited to 4kHz and peak voltage varying between + 5V and - 5V is sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits. The number of quantization levels required to reduce the quantization noise by a factor of 4 would be |
| Answer» To reduce quantization noise by 4 No. of levels should increase by 2 Thus 512 levels are required. | |
| 386. |
The fourier transform of a voltage signal () is X(). The unit of X() is __________ . |
| Answer» F[x(t)] = X(f) X(f) = x(t).e- jpft.dt From above expression, the unit of x(t) x e-j2pft is volts and that of dt is seconds. | |
| 387. |
A diode detector load consists of 0.01 μF capacitor in parallel with a 5 resistor. The maximum depth of modulation without diagonal clipping at modulating frequency of 1000 H and 10 kH is |
| Answer» At f = 1 KHz Similarly for 10 kHz, | |
| 388. |
The given characteristic polynomial + + 2 + 2 + 3 = 0 has |
| Answer» By Routh's Array When ε -> 0, two sign changes in the first column, so there are two roots in RHS plane. | |
| 389. |
In -well CMOS fabrication substrate is: |
| Answer» Lightly doped p-type. | |
| 390. |
The transfer function of a zero-order-hold system is : |
| Answer» . | |
| 391. |
The amplitude modulated wave is given by __________ = 25(1 + 0.7 cos 5000 - 0.3 cos 1000) x sin 5 x 10 The amplitude of carrier and sideband frequencies in magnitudes are |
| Answer» Modulation index is 0.7 and 0.3 so that amplitude of signal is 0.7 x 25 = 17.5 and 0.3 x 25 = 7.5 On modulation the sideband amplitude is . Hence 8.75 and 3.75 | |
| 392. |
A signal flow graph of a system is given belowThe set of equations that correspond of this signal flow graph is |
| Answer» . | |
| 393. |
(11) + (11) = (11) |
| Answer» (11)2 + (11)3 = 3 + 4 = 7 and (11)6 = 61 + 60 = 6 + 1 = 7. | |
| 394. |
If [] = 3δ[] + δ[ - 1], [] = 2δ[] + δ[ - 2][] = δ[] - 3δ[ - 1] + 7δ[ - 4] + 6δ[ - 6][] = [] * [] + 3[n] * [] has value |
| Answer» h1[n] * h2[n] + h3[n] * h1[n] = h1[n] * (h2[n] + h3[n]) Distributive property for convolution, (h2[n] + h3[n]) = 3δ(n) - 3δ(n - 1) + δ(n - 2) + 7δ(n -4) + 6δ(n - 6) then h1(n) * (h2(n) + h3(n)) {9, - 6, 0, 1, 21, 7, 18, 6}. | |
| 395. |
The feedback control system shown in the given figure represents |
| Answer» j gives type of system j = 3 ∴ given system is type 3 system. | |
| 396. |
If the -transform of () is given by then the filter it represents is __________ . |
| Answer» i.e. we can see that According to the property, the filter is an all-pass filter. | |
| 397. |
Find the voltage across 64 Ω resistor |
| Answer» Voltage across 64 Ω is = 1x=V. | |
| 398. |
If (Z) = + , then is given by |
| Answer» | |
| 399. |
The capacitors C, C, C and C have a capacitance of 4 μF each and the capacitor C has a capacitance of 10 μF. The effective capacitance (in μF) between the points X and Y will be |
| Answer» Converting into T network | |
| 400. |
(9) x (9) x(9) = (?) |
| Answer» (9 x 9 x 9)15 = (729)15 = (2D9)16 . | |