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.
| 301. |
For a four point DFT, Linear transformation matrix will be: |
| Answer» | |
| 302. |
In the circuit, the value of the voltage source E |
| Answer» 13 + 4 + E + 5 = 0 ∴ E = - 22 V. | |
| 303. |
Skin depth depends on |
| Answer» | |
| 304. |
The time constant of the network shown in figure is |
| Answer» Req = (R || 2R) + R . | |
| 305. |
For the 8085 assembly language program given below, the content of the accumulator after the execution of the program is |
| Answer» A → 0100 0101 B → 0100 0101 Carry flag → 0 RAR will A → 0010 0010 XRA B 0110 0111 67. | |
| 306. |
. The signal is __________ . |
| Answer» for ∴ for periodicity should be a rational number. But our case not a rational number. Hence non periodic. | |
| 307. |
Conversion rate, for a successive approximation ADC which uses 2 MH clock and 5 bit binary ladder containing 8 VV reference voltage is __________ . |
| Answer» Conversion rate = = = 400000 conversions/sec. | |
| 308. |
V() across R for > 0 is given by, |
| Answer» Time constant t = RC Req for t > 0 is; R1 = R2 = C = 1F ∴ t = 1/2 ∴ V(t) = VF - (VF - Vi)e-t/z VF = VR2(t = ∞) = 2 V Vi = VR2 (t = 0+) = 1 V V(t) = 2 - (2 - 1)e-t/1/2 V(t) = 2 - e-2t. | |
| 309. |
The rms value of the periodic waveform shown in fig. is |
| Answer» | |
| 310. |
The logic circuit given below converts a binary code , into |
| Answer» Let y1 = 1, y2 = 0, y3 = 1 then x1 = 1 x2 = y1 ⊕ y2 ⇒ 1 ⊕ 0 ⇒ 1 x3 = x2 ⊕ y3 = 1 ⊕ 0 = 1 which is gray code conversion. | |
| 311. |
A transmission line terminates in two branches, each of length λ/4, as shown. The branches are terminated by 50Ω loads. The lines are lossless and have the characteristics impedances shown. Determine the impedance Z as seen by the source |
| Answer» . | |
| 312. |
The Nyquist plot of |
| Answer» | |
| 313. |
The counter shown in figure is built using 4 negative edge triggered toggle FFs. The FFs can be set synchronously when R = 0. The combinational logic required to realize a modulo 13 counter is |
| Answer» According to figure, output of combinational logic is applied to R of all FF N = 13 = (1101)2 The logic required is Q4 Q3 Q2 Q1 The gate used is a NAND. Y = Q4 + Q3 + Q2 + Q1 . | |
| 314. |
A plane wave having the electric field component V/m and travelling in free space in incident normally on a lossless medium with = and = 90 which occupies the region ≥ 0. The reflected magnetic field component is given by |
| Answer» ∴ r = - Ve will change direction and M will not change its direction Direction of for incident field is in x, so direction of in the reflected field is also is in x. . | |
| 315. |
I() and I() are the Laplace transforms of () and () respectively. The equations for the circuit shown in above figure when switch S is closed at = 0; are |
| Answer» For t < 0, steady state is reached iL = 2A L = 2 VC = 8 V Using Laplace transformation for t = 0+ Apply KVL to mesh 1 Apply KVL to mesh 2 . | |
| 316. |
The residues of a complex function at its poles are |
| Answer» . | |
| 317. |
The Z-inverse of the given Z-transform is __________ |
| Answer» value of x(n) = d(n) impulse response. | |
| 318. |
Find the Fourier transform of the half cosine pulse as shown below : |
| Answer» The given signal can be expressed as multiplication of x1(t) and x2(t) as shown below. where A = 2, T/2 = 0.25 => T = 0.5 ∴ x(t) = x1(t) x x2(t) => X(f) = X1(f) * X2(f) Now X1(f) = [δ(f - f0) + δ(f + f0)] = [sin c [T(f - f0)] + sin c[T(f + f0)]] Now, A = 2, T = 0.5 and f0 = = 1 => X(f) = 0.5[sin c (0.5(f - 1)) + sin c (0.5(f + 1))]. | |
| 319. |
The - relation of a capacitor is = 1 + + . Find the amount of energy required to charge this capacitor from () = 0 to () = 5 C. |
| Answer» | |
| 320. |
V-I relationship of a network N is given by, V = 8 - 4I.The equivalent circuit to the network N is shown belowThe values of I and R are |
| Answer» Given equivalent circuit is Norton's equivalent circuit V = 8 - 4I Put V = 0I = IN = 2 A Put I = 0 V = VTH = 8 V . | |
| 321. |
If the open-loop transfer function is a ratio of a numerator polynomial of degree '' and a denominator polynomial of degree '', then the integer ( - ) represents the number of: |
| Answer» Asymptotes. | |
| 322. |
The expression given is F(A, B, C, D, E, F, G, H) = Σ(33, 48, 102, 158, 222, 229, 233, 243, 255); then the __________ and __________ minterms are grouped together into dual and variable __________ is reduced using the dual |
| Answer» We use Q-McCluskyey method of redaction We can observe that minterms 158 and 222 can be grouped to reduce the variable 'B'. | |
| 323. |
In = the value of is |
| Answer» = (10)2 + (10)3 + ... + (10)8 + (10)9 = (2)10 + (3)10 + .... + (8)10 + (9)10 = 2 + 3 + ... + 8 + 9 = 44. | |
| 324. |
From the above statements which are the false statements? |
| Answer» where r is charge density and RH = + ve, then the material is p - type = - ve, then the material is n - type RH ∝ VH where VH is Hall voltage. | |
| 325. |
In a series operated thyristor configuration, each thyristor should support 100 V. However, the three thyristors only support voltage of 100 V, 90 V and 85 V respectively. Hence, the string efficiency in the case is : |
| Answer» String efficiency = = = 0.9167 = 91.67%. | |
| 326. |
A uniform plane wave in the free space is normally incident on an infinitely thick dielectric slab (dielectric constant ε = 9). The magnitude of the reflection coefficient is |
| Answer» Thus reflection coefficient = = - 0.5 Thus magnitude is 0.5 . | |
| 327. |
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. If the bits 0 and 1 are transmitted using bipolar pulses, the minimum bandwidth required for distortion free transmission is |
| Answer» According to Nyquist Criteria, minimum sampling frequency = 2 x 4 k = 8 kHz But each cycle can accommodate two bits Thus minimum frequency = 4 kHz. | |
| 328. |
The question below consists of a pair of related words followed by four pairs of words. Select the pair that best expresses the relation in the original pair.Unemployed: Worker |
| Answer» A worker who is inactive or not working is termed as unemployed, similarly land which is inactive or not in use is called Fallow. | |
| 329. |
An FM transmitter delivers 80W to a load of 30 W when no modulation is present. The carrier is now frequency modulated by a single sinusoidal signal and the peak frequency deviation is so adjusted to make the amplitude of the second sideband is zero in the given output.J(0, 0) = 1, J(2, 4) = 0, J(3.8) = - 0.4, J(5.1)= - 0.1J(2, 4) = 0.52, J((3.8) = 0, J(5.1) = - 0.33, J(2.4) = 0.1J(3.8)= 0.41, J(5.1) = 0Power in the first order sidebands is : |
| Answer» Power in first order side bands = 2(J1(β))2 x 100 = 21.78 W. | |
| 330. |
Figure below shows D type Flip-Flops connected as shift register, then outputs Q, Q, Q, Q are given by: |
| Answer» Here DA = QD ⊕ QC; DB = QA; DC = QB; DD = QC For the given input string, outputs are QD, QC, QB, QA. Therefore right most bit is QA and left most bit is QD. Now we determine the string sequence according to the above mentioned equations. (i) (ii) (iii) | |
| 331. |
In a certain network not containing any dependent sources, number of nodes = 4 and number of branches is equal to 6. |
| Answer» No. of independent loop currents is 6 - 4 + 1 = 3. | |
| 332. |
A 4 cm by 3 cm rectangular waveguide with air cavity, is operating at 12 GHz with TE mode. The cut-off frequency is: |
| Answer» . | |
| 333. |
For a bipolar transistor, avalanche multiplication factor depends on __________ . |
| Answer» Avalanche multiplication factor where, βVCBO = collector - emitter breakdown voltage with open circuited base. | |
| 334. |
Given L[] = and L[] = also = , where, = then, L[] = __________ . |
| Answer» Now, x = ∴ log x = ejp x log S log x = y, x = ey log x = log Sejp x = Sejp x = S-1 ∴ ∴ ∴ L[f(t)] = L[f(t)] x L[g(t)]. | |
| 335. |
The voltage reflection coefficient on a 50 Ω line is 0.7∠30°.The load impedance is |
| Answer» . | |
| 336. |
For finding the square root of N using Newton's iterative formula, the correct expression is: |
| Answer» f(xn) = (x2n - N). | |
| 337. |
V at < T, = ∞ |
| Answer» At t = 0, C1, C2, C3 acts as short. ∴ entire voltage will appear cross R. At t increases the voltage across the resistance decrease exponentially. T = RCnet = 10 x 2.5 x 10-6 = 25 μ sec Initial voltage VR = V0e-t/RC at t = 0 = 100 = 100 e-t/25x10-6 V. VR = 100 e-40x104t V. At t = ∞, VR = 0 The charge C0 will be distributed in C1, C2, C3 and no current will flow. | |
| 338. |
The function shown in figure, can represent a probability density function for A __________ . |
| Answer» The condition for the probability density of function d(x) is f(x)dx = 1 and is always non negative or the area under the curve f(x) is unity. ⇒ A x 6 + x 2 x 2A = 1 or 8A = 1, or A = . | |
| 339. |
For the circuit shown in figure below, the initial conditions are zero, its transfer function H() = is : |
| Answer» . | |
| 340. |
For a lossless transmission line, the propagation constant is given as γ = __________ . |
| Answer» For lossless line, a = 0, R = 0 and G = 0 ∴ | |
| 341. |
Consider the pulse shape () as shown. The impulse response () of the filter matched to this pulse is |
| Answer» h(t) = s(T - t). | |
| 342. |
If each stage has a gain of 10 dB and noise figure of 10 dB, then the overall noise figure of a two stage cascade amplifier will be |
| Answer» Overall noise figure F of the system is F = F1 + F1 = F2 = 10 dB = 10 in absolute units g = 10 dB = 10 in absolute units ∴ F = 10 + = 10.9 . | |
| 343. |
A rectangular waveguide of internal dimensions ( = 4 cm and = 3) is to be operated in TE mode. The minimum operating frequency is |
| Answer» . | |
| 344. |
Using 4-bit numbers ( = 4) if = 0011, how is expressed in 2' complement? |
| Answer» In 2's complement k + 11012 = 2n = 16 So -k is represented by 11012 . | |
| 345. |
The effective length of an 1/2 antenna (Given R = 73 Ω and (A) = 0.34λ and η = 120) is __________ . |
| Answer» = le = 0.5133λ. | |
| 346. |
Find , |
| Answer» Z11(s) = Z1(s) + Z2(s) = 1 Ω. | |
| 347. |
The transfer function of a system is . The characteristic equation of the system is : |
| Answer» Characteristic equation = 1 + GH = 0 (s + 1)2(s + 2) + (2s2 + 6s + s) = 0 Apply R-H. | |
| 348. |
With respect to the Miller integrator sweep shown below, the sweep rate will be |
| Answer» Sweep rate = . | |
| 349. |
The trigonometric Fourier series for the waveform shown below contain |
| Answer» Since given waveform is even, thus it will have only cosine terms. Function f(t) has more negative value then positive, so it will have negative DC Component. | |
| 350. |
For a class B amplifier using a supply of V = 30 V and driving a load of 16 Ω the maximum efficiency will be __________ . |
| Answer» The maximum output power is | |