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.
| 201. |
An astable multivibrator circuit using IC 555 timer is shown below. Assume that the circuit is oscillating steadilyThe voltage V across capacitor varies between |
| Answer» Charging and discharging level of capacitor will be the voltage across it. This is equal to Vcc and Vcc. Thus 3V to 6V is the voltage VC across the capacitor. | |
| 202. |
The current I for the network |
| Answer» . | |
| 203. |
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 probability of bit error is |
| Answer» Probability of error = 0.5e-107 x 10 -6 = 0.5e-10. | |
| 204. |
In a synchro error detector, the output voltage is proportional to [ω()], where ω() is the rotor velocity and equals : |
| Answer» In a synchro detector, output voltage is proportional to ω(t) v0 a ω(t) So, n must be equal to one. | |
| 205. |
A system with transfer function has an output for the input signal . Then, the system parameter '' is |
| Answer» |y(t)| = |M(ω)||x(t)|. ω2 + p2 = ω2p2, 4 + p2 = 4p2 3p2 = 4, . | |
| 206. |
The current (), though a 10 Ω resistor in series with an inductance, is given by () = 3 + 4 sin (10 + 45°) + 4 sin (300 + 60°) Amperes. The RMS value of the current and the power dissipated in the circuit are |
| Answer» Irms = = 6.4 Power P = I2R = 25 x 10 = 41 x 10 = 410 W. | |
| 207. |
The differential equation for the current () in the circuit of figure is |
| Answer» Apply KVL to the loop Differentiate w.r.t. t . | |
| 208. |
is __________ . |
| Answer» E = e-4tdt which is infinity Power of the signal This is of ∞/∞ form, we make use of L' hospital rule e-2t is neither an energy signal nor a power signal. | |
| 209. |
The order of the differential equation is |
| Answer» The order of differential equation is two. | |
| 210. |
Consider an incremental vector element of length ΔL and angle between an electric field E and ΔL s cos θ. Then in which direction should ΔL be placed to be obtain maximum value of Δ? |
| Answer» Now E is a definite value at the point which it is present and is independent of direction of ΔL. The magnitude of ΔL is constant and direction is along the unit vector. ∴ Maximum value of Δv will occur at cos θ = - 1 or ΔL points in opposite direction to E. | |
| 211. |
A MOD - 2 counter, the FF has a = 60 nsec. The NOR gate has a of 30 nsec. The clock frequency is |
| Answer» Clock period should be greater than or equal tpd MOD-12 → X FF = 4 x 60 = 240 n sec tpd of NOR gate = 30 n sec Total = 240 + 30 ⇒ 270 n sec . | |
| 212. |
Evaluate the power stored in inductor and capacitor |
| Answer» Energy stored in inductor L is as inductor will act as a short circuit for dc. ∴ EL Energy stored in capacitor C is Voltage across capacitor C is 5 V (i.e. voltage across shunt resistor R) ∴ EC | |
| 213. |
If the Laplace transform of the voltage transform of the voltage across a capacitor of value F is the value of the current through the capacitor at = 0 is |
| Answer» capacitor = ∴ I(s) = vc(s)/c(s) . | |
| 214. |
The switching expression corresponding to :(A, B, C, D)= ∑(0, 3, 4, 7, 8) + (10, 11, 12, 14, 15) is: |
| Answer» | |
| 215. |
Given the function F(A, B, C, D) = (0, 1, 2, 3, 4, 6, 8, 9, 10, 11, 12). The minimized form of the function can be given by : |
| Answer» Given S(A, B, C, D) = m(0, 1, 2, 3, 4, 6, 8, 9, 10, 11, 12) S = (B + D)(A + B + C) . | |
| 216. |
For the network shown, determine , and A |
| Answer» = 24.04 μA IE = (1 + β)IB = 2.428 mA; = 10.7 Ω (i) zi = RB||Bre = = 0.937 K = 1.067 K (ii) Z0 = RC = 3 kΩ (iii) = -280.37 ∴ Zi = 1.067 KΩ; Z0 = 3 KΩ; Av = -280.37 . | |
| 217. |
A germanium diode has a reverse saturation current of 30 μA at 125°C and forward current of 1 A at a forward potential of 0.8 V and the reverse potential is 0.6 V What are its dynamic forward and reverse resistance at this temperature. |
| Answer» We have, I = I0•(eV/nVT -1) ∴ For forward voltage greater than few teeth's of volts, I >> I0 ∴ ∴ Forward dynamic resistance, = 34.3 Ω Reverse dynamic resistance = = 20 kΩ. | |
| 218. |
A radio wave is incident on a layer of ionosphere at an angle of 30° with the vertical. If the critical frequency is, 1.2 MHz, the maximum usable frequency is |
| Answer» fmuf = for sec i = 1.2 x 106 x sec (30°) = 1.386 MHz. | |
| 219. |
Fourier transform of signal obtained after taking difference between step function and its reverse version is given as __________ . |
| Answer» u(t) - u(-t) = sgn (t) | |
| 220. |
{ () } is a real-valued periodic sequence with a period N. () and X() form N-point. Discrete Fourier Transform (DFT) pairs. The DFT Y() of the sequence. () = () X ( + ) is |
| Answer» Direct property Convolution in time domain ↔ Multiplication in frequency domain. | |
| 221. |
The signal x(t) is described by Two of the angular frequencies at which its fourier transform becomes zero are |
| Answer» X(jω) = 1.e-jωt = At ω = 0 X(j0) = 2 At ω = p X(jp) = 0 ω = 2p X(j2p) = 0 | |
| 222. |
(29) ÷ (29) x (29) = (23) Find |
| Answer» (29)13 ÷ (29)12 x (29)x = (23)16 => (35)10 , (33)10 x (29)x = (35)10 => x(29)X = 35 ∴ (29)x = (33)10 => X = 12. | |
| 223. |
If the scattering matrix [S] of a two port network is then the network is |
| Answer» S11 ≠ S22 ≠ lossy [S]1 = [S] Reciprocal. | |
| 224. |
Consider the frequency modulated signal 10 cos (2 x 10 + 5 sin (2 x 1500) + 7.5 sin (2 x 1000) with carrier frequency of 10Hz. The modulation index is |
| Answer» Total frequency deviation = 5 x 1500 + 7.5 x 1000 = 15000 Thus modulation index = . | |
| 225. |
An air-filled rectangular waveguide has dimensions 8 x 10. The group velocity is |
| Answer» =2.6 x 108 m/s. | |
| 226. |
A function n(x) satisfied the differential equation where L is a constant. The boundaryconditions are: (0) = K and (∞) = 0. The solution to this equation is |
| Answer» n(∞) = 0, ∴ B = 0, n(0) = k ∴ B = k. | |
| 227. |
If over the path shown in the figure is |
| Answer» . | |
| 228. |
50 independent low pass signals of bandwidth W, 2W, 3W, W, 2W, 3W, W, 2W, 3W H are to be time division multiplexer on a common channel using PAM. To achieve this, the minimum transmission capacity of the channel should be __________ . |
| Answer» The commulator in TDM has to rotate at a speed twice that of highest bandwidth 3W x 2 = 6W Hz. ∴ Channel capacity = 2B = 2(6 W) = 12 W Hz. | |
| 229. |
Find the value of K for which V = 0 |
| Answer» | |
| 230. |
Two coils X of 1000 turns and Y of 2000 turns are placed such that 60% of the flux produced by coil X links coil Y. A current of 1 A in coil X produces 0.1 wb flux. The mutual inductance between coils is |
| Answer» L2 = 0.1 X 4 = 0.4 M = kL1L2 = 0.6 0.4 X 0.1 = 0.6 X 0.2 = 0.12 . | |
| 231. |
A delta modulation system is designed to operate at 3 times the Nyquist rate for a signal with a 3 KHz bandwidth. The quantization step size is 250 mv. The maximum amplitude of a 1 KHz input sinusoidal for which the delta modulator does not show slope overload is : |
| Answer» . | |
| 232. |
Values of , , V and at = 0 are given by, |
| Answer» For t = 0 + iL(0-) = = 6 A VC(0-) = 0 V For t = 0+ iL(0+) = 6 VC(0+) = 0 Apply KCL at node V 6 - V = (6 + v)2 6 - V = 12 + 2V - 6 = 3V V = VL = - 2V At t = 0+, = -2 ∴ ∴ ∴ iL = 6 Amp VC = 0 V . | |
| 233. |
A continuous time LTI system is described byAssuming zero initial conditions, the response () of the above system for the input () = () is given by |
| Answer» y(t) = (e-t - e-3t) u(t). | |
| 234. |
Two discrete time systems with impulse responses h[] = δ[ - 1] and h[] = δ[ - 2] are connected in cascade. The overall impulse response of the cascaded system is |
| Answer» h1(n) → delay by 1, h2(n) → delay by 2, h1(n) * h2(n) → delay by 3. | |
| 235. |
The magnitude plot of a composite signal () = + is |
| Answer» x(t) = ej2 . 5t (e-j0.5t + e-j0 . 5t) x(t) = 2ej2.5t cos (0.5)t The magnitude of x(t) is |x(t)| = 2|cos (0.5)t|. | |
| 236. |
A message signal given by is amplitude modulated with a carrier of frequency ω to generate = [1 + ] cos ω |
| Answer» . | |
| 237. |
A 16 bit counter type ADC uses 1 MH clock then its maximum conversion time is __________ . |
| Answer» Maximum conversion time of counter type ADC = (216 - 1) x 1 μs = 65.5 ms » 66 ms. | |
| 238. |
A first order system will never be able to give a __________ response. Choose the correct option |
| Answer» The ideal response of a Band pass filter BRF and APF is as shown So for BPF we require 0 at ω = 0, p i.e. at least 2 zeroes are required in the Transform. Hence order has to be atleast 2. Similarly for band reject filter order should be at least 2. Whereas APF has a constant amplitude. Hence Zeroes in the function need not matter. APF function is given as Hence first order can provide APF. | |
| 239. |
The analog signal given below is sampled by 600 samples per second() = 2 sin 480 + 3 sin 720The folding (maximum) frequency is __________ . |
| Answer» Folding frequency (max.). | |
| 240. |
In the circuit shown, the power supplied by the voltage source |
| Answer» KVL → 10 = 2(i + 3) + 2(i + 2) 10 = 10 + 4i ⇒ i = 0 ∴ Power = 0 | |
| 241. |
A sequential multiplexer is connected as shown in figure. Each time the multiplexer receives the clock, it switches to the next channel from 6 to 1. If the input signals are :A = 10 cos 2(4 x 10t)B = 15 cos 2(5 x 10t)C = 20 cos 2(6 x 10t)D = 10 cos 2(10 x 10t)The minimum clock frequency should be __________ kHz. |
| Answer» Minimum clock freq. = 2fm Where fm is highest freq. component = 2 x 10 kHz = 10 kHz. | |
| 242. |
The Boolean function realized by the logic circuit shown is |
| Answer» F = Σm (2, 3, 5, 7, 8, 9, 15). | |
| 243. |
() for > 0 is given by |
| Answer» The time constant t is given by As Req = 4 || 4 = 2 Ω ∴ i1(t) = iF - (iF - i1)e-t/t ifinal = = 2.5 A ∴ i1(t) = 2.5 - (2.5 - 1.25)e-t/1/2 i1(t) = 2.5 - 1.25e-2t. | |
| 244. |
Find I in 4 Ω resistor. |
| Answer» Apply KVL to first to first loop 6i1 - 2i2 = 10 (4 + R)i2 - 2i1 - 2i3 = 0 4i3 - 2i2 = 0 but i3 = 0.5 A 2 - 2i2 = 0 ∴ i2 = 1 A, 6i1 - 2 = 10 ∴ i1 = 2 A, (4 + R) - 2 X 2 -1 = 0 ∴ R = 1 Ω Using Nodal analysis for loop 2 At node A, VA - 20 + VA + VA - VB = 0 3VA - VB - 20 = 0 3VA - VB = 20 ...(i) At node B, 2VB - 2VA + 2VB + VB = 0 5VB - 2VA = 0 ...(ii) Multiplying (i) by 2 and (ii) by 3 6VA - 2VB = 40 - 6VA - 15VB = 0 13VB = 40 VB ≅ 3 V ∴ | |
| 245. |
A white noise () with two-sided power spectral density 1 x 10 W/Hz is input to a filter whose magnitude squared response is shown below.The power of the output process Y() is given by |
| Answer» Since bandwidth is 10 kHz, thus output power is 10 x 10-11 x 10 x 103 = 1 x l0-6 W. | |
| 246. |
A short vertical grounded antenna is required to radiate at 1 MHz. The effective height of the antenna is 30 m. The calculated value of radiation resistance is |
| Answer» Rr = 80p2. | |
| 247. |
A certain JK FF has = 12 nsec. The largest MOD counter that can be constructed from these FFs and still operate up to 10 MHz is |
| Answer» f0 = 107 Hz ⇒ 8.33 n = 8. | |
| 248. |
A second-order system has a transfer function is given by . If the system, initially at rest is subjected to a unit step input at = 0, the second peak in response will occur at : |
| Answer» Peak time . | |
| 249. |
Value of Z for maximum power transfer is |
| Answer» The value of load for maximum power transfer is given by complex conjugate of ZAB. ZAB = R + jXL = R + jωL. ∴ ZL for maximum power transfer is given by, ZL = R - jωL. | |
| 250. |
Consider an angle modulated signal() = 6cos[2 x 10 + 2sin(8000) + 4cos(8000)] VThe average power of () is |
| Answer» . | |