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.
| 251. |
If the closed-loop transfer function () of a unity negative feedback system is given by then the steady state error for a unit ramp input is : |
| Answer» . | |
| 252. |
In 8085 microprocessor system, can an input and output port have same port address? |
| Answer» Since microprocessor differentiates between an input or output operation on a port depending upon control signals RD and WR . Therefore in a microprocessor system same port address can be given to two different input and output ports. | |
| 253. |
5 skilled workers can build a wall in 20 days; 8 semi-skilled worker can build a wall in 25days; 10 unskilled workers can build a wall in 30 days, If a team has 2 killed, 6 semi-skilled and 5 unskilled workers, how long will it take to build the wall? |
| Answer» Given, 5 skilled workers can build a wall in 20 days i.e., 1 skilled worker can the wall build in 100 days ∴ The capacity of each skilled worker is 8 semi-skilled workers can build a wall in 25 days i.e., 1 semi-skilled worker can build a wall in 200 days ∴ The capacity of each semi-skilled worker is Similarly, The capacity of 1 unskilled worker is Now, The capacity of 2 skilled + 6 semi-skilled + 5 unskilled workers is ∴ The required numbers of days is 15. | |
| 254. |
The fourier transform of the half cosine pulse as shown below is __________ |
| 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)] where and X2(f) = T . sin c[fT] ⇒ X(f) = [δ(f - f0) + δ(f + f0)] *T. sinc(fT) = [sinc[T(f - f0)] + sinc [T(f + f0)]] Now, A = 2, T = 0.5 and ⇒ X(f) = 0.5[sin c(0.5(f - 1)) + sinc(0.5(f + 1))]. | |
| 255. |
A lossless transmission line is 50 cm long and operates at a frequency of 500 MHz. Inductance and capacitances present in a line are 0.5 μH/m and 200 pF/m. Phase constant and phase velocity are respectively. |
| Answer» β = ωLC = 106 x 2p x 500 x | |
| 256. |
The wavelength of a wave with propagation constant (0.1 + 0.4)m is |
| Answer» γ = a + jβ ∴ β = 0.4p | |
| 257. |
For the network shown below, determine and , if, = ∞, C = 36 P, C = 4 P, C = 1 P, C = 8 P, C = 6 P where C = wiring capacitance at output side, C = wiring capacitance at input side |
| Answer» Equivalent is RTi = RS || R1 || R2 || Ri = 0.531 KΩ Ci = Cwi + Cbe + (1 + AV)Cbc = 6pf + 36pf + (1 - c - 90) 4 pf = 460 pf RT2 = RC || RL = 4 K || 2.2 K = 1.419 KW C0 = Cwo + Cce + CMO | |
| 258. |
Complex pole in z-plane is as shown below. This is the pole diagram for __________ . |
| Answer» Decaying sinusoidal sequence. | |
| 259. |
Four messages band limited to W, W, 2W and 3W respectively are to be multiplex using Time Division Multiplexing (TDM). The minimum bandwidth required for transmission of this TDM signal is |
| Answer» In TDM minimum bandwidth is twice the maximum frequency present. Thus, 6W is the answer. | |
| 260. |
When transient and steady state response require improvement; __________ is used. |
| Answer» When improvement in transient state response is required Lead compensator is used. When improvement in steady state response is required lag compensator is used. Use of lag-lead compensator increases the low frequency gain which improves the steady state while at the same time it increases bandwidth of the system making the system response very fast. | |
| 261. |
For the collector-feedback amplifier shown below the voltage gain A will be __________ . |
| Answer» = - 103.07 RF -> feedback resistor. | |
| 262. |
Narrow band FM signal can be represented as |
| Answer» General expression for narrowband FM is A cos (2pfct) - βA sin (2pfct) sin (2pfmt). | |
| 263. |
2's complement representation of a 16 bit number (one sign bit and 15 magnitude bits) is FFFF. Its magnitude in decimal representation is |
| Answer» First bit is sign bit, hence its magnitude its → 2s complement ⇒ Sign bit is part of magnitude, so it may be 1. | |
| 264. |
The circuit shown in the figure converts |
| Answer» Let input is 1 1 0 1 y1 = x1 y2 = x2 ⊕ x1 ⇒ 1 ⊕ 1 = 0 y3 = x3 ⊕ x2 = 0 ⊕ 0 = 0 y4 = x4 ⊕ x3 = 1 ⊕ 0 = 1 if input is gray then y1 y2 y3 y4 will be Binary. | |
| 265. |
Using 4-bit numbers ( = 4) if = (0011) how is expressed in 2's complement. |
| Answer» In 2's complement Simply find 2s complement ⇒ (0011)2 ⇒ 1100 + 1 ⇒ (1101)2 . | |
| 266. |
System is : |
| Answer» ROC is towards infinity hence the signal is causal. Now the ROC does not contain the unit circle hence the system is unstable. | |
| 267. |
In analog signal given below is sampled at such a rate that the samples of signal can be given to a collection of buffers having capacity of either 24 samples/sec. or 25 samples/sec. The sampling rate is minimum such that it fully utilizes the collection of both types of buffers = 2 sin 480 + 3 sin 720 The folding (maximum) frequency is |
| Answer» LCM of 24 and 25 is 600 which will give maximum utilization of the buffers ∴ Sampling frequency is 600 ∴ Folding frequency = 600/2 = 300. | |
| 268. |
The system with given T.F. to be stable will be |
| Answer» ∴ Pole zero plot diagram is ROC should contain unit circle. Hence system is noncausal. | |
| 269. |
For a Gaussian process, auto-correlation also implies |
| Answer» Statistical independence: Suppose that Aj and Bk are the possible outcomes of two successive experiments or the joint outcome of a single experiment. And suppose that it turns out that the probability of the occurrence of outcome Bk simply does not depend at all on which outcome Aj accompanies it. Then we say that the outcomes Aj and Bk are independent. Auto correlation also implies comparison of two signals of same source. Hence for random Gaussian process. Auto correlation also implies statistical independence. | |
| 270. |
Laplace transform of { cos } is: |
| Answer» x(t) x(s) . | |
| 271. |
The fourier transform of a double sided exponential function is |
| Answer» Consider general expression X(jω) = ea|t| e- jωt dt = eat e-jωt dt + e-ate-jωt dt X(jω) = e(a - jω)t dt + e -(a + jω)t dt Since ∴ . | |
| 272. |
If A = then |A| will be |
| Answer» An = ? Every n x n matrix satisfy its characteristic equation |A - λI| = 0 λ -> eigen vector A - λI = |A - λI| = = 0 ∴ 1 =, ∴ f(A) = An = β0I + β1A Replace A by 1, I by 1 f(λ) = λn = β0 + β1λ Differentiate w.r.t. λ nλn - 1 = β1 β1 = ∴ β0 = λn - β x λ β1 = β0 = - = [1 - n] ∴ An = [1 - n]+ . 2n ∴ An = = ∴ A50 = |A50| = (1 - 502). | |
| 273. |
The probability density function of a random variable is as shown. The mean of the distribution is: |
| Answer» Mean of the distribution = A = 1/5 then find mean. | |
| 274. |
Given = GEF + KHIJ + LMON + TUWXYZ and Then is equivalent to |
| Answer» (UVW ⊕ WVU) = 1 = KJ(XY ⊕ XY) ∴ KJ = 1 NAND B ∴ HI = 1 Since f = GEF + 1 + LMON + TUVWXYZ = 1. | |
| 275. |
The value of the integral of the function g() = 4 + 10 along the straight line segment from the point (0, 0) to the point (1, 2) in the x-y plane is |
| Answer» (4x3 + 10y4) y = 2x ⇒[4x3 + 10(2x)4]dx = 33 | |
| 276. |
A communication channel with AWGN operating at a signal to noise ratio SNR >> 1 and bandwidth B has capacity C. If the SNR is doubled keeping B constant, the resulting capacity C is given by |
| Answer» C = B log2[1 + SNR] C = B log2[SNR] C' = B log2[2SNR] = B log2 SNR + B log22 C' = C1 + B. | |
| 277. |
The eigen values of a skew symmetric matrix are |
| Answer» For a real skew symmetric matrix the non-zero eigen values are all pure imaginary and thus occurs in complex conjugate pair. | |
| 278. |
A uniform plane magnetic wave incident normally on a plane surface of a dielectric material is reflected and the percentage of reflected power is 75% What is VSWR? |
| Answer» | |
| 279. |
Silicon is doped with boron to a concentration of 4 x 10 atoms/cm. Assuming the intrinsic carrier concentration of silicon to be 1.5 x 10/cm and the value of to be 25mV at 300K. Compared to undoped silicon, the Fermi level of doped silicon |
| Answer» Use EF - Ev = Since it is doped with acceptor impurity, Fermi level will shift down. | |
| 280. |
The Fourier series of a real periodic function has only Which of the above statements are correct? |
| Answer» , then it has 0 and HZ frequency component. | |
| 281. |
The circuit shown in the figure has 4 boxes each described by inputs P, Q, R and outputs Y, Z withY = P ⊕ Q ⊕ R, Z = RQ + R + Q The circuit is a |
| Answer» Let P = 1101 Q = 1101 Yn = Pn ⊕ Qn ⊕ Rn Z = Rn Qn + Pn Rn + Qn Pn Constructing truth table So that, Rn + 1 = Zn 1 ≥ n ≥ 3 Z4 = R5(MSB) Hence, output is 00010 which show that it is a 4 bit subtractor giving P - Q. | |
| 282. |
A switch tail ring counter is made by using a single D FF. The resulting circuit is a |
| Answer» In both case input will always opposite to each other hence it is D FF. | |
| 283. |
A transmitter radiates a power 20 KW and its base current is 18A. The radiation resistance of the antenna is |
| Answer» . | |
| 284. |
The Thevenin resistance of the circuit is |
| Answer» Diode is the non-linear element Find RTH across diode short voltage supplies. RAB= RTH = 3K + (2K || 1K) . | |
| 285. |
The electric field of uniform plane wave is given by = 20 sin (2 x 10- ) + 20 cos (2 x 10 - ) . The polarization of wave is |
| Answer» The given two components are equal in amplitude, perpendicular and out of phase by 90°. ∴ The wave is circularly polarized. Now we consider z = 0 and put t = 0 and ∴ At t = 0, E = 10ay At t ∴ Left circularly polarized. | |
| 286. |
The Z-transform of a particular signal is given as X(Z) = where This system after practical implementation will be |
| Answer» Let x = zx + x2 = 1 Substitute in the equation given, we get Hence we have pole at Z = 0.8 for stability the ROC should be away from Z = 0.8 towards infinity. | |
| 287. |
The JFET in the circuit shown in figure has an I = 10 A and V = -5 V. The value of the resistance R for a drain current I = 6.4 A is (select the nearest value) |
| Answer» 6.4 x 10-3 = 10 x 10-3 = 0.8 = 0.8 - 1 = - 0.2 VGS = -5 x 0.2 = -1 V IDSRS = -VGS = - (-1) = 1 = 0.156 x 103 = 156 Ω. | |
| 288. |
In the circuit closed for a long time and steady state is reached. is opened at = 0 Determine current through resistor. |
| Answer» Steady state with s closed. C is open Is = 4. This source current is divided between two resistors Ω and 1 Ω. Across 1 Ω, C is open, I in 1 Ω is = 1 A. Hence voltage across C is 1 V. The initial charge on C is V0 = 1 V. At t = 0+, s is opened bringing R = 1 Ω into the circuit. C is replaced by 1 V source with positive polarity left. The circuit has two sources Is = 4 and V0 = 1. The currents in the resistor R = and R = 1 can be determined by superposition. For R = Due to Is only (short V0) I = Is x= 3 Due to V0 only (open Is) I == Downwards Adding, current through R = is 3= . | |
| 289. |
The depth of penetration at 4 MHz is 10 cm, then at 2.5 MHz it is |
| Answer» Depth of penetration δ2 = 12.64 cm. | |
| 290. |
In a J-K FF we have J = and K = 1. Assuming the FF was initially cleared and then clocked for 6 pulses, the sequence at the Q output will be |
| Answer» | |
| 291. |
The V is given by |
| Answer» Using Nodal analysis ...(i) ...(ii) ∴ V2 = 2Vx ∴ V2 = 2V ...(iii) ...(iv) Put V2 = 2V Put in equation (iii), Vs = 3V1 - V2 Vs = 3 x (+5V) - 2V Vs = 15V - 2V = 13V. | |
| 292. |
The current in the circuit shown in |
| Answer» | |
| 293. |
The magnitude plot of a composite signal = + is __________ . |
| Answer» x(t) = ej3.5t(e-j05t + ej0.5t) x(t) = 2ej3.5t cos(0.5)t The magnitude of x(t) is |x(t)| = 2|cos(0.5)t|. | |
| 294. |
The expression given is F(A, B , C, D, E, F, G, H) = (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 Quine - McCluskey method of reduction : We can observe that minterms 158 and 222 can be grouped together to reduce the variable 'B'. | |
| 295. |
Calculate () for ≥ 0 assuming the switch has been in position A for a long time. At = 0, the switch is moved to position B. |
| Answer» | |
| 296. |
A source produces 4 symbols with probabilities , , , For this source, a practical coding scheme has an average codeword length of 2 bits/symbol. The efficiency of the code is |
| Answer» Total information for L message is Efficiency =. | |
| 297. |
A generator of 50 Ω internal impedance and operating at 1 GHz feeds a 75 Ω load via a coaxial line of characteristic impedance 50 Ω. The VSWR on the feed line is |
| Answer» ZL = 75 Ω Zo = 50 Ω ∴ Reflection coefficient: ∴ . | |
| 298. |
Power gain of antenna equals directive gain in VHF and UHF range if and only if efficiency of the antenna is __________ . |
| Answer» Directive gain and power gain are identical except that power gain takes into account the antenna losses. It may be written as Gp = ηGd η => efficiency If η = 1 , GP = Gd . | |
| 299. |
For the given phase lead network, the maximum possible phase lead is |
| Answer» The maximum phase lead is given by, sin φm = Where Hence sin φm = Hence φm = 30°. | |
| 300. |
A second order system has the transfer function . With as the unit step function, the response () of the system is represented by |
| Answer» Take Inverse L.T. and then observe Plot. | |