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.
| 151. |
The impulse response and the excitation function of a linear time invariant causal system are shown in figure () and () respectively. The output of the system at = 2 sec is equal to |
| Answer» The area of the shaded region . | |
| 152. |
The specific gravity of tungsten is 13.8 and its atomic weight is 184.0 Assume that there are two free elements per atom. Then the Fermi level or characteristic energy for the crystal in V will be |
| Answer» A quantity of any substance equal to its molecular weight in grams is a mole of that substance contains the same number of molecules as one mole of any other substance. This Avogadro's number and equals 6.02 x 1023 molecules per mole. Thus = 9.03 x 1028 electrons/m3 Since for tungsten the atomic and the molecular weights are the same. Therefore for tungsten EF = 3.64 x 10-19(η)2/3 = 3.64 x 10- 19(9.03 x 1028)2/3 EF = 7.326 eV. | |
| 153. |
An air-filled rectangular waveguide has dimensions 8 x 10. The phase velocity of guided wave at frequency of 3 GH for TE mode is |
| Answer» ∴ VP = 3.46 x 108 m/s. | |
| 154. |
A lossless transmission line with air dielectric is 6 m long. What is the line length at 30 MHz? |
| Answer» V = 3 x 108 m/s for lossless line ∴ ∴ Line length = = 0.6 wave length long. | |
| 155. |
Given that () is the one-sided Laplace transform of (), the Laplace transform of ()d is |
| Answer» Since it is one-sided Laplace transform. | |
| 156. |
The antenna current of an AM broad cast transmitter, modulated to a depth of 40% by an audio sine wave is 11 amperes. It increased to 12 ampere as a result of simultaneous modulation by another sine wave. The modulation index due to this second sine wave is : |
| Answer» μ2 = 0.57057, μ12 + 0.16 = 0.57057 ⇒ μ1 = 0.64 . | |
| 157. |
If the system T.F. is The differential equation representing the system is |
| Answer» (s3 + 2s2 + 5s + 6)y(s) = (s2 + 4) x (s) s3y(s) + 2s2y(s) + 5sy(s) + 6y(s) = s2 x (s) + 4 x (s) Replacing s by and y(s) by y(t) and x(s) by x(t) we get y(t) + 2y(t) + 5y(t) + 6y(t) = x(t) + 4x(t) + 2 + 5 + 6 = + 4x This is required differential equation. | |
| 158. |
In the circuit below, delay of the EXOR and ANDLogic gates are 20 μ and 10 μ respectively, then the wave and (with initial condition = 1) will be : |
| Answer» Since y is at logic '1' initially, y' must be at logic '1'. Also, one input of the EXOR gate is tied to zero therefore 'f' will be directly transferred to the output. The logic gates will introduce gate delays which will lead to the answer . | |
| 159. |
The antenna current of an AM transmitter is 8 A, when only the carrier is sent, but it increases to 8.93 A when the carrier is modulated by a single sine wave. Determine the antenna current when the percent of modulation changes to 0.8 |
| Answer» Ic = 8 A m = 0.8 = 8 x 1.149 = 9.19 A. | |
| 160. |
A noiseless 3 KH channel can transmit binary at the rate __________ . |
| Answer» Maximum data rate for noiseless channel is given by Nyquist Theorem = 2H log2 V bits/sec V - discrete levels (Here Binary i.e. 2) H - Bandwidth ∴ Maximum data rate = 2H log2 V = 2(3 KHz) log2 2 = 2 x 3 x 103 x 1 = 6 x 103 = 6000 bps. | |
| 161. |
Find V V V |
| Answer» Apply Mesh analysis Apply KVL to mesh 2 -2 + Vy + 1 = 0 ∴ Vy = 1 Apply KVL to mesh 1 ∴ -8 + Vx + 2 = 0 ∴ Vx = 6 V Apply KVL to mesh 3 -1 + Vz -2 = 0 ∴ Vz = 3 V. | |
| 162. |
The electric field of uniform plane wave is given by = 20 sin (2 x 10- ) + 20 cos (2 x 10 - ) . Find phase velocity |
| Answer» = 2 x 108 m/s. | |
| 163. |
The z-transform of a signal is given by . Its final value is : |
| Answer» Final value is . | |
| 164. |
The residue of the function at = 2 is |
| Answer» Residue at z = 2 . | |
| 165. |
The number of variables in a function is odd and the number of minterms with even number of 0's are same as the number of minterms with odd number at 1's, the expression formed can be represented in __________ as well as __________ form. |
| Answer» With odd number of variables (n) Let n = 3, by using k map A (B ⊕ C) + A(BC + B C) AD + AD = A ⊕ D D = B⊕C D = B⊕C This can be represented by EX-NOR of EX-OR expressions. EX-NOR means equivalence. | |
| 166. |
In a number system with radix , the decimal value of 110 is equal to 4. The decimal values of 111 and respective are equal to |
| Answer» (110)r = 4r ⇒ r2 + r = 4r r(r - 3) = 0 ⇒ r ≠ 0, r = 3 (lll)r = 32 + 3 + 1 = 13. | |
| 167. |
Consider the signal flow graph shown in figure below. The gain is : |
| Answer» P1 = abcd L1 = be, L2 = cf, L3 = dg Δ1 = 1 . | |
| 168. |
During transmission over a communicate channel, bit errors occur independently with probability . If a block of 3 bits are transmitted, the probability of at most one bit error is equal to |
| Answer» Probability of no-error + probability one error . | |
| 169. |
If [] () with ROC = R and [] () with ROC = R then ROC for [] + [] will be __________ (where R → Region of convergence and [] and [] are causal). |
| Answer» R1 ∩ R2, Suppose R1 for x1(n) be Then ROC for x1(n) + x2(n) R1 ∩ R2 (R1 < R2) Then ROC for x1(n) + x2(n) R1 ∩ R2(R1 < R2) | |
| 170. |
Consider a unity feedback control system with open-loop transfer function The steady state error of the system due to a unity step input is : |
| Answer» . | |
| 171. |
The analog signal given below is sampled at 1200 samples per second. = 2 sin 120 + 3 sin 240 + 4 sin 360 + 5 sin 480 + 6 sin 600 + 7 sin 720 . The folding (maximum) frequency is __________ . |
| Answer» Folding frequency (max.) = = 600 Hz. | |
| 172. |
The V at terminals A and B is equal to, |
| Answer» Apply KVL to loop 1 5V - 3I1 - 2(I1 - I2) - 5V = 0 ∴ I2 = 0 5 - 3I1 - 2I1 - 5V = 0 ∴ I1 = 0 ∴ VAB = 5V. | |
| 173. |
If , then has |
| Answer» . | |
| 174. |
If . , then [()] is |
| Answer» ⇒ F{Y(t)} = Y(f) = X(f) x sin (t) ⇌ By duality :⇌ sgn + (- f) ⇒ ⇌ jsgn (- f) ∴⇌ - j sgn (f) ∵ sgn(- f) = - sgn(f) ⇒ Y(f) = X(f). [- j sgn(f)]. | |
| 175. |
The two numbers represent in signed 2's complement form are P = 11101101 and Q = 111001110. If Q subtracted from P, the value obtained in signed 2's complement form is |
| Answer» P = 11101101 = - 25 Q = 11100110 = - 18 Thus P - Q = - 25 - (- 18) = - 7 Thus '- 7's two complement representation is 11111001. | |
| 176. |
The transfer function of an RLC circuit is given by quality factor (Q-factor) of this circuit is : |
| Answer» . | |
| 177. |
Consider the circuit shown below of 2 : 1 MUX is given by the function = + Then is : |
| Answer» Since g = ac + bc i.e. when c = 0, g = b and c = 1, g = a Now we evaluate f = (E)(X) + (0)(X) = EX f = (X)(Y) + (EX)(Y) = XY + EXY. | |
| 178. |
For the gear train, If θ = 2 . clockwise and the torque of gear 1 is 10 lb-ft, what is the displacement and torque of gear 7. |
| Answer» Number of teeths on any gear is proportional to its radius, i.e. and also displacement on all gear must be same => r1θ1 = r7θ7 => => ∴ ∴ rad CW Now, work done by one gear is equal to other, ∴ T1θ1 = T7θ7 => . | |
| 179. |
The minimal function that can detect a "divisible by 3" 8421 BCD code digit (representation DDDD) is given by |
| Answer» Now the problem can be solved using k-map method. The function can be expressed as Σm(0, 3, 6, 9) + d(10, 11, 12, 13, 14, 15) ∴ K-map is ∴ The function is D8D4D2D1 + D4D2D1 + D4D2D1 + D8D1 . | |
| 180. |
Given digits 2, 2, 3, 3, 3, 4, 4, 4, 4 how many distinct 4 digit numbers greater than 3000 can be formed? |
| Answer» The given digits are 2, 2, 3, 3, 3, 4, 4, 4, 4 we have to find the numbers that are greater than 300 ∴ The first digit can be 3 or 4 but not 2. Now, let us fix the first, second and third digits as 3, 2, 2, then the fourth place can be filled in 3 ways. ∴ The number of ways is 3 similarly, we fix first third and fourth place as 3, 2 and 2 respectively (4) so the second place can be filled in 3 ways again, The number of ways is 3 Now, we fix first, second and fourth, previous cases and we obtain the same result. ∴ The number of ways is 3 so, the total number of ways is 9 similarly this can done by fixing the numbers as 3 and 4 (instead of 2) and thereby we obtain the a ways each The number of numbers starting with 3 is 27 Similarly by taking 4 as the first digit we get 27 numbers ∴ The number of numbers that are greater than 3000 is 27 + 27 = 54 But, 3222, 4222, is not possible as there are only two 2's, 3333 is not possible as there are only three 3's ∴ The total number of numbers that are greater than 3000 is 54 - 3 = 51. | |
| 181. |
The number of variables in a function is odd and the number of minterms with even number of 0's are same as the number of minterms with odd number at 1 's, the expression formed can be represented in __________ as well as __________ form. |
| Answer» With odd number of variables (n) terms with even number of 0's and terms with odd number of 1 's are the same. Thus the k-map for the above condition with minterms represented by terms with even number of 0's is for n = 3 ∴ The expression can be represented by equivalence (EX NOR) or (EX OR) expressions. | |
| 182. |
Find the power absorbed or delivered by each source: |
| Answer» Source a : Power : 8 x (4) = - 32 W. Source A has delivered the power as it is. - ve. Source b: Source B has absorbed power of + 12 W as power is positive Source c: Source C has absorbed power of 5 x 16 = 80 W Source d: Delivers power of 3 x ix x (5) = 3 x 4 x (- 5) = -60 W Total power of the network is - 32 + 12 + 80 - 60 = 0 Watts. | |
| 183. |
For the network shown below, if V = 12 V and V = - 2 V value of R will be |
| Answer» VGS = VG - IDRS - 2 = 5.44 - 2.22 X 10-3 X RS RS = 3.35 KΩ. | |
| 184. |
In the circuit shown, all the transmission line sections are lossless. The Voltage Standing Wave Ration (VSWR) on the 60W line is |
| Answer» | |
| 185. |
If for certain (), the Fourier transform is: Then for (2) the Fourier transform will be: |
| Answer» Using scaling property . | |
| 186. |
A parallel plate capacitor has surface density of C/m. The dielectric inside capacitor is air. The electric field intensities outside and inside the capacitor are |
| Answer» In case of capacitor, the charges on two plates of capacitor are of opposite sign. We know that electric field due to infinite sheet charge with density rs is Outside the capacitor, electric field intensities due to upper plate and lower plate cancel each other. ∴ Eoutside = 0 Inside the capacitor, the electric field intensities due to upper and lower plate add together ∴ | |
| 187. |
One of a loss-less transmission line having the characteristic impedance of 75 Ω and length of 1cm is short-circuit. At 3 GHz, the input impedance at the other end of the transmission line is |
| Answer» X = 10 cm, ZL = 0, βl = Thus Zin is Z0which is inductive. | |
| 188. |
Total electric field intensity is given as (V/m). The maximum value of will occur; if and only if |
| Answer» To find value of y at which E will be maximum, differentiable ET with respect to y and set it equal to zero. | |
| 189. |
In the signal flow graph of figure, equals. |
| Answer» P1 = 5 x 2 x 1 = 10 L1 = - 4 . | |
| 190. |
Find resistance R to bring transistor to threshold of saturation (V = 0), V = 0.7 V, = 0.96 |
| Answer» = 0.07 A = 24 = 3 mA = 1433.33 Ω ≅ 1.5 KΩ. | |
| 191. |
The impulse response of an initially relaxed linear system is () . To produce a response U(), the input must be equal to __________ . |
| Answer» Given, h(t) = e-2tu(t) H(s) = y(t) = te-2t, u(t) y(s) = Using Laplace Transform ∴ x(t) = e-2tu(t) Input must be equal to e-2tu(t). | |
| 192. |
The continuous time system with impulse response () () is stable, if __________ ( is even). |
| Answer» h(t) = consider ax = x ax = x2 a = x For h(t) to be bounded, exponential term should have a negative power. i.e. e-t will have response as shown. Hence should be < 0 ∴ a < 0. | |
| 193. |
A sphere with charge Q was placed in, but not touching a larger hollow sphere. If there is any charge initially on the outer sphere; to remove it the outer sphere was earthed momentarily, and then the inner sphere was removed. The charge remaining on the outer sphere was then measured. This charge was found to be __________ to the charge on the inner sphere. |
| Answer» This is true for all sizes of the spheres and for all types of dielectric media between the spheres. | |
| 194. |
At = ∞; V and V will be |
| Answer» When steady state is reached again; then C is open Now VR2 = 2V . | |
| 195. |
For the asymptotic Bode magnitude plot shown below, the system transfer function can be |
| Answer» . | |
| 196. |
A square in which each (, ) element of matrix is equal to the negative of the conjugate complex of (, ) element, is called |
| Answer» When aij = - . This matrix is called skew hermitian matrix. | |
| 197. |
Mid-band gain = |
| Answer» . | |
| 198. |
The characteristic equation of a feedback control system is given by + 5 + (K + 6) + K = 0where K > 0 is a scalar variable parameter. In the root-loci diagram of the system the asymptotes of the root-locus for large values of meet at a point in the -plane, whose coordinates are : |
| Answer» Apply R-H, and find ω by auxiliary equation. | |
| 199. |
The input 3-dB frequency is : |
| Answer» Input 3 dB frequency C = 0.40 μF. | |
| 200. |
If 137 + 276 = 435 how much is 731 + 672? |
| Answer» Given, 137 + 276 = 435 Adding units digits i.e., 7 + 6 = 13, but given as 5, which is 13 - 8 and also 1 is carry forwarded to the tens digit. i.e., + 1 Here, 7 + 3 + 1 = 1 i.e., 11 - 8 = 3 and 1 is carry forwarded to hundred digits Now, the sum of digits in hundred's place is 1 + 1 + 2 = 4 ie., Using the same logic, we have Using digits sum 1 + 2 = 3 Tens digits sum = 3 + 7 = 10 i.e., 10 - 2 and + 1 carry forward Hundreds digits sum = 1 + 7 + 6 = 14 i.e., 14 - 8 = 6 and one carry forward. | |