#1
3rd August 2014, 12:11 PM
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Maharashtra State Electricity Distribution Company Limited Assistant Engineer papers
Will you please sharewith me the Maharashtra State Electricity Distribution Company Limited Assistant Engineer papers?
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#2
3rd August 2014, 01:54 PM
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Re: Maharashtra State Electricity Distribution Company Limited Assistant Engineer pap
As you want to get the Maharashtra State Electricity Distribution Company Limited Assistant Engineer papers so here it is for you: 1. Find the Fourier sine transform of f(x), where f(x) = f(x) 1,0<x<a { 0, x>a (a) √2/p (cos st / s) (b) √2/p (1 - cos as / s) (Ans) (c) √2 (1 - cos as) (d) None of these 2. A random variable X with uniform density in the interval 0 to 1 is Quantized as follows: If 0 ≤ X ≤ 0.3, xq = 0 If 0.3 ≤ X ≤ 1, xq = 0.7 Where Xq is the quatized value of x The root mean square value of the quantization noise is (a) 0.573 (b) 0.198 (ans) (c) 2.205 (d) 0.266 Solution : Since it is uniform as xq = 0 in the range 0≤x≤0.3 xq = 0.7 in the range 0.3≤x≤1 The square mean value is ¥ s2 = ∫ (x - xq)2 f (x) dx -¥ 1 = ∫ (x - xq)2 f (x) dx 0 0.3 0.1 = ∫ (x - 0)2 f (x) dx + ∫ (x - 0.7)2 f (x) dx 0 0.3 0.3 1 = [x3/3] + [x3/3 + 0.49 x - 1.4] 0 0.3 or s2 = 0.039 The root mean square quantization noise RMS = √s2 = √0.039 = 0.198 3. Choose the correct one from among the alternatives A, B, C, D after matching an item from Group 1 with the most appropriate item in Group 2. Group 1 Group 2 1 : FM P : Slope overload 2 : DM Q : m Law 3 : PSK R : Envelope detector 4 : PCM S : Capture effect T : Hilbert transform U : Matched filter (a) 1 - T, 2 - P, 3 - U, 4 - S (b) 1 - S, 2U, 3 - P, 4 - T (c) 1 - S, 2 - P, 3U, 4 - Q (ans) (d) 1 - U, 2 - R, 3 - S, 4 - Q Solution : FM --- Capture effect --- Receives only strong signal DM ---- Slop over load Noise PSK --- Matched filter PCM - m law - Non linear quantization by using Companding with a law V = log (1 + m |M|) log (1 +m ) 4. There analog signals, having bandwidth 1200 Hz, 600 Hz and 600 Hz, are sampled at their respective Nyquist rates, encoded with 12 bit words, and time division multiplexed. The bit rate for the multiplexed signal is (a) 115.2 kbps (b) 28.8 kbps (c) 57.6 kbps (ans) (d) 38.4 kbps Solution : The three analog Signals having BW 1200 Hz, 600Hz and 600 Hz are sampled at their respective Nyquist rate i.e. at 2400, 1200, 1200 sample/sec respectively. The total of (2400 + 1200 + 1200) = 4800 sample/sec The Bit rate = n. fs = (4800 sample/sec) x 12 = 57.6 Kbps Where n = number of bit in a symbol 5. Find the correct match between group 1 and group 2. Ground I P - [1 + km (t)] A sin (wct) Q - km (t) A sin (wct) R - A sin [ w'c + k]'-¥ m (t) dt S - A sin [wct + k '∫-¥ m (t) dt] Solution : Group II W - Phase modulation X - Frequency modulation Y - Amplitude modulation Z - DSB-SC modulation P Q R S (a) Z Y X W (b) W X Y Z (c) X W Z Y (d) Y Z W X (ans) Solution : The correct match is given below [1 + km (t)] A sin (wct) Amplitude modulation km (t) A sin (wct) DB-SC modulation A sin [w'c + k]'-¥ m (t) dt Phase modulation A sin [wct + k '∫-¥ m (t) dt] Frequency modulation 6. Which of the following analog modulation scheme requires the minimum transmitted power and minimum channel bandwidth? (a) VSB (b) DSB-SC (c) SSB (ans) (d) AM Solution : VSB → fm + fc DBS - SC → 2 fm SSB → fm AM → 2 fm Thus SSB has minimum bandwidth and it required minimum power i.e. 17% as compared to AM. 7. A device with input x(t) and output y(t) is characteristic by : y (t) = x2(t). An FM signal with frequency deviation of 90 KHz and modulation signal bandwidth of 5 KHz is applied to this device. The bandwidth of the output signal is (a) 370 KHz (ans) (b) 190 KHz (c) 380 KHz (d) 95 KHz Solution : In present case ∆f = 90; fm = 5 β = [∆f / fm] = [90/5] = 18 FM equation A cos [wct + β = sin wmt] = A cos [wct + 18 sin wmt] y(t) = x2 (t) = A2 cos2 [wct + 18 Sin wmt] Note : Cos2 q = [1 + Cos2q] / 2 If there is change in frequency the modulation index also changes in same ratio y(t) = A2 [(1/2) + (1/2) Cos {2wct + 36Sin wmt}] y(t) = [(A2/2) + (A2/2) Cos {2wct + 36Sin wmt}] After the device, β(new) = 36 = [∆f(new) / fm] ∆f(new) = 36 x 5 = 180 By carson's rule Bandwidth = 2(∆f + fm) = 2 (180 + 5) Bandwidth = 370 kHz 9. A carrier is phase modulated (PM) with frequency deviation of 10 KHz by a single tone frequency of 1 KHz. If the single tone frequency is increased to 2 KHz, assuming that phase deviation remains unchanged, the bandwidth of the PM signal is (a) 21 kHz (b) 22 kHz (c) 42 kHz (d) 44 kHz (ans) Solution : ∆f = 10 KHz fm(new) = 2 KHz fm = 1 KHz By carson's Rule BW = 2 (∆f + fm) = 2 (10 + 1) = 22 KHz ∆ f(new) = 2 x 10 = 20 BW(new) = 2 (20 + 2) = 44 kHz 10 If A and B be the set and Ac and Bc denote the complements of the sets. A and B, then set (A - B) È (B - A) È (A Ç B) is equal to (a) A È B (Ans) (b) Ac È Bc (c) A Ç B (d) Ac Ç Bc 11 Let G = G(V, E) has five vertices, then the maximum number of m of edges in E, if G is a multigraph ? (a) 5 (b) 2 (c) 10 (d) Finite or infinite (Ans) 12 How many straight line can be drawn through 10 points on a circle ? (a) 10 (b) 20 (c) 45 (Ans) (d) Infinite 13 . The Fourier transform of unit step function u(t) is (a) 1 (b) pd(w) (c) pd(w) - 1/jw (Ans) (d) pd(w) + 1/jw 14. The value of the integral ∫ e-2(x - t) d(t - 2) dt is -∞ (a) e-2(x - 2) (Ans) (b) e2(x - 2) (c) e-2(x + 2) (d) e2(x + 2) 15. The uint of Ñ x H is (a) A (b) A/m (c) A/m2 (Ans) (d) A-m |
#3
18th May 2015, 03:40 PM
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Re: Maharashtra State Electricity Distribution Company Limited Assistant Engineer pap
I want to apply for the Assistant Engineer exam. Will you please forward me question paper of the Assistant Engineer (Civil Engineering) exam of Maharashtra State Electricity Distribution Company Limited?
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#4
18th May 2015, 03:40 PM
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Re: Maharashtra State Electricity Distribution Company Limited Assistant Engineer pap
As per your request, I am forwarding you question paper for the Assistant Engineer Civil Engineering exam conducted by the Maharashtra State Electricity Distribution Company Limited (MAHAGENCO). 1. Which of the following is/are the assumption(s) of Bernaulli’s equation? 1] There is loss of liquid while flowing. 2] There is no external force except the gravity acts on the liquid. 3] The velocity of energy of liquid particle, across any cross-section of pipe is uniform. 4] Both [2] and [3] 2. In constructions, why are the lintels preferred to arches? 1] Arches will not long last 2] Arches require more head room to span the open as like doors, windows etc. 3] Arches require strong abutments to withstand arch thrust 4] Both [2] and [3] 3. What is called a ‘level line’? 1] The line parallel to the mean sphireodal surface of earth 2] The line is horizontal 3] The line passing through the centre of cross-hairs and the centre of the eye piece 4] The line passing through the objective lens and the eye piece of a dumpy or tilting level APTITUDE 1. 1. Find the amount obtained by investing Rs. 24, 000 at 18% per annum simple interest for five years? 1] Rs. 21, 600 2] Rs. 44, 000 3] Rs. 48, 000 4] Rs. 45, 600 2. A number when increased by 30% becomes 78. What is the number? 1] 60 2] 70 3] 40 4] 48 3. A two-digit number is such that twice the ten’s digit add to eleven times the units digit is equal to the number itself. What is the number? 1] 48 2] 86 3] 73 4] 54 Direction for question 4 and 5: Fill in the blank in the given sentence so as to make sense. Select the correct word from the answer choices and mark its number as the answer. 4. A welcoming party was _________ the day after the new teach arrived. 1] conducted 2] thrown 3] initiated 4] organised 5. The store _________ medicines as well as cosmetics. 1] stocks |