Ece Practice Book PDF

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  ECE 

GATE Practice Book 

󰁛󰀱󰁝 

GATE ELECTRONICS COMMUNICATION ENGINEERING (ECE)

PRACTICE BOOK  

󰁐󰁲󰁡󰁣󰁴󰁩󰁣󰁥 󰁳󰁵󰁢󰁪󰁥󰁣󰁴 󰁷󰁩󰁳󰁥 󰁱󰁵󰁥󰁳󰁴󰁩󰁯󰁮󰁳.

 

󰁍󰁯󰁲󰁥 󰁴󰁨󰁡󰁮 󰀱󰀰󰀰 󰁱󰁵󰁥󰁳󰁴󰁩󰁯󰁮󰁳 󰁱󰁵 󰁥󰁳󰁴󰁩󰁯󰁮󰁳 󰁰󰁥󰁲 󰁳󰁵󰁢󰁪󰁥󰁣󰁴 󰁷󰁩󰁴󰁨 󰁤󰁥󰁴󰁡󰁩󰁬󰁥󰁤 󰁳󰁯󰁬󰁵󰁴󰁩󰁯󰁮󰁳.

 

󰁏󰁮󰁬󰁩󰁮󰁥 󰁳󰁵󰁰󰁰󰁯󰁲󰁴, 󰁤󰁯󰁵󰁢󰁴 󰁳󰁥󰁳󰁳󰁩󰁯󰁮 󰁡󰁮󰁤 󰁤󰁩󰁳󰁣󰁵󰁳󰁳󰁩󰁯󰁮󰁳 󰁷󰁩󰁴󰁨 󰁦󰁡󰁣󰁵󰁬󰁴󰁹 󰁡󰁴 󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭󰀯󰁥󰁬󰁥󰁣 󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲 󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭󰀯󰁥󰁬󰁥󰁣󰁴󰁲󰁯󰁮󰁩󰁣󰁳. 󰁴󰁲󰁯󰁮󰁩󰁣󰁳.

Register at www.engineersinstitute.com & get 3 Full length GATE-Mock Test  

󰁐󰁲󰁥󰁰󰁡󰁲󰁥󰁤 󰁢󰁹󰀺

Er. KUNAL SRIVASTAVA  AIR-1 ESE 2012, AIR-44 GATE 2013

󰁐󰁵󰁢󰁬󰁩󰁳󰁨󰁥󰁤 󰁢󰁹󰀺 󰁅󰁎󰁇󰁉󰁎󰁅󰁅󰁒󰁓 󰁉󰁎󰁓󰁔󰁉󰁔󰁕󰁔󰁅 󰁏󰁆 󰁉󰁎󰁄󰁉󰁁󰀭󰁅.󰁉.󰁉. 󰂩 󰁁󰁌󰁌 󰁒󰁉󰁇󰁈󰁔 󰁒󰁅󰁓󰁅󰁒󰁖󰁅󰁄  󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭   󰀲󰀸󰁂󰀯󰀷 󰁊󰁩󰁡󰁳󰁡󰁲󰁡󰁩 󰁎󰁥󰁡󰁲 󰁉󰁉󰁔 󰁈󰁡󰁵󰁺󰁫󰁨󰁡󰁳 󰁎󰁥󰁷󰁤󰁥󰁬󰁨󰁩󰀭󰀱󰀱󰀰󰀰󰀱󰀶 󰁰󰁨󰀮 󰀰󰀱󰀱󰀭󰀲󰀶󰀵󰀱󰀴󰀸󰀸󰀸󰀮 󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭

 

  ECE 

GATE Practice Book 

󰁛󰀲󰁝 

󰂩 2013 By Engineers Institute of India ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmitted, stored or used in any form or by any means graphic, electronic, or mechanical & chemical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems.

Engineers Institute of India

28-B/7, Jia Sarai, Near IIT Hauz Khas New Delhi-110016 Tel: 011-26514888 For publication information, visit www.engineersinstitute.com/publication ISBN: 978-93-5137-760-3 Price: Rs. 299

 

󰁐󰁵󰁢󰁬󰁩󰁳󰁨󰁥󰁤 󰁢󰁹󰀺 󰁅󰁎󰁇󰁉󰁎󰁅󰁅󰁒󰁓 󰁉󰁎󰁓󰁔󰁉󰁔󰁕󰁔󰁅 󰁏󰁆 󰁉󰁎󰁄󰁉󰁁󰀭󰁅.󰁉.󰁉. 󰂩 󰁁󰁌󰁌 󰁒󰁉󰁇󰁈󰁔 󰁒󰁅󰁓󰁅󰁒󰁖󰁅󰁄  󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭   󰀲󰀸󰁂󰀯󰀷 󰁊󰁩󰁡󰁳󰁡󰁲󰁡󰁩 󰁎󰁥󰁡󰁲 󰁉󰁉󰁔 󰁈󰁡󰁵󰁺󰁫󰁨󰁡󰁳 󰁎󰁥󰁷󰁤󰁥󰁬󰁨󰁩󰀭󰀱󰀱󰀰󰀰󰀱󰀶 󰁰󰁨󰀮 󰀰󰀱󰀱󰀭󰀲󰀶󰀵󰀱󰀴󰀸󰀸󰀸󰀮 󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭

 

  ECE 

GATE Practice Book 

This book is dedicated to all Electronics C Communication ommunication Engineers preparing for GATE & Public sector e examination xaminations s. xaminations.

󰁐󰁵󰁢󰁬󰁩󰁳󰁨󰁥󰁤 󰁢󰁹󰀺 󰁅󰁎󰁇󰁉󰁎󰁅󰁅󰁒󰁓 󰁉󰁎󰁓󰁔󰁉󰁔󰁕󰁔󰁅 󰁏󰁆 󰁉󰁎󰁄󰁉󰁁󰀭󰁅.󰁉.󰁉. 󰂩 󰁁󰁌󰁌 󰁒󰁉󰁇󰁈󰁔 󰁒󰁅󰁓󰁅󰁒󰁖󰁅󰁄  󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭   󰀲󰀸󰁂󰀯󰀷 󰁊󰁩󰁡󰁳󰁡󰁲󰁡󰁩 󰁎󰁥󰁡󰁲 󰁉󰁉󰁔 󰁈󰁡󰁵󰁺󰁫󰁨󰁡󰁳 󰁎󰁥󰁷󰁤󰁥󰁬󰁨󰁩󰀭󰀱󰀱󰀰󰀰󰀱󰀶 󰁰󰁨󰀮 󰀰󰀱󰀱󰀭󰀲󰀶󰀵󰀱󰀴󰀸󰀸󰀸󰀮 󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭

󰁛󰀳󰁝 

 

  ECE 

GATE Practice Book 

󰁛󰀴󰁝 

ion is one of the most p prresti tig gious competitive examination GATE exa ination gradua uate te engi engine neer ers. s. Ov Over er th thee past past few few y ars, it has become conducted or grad more com etitive as a number of aspirants are incr asingly becoming M.Tec ech h & go gove vern rnme ment nt jo jobs bs du duee to to dec decll ne in other career inte intere rest sted ed in M.T options. In my opi ion, GATE exam test candidates’ basics understanding of bility ty to apply pply th thee same same in a nume numerical approach. A concepts a d abili with the the sy sylla lab bus not ju just mu mu ging up concepts. candidate is supposed to s artly deal wi Thorou Tho rough gh under understa standi nding ng wit critical analysis of topics and abilit y to xpress clearly are some of the pre-requisites to crack this exam. The questioning & examination pattern has changed over the past few y ars, ars, as as num numer eric ical al ans answe werr type type qu ques estio tions ns p ay a major role to score a good rank. Keeping in mind, the difficulties of an average student, we have composed this booklet. We are thankf l to Er. KUNAL SRIVASTAVA (AIR 1 SE 2012, AIR 44 GATE 2013 in ECE) for preparing this booklet.  In Electronics Communicati n Engineering, over 2,50,000 candidates ppeared in GATE 2013 with 36,394 finally qualifie lified. d. The The cut cut-o -off ff mar marks ks for for gen gener eral al cat categ eg ry is 25. For more details log onto our institute’s website http://www.engineersinstitute.com/   lectronics  Engi gine neer er Institute of India  Esta Establ blis ishe hed d in 20 2006 06 by a tea tea of IES and GATE toppers, we at En orous class lassees and and pro proper per g gu uid idaance nce tto o eng engin ineeering students over have consistently provided ri oro ccomplish mplishing ing their their dreams dreams.. We believ believee i providing examthe nation in successfully acco oriented teaching methodology with updated study material and  test series so that our compet etiti ition on.. The The facu faculty lty at EII are are a te m of experienced students stay ahead in the comp professionals who have gui ed thousands to aspirants over the years. They are readily ses to assi assist st stud studen ents ts and and we main mainta tain in a h hea ea thy student-faculty avail vailaable be befo fore re an and d after fter cla ses toppers assoc associate iate with with us for for contributin contributing towards our goal ratio. Many current and past ear toppers of provid providing ing quali quality ty educa educatio tion and share their success with the future as irants. Our results EII are currentl tly y working in various government speak for themselves. Past tudents of EII departments and PSU’s and ursuing higher specializations at IISc, IITs , NITs & reputed institutions.

 Best wishes for future career R.K. Rajesh Director Engineers Institute of India [email protected]  

󰁐󰁵󰁢󰁬󰁩󰁳󰁨󰁥󰁤 󰁢󰁹󰀺 󰁅󰁎󰁇󰁉󰁎󰁅󰁅󰁒󰁓 󰁉󰁎󰁓󰁔󰁉󰁔󰁕󰁔󰁅 󰁏󰁆 󰁉󰁎󰁄󰁉󰁁󰀭󰁅.󰁉.󰁉. 󰂩 󰁁󰁌󰁌 󰁒󰁉󰁇󰁈󰁔 󰁒󰁅󰁓󰁅󰁒󰁖󰁅󰁄  󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭   󰀰󰀱󰀱󰀭󰀲󰀶󰀵󰀱󰀴󰀸󰀸󰀸󰀮󰀮 󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥 󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭 󰀲󰀸󰁂󰀯󰀷 󰁊󰁩󰁡󰁳󰁡 󰁊󰁩󰁡󰁳󰁡󰁲󰁡󰁩 󰁲󰁡󰁩 󰁎󰁥󰁡󰁲 󰁉󰁉󰁔 󰁈 󰁈󰁡󰁵󰁺󰁫 󰁡󰁵󰁺󰁫󰁨󰁡󰁳 󰁎󰁥󰁷󰁤󰁥󰁬󰁨󰁩󰀭󰀱󰀱󰀰󰀰󰀱󰀶 󰁰󰁨󰀮 󰀰󰀱󰀱󰀭󰀲󰀶󰀵󰀱󰀴󰀸󰀸󰀸

 

  ECE 

GATE Practice Book 

󰁛󰀵󰁝 

CONTENTS 1. 

2. 

3. 

4. 

5. 

6. 

7. 

ANALOG ELECTRONICS.......................... ELECTRONICS................................................... ................................. ........ 01-70 Questions ………………………………………………………..

01-38

Solutions ……………………………………………………......

39-70

COMMUNICATION ................................................................... 71-106 Questions ………………………………………………………..

71-87

Solutions ……………………………………………………......

88-106

CONTROL SYSTEM SYSTEM ........................... ...................................................... ....................................... ............ 107-152 Questions ………………………………………………………..

107-132

Solutions ……………………………………………………......

133-152

DIGITAL ELECTRONICS ELECTRONICS ............................................ ........................................................... ............... 153-188 Questions ………………………………………………………..

153-173

Solutions ……………………………………………………......

174-188

ELECTRONICS DEVICES AND CIRCUITS (EDC) ..................... Questions ………………………………………………………..

189-206

Solutions ……………………………………………………......

207-220

ELECTROMAGNETIC ELECTROMAGNET IC THEORY-EMT....................... THEORY-EMT......................................... .................. 221-248 Questions ………………………………………………………..

221-236

Solutions ……………………………………………………......

237-248

NETWORK THEORY............................... THEORY........................................................ ................................... .......... 249-288 Questions ……………………………………………………….. Solutions ……………………………………………………......

8. 

189-220

249-273 274-288

SIGNAL & SYSTEM SYSTEM ............................... ........................................................ ................................... .......... 289-330 Questions ………………………………………………………..

289-311

Solutions ……………………………………………………......

312-330

󰁐󰁵󰁢󰁬󰁩󰁳󰁨󰁥󰁤 󰁢󰁹󰀺 󰁅󰁎󰁇󰁉󰁎󰁅󰁅󰁒󰁓 󰁉󰁎󰁓󰁔󰁉󰁔󰁕󰁔󰁅 󰁏󰁆 󰁉󰁎󰁄󰁉󰁁󰀭󰁅.󰁉.󰁉. 󰂩 󰁁󰁌󰁌 󰁒󰁉󰁇󰁈󰁔 󰁒󰁅󰁓󰁅󰁒󰁖󰁅󰁄  󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭   󰀲󰀸󰁂󰀯󰀷 󰁊󰁩󰁡󰁳󰁡󰁲󰁡󰁩 󰁎󰁥󰁡󰁲 󰁉󰁉󰁔 󰁈󰁡󰁵󰁺󰁫󰁨󰁡󰁳 󰁎󰁥󰁷󰁤󰁥󰁬󰁨󰁩󰀭󰀱󰀱󰀰󰀰󰀱󰀶 󰁰󰁨󰀮 󰀰󰀱󰀱󰀭󰀲󰀶󰀵󰀱󰀴󰀸󰀸󰀸󰀮 󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭

 

  ECE 

GATE Practice Book 

󰁛󰀶󰁝 

1. ANALOG ELECTRONICS (1.) 

16

3

15

3

Consider a Si p.n diode with doping concentrations,  N a = 10 / cm and N d  = 10 / cm . If

ni = 1.5 ×1010 / cm   3 and zero bias junction capacitance = 0.5pF. The junction capacitance at an applied reverse bias of 2V is: (a.) 0.5pF

(b.) 0.245pF 17

(c.) 0.168pF 3

16

(d.) 2.0698pF

3

(2.) 

If at the doping levels of N D = 10  /cm . And NA = 10  /cm  and at a reverse bias of 1V, a Si diode shows a junction capacitance of 0.5pF, find the zero bias junction. (a.) 0.76pF (b.) 1.16pF (c.) 0.86pF (d.) 1.31pF

(3.) 

Consider a silicon p – n diode with reverse saturation current, Io = 10 A. Find the power

-13

dissipated in the 2k Ω resistor.

(a.) 20mW

(b.) 19.2mW

(c.) 8mW

(d.) 9.6mW

2 Linked Questions:  Questions:  (4.) 

Find the power in the 2k Ω resistor if the diode has a cut in voltage = 0.7V and d.c. forward resistance = 10Ω 

(a.) 8mW (5.)  (6.) 

(b.) 16.7mW

(c.) 9.15mW

(d.) 19.1mW

Find the power consumed in the diode in the previous question (a.) 1.5mW (b.) 9.15mW (c.) 1.4mW (d.) 0.046mW If in the following circuit, diode has a power rating of 1mW, with a cut in voltage = 0.7V and zero forward resistance, find which of the following values of “R” will meet the requirement.

(a.) 3k Ω 

(b.) 7k Ω 

(c.) 3.97k Ω 

(d.) 6.5k Ω 

󰁐󰁵󰁢󰁬󰁩󰁳󰁨󰁥󰁤 󰁢󰁹󰀺 󰁅󰁎󰁇󰁉󰁎󰁅󰁅󰁒󰁓 󰁉󰁎󰁓󰁔󰁉󰁔󰁕󰁔󰁅 󰁏󰁆 󰁉󰁎󰁄󰁉󰁁󰀭󰁅.󰁉.󰁉. 󰂩 󰁁󰁌󰁌 󰁒󰁉󰁇󰁈󰁔 󰁒󰁅󰁓󰁅󰁒󰁖󰁅󰁄  󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭   󰀲󰀸󰁂󰀯󰀷 󰁊󰁩󰁡󰁳󰁡󰁲󰁡󰁩 󰁎󰁥󰁡󰁲 󰁉󰁉󰁔 󰁈󰁡󰁵󰁺󰁫󰁨󰁡󰁳 󰁎󰁥󰁷󰁤󰁥󰁬󰁨󰁩󰀭󰀱󰀱󰀰󰀰󰀱󰀶 󰁰󰁨󰀮 󰀰󰀱󰀱󰀭󰀲󰀶󰀵󰀱󰀴󰀸󰀸󰀸󰀮 󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭

 

  ECE 

GATE Practice Book 

SOLUTION (1.) 

ANS: (b) EXP:

C  jo

C  j =

1 + V  R V bi V bi = (2.) 

kT  q

ni

2

ANS: (a) EXP:

c j =

C  jo

1+ (3.) 

   N a .N d 

ln  

V  R

 

V bi

ANS: (d) EXP: Using diode equation and K.V.L V     0.026  − 1 + V  D ⇒ 5  = 2 × 10 × 10 e     hit and tria iall) ⇒ V D  = 0.62V  (by hit  D

3

−13

∴ I D =  2.2mA ∴ Power = ( 2.2mA) 2 × 2k Ω = 9.68mW    (4.) 

ANS: (c) EXP: Replace the diode by its piecewise linear model

∴ I D =

5V − 0.7V 

= 2.14mA 2.01k Ω ∴ Power = 9.15mW    

󰁐󰁵󰁢󰁬󰁩󰁳󰁨󰁥󰁤 󰁢󰁹󰀺 󰁅󰁎󰁇󰁉󰁎󰁅󰁅󰁒󰁓 󰁉󰁎󰁓󰁔󰁉󰁔󰁕󰁔󰁅 󰁏󰁆 󰁉󰁎󰁄󰁉󰁁󰀭󰁅.󰁉.󰁉. 󰂩 󰁁󰁌󰁌 󰁒󰁉󰁇󰁈󰁔 󰁒󰁅󰁓󰁅󰁒󰁖󰁅󰁄  󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭   󰀲󰀸󰁂󰀯󰀷 󰁊󰁩󰁡󰁳󰁡󰁲󰁡󰁩 󰁎󰁥󰁡󰁲 󰁉󰁉󰁔 󰁈󰁡󰁵󰁺󰁫󰁨󰁡󰁳 󰁎󰁥󰁷󰁤󰁥󰁬󰁨󰁩󰀭󰀱󰀱󰀰󰀰󰀱󰀶 󰁰󰁨󰀮 󰀰󰀱󰀱󰀭󰀲󰀶󰀵󰀱󰀴󰀸󰀸󰀸󰀮 󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭

󰁛󰀷󰁝 

 

  ECE 

(5.) 

GATE Practice Book 

ANS: (a) EXP: Power consumed in diode = sum of powers consumed in the Vγ  and rf .

= Vγ  .I D + ( I D )2 .rf     

= 1.54mW 

(6.) 

Mind that 1.54mW + 9.15mW = 10.69mW = power delivered by the 5V source ANS: (b) EXP: The power consumed in the diode must not exceed 1mW.

∴ I D ≤

∴ R ≥

1mW  0.7V 

= 1.43mA  

10V − 0.7V 

1.43mA thus, R ≥ 6.51k Ω  

= 6.51k Ω

 

󰁐󰁵󰁢󰁬󰁩󰁳󰁨󰁥󰁤 󰁢󰁹󰀺 󰁅󰁎󰁇󰁉󰁎󰁅󰁅󰁒󰁓 󰁉󰁎󰁓󰁔󰁉󰁔󰁕󰁔󰁅 󰁏󰁆 󰁉󰁎󰁄󰁉󰁁󰀭󰁅.󰁉.󰁉. 󰂩 󰁁󰁌󰁌 󰁒󰁉󰁇󰁈󰁔 󰁒󰁅󰁓󰁅󰁒󰁖󰁅󰁄  󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭   󰀲󰀸󰁂󰀯󰀷 󰁊󰁩󰁡󰁳󰁡󰁲󰁡󰁩 󰁎󰁥󰁡󰁲 󰁉󰁉󰁔 󰁈󰁡󰁵󰁺󰁫󰁨󰁡󰁳 󰁎󰁥󰁷󰁤󰁥󰁬󰁨󰁩󰀭󰀱󰀱󰀰󰀰󰀱󰀶 󰁰󰁨󰀮 󰀰󰀱󰀱󰀭󰀲󰀶󰀵󰀱󰀴󰀸󰀸󰀸󰀮 󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭

󰁛󰀸󰁝 

 

  ECE 

GATE Practice Book 

󰁛󰀹󰁝 

2. COMMUNICATION (1.) 

For an A.M. signal  x(t ) = Ac   1 + (a.) 25%

(b.) 20%

1 2

 

6

sin 10t . cos 10 t  find the power efficiency? (c.) 70.7%

(d.) 56.56%

(2.) 

For an A.M signal  x (t ) = 10   1 + 0.5 cos( 2π .103.t )   cos 10 6 t  find the upper side band power

(3.) 

and modulated signal bandwidth. (a.) 3.125W, 2kHz (b.) 6.25W, 2kHz (c.) 3.125W, 1kHz (d.) 6.25W, 1kHz In commercial T.V, picture signals are transmitted by …………….. and speech signals by ……………… modulation. (a.) S.S.B., F.M (b.) F.M, V.S.B (c.) F.M., S.S.B (d.) V.S.B., F.M.

2 Linked Questions  Questions  (4.) 

A signal by  f (t ) = 2 cos(1002π t ) + 20 cos(1000π t ) + 2 cos(998π t ) has been radiated with an antenna of resistance 10Ω. Find modulation index and total power transmitted. (a.) µ = 0.1 (b.) µ = 0.1 (c.) µ = 0.1 (d.) µ = 0.2 Pt = 201W Pt = 201W Pt = 20.1W Pt = 20.4W

 

(5.) (6.) 

The bandwidth and the power of the upper sideband in the previous question are (a.) 2Hz, 0.2W (b.) 4Hz, 0.4W (c.) 2Hz, 0.4W (d.) 4Hz, 0.2W An A.M. transmitter current is given by 10A with 40% single tone modulation, Find A.M. transmitter current with 80% single tone modulation. (a.) 11.05A (b.) 8.18A (c.) 9.045A (d.) 12.22A

2 Linked Questions  Questions  6

A carrier signal c(t) = 10. cos(2π  × 10 t )  is simultaneously amplitude modulated with two message signals of frequencies 1kHz and 2kHz with modulation indices of 0.4 and 0.3 respectively (7.)  (8.) 

(9.) 

Find overall modulation index and total power transmitted if antenna resistance = 1Ω  (a.) 0.5, 56.25W (b.) 0.5, 62.25W (c.) 0.7, 56.25W (d.) 0.7, 62.25W Find the bandwidth and power in the two sidebands corresponding to 2kHz modulating frequency. (a.) 4kHz, 2.25W (b.) 3kHz, 4W (c.) 3kHz, 2.25W (d.) 4kHz, 4W An amplitude modulated signal is plotted.

Find the modulation index and amplitude of the carrier wave (a.) 0.6, 60V (b.) 0.4, 50V (c.) 0.4, 60V

(d.) 0.6, 50V

󰁐󰁵󰁢󰁬󰁩󰁳󰁨󰁥󰁤 󰁢󰁹󰀺 󰁅󰁎󰁇󰁉󰁎󰁅󰁅󰁒󰁓 󰁉󰁎󰁓󰁔󰁉󰁔󰁕󰁔󰁅 󰁏󰁆 󰁉󰁎󰁄󰁉󰁁󰀭󰁅.󰁉.󰁉. 󰂩 󰁁󰁌󰁌 󰁒󰁉󰁇󰁈󰁔 󰁒󰁅󰁓󰁅󰁒󰁖󰁅󰁄  󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭   󰀲󰀸󰁂󰀯󰀷 󰁊󰁩󰁡󰁳󰁡󰁲󰁡󰁩 󰁎󰁥󰁡󰁲 󰁉󰁉󰁔 󰁈󰁡󰁵󰁺󰁫󰁨󰁡󰁳 󰁎󰁥󰁷󰁤󰁥󰁬󰁨󰁩󰀭󰀱󰀱󰀰󰀰󰀱󰀶 󰁰󰁨󰀮 󰀰󰀱󰀱󰀭󰀲󰀶󰀵󰀱󰀴󰀸󰀸󰀸󰀮 󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭

 

  ECE 

GATE Practice Book 

SOLUTION

(1.) 

ANS: b EXP: 2

η  =

 µ 

2 +  µ 2

(2.)  (3.) 

ANS: a ANS: d

(4.) 

ANS: c EXP:

P=

 Ac

2

2 R

(5.) 

ANS: a

(6.) 

ANS: a

(7.)  (8.)  (9.) 

ANS: a ANS: a  a  ANS: d

 

2

 1+

 µ 

2

 

watt  

󰁐󰁵󰁢󰁬󰁩󰁳󰁨󰁥󰁤 󰁢󰁹󰀺 󰁅󰁎󰁇󰁉󰁎󰁅󰁅󰁒󰁓 󰁉󰁎󰁓󰁔󰁉󰁔󰁕󰁔󰁅 󰁏󰁆 󰁉󰁎󰁄󰁉󰁁󰀭󰁅.󰁉.󰁉. 󰂩 󰁁󰁌󰁌 󰁒󰁉󰁇󰁈󰁔 󰁒󰁅󰁓󰁅󰁒󰁖󰁅󰁄  󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭   󰀲󰀸󰁂󰀯󰀷 󰁊󰁩󰁡󰁳󰁡󰁲󰁡󰁩 󰁎󰁥󰁡󰁲 󰁉󰁉󰁔 󰁈󰁡󰁵󰁺󰁫󰁨󰁡󰁳 󰁎󰁥󰁷󰁤󰁥󰁬󰁨󰁩󰀭󰀱󰀱󰀰󰀰󰀱󰀶 󰁰󰁨󰀮 󰀰󰀱󰀱󰀭󰀲󰀶󰀵󰀱󰀴󰀸󰀸󰀸󰀮 󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭

󰁛󰀱󰀰󰁝 

 

  ECE 

GATE Practice Book 

󰁛󰀱󰀱󰁝 

3. CONTROL SYSTEM (1.)  (2.) 

The transfer function of a system in Laplace transform of (a.)  Impulse response (b.) Step response (c.) Ramp response (d.) None of these The open loop D.C. gain of a unity negative feedback system with closed loop transfer

s+4 function (a.)  (3.) 

4

4

(b.)

13  

(c.) 4



  −3t 

If unit step response of a system is C (t ) = 1 − e (a.) 

(4.) 

s + 7 s + 13 is: 2

3

(b.)

s (s + 3)  

3

s+6



5

1 −3  − .e t + .e t   .u (t )  

(d.) ( s + 3)

 

5 − 1 −3   (b.)  2 − .e t + .e t   .u (−t )   

  2 2  5 − t 1 −3t    (c.)  2 − .e + .e  .u (t )   2 2    

2 2   5 − t 1 −3t    (d.)  2 + .e − .e  .u (t )    2 2  



Find the inverse Laplace transform t ransform

F (s) =

( (c.)  ( −e (a.)  e

s +6 2

( s + 2)( s + 2 s + 1)

−2 t



) .u(t )   − t.e ) .u (t )    

− e t + t.e

−2 t

+e

−t

 

− t 

− t 

Find the inverse Laplace transform t ransform of F ( s) = (a.)   e



−3t

. cos 4t + 3 .e 2

(c.)  cos 3t.u (t );

2

 

3 2 3

sin 3t.u (t )  

(d.) ( e

−2 t

− e t + t2





 3 −3  −3 (d.)  e t . cos 4t + .e t . sin 4t  .u (t )    2  s 2

s −9

; and f 2 ( s ) =

   

− t 

 

(b.) cos h3t.u (t ));;

sin h3t.u (t ) 

) .u (t )    .e ) .u (t )    − t 

−3 t −3t  (b.)  e . cos 4t + 2 .e . sin 4t  .u (t )    3  

 2 −3 t  −3 e . cos 4t + .e t . sin 4t  .u (t )   3   

(a.)  cos h3t.u (t ); );

+ e t + t.e

2

. sin 4t  .u (t )  

Find the inverse Laplace transform t ransform of  f1 ( s) =

−2 t

s + 6s + 25

−3t 



(b.) ( e

s+9

(c.)   (7.) 



 

2

s( s + 4s + 3)

(a.)   2 +

(6.) 

3

The inverse Laplace transform of

F ( s) =

(5.) 

≥ 0 , its transfer function is, for t   ≥

(c.) 3

( s + 3)  

(d.) 13

(d.) cos 3t.u (t ); );

2 2

s −9

2 3 2 3

 

sin h3t.u (t )   sin 3t.u (t )   

󰁐󰁵󰁢󰁬󰁩󰁳󰁨󰁥󰁤 󰁢󰁹󰀺 󰁅󰁎󰁇󰁉󰁎󰁅󰁅󰁒󰁓 󰁉󰁎󰁓󰁔󰁉󰁔󰁕󰁔󰁅 󰁏󰁆 󰁉󰁎󰁄󰁉󰁁󰀭󰁅.󰁉.󰁉. 󰂩 󰁁󰁌󰁌 󰁒󰁉󰁇󰁈󰁔 󰁒󰁅󰁓󰁅󰁒󰁖󰁅󰁄  󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭   󰀲󰀸󰁂󰀯󰀷 󰁊󰁩󰁡󰁳󰁡󰁲󰁡󰁩 󰁎󰁥󰁡󰁲 󰁉󰁉󰁔 󰁈󰁡󰁵󰁺󰁫󰁨󰁡󰁳 󰁎󰁥󰁷󰁤󰁥󰁬󰁨󰁩󰀭󰀱󰀱󰀰󰀰󰀱󰀶 󰁰󰁨󰀮 󰀰󰀱󰀱󰀭󰀲󰀶󰀵󰀱󰀴󰀸󰀸󰀸󰀮 󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭

 

  ECE 

GATE Practice Book 

SOLUTION (1.)  (2.) 

ANS: (a) ANS: (a) ANS: (b) ANS:  (b) EXP:   EXP:

G( s)

1 + G ( s).H ( s) (3.) 

s+4

=

 

2

s + 7 s + 13

Put H(s) = 1, find G(s) and put s = 0 (for d.c.) ANS: (b) ANS:  (b) EXP:   EXP: Impulse response, h(t) =

d  dt 

c(t )  

 H (s) = L [ h(t ) ]   (4.) 

ANS: (c) ANS: (c) EXP:   EXP:

F ( s) =

s+6 s ( s + 1) 1)( s + 3)

=

A s

+

B s +1

+

C s+3

 A = 6 = 2 3 −5 5  B = = −1× 2 2 3 1 C  = = −3 × −2 2 (5.) 

 

 

ANS: (a) ANS: (a) EXP:   EXP:

F ( s) =  A =  B =

C  =

1 ( s + 2)( s + 1)

1

=

 A

( s + 2)

+

B

( s + 1)

=1

( −1) 2 −1

( −1 + 2) 1 ( −1 + 2)

∴ F (s) =

2

+



(s + 1) 2  

= −1

=1

1 ( s + 2)



1 ( s + 1)

+

1 ( s + 1) 2

∴  f (t ) = ( e−2 t − e −t + t.e −t  ) .u (t )

 

󰁐󰁵󰁢󰁬󰁩󰁳󰁨󰁥󰁤 󰁢󰁹󰀺 󰁅󰁎󰁇󰁉󰁎󰁅󰁅󰁒󰁓 󰁉󰁎󰁓󰁔󰁉󰁔󰁕󰁔󰁅 󰁏󰁆 󰁉󰁎󰁄󰁉󰁁󰀭󰁅.󰁉.󰁉. 󰂩 󰁁󰁌󰁌 󰁒󰁉󰁇󰁈󰁔 󰁒󰁅󰁓󰁅󰁒󰁖󰁅󰁄  󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭   󰀲󰀸󰁂󰀯󰀷 󰁊󰁩󰁡󰁳󰁡󰁲󰁡󰁩 󰁎󰁥󰁡󰁲 󰁉󰁉󰁔 󰁈󰁡󰁵󰁺󰁫󰁨󰁡󰁳 󰁎󰁥󰁷󰁤󰁥󰁬󰁨󰁩󰀭󰀱󰀱󰀰󰀰󰀱󰀶 󰁰󰁨󰀮 󰀰󰀱󰀱󰀭󰀲󰀶󰀵󰀱󰀴󰀸󰀸󰀸󰀮 󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭

󰁛󰀱󰀲󰁝 

 

  ECE 

(6.) 

GATE Practice Book 

ANS: (a) ANS: (a) EXP:   EXP:

F ( s) =

=

s+9 2

s + 6 s + 25

=

s+3 2

( s + 3) + ( 4)

 

−3

2

=   e t . cos 4t + (7.) 

s + 3+ 6 2

( s + 3) + ( 4) +

3 2

3 2

×

2

4 2

( s + 3) + ( 4)

2

−3 .e t . sin 4t .u (t )  

ANS: (b) ANS:  (b)

󰁐󰁵󰁢󰁬󰁩󰁳󰁨󰁥󰁤 󰁢󰁹󰀺 󰁅󰁎󰁇󰁉󰁎󰁅󰁅󰁒󰁓 󰁉󰁎󰁓󰁔󰁉󰁔󰁕󰁔󰁅 󰁏󰁆 󰁉󰁎󰁄󰁉󰁁󰀭󰁅.󰁉.󰁉. 󰂩 󰁁󰁌󰁌 󰁒󰁉󰁇󰁈󰁔 󰁒󰁅󰁓󰁅󰁒󰁖󰁅󰁄  󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭   󰀲󰀸󰁂󰀯󰀷 󰁊󰁩󰁡󰁳󰁡󰁲󰁡󰁩 󰁎󰁥󰁡󰁲 󰁉󰁉󰁔 󰁈󰁡󰁵󰁺󰁫󰁨󰁡󰁳 󰁎󰁥󰁷󰁤󰁥󰁬󰁨󰁩󰀭󰀱󰀱󰀰󰀰󰀱󰀶 󰁰󰁨󰀮 󰀰󰀱󰀱󰀭󰀲󰀶󰀵󰀱󰀴󰀸󰀸󰀸󰀮 󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭

󰁛󰀱󰀳󰁝 

 

  ECE 

GATE Practice Book 

󰁛󰀱󰀴󰁝 

4. Digital Electronics (1.) 

Assuming that initially Qn = 0, now the clock pulses are given, fine the resulting sequence at

Qn .

(2.) 

(a.) 0, 0, 0, 0… (b.) 1, 1, 1, 1… (c.) 0, 1, 0, 1, 0, 1… (d.) 1, 0, 1, 0… Fine the output frequency of the circuit.

(a.) 49.76 MHz (b.) 995.22 Hz (c.) 6.25 KHz (d.) 7.961 KHz

󰁐󰁵󰁢󰁬󰁩󰁳󰁨󰁥󰁤 󰁢󰁹󰀺 󰁅󰁎󰁇󰁉󰁎󰁅󰁅󰁒󰁓 󰁉󰁎󰁓󰁔󰁉󰁔󰁕󰁔󰁅 󰁏󰁆 󰁉󰁎󰁄󰁉󰁁󰀭󰁅.󰁉.󰁉. 󰂩 󰁁󰁌󰁌 󰁒󰁉󰁇󰁈󰁔 󰁒󰁅󰁓󰁅󰁒󰁖󰁅󰁄  󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭   󰀲󰀸󰁂󰀯󰀷 󰁊󰁩󰁡󰁳󰁡󰁲󰁡󰁩 󰁎󰁥󰁡󰁲 󰁉󰁉󰁔 󰁈󰁡󰁵󰁺󰁫󰁨󰁡󰁳 󰁎󰁥󰁷󰁤󰁥󰁬󰁨󰁩󰀭󰀱󰀱󰀰󰀰󰀱󰀶 󰁰󰁨󰀮 󰀰󰀱󰀱󰀭󰀲󰀶󰀵󰀱󰀴󰀸󰀸󰀸󰀮 󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭

 

  ECE 

(3.) 

GATE Practice Book 

Find the expression for the output logic function.

(a.) AB + C(D + E) (b.) (A+B) .C + (D + E) (c.) AB + C. (D+E) (d.) AB + C. ( D + E )

 

󰁐󰁵󰁢󰁬󰁩󰁳󰁨󰁥󰁤 󰁢󰁹󰀺 󰁅󰁎󰁇󰁉󰁎󰁅󰁅󰁒󰁓 󰁉󰁎󰁓󰁔󰁉󰁔󰁕󰁔󰁅 󰁏󰁆 󰁉󰁎󰁄󰁉󰁁󰀭󰁅.󰁉.󰁉. 󰂩 󰁁󰁌󰁌 󰁒󰁉󰁇󰁈󰁔 󰁒󰁅󰁓󰁅󰁒󰁖󰁅󰁄  󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭   󰀲󰀸󰁂󰀯󰀷 󰁊󰁩󰁡󰁳󰁡󰁲󰁡󰁩 󰁎󰁥󰁡󰁲 󰁉󰁉󰁔 󰁈󰁡󰁵󰁺󰁫󰁨󰁡󰁳 󰁎󰁥󰁷󰁤󰁥󰁬󰁨󰁩󰀭󰀱󰀱󰀰󰀰󰀱󰀶 󰁰󰁨󰀮 󰀰󰀱󰀱󰀭󰀲󰀶󰀵󰀱󰀴󰀸󰀸󰀸󰀮 󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭

󰁛󰀱󰀵󰁝 

 

  ECE 

GATE Practice Book 

SOLUTION (1.) 

(2.) 

ANS: c  c  EXP:   EXP: The input changes as 11 ANS: b  b 

 01



 11



 01, use JKFF truth table.



EXP: The frequency of Hartley’s oscillator = 7961.8Hz. The output of mod – 8 counter

= (3.) 

7961.8 Hz 8

 

ANS: d  d  EXP: NMOS  Parallel  OR NMOS  series  AND Reverse for PMOS Put complement on the final expression →







󰁐󰁵󰁢󰁬󰁩󰁳󰁨󰁥󰁤 󰁢󰁹󰀺 󰁅󰁎󰁇󰁉󰁎󰁅󰁅󰁒󰁓 󰁉󰁎󰁓󰁔󰁉󰁔󰁕󰁔󰁅 󰁏󰁆 󰁉󰁎󰁄󰁉󰁁󰀭󰁅.󰁉.󰁉. 󰂩 󰁁󰁌󰁌 󰁒󰁉󰁇󰁈󰁔 󰁒󰁅󰁓󰁅󰁒󰁖󰁅󰁄  󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭   󰀲󰀸󰁂󰀯󰀷 󰁊󰁩󰁡󰁳󰁡󰁲󰁡󰁩 󰁎󰁥󰁡󰁲 󰁉󰁉󰁔 󰁈󰁡󰁵󰁺󰁫󰁨󰁡󰁳 󰁎󰁥󰁷󰁤󰁥󰁬󰁨󰁩󰀭󰀱󰀱󰀰󰀰󰀱󰀶 󰁰󰁨󰀮 󰀰󰀱󰀱󰀭󰀲󰀶󰀵󰀱󰀴󰀸󰀸󰀸󰀮 󰁷󰁷󰁷.󰁥󰁮󰁧󰁩󰁮󰁥󰁥󰁲󰁳󰁩󰁮󰁳󰁴󰁩󰁴󰁵󰁴󰁥.󰁣󰁯󰁭

󰁛󰀱󰀶󰁝 

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