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GATE CLOUD SIGNALS & SYSTEMS Sample Chapter GATE CLOUD Signals & System by Kanodia Sample Chapter GATE CLOUD Signals & System by Kanodia GATE CLOUD SIGNALS & SYSTEMS R. K. Kanodia Ashish Murolia JHUNJHUNUWALA Jaipur Sample Chapter GATE CLOUD Signals & System by Kanodia GATE CLOUD Signals & Systems, 1e R. K. Kanodia, Ashish Murolia AA1213 Copyright by Jhunjhunuwala Information contained in this book has been obtained by author, from sources believes to be reliable. However, neither Jhunjhunuwala nor its author guarantee the accuracy or completeness of any information herein, and Jhunjhunuwala nor its author shall be responsible for any error, omissions, or damages arising out of use of this information. This book is published with the understanding that Jhunjhunuwala and its author are supplying information but are not attempting to render engineering or other professional services. JHUNJHUNUWALA B-8, Dhanshree Tower Ist, Central Spine, Vidyadhar Nagar, Jaipur 302023 Ph : +91-141-2101150. www.jhunjhunuwala.com email : info@jhunjhunuwala.com Printed By: Jaipur Printing Centre, Jaipur Sample Chapter GATE CLOUD Signals & System by Kanodia Preface to First Edition GATE Question Cloud caters a versatile collection of Multiple Choice Questions to the students who are preparing for GATE(Gratitude Aptitude Test in Engineering) examination. This book contains over 1500 multiple choice solved problems for the subject of Signals & Systems, which has a significant weightage in the GATE examinations of EC, EE & IN branches. The GATE examination is based on multiple choice problems which are tricky, conceptual and tests the basic understanding of the subject. So, the problems included in the book are designed to be as exam-like as possible. The solutions are presented using step by step methodology which enhance your problem solving skills. The book is categorized into eleven chapters covering all the topics of syllabus of the examination. Each chapter contains : Exercise 1 : Theoretical & One line Questions Exercise 2 : Level 1 Exercise 3 : Level 2 Exercise 4 : Mixed Questions taken form previous examinations of GATE & IES. Detailed Solutions to Exercise 2, 3 & 4 Summary of useful theorems Although we have put a vigorous effort in preparing this book, some errors may have crept in. We shall appreciate and greatly acknowledge the comments, criticism and suggestion from the users of this book which leads to some improvement. Wish you all the success in conquering GATE. Authors Sample Chapter GATE CLOUD Signals & System by Kanodia Sample Chapter GATE CLOUD Signals & System by Kanodia Sample Chapter GATE CLOUD Signals & System by Kanodia GATE CLOUD DIGITAL ELECTRONICS R. K . Kanodia & Ashish Murolia GATE CLOUD is an exclusive series of books which offers a completely solved question bank to GATE aspirants. The book of this series are featured as Over 1300 Multiple Choice Questions with full & detailed explanations. Questions are graded in the order of complexity from basic to advanced level. Contains all previous year GATE and IES exam questions from various branches. Each question is designed to GATE exam level. Upcoming titles in this series & Circuit Analysis (For EC, EE & IN branches) & Analog Circuit and Devices (For EC, EE & IN branches) & Control Systems (For EC, EE & IN branches) Sample Chapter GATE CLOUD Signals & System by Kanodia CHAPTER 1 CONTINUOUS TIME SIGNALS Sample Chapter GATE CLOUD Signals & System by Kanodia EXCERCISE 1.1 MCQ 1.1.1 The graphical representation of a signal in the time domain is known as (A) frequency (B) waveform (C) frequency spectrum MCQ 1.1.2 A continuous-time signal is a signal in which the independent variable is (A) discrete (B) continuous (C) (A) or (B) MCQ 1.1.3 (D) none of the above (D) none of the above Digital signals are those signal which (A) do not have a continuous set of values (B) have values at discrete instants (C) can utilize decimal or binary system (D) are all of the above MCQ 1.1.4 A deterministic signal is the signal which (A) can not be represented by a mathematical expression (B) has no uncertainty (C) has uncertainty (D) none of the above MCQ 1.1.6 A random signal is the signal which (A) has uncertainty (B) (C) is a completely specified function of time MCQ 1.1.5 (D) none of the above has no uncertainty Speech signals and the sine wave respectively are the example of (A) deterministic signal, random signal. (B) both random signals (C) both deterministic signals (D) random signal, deterministic signals MCQ 1.1.7 Which of the following is a periodic signal ? (B) x (t) = Ae j t (A) x (t) = At 2 (C) x (t) = Ae t (D) x (t) = Au (t) Sample Chapter GATE CLOUD Signals & System by Kanodia Page 4 MCQ 1.1.8 Continuous Time Signals The sum of two periodic signals having periods T1 and T2 is periodic only if the ratio of their respective periods (T1 /T2) is (A) an irrational number (B) a rational number (C) an odd number MCQ 1.1.9 Chapter 1 (D) an even number A continuous-time signal x (t) is said to be periodic with a fundamental period T0 , where T0 is the (A) smallest positive integer satisfying the relation x (t) = x (t + mT0) for any t and any m . (B) positive constant satisfying the relation x (t) = x (t + mT0) for every t and any integer m . (C) largest positive constant satisfying the relation x (t) = x (t + mT0) for any t and any integer m (D) smallest positive constant satisfying the relation x (t) = x (t + mT0) for every t and any integer m MCQ 1.1.10 Sine waves, cosine waves, square waves and triangular waves are the examples of (A) non-deterministic functions (B) multiple frequency functions (C) periodic functions MCQ 1.1.11 A signal is given by x (t) = 2 cos ( t) sin2 ( t) + 2 cos ( t) + sin (t) + sin2 ( t). The odd component of x (t) is (A) cos ( t) sin2 ( t) (B) sin ( t) (C) sin2 ( t) MCQ 1.1.12 (D) all of the above (D) cos ( t) f (t) is even while g (t) is odd. If x (t) = f (t) + g (t) and y (t) = f (t) g (t) then x (t) and y (t) are respectively (A) neither, even (B) odd, even (C) neither, odd MCQ 1.1.14 Signal x (t) = 5 sin 20 t (A) is an even signal (B) is an odd signal (C) has even and odd parts MCQ 1.1.13 (D) even, odd (D) none of the above Which of the following statements is not true ? 1. The product of two even signals in an even signal 2. The product of two odd signals in an odd signal. 3. The product of even and odd signals in an even signal. 4. The product of even and odd signal is an odd signal. (A) 2 and 3 (B) 1 only (C) 3 only (D) 4 only Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 MCQ 1.1.15 Continuous Time Signals x (t) = 5 sin (10 t + 30c) (A) is an odd signal (B) is an even signal (C) has an even part as well as an odd part (D) none of the above MCQ 1.1.17 The signal x (t) = 10e j10 t is (A) an energy signal (B) a power signal (C) neither energy nor power signal MCQ 1.1.16 (D) both energy and power signal Signal e 2t u (t) is (A) a power signal (B) an energy signal (C) neither an energy signal nor a power signal (D) none of the above MCQ 1.1.21 (D) both (B) and (C) A signal is a power signal if it has (A) infinite energy (B) infinite power (D) both (A) and (C) The signal A cos ( 0 t + ) is (A) a periodic signal (B) a power signal (C) both periodic and power signals MCQ 1.1.20 (B) finite energy (C) finite power MCQ 1.1.19 A signal is an energy signal if it has (A) infinite energy (C) zero average power MCQ 1.1.18 (D) a energy signal Which of the following is an energy signal ? (B) x (t) = A sin 0 t (A) x (t) = A cos 0 t (C) x (t) = Ae j t (D) x (t) = e at u (t) 0 MCQ 1.1.22 Which of the following statement are true ? 1. Most of the periodic signals are energy signals. 2. Most of the periodic signals are power signals. 3. For energy signals, the power is zero. 4. For power signals, the energy is zero. (A) 1, 2 and 3 only (B) 1 only (C) 1 and 2 only (D) 1, 2, 3, and 4 Sample Chapter GATE CLOUD Signals & System by Kanodia Page 5 Page 6 MCQ 1.1.23 Continuous Time Signals A complex valued signal x (t) = xR (t) + jxI (t) has conjugate symmetry if (A) xR (t) is odd while xI (t) is even (B) xR (t) and xI (t) are both odd (C) xR (t) is even while xI (t) is odd MCQ 1.1.24 Chapter 1 (D) xR (t) and xI (t) are both even A signal x (t) has energy Ex , then energy of the signal x (at) is given by (B) Ex / a (A) Ex / a 2 2 (C) Ex a (D) a Ex MCQ 1.1.25 The value of # 2 cos t ( ) d is (A) 0 (B) /2 (C) 1 (D) 2 3 MCQ 1.1.26 If (t) is the unit impulse function, then # x (t) (t) dt equals to 3 (A) x (t) (C) x (3) MCQ 1.1.27 (B) x (0) (D) x (1) For unit impulse function (t), which of the following relation holds true ? (A) ( t) = a t k (B) ( t) = (t2) 2 (D) ( t) = 2 (t) (C) ( t) = (t) MCQ 1.1.28 The function f (t) = t (t) will be equal to (A) t (B) 3 (C) 1 MCQ 1.1.29 (D) 0 The unit impulse is defined as, 3, t = 0 (B) (t) = ) 0, t ! 0 (A) (t) = 3, t = 0 (C) (t) = 3, t = 0 and MCQ 1.1.30 +3 # (t) dt = 0 3 3, t = 0 (D) (t) = ) and 0, t ! 0 +3 # (t) dt = 1 3 If x (t) is a continuous time signal and (t) is a unit impulse signal then value of integral # x (t) (t t ) is equal to 3 3 0 (A) x (t) (C) (t) MCQ 1.1.31 (B) x (t 0) (D) 1 A weighted impulse function (at) has (A) unit area and unit amplitude (B) infinite area and finite amplitude (C) finite area and infinite amplitude (D) infinite area and infinite amplitude Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 MCQ 1.1.32 Continuous Time Signals Page 7 Unit step signal u (t) is (A) an energy signal (B) a power signal (C) neither power signal nor energy signal (D) both MCQ 1.1.33 A unit step function is given by 1, (A) u (t) = * 0, 1, (B) u (t) = ) 0, t=0 t!0 t, (C) u (t) = * 0, MCQ 1.1.34 t$0 t<0 t$0 t<0 1, (D) u (t) = ) 0, t>0 t<0 Match List I with List II and choose the correct answer using the codes given below the lists : List I (Signal) List II (Nature) 1. Sample values are unpredictable Q. Unit Step signal 2. Has only one non-zero value R. Random noise signal 3. Amplitude decreases as time increases S. Decaying exponential Codes : P Q R (A) 3 2 4 (B) 2 4 1 (C) 1 2 3 (D) 2 1 4 4. Has only two possible values P. MCQ 1.1.35 Unit Impulse signal S 1 3 4 3 A unit ramp function is defined as 1, (A) r (t) = ) 0, (B) r (t) = * t, (C) r (t) = * 0, MCQ 1.1.36 t=0 t!0 t$0 t<0 1, (D) r (t) = * 0, t$0 t<0 t$0 t<0 The differentiation of a unit step signal is, (A) an impulse signal (B) a ramp signal (C) an exponential signal MCQ 1.1.37 t + 1, 0, (D) a parabolic signal In terms of unit-step function, signum function is given as (A) sgn (t) = u (t) (B) sgn (t) = 2u (t) (C) 2sgn (t) = u (t) (D) sgn (t) = 2u (t) 1 Sample Chapter GATE CLOUD Signals & System by Kanodia Page 8 MCQ 1.1.38 MCQ 1.1.39 Continuous Time Signals The signum function is defined as 1, t > 0 (A) sgn (t) = ) 0, t < 0 0, t > 0 (C) sgn (t) = ) 1, t < 0 1, t > 0 (B) sgn (t) = ) 1, t < 0 1, t > 0 (D) sgn (t) = ) 1, t<0 Differentiation of signum function will be (A) 1 (t) 2 (D) 2u (t) The sinc function f (t) is defined as (A) f (t) = sin t t (B) f (t) = sin t t (C) f (t) = sin t t MCQ 1.1.41 (B) (t) (C) 2 (t) MCQ 1.1.40 Chapter 1 (D) f (t) = sin t t The mathematical expression for the signal x (t) shown in figure is given by (A) u (t 0.5) + u (t + 0.5) (B) u (t + 0.5) u (t 0.5) (C) u (t 0.5) u (t 0.5) (D) u (t + 0.5) + u (t 0.5) *********** Sample Chapter GATE CLOUD Signals & System by Kanodia EXCERCISE 1.2 MCQ 1.2.1 What is the period of a signal x (t) = 3 sin (4 t) + 7 cos (3 t) ? (A) 2 sec (B) 4 sec (C) 12 sec MCQ 1.2.2 (D) x (t) is not periodic The period of a signal x (t) = 3 sin (4 t) + 7 cos (10t) is (A) 10 sec (B) 5 sec (C) 6 sec MCQ 1.2.3 (D) x (t) is not periodic Consider the following continuous time signals x1 (t) = 6 sin (8 t) + 14 cos (6 t) x2 (t) = 6 sin (8 t) + 14 cos (20t) Which of the following statement regarding the periodicity of the signals is true ? (A) x1 (t) is periodic, x2 (t) is aperiodic (B) Both x1 (t) and x2 (t) are periodic (C) x1 (t) is aperiodic, x2 (t) is periodic (D) Both x1 (t) and x2 (t) are aperiodic MCQ 1.2.4 What is the period of the signal x (t) = sin b 2 t l cos b 4 t l ? 5 3 (A) 13 sec (C) 15 sec MCQ 1.2.5 (B) 91 sec (D) x (t) is aperiodic Match List I (Signal) with List II (Period of the signal) and select the answer using the codes given below List I (Signals) List II (Period of the signal) P. f1 (t) = sin b 2 l t 3 1. 15 Unit Q. f2 (t) = sin b 2 t l cos b 4 t l 5 3 2. 3 Unit R. f3 (t) = sin 3t 3. aperiodic S. f4 (t) = f1 (t) 2f3 (t) 4. 2 /3 unit Sample Chapter GATE CLOUD Signals & System by Kanodia Page 10 Continuous Time Signals Codes : P (A) 1 (B) 3 (C) 1 (D) 2 MCQ 1.2.6 Q 4 2 2 1 (D) 5 sec Consider the signals x1 (t) = 5 cos (4t + ), x2 (t) = e j ( t 1) and x 3 (t) = 6cos (2t )@2 3 3 Which signals is/are aperiodic (B) x2 (t) and x 3 (t) (A) x 3 (t) only (C) x2 (t) only MCQ 1.2.9 (D) none of these The period of the signal g (t) = 2 cos (10t + 1) + sin (4t 1) is equal to (A) 10 sec (B) sec (C) 2 sec MCQ 1.2.8 S 2 4 4 3 Which of the following signal is not periodic? (A) sin (10t) (B) 2 cos (5 t) (C) sin (10 t) u (t) MCQ 1.2.7 R 3 1 3 4 Chapter 1 (D) none of above t, 0 # t < 1 Consider a signal g (t) defined as g (t) = * . The odd part of g (t) is 0, elsewhere Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 MCQ 1.2.10 Continuous Time Signals Page 11 A signal g (t) is defined as t, 0 # t < 1 . g (t) = * 0, elsewhere The even part of the signal g (t) is Zt/2, ] (A) ge (t) = [t/2, ]0, \ Z 2t, ] (C) ge (t) = [2t, ]0, \ MCQ 1.2.11 1 # t < 0 0#t<1 elsewhere 1 # t < 0 0#t<1 elsewhere Z t/2, ] (B) ge (t) = [t/2, ]0, \ Z2t, ] (D) ge (t) = [2t, ]0, \ 1 # t < 0 0#t<1 elsewhere 1 # t < 0 0#t<1 elsewhere A CT signal is defined as 2, t > 0 x (t) = * 0, t < 0 The odd part of x (t) is an unit (A) step function (C) impulse function MCQ 1.2.12 (B) signum function (D) ramp function The odd part of a unit step signal is Sample Chapter GATE CLOUD Signals & System by Kanodia Page 12 MCQ 1.2.13 Continuous Time Signals Chapter 1 A signal x (t) is shown in figure below The odd part of the signal g (t) = x ^t 3 h + x ^t + 3 h will be 4 4 MCQ 1.2.14 If xe (t) and xo (t) are the even and odd part of a signal x (t), then which of the following is true? (A) xo (0) = 0 (B) xe (0) = x (0) (C) xo (0) = xe (0) = 0 (D) Both (A) and (B) Statement For Q. 15 & 16 : The figure shows parts of a signal x (t) and its odd part xo (t), for t $ 0 only, that is x (t) and xo (t) are not given for t < 0 . Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 Continuous Time Signals MCQ 1.2.15 The complete odd part xo (t) of the signal will be MCQ 1.2.16 The complete even part xe (t) of the signal x (t) is Sample Chapter GATE CLOUD Signals & System by Kanodia Page 13 Page 14 MCQ 1.2.17 Continuous Time Signals A signal x (t) is shown in figure below The odd part of signal x (t) is MCQ 1.2.18 Two signals g1 (t) and g2 (t) are shown in the following figures Which of the following statement is true ? (A) g1 (t) is a power signal, g2 (t) is an energy signal. (B) g1 (t) is an energy signal, g2 (t) is a power signal. (C) Both g1 (t) and g2 (t) are power signals. (D) Both g1 (t) and g2 (t) are energy signals. Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 Chapter 1 MCQ 1.2.19 Continuous Time Signals The average power (Pg) and energy (Eg) of the signal g (t) shown in figure are (A) Pg = 25, Eg = 150 (B) Pg = 0, Eg = 150 (C) Pg = 25, Eg = 3 MCQ 1.2.20 (D) Pg = 25, Eg = 50 The energy and average power of a signal x (t) as shown in figure are respectively : (A) 100, 0 (B) 3, 25 (C) 50, 0 MCQ 1.2.21 (D) 3, 12.5 The energy of the signal shown in figure is (A) A2 /2 (B) A2 (C) A2 /4 MCQ 1.2.22 (D) None of above The power and rms value of a voltage signal x (t) = 20 cos (5t) cos (10t) V are respectively : (A) 200 W, 14.14 volt (B) 100 W, 7.07 volt (C) 100 W, 10 volt MCQ 1.2.23 Page 15 (D) 200 W, 10 volt The signal x (t) = e ja2t + 4 k is (A) a power signal (B) an energy signal (C) neither a power nor an energy (D) none of above Sample Chapter GATE CLOUD Signals & System by Kanodia Page 16 MCQ 1.2.24 Continuous Time Signals Chapter 1 The power of a periodic signal shown in figure is (A) 56 unit (C) 11.2 unit MCQ 1.2.25 (B) 8 unit (D) 32 unit A signal x (t), defined over the range 3 # t # 3 , has energy equal to 12 units. Match List I (signal) with List II (Energy of the signal) and select correct answer using the codes given below List I (Signal) List II 2x (t) 1. 48 unit Q. x (3t) 2. 12 unit R. x (t 4) (Energy) 3. 4 unit P. S. 4. 24 unit 2x (2t) Codes: (A) (B) (C) (D) MCQ 1.2.26 P 1 4 1 4 Q 3 3 4 1 R 2 1 3 2 S 4 2 2 3 Consider the following statements regarding a signal x (t) = e t . 1. x (t) is an energy signal 2. x (t) is an odd signal 3. x (t) is an even signal 4. x (t) is neither even nor odd. Which of the above statement is/are true? (A) only 4 (B) 1 and 3 (C) 1 and 4 (D) 1 and 2 Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 MCQ 1.2.27 Continuous Time Signals Consider the signals x1 (t), x2 (t) and y (t) as shown in below : Which of the following relation is true ? (B) y (t) = x1 (t) + x2 (t) (A) y (t) = x1 (t) x2 (t) (C) y (t) = x1 (t) x2 (t) MCQ 1.2.28 (D) none of above Two CT signals f (t) and g (t) are shown in following figure : The plot for a signal x (t) = f (t) g (t 1) will be Sample Chapter GATE CLOUD Signals & System by Kanodia Page 17 Page 18 MCQ 1.2.29 Continuous Time Signals Chapter 1 A continuous time signal is given as t + 1, 1 # t # 0 g (t) = *1, 0#t<2 elsewhere 0, The correct expression for g (2t) is Zt 0. 5 # t # 0 ] 2 + 1, (A) g (2t) = [t, 0#t<1 ] 0, elsewhere \ t + 1, 0.5 # t # 0 (C) g (2t) = *1, 0#t<1 elsewhere 0, 2t + 1, (B) g (2t) = *2, 0, 0.5 # t # 0 0#t<1 elsewhere 2t + 1, (D) g (2t) = *1, 0, 0.5 # t # 0 0#t<1 elsewhere MCQ 1.2.30 Consider a signal g (t) defined as following Z 1 # t # 0 ]t + 1, ]1, 0#t#2 g (t) = [ 2#t#3 t + 3, ] ]0, elsewhere The waveform of\signal g (t/2) is MCQ 1.2.31 Two signals f (t) and g (t) are shown in the figure below Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 Continuous Time Signals Page 19 Which of the following is the correct expression of f (t) ? (A) f (t) = g (t) + g (t + 2) + g (t + 3) (B) f (t) = g (t) + g (t 2) + g (t 3) (C) f (t) = g (t) + g (t/2) + g (t/3) (D) f (t) = g (t) + g (2t) + g (3t) MCQ 1.2.32 Consider a unit triangular function (t) and a unit rectangular function (t) as shown in figure Which of the following waveform is correct for g (t) = 3 ^2t/3h + 3 ^t/3h MCQ 1.2.33 Time compression of a signal (A) Reduces its energy (B) increases its energy (C) does not effect the energy (D) none of above. Sample Chapter GATE CLOUD Signals & System by Kanodia Page 20 MCQ 1.2.34 Continuous Time Signals A CT signal is shown below The plot of signal g (t + 2) is MCQ 1.2.35 Consider the signal x (t) and y (t) shown is figures Which of the following is correct statement ? (A) y (t) is amplitude scaled version of x (t) (B) y (t) is time scaled version of x (t) by a factor of 2. (C) y (t) is time advanced version of x (t) by 2 units. (D) y (t) is time delayed version of x (t) by 2 units. Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 Chapter 1 MCQ 1.2.36 Continuous Time Signals The plot of a signal x (t) is shown in figure If x (t) is delayed by 3 sec, then plot will be Statement For Q. 37 & 38 Consider the signal g (t) as shown in figure MCQ 1.2.37 Plot for signal g (t 2) will be Sample Chapter GATE CLOUD Signals & System by Kanodia Page 21 Page 22 Continuous Time Signals Chapter 1 MCQ 1.2.38 Plot for signal g ( t + 1) will be MCQ 1.2.39 If the energy of a signal x (t) is Ex then what will be the energy for a signal x (at b)? (A) Ex a (C) 1 Ex + b a MCQ 1.2.40 (B) b b l Ex a (D) b 1 + b l Ex a Consider a signal f (t) as shown is figure Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 Continuous Time Signals The plot of signal f (4 2t) is MCQ 1.2.41 If plot of a signal f (t) is shown in figure below Then the plot of signal f ( t 3) will be Sample Chapter GATE CLOUD Signals & System by Kanodia Page 23 Page 24 MCQ 1.2.42 Continuous Time Signals Chapter 1 A signal x (t) and its transformed signal y (t) are shown in figure(A) and figure(B) respectively If y (t) = x (at + b), then values of a and b are respectively (A) 3, 2 (B) 3 , 6 (C) 3, 6 MCQ 1.2.43 (D) 2 , 3 Consider two signals x1 (t) and x2 (t) as shown below Which of the following procedure is correct to obtain x2 (t) from x1 (t) ? (A) First compress x1 (t) by a factor of 3, then shift to the right by 6 time units. (B) First expand x1 (t) by a factor of 6, then shift to the right by 3 time units. (C) First compress x1 (t) by a factor of 3, then shift to the right by 2 time units. (D) First shift x1 (t) to the right by 2 time units then expand by a factor of 3. Statement For Q. 44 & 45 The plot of a signal x (t) is shown in figure Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 Continuous Time Signals MCQ 1.2.44 Plot for the signal x1 (t) = x [0.5 (t 2)] will be MCQ 1.2.45 Plot for the signal x2 (t) = x ( 0.5t 1) will be Statement For Q. 46 & 47 Consider two CT signal x (t) and y (t) shown in figure below Sample Chapter GATE CLOUD Signals & System by Kanodia Page 25 Page 26 MCQ 1.2.46 Continuous Time Signals Chapter 1 Which of the following relation is true ? (A) y (t) = x (2t 8) (B) y (t) = x (2t 4) (C) y (t) = x a t 2k 2 (D) y (t) = x a t 4 k 2 MCQ 1.2.47 The sketch of signal x (2 t) will be MCQ 1.2.48 Consider two signals x (t) and y (t) shown in figure below 1.4 If y (t) = Ax d t t 0 n then, the values of A , t 0 and W are respectively. W (A) 2, 0, 2 (B) 2, 1, (C) 2, 0, (D) 2, 1, 2 1 2 Sample Chapter GATE CLOUD Signals & System by Kanodia 1 2 Chapter 1 MCQ 1.2.49 Continuous Time Signals Page 27 A signal x (t) is shown in the following figure The plot for a transformed signal y (t) = 6x b t 1 l will be 2 MCQ 1.2.50 A signal x (t) is transformed into another signal y (t) given as y (t) = x a1 t k 2 The waveform of the original signal x (t) is Sample Chapter GATE CLOUD Signals & System by Kanodia Page 28 Continuous Time Signals Chapter 1 3 MCQ 1.2.51 If (t) is an unit impulse function, then the value of integral equals to (A) 1 2 MCQ 1.2.52 MCQ 1.2.54 (2t 2) dt 3 (D) 1 For an unit impulse function (t), which of the following is true? (B) [a (t t 0)] = a (t t 0) (A) [a (t t 0)] = 1 (t) a (C) [a (t t 0)] = 1 (t t 0) a MCQ 1.2.53 t (B) 1 e (C) 1 2e #e (D) [a (t t 0] = a (t) If (t) is an unit impulse function then which of the following waveform represents a signal g (t) = 6 (3t + 9)? What is the numerical value of the following integral 3 x (t) = # (t + 5) cos ( t) dt 3 (A) 1 (B) 1 (C) 0 (D) 5 Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 MCQ 1.2.55 MCQ 1.2.56 Continuous Time Signals If r (t) is a unit ramp function, then plot for signal r ( t + 2) will be Consider three signals x1 (t) = u (t) u (t 1), x2 (t) = r (t) r (t 2) and x 3 (t) = (1 + e 6t) u (t) where u (t) and r (t) are unit-step function and unit-ramp function respectively. Which of the above signals have finite energy? (A) x1 (t) and x 3 (t) (B) x1 (t) only (C) x2 (t) and x 3 (t) MCQ 1.2.57 Page 29 (D) x2 (t) only For a signal x (t) = u (t + 2) 2u (t) + u (t 2) the waveform is Sample Chapter GATE CLOUD Signals & System by Kanodia Page 30 Continuous Time Signals Chapter 1 MCQ 1.2.58 Which of the following is correct waveform of a signal x (t) given as below x (t) = u (t + 3) + 2u (t + 1) 2u (t 1) + u (t 3) MCQ 1.2.59 Consider a signal x (t) which is a linear combination of ramp signals given as x (t) = r (t + 2) r (t + 1) r (t 1) + r (t 2) The correct waveform of x ^ t h is *********** Sample Chapter GATE CLOUD Signals & System by Kanodia EXCERCISE 1.3 MCQ 1.3.1 The period of signal x (t) = 14 + 50 cos 60t is (A) sec 30 (C) 1 sec 60 MCQ 1.3.2 (B) 60 sec (D) Not periodic The period of signal x (t) = 10 sin 5t 4 cos 7t is (A) 24 35 (C) 2 MCQ 1.3.3 (B) 4 35 (D) Not periodic The period of signal x (t) = 5t 2 cos 5000 t is (A) 0.96 ms (C) 0.4 ms MCQ 1.3.4 (B) 1.4 ms (D) Not periodic The period of signal x (t) = 4 sin 3t + 3 sin t is (B) 2 sec 3 (D) Not periodic (A) 2 sec 3 (C) 2 sec Statement for Q. 5 & 6 Consider the signal shown below Sample Chapter GATE CLOUD Signals & System by Kanodia Page 32 Continuous Time Signals MCQ 1.3.5 The even part of signal is MCQ 1.3.6 The odd part of signal is MCQ 1.3.7 Consider the function x (t) shown in figure Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 Chapter 1 Continuous Time Signals Page 33 The even part of x (t) is MCQ 1.3.8 The signal x (t) = e 4t u (t) is a (A) power signal with power Px = 1/4 (B) power signal with power Px = 0 (C) energy signal with energy Ex = 1/4 (D) energy signal with energy Ex = 0 4 MCQ 1.3.10 The signal x (t) = e j (2t + ) is a (A) power signal with Px = 1 (B) power signal with Px = 2 (C) energy signal with Ex = 2 MCQ 1.3.9 (D) energy signal with Ex = 1 The raised cosine pulse x (t) is defined as 1 (cos t + 1), # t # x (t) = * 2 0, otherwise The total energy of x (t) is (A) 3 4 (B) 3 8 (C) 3 (D) 3 2 Statement for Q. 11 -14 : Consider the six signals shown in figure below. Sample Chapter GATE CLOUD Signals & System by Kanodia Page 34 MCQ 1.3.11 Continuous Time Signals The signal f 1 (t) can be expressed as (A) x (t 1) + y (t + 1) (C) x (t + 1) + y (t + 1) MCQ 1.3.12 (B) x (t 1) + y (t 1) (D) x (t + 1) + y (t 1) The signal f 2 (t) can be expressed as (A) x (t 1) + y (t + 1) (C) x (t + 1) + y (t + 1) MCQ 1.3.13 (B) x (t 1) + y (t 1) (D) x (t + 1) + y (t 1) The signal f 3 (t) can be expressed as (A) x (t 0.5) + y (t + 0.5) (C) x (t 0.5) + y (t 0.5) MCQ 1.3.14 (B) x (t + 0.5) + y (t + 0.5) (D) x (t + 0.5) + y (t 0.5) The signal f 4 (t) can be expressed as (A) 1.5x (2t 2) (B) 1.5x ( t 1 ) 2 (C) 1.5x (2t 1) t (D) 1.5x ( 2 1) Statement for Q. 15-19 : The signal x (t) is depicted in figure below : Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 Chapter 1 MCQ 1.3.15 Continuous Time Signals Page 35 The trapezoidal pulse y (t) is related to the x (t) as y (t) = x (10t 5). The sketch of y (t) is MCQ 1.3.16 The trapezoidal pulse x (t) is time scaled producing y (t) = x (5t). The sketch for y (t) is MCQ 1.3.17 The trapezoidal pulse x (t) is time scaled producing y (t) = x a t k. The sketch for 5 y (t) is Sample Chapter GATE CLOUD Signals & System by Kanodia Page 36 MCQ 1.3.18 Continuous Time Signals The trapezoidal pulse x (t) is applied to a differentiator, defined by y (t) = The total energy of y (t) is (A) 0 (B) 1 (C) 2 MCQ 1.3.19 (D) 3 The total energy of x (t) is (A) 0 (B) 13 (C) 13/3 MCQ 1.3.20 (D) 26/3 Consider the two signal shown in figure below. The signal y (t) can be represented as (A) 2x b 1 t + 2 l + 2 2 (B) 2x (2t 2) 2 (C) 2x ( 2t + 2) + 2 MCQ 1.3.21 Chapter 1 (D) 2x b 1 t + 4 l + 2 2 The numerical value of integral # 8 1 [ (t + 3) 2 (4t)] dt is (A) 1 2 (B) 1 2 (C) 2 (D) 2 Sample Chapter GATE CLOUD Signals & System by Kanodia dx (t) . dt Chapter 1 Continuous Time Signals Page 37 MCQ 1.3.22 The graph of function x (t) = 2 (2t) + 6 (3 (t 2)) is MCQ 1.3.23 The function # 3 3 x ( ) [ ( 2) + ( + 2)] d is equal to x (2) + x ( 2) 2 (A) x (2) + x ( 2) (C) 2x (2) + 2x ( 2) MCQ 1.3.24 (B) (D) None of these The value of the function # 3 3 (at b) sin2 (t 4) dt where a > 0 , is (A) 1 (C) 0 MCQ 1.3.25 sin2 ^ a 4h b (B) b (D) b sin2 ^ a 4h a Consider the function x (t) = u bt + 1 l ramp b 1 t l. The graph of x (t) is 2 2 Sample Chapter GATE CLOUD Signals & System by Kanodia Page 38 MCQ 1.3.26 MCQ 1.3.27 Continuous Time Signals Consider the signal x (t) = rect (t) tri (t). The graph of x (t) is A signal is defined as x (t) = 4tri (t). The value of x ( 1 ) is 2 (A) 1/2 (B) 1 (C) 2 MCQ 1.3.28 (D) 0 Consider the signal x (t) = 3tri b 2t l + 3rect a t k. The graph of x (t) is 3 3 Statement for Q. 29 - 30 : Let the CT unit impulse function be defined by (x) = lim b 1 l tri a x k, a > 0 "0 The function (x) has an area of one regardless the value of Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 Chapter 1 MCQ 1.3.29 Continuous Time Signals What is the area of the function (4x) ? (A) 1 (B) 1 4 (C) 4 MCQ 1.3.30 (D) 2 What is the area of the function ( 6x) ? (A) 1 (B) 1/6 (C) 4 MCQ 1.3.31 Page 39 (D) 2 t 3 A signal x (t) is defined as x (t) = 2tri [2 (t 1)] + 6rect ( 4 ). The value of x ( 2 ) is (A) 4 (B) 5 (C) 6 (D) 7 MCQ 1.3.32 A function is defined as x (t) = 1 + sgn (4 t). The graph of x (t) is MCQ 1.3.33 Consider the voltage waveform shown below The equation for v (t) is (A) u (t 1) + u (t 2) + u (t 3) (B) u (t 1) + 2u (t 2) + 3u (t 3) (C) u (t 1) + u (t 2) + u (t 2) (D) u (t 1) + u (t 2) + u (t 3) 3u (t 4) Sample Chapter GATE CLOUD Signals & System by Kanodia Page 40 MCQ 1.3.34 Continuous Time Signals Chapter 1 Consider the following function for the rectangular voltage pulse shown below (1) v (t) = u (a t) # u (t b) (2) v (t) = u (b t) # u (t a) (3) v (t) = u (t a) u (t b) The function that describe the pulse are (A) 1 and 2 (B) 2 and 3 (C) 1 and 3 MCQ 1.3.35 MCQ 1.3.36 (D) all A signal is described by x (t) = r (t 4) r (t 6), where r (t) is a ramp function starting at t = 0 . The signal x (t) is represented as For the waveform shown in figure the equation is (A) 3tu (t) + 1.5 (t 2) u (t 1) + 1.5 (t 3) u (t 3) (B) 3 (2 t) u (t) + 1.5 (t 2) u (t 1) + 1.5 (t 3) u (t 3) Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 Continuous Time Signals Page 41 (C) 3 (1 t) u (t) + 1.5tu (t 1) + 1.5 (t 2) u (t 3) (D) None of these MCQ 1.3.37 For the signal x (t) = u (t + 1) 2u (t 1) + u (t 3), the correct wave form is MCQ 1.3.38 For the signal x (t) = u (t) + u (t + 1) 2u (t + 2), the correct waveform is MCQ 1.3.39 For the signal x (t) = 2 (t 1) u (t 1) 2 (t 2) u (t 2) + 2 (t 3) u (t 3) the correct waveform is Sample Chapter GATE CLOUD Signals & System by Kanodia Page 42 Continuous Time Signals Chapter 1 MCQ 1.3.40 For the signal x (t) = (t + 1) u (t 1) tu (t) u (t 2) the correct waveform is MCQ 1.3.41 Consider the two signal shown in figure The signal y (t) can be explained as (A) x ( 1 t 1) + x ( 2 t 5 ) + x (t 3) + x (2t 7) 2 3 3 3 (B) x (2t + 1) + x ( 2 t + 5 ) + x (t + 3) + x (2t + 7) 3 (C) x ( 1 t + 1) + x ( 2 t + 5 ) + (t + 3) + x (2t + 7) 2 3 3 3 (D) x (2t 1) + x ( 2 t 5 ) + x (t 3) + x (2t 7) 3 Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 Continuous Time Signals Page 43 Statement for Q. 42-43 : Consider the triangular pulses and the triangular wave of figure MCQ 1.3.42 The mathematical function for x 1 (t) is (A) 2tu (t) 4 (t + 1) u (t 1) + 2 (t + 2) u (t 2) (B) 2tu (t) 4 (t 1) u (t 1) + 2 (t 2) u (t 2) (C) 2tu (t) 4 (t 1) u (t + 1) + 2 (t 2) u (t + 2) (D) None of the above MCQ 1.3.43 The mathematical function for waveform x (t) is 3 (A) 3 / x (t + 2k) (B) / x (t 2k) (D) 1 k=0 3 (C) / x (t 2k) 1 k = 3 3 1 k=0 / x (t + 2k) 1 k = 3 Here, T0 = 2 , therefore x (t) = 3 / x (t 2k) 1 k = 3 *********** Sample Chapter GATE CLOUD Signals & System by Kanodia EXCERCISE 1.4 MCQ 1.4.1 IES EC 2009 A function of one or more variable which conveys information on the nature of physical phenomenon is called (A) Noise (B) Interference (C) System MCQ 1.4.2 GATE IN 2006 (D) Signal The Fourier series for a periodic signal is given as x (t) = cos (1.2 t) + cos (2 t) + cos (2.8 t) The fundamental frequency of the signal is (A) 0.2 Hz (B) 0.6 Hz (C) 1.0 Hz MCQ 1.4.3 GATE IN 2007 (D) 1.4 Hz Consider the periodic signal x (t) = (1 + 0.5 cos 40 t) cos 200 t , where t is in seconds. Its fundamental frequency, in Hz, is (A) 20 (B) 40 (C) 100 MCQ 1.4.4 GATE IN 2009 (D) 200 The fundamental period of x (t) = 2 sin 2 t + 3 sin 3 t , with t expressed in seconds, is (A) 1 s (B) 0.67 s (C) 2 s MCQ 1.4.5 IES EC 1999 (D) 3 s The period of the function cos [ /4 (t 1)] is (A) 1/8 s (B) 8 s (C) 4 s MCQ 1.4.6 IES EC 2001 (D) 1/4 s If x1 (t) = 2 sin t + cos 4 t and x2 (t) = sin 5 t + 3 sin 13 t , then (A) x1 and x2 both are periodic (B) x1 and x2 both are not periodic (C) x1 is periodic, but x2 is not periodic (D) x1 is not periodic, but x2 is periodic MCQ 1.4.7 IES EC 2003 The sum of two or more arbitrary sinusoids is (A) Always periodic (B) Periodic under certain conditions Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 Continuous Time Signals Page 45 (C) Never periodic (D) Periodic only if all the sinusoids are identical in frequency and phase MCQ 1.4.8 IES EC 2004 Which one of the following must be satisfied if a signal is to be periodic for 3 < t < 3 ? (B) x (t + T0) = dx (t) /dt (A) x (t + T0) = x (t) (C) x (t + T0) = MCQ 1.4.9 IES EC 2007 T0 #t x (t) dt (D) x (t + T0) = x (t) + kT0 Consider two signals x1 (t) = e j 20t and x2 (t) = e( 2 + j) t . Which one of the following statements is correct ? (A) Both x1 (t) and x2 (t) are periodic (B) x1 (t) is periodic but x2 (t) is not periodic (C) x2 (t) is periodic but x1 (t) is not periodic (D) Neither x1 (t) nor x2 (t) is periodic MCQ 1.4.10 IES EC 2008 Which one of the following function is a periodic one ? (A) sin (10 t) + sin (20 t) (B) sin (10t) + sin (20 t) (C) sin (10 t) + sin (20t) MCQ 1.4.11 GATE EE 2010 The period of the signal x (t) = 8 sin `0.8 t + j is 4 (A) 0.4 s (B) 0.8 s (C) 1.25 s MCQ 1.4.12 IES EC 2009 (D) 2.5 s A signal x1 (t) and x2 (t) constitute the real and imaginary parts respectively of a complex valued signal x (t). What form of waveform does x (t) possess ? (A) Real symmetric (B) Complex symmetric (C) Asymmetric MCQ 1.4.13 IES EC 1991 (D) sin (10t) + sin (25 t) (D) Conjugate symmetric If from the function f (t) one forms the function, (t) = f (t) + f ( t), then (t) is (A) even (B) odd (C) neither even nor odd MCQ 1.4.14 IES EC 2001 (D) both even and odd The signal x (t) = A cos ( t + ) is (A) an energy signal (B) a power signal (C) an energy as well as a power signal (D) neither an energy nor a power signal MCQ 1.4.15 IES EC 2007 Which one of the following is the mathematical representation for the average power of the signal x (t) ? T T (A) 1 # x (t) dt (B) 1 # x2 (t) dt T0 T0 (C) 1 T T/2 # T/2 x (t) dt (D) lim 1 T"3T Sample Chapter GATE CLOUD Signals & System by Kanodia T/2 # T/2 x 2 (t) dt Page 46 MCQ 1.4.16 IES EC 2007 Continuous Time Signals Which one of the following is correct ? Energy of a power signal is (A) finite GATE EC 2005 (B) zero (C) infinite MCQ 1.4.17 Chapter 1 (D) between 1 and 2 The power in the signal s (t) = 8 cos (20 ) + 4 sin (15 t) is 2 (A) 40 (B) 41 (C) 42 (D) 82 MCQ 1.4.18 Which of the following is true ? GATE EE 2006 (A) A finite signal is always bounded (B) A bounded signal always possesses finite energy (C) A bounded signal is always zero outside the interval [ t0, t0] for some t0 (D) A bounded signal is always finite MCQ 1.4.19 If a signal f (t) has energy E , the energy of the signal f (2t) is equal to GATE EC 2001 (A) 1 (B) E/2 (C) 2E (D) 4E MCQ 1.4.20 If a function f (t) u (t) is shifted to right side by t 0 , then the function can be expressed IES EC 2001 as (A) f (t t 0) u (t) (C) f (t t 0) u (t t 0) MCQ 1.4.21 (B) f (t) u (t t 0) (D) f (t + t 0) u (t + t 0) If a plot of signal x (t) is as shown in the figure IES EC 1999 then the plot of the signal x (1 t) will be Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 Continuous Time Signals Page 47 MCQ 1.4.22 A signal v [n] is defined by IES EC 2005 Z1 ; n=1 ] ] v [n] = [ 1 ; n = 1 ]0 ; n = 0 and n > 1 \ Which is the value of the composite signal defined as v [n] + v [ n] ? (A) 0 for all integer values of n (B) 2 for all integer values of n (C) 1 for all integer values of n (D) 1 for all integer values of n MCQ 1.4.23 Which one of the following relations is not correct ? IES EC 2011 (A) f (t) (t) = f (0) (t) (C) MCQ 1.4.24 GATE EC 2006 # 3 3 ( ) d = 1 (B) # f (t) ( ) d = 1 3 3 (D) f (t) (t ) = f ( ) (t ) The Dirac delta function (t) is defined as 3 1 t=0 (A) (t) = ) (B) (t) = ) 0 otherwise 0 Sample Chapter GATE CLOUD Signals & System by Kanodia t=0 otherwise Page 48 Continuous Time Signals 1 t=0 (C) (t) = ) and 0 otherwise MCQ 1.4.25 # (t) dt = 1 3 t=0 (D) (t) = ) and 0 otherwise Chapter 1 # (t) dt = 1 3 3 3 3 Let (t) denote the delta function. The value of the integral GATE EC 2001 (A) 1 MCQ 1.4.26 (D) The Integral GATE IN 2010 2 at k 6 sin (t) dt evaluates to 6 3 # 3 (A) 6 (B) 3 (C) 1.5 MCQ 1.4.27 3 (B) 1 (C) 0 (t) cos b 3t l dt is 2 3 # (D) 0 The integral GATE IN 2011 (A) # 3 2 t 2 /2 3 1 e 1/8 8 2 (C) 1 2 1 e 1/2 2 te (1 2t) dt is equal to (B) 1 e 1/8 4 2 (D) 1 MCQ 1.4.28 Double integration of a unit step function would lead to IES EC 1995 (A) an impulse (B) a parabola (C) a ramp (D) a doublet MCQ 1.4.29 The function x (t) is shown in the figure. Even and odd parts of a unit step function GATE EC 2005 u (t) are respectively, (A) 1 , 1 x (t) 22 (B) 1 , 1 x (t) 22 (C) 1 , 1 x (t) 2 2 (D) 1 , 1 x (t) 2 2 Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 MCQ 1.4.30 Continuous Time Signals Page 49 The expression for the wave form in terms of step function is given by IES EC 1991 (A) v (t) = u (t 1) u (t 2) + u (t 3) (B) v (t) = u (t 1) + u (t 2) + u (t 3) (C) v (t) = u (t 1) + u (t 2) u (t 3) (D) v (t) = u (t 1) + u (t 2) + u (t 3) 3u (t 4) MCQ 1.4.31 IES EC 1991 The impulse train shown in the figure represents the second derivative of a function f (t). The value of f (t) is (A) tu (t 1) tu (t 2) + tu (t 3) + tu (t 4) tu (t 5) + 2tu (t 6) tu (t 7) (B) tu (t 1) tu (t 2) tu (t 3) tu (t 4) + tu (t 5) (C) tu (t 3) + tu (t 4) + 2tu (t 6) (D) tu (t + 1) + tu (t + 2) + tu (t + 3) + tu (t + 4) + tu (t + 5) + 2tu (t + 6) + tu (t + 7) MCQ 1.4.32 IES E C 1997 Match List I with List II and select the correct answer using the codes given below the Lists: Sample Chapter GATE CLOUD Signals & System by Kanodia Page 50 Continuous Time Signals Codes : A (A) 1 (B) 3 (C) 4 (D) 4 MCQ 1.4.33 B 2 4 3 3 C 3 1 2 1 Chapter 1 D 4 2 1 2 Consider the following waveform diagram IES EC 2004 Which one of the following gives the correct description of the waveform shown in the above diagram ? (B) u (t) + (t 1) u (t 1) (A) u (t) + u (t 1) (C) u (t) + u (t 1) + (t 2) u (t 2) (D) u (t) + (t 2) u (t 2) Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 MCQ 1.4.34 GATE EE 1994 Continuous Time Signals Match the waveforms on the left-hand side with the correct mathematical description listed on the right hand side. (A) P-1, Q-3, R-4, S-2 (B) P-3, Q-6, R-4, S-1 (C) P-1, Q-6, R-2, S-4 MCQ 1.4.35 Page 51 (d) P-2, Q-3, R-4, S-1 In the graph shown below, which one of the following express v (t) ? IES EC 2005 (A) (2t + 6) [u (t 3) + 2u (t 4)] (B) ( 2t 6) [u (t 3) + u (t 4)] (C) ( 2t + 6) [u (t 3) + u (t 4)] (D) (2t 6) [u (t 3) u (t 4)] *********** Sample Chapter GATE CLOUD Signals & System by Kanodia SOLUTIONS 1.1 Sample Chapter GATE CLOUD Signals & System by Kanodia SOLUTIONS 1.2 SOL 1.2.1 Option (A) is correct. T1 = 2 = 1 4 2 Period of cos (3 t), T2 = 2 = 2 3 3 T1 = 1/2 = 3 (rational) Ratio, T2 2/3 4 So, the signal x (t) is periodic. Period of x (t), T = LCM (T1, T2) = LCM b 1 , 2 l = 2 sec 23 Alternate Method : T1 = m n T2 Period of sin (4 t), Fundamental period of x (t) T = nT1 = mT2 T1 = 3 = m Here n T2 4 Thus m = 3, n = 4 Period of x (t), or SOL 1.2.2 T = nT1 = 4 # 1 = 2 sec 2 T = mT2 = 3 # 2 = 2 sec 3 Option (D) is correct. T1 = 2 = 1 4 2 Period of cos (10t), T2 = 2 = 10 5 T1 = 1/2 = 5 (not rational) Here T2 /5 2 Since the ratio T1 /T2 is not rational, x (t) is not periodic. Period of sin (4 t), SOL 1.2.3 Option (A) is correct. For x1 (t) : Period of sin (8 t), Period of cos (6 t), T1 = 2 = 8 T2 = 2 = 6 1 4 1 3 Sample Chapter GATE CLOUD Signals & System by Kanodia Page 54 Continuous Time Signals Chapter 1 T1 = 1/4 = 3 (rational) T2 1/3 4 Ratio T1 /T2 is a rational number, therefore x1 (t) is a periodic signal. For x2 (t) : Period of sin (8 t), T1 = 2 = 1 8 4 Period of cos (20t), T2 = 2 = 20 10 T1 = 1/4 = 5 (not rational) Check for periodicity T2 /10 2 Ratio T1 /T2 is not rational, therefore x2 (t) is not periodic. Now SOL 1.2.4 Option (C) is correct. sin A cos B = x (t) = sin ;b 2 l tE cos ;b 4 l tE 5 3 = 1 ;sin b 2 4 l t + sin b 2 + 4 l tE 2 5 3 5 3 = 1 ;sin b 14 l t + sin b 26 l tE 2 15 15 = x1 (t) + x2 (t) 2 Period of x1 (t), T1 = = 15 7 (14 /15) 2 15 Period of x2 (t), T2 = = (26 /15) 13 T1 = 15/7 = 13 = m (rational) n T2 7 15/13 Here m = 13 and n = 7 . Let period of x (t) is T , then T = mT2 = nT1 Thus, T = 13 # 15 = 15 sec 13 or T = 7 # 15 = 15 sec 7 Alternate Method : Period of x (t), T = LCM (T1, T2) T = LCM b 15 , 15 l 7 13 = 15 sec SOL 1.2.5 Option (D) is correct Period of f1 (t), T1 = 2 = 3 unit 2 /3 f2 (t) can be written as f2 (t) = 1 ;sin b 2 4 l t + sin b 2 + 4 l tE 2 5 3 5 3 = 1 ;sin b 14 l t + sin b 26 l tE 2 15 15 Let f2 (t) = f21 (t) + f22 (t) Sample Chapter GATE CLOUD Signals & System by Kanodia 1 [sin (A B) + sin (A + B)] 2 Chapter 1 Continuous Time Signals 2 = 15 7 (14 /15) 2 Period of f22 (t), T22 = = 15 (26 /15) 13 T21 = 15/7 = 13 (rational) Ratio, T22 7 15/13 So, f2 (t) is periodic. Period of f21 (t), T21 = T2 = LCM (T21, T22) = LCM b 15 , 15 l = 15 sec 7 13 Period of f3 (t), T3 = 2 unit 3 f4 (t) = f1 (t) 2f3 (t) Ratio T1 = 3 = 9 (not rational) T3 2 /3 2 Therefore f4 (t) is aperiodic. Codes, P " 2, Q " 1, R " 4, S " 3 Period of f2 (t), SOL 1.2.6 Option (C) is correct Signal sin (10 t) u (t) is not periodic as it is defined for t > 0 only. SOL 1.2.7 Option (B) is correct. Let, g (t) = 2 cos (10t + 1) + sin (4t 1) 1 44 g (t) 44 3 1 44 g (t) 44 3 42 4 2 1 2 Period of g1 (t), T1 = 2 = sec 10 5 Period of g2 (t), T2 = 2 = sec 2 4 T1 = /5 = 2 (rational) Ratio, T2 /2 5 Therefore, g (t) is periodic Period of g (t), SOL 1.2.8 T = LCM (T1, T2) = LCM a , k = sec 52 Option (D) is correct. All the given signals are periodic. Period of x1 (t), T1 = 2 = 2 4 Period of x2 (t), T2 = 2 = 2 Period of x 3 (t), T3 = 2 = 2 4 None of the above signals is aperiodic. SOL 1.2.9 Option (C) is correct. Odd part of g (t), go (t) = 1 [g (t) g ( t)] 2 Sample Chapter GATE CLOUD Signals & System by Kanodia Page 55 Page 56 Continuous Time Signals t, g ( t) = * 0, t, =* 0, Zt/2, ] go (t) = [t/2, ]0, \ So, SOL 1.2.10 0 # t < 1 elsewhere 1 1 t # 0 elsewhere 1 # t < 0 0#t<1 elsewhere Option (B) is correct. t, g ( t) = * 0, 1 # t < 0 elsewhere Even part ge (t) = 1 [g (t) + g ( t)] 2 Z t/2, 1 # t < 0 ] = [t/2, 0#t<1 ]0, elsewhere \ Graphically : SOL 1.2.11 Option (B) is correct. Odd part of x (t), xo (t) = 1 [x (t) x ( t)] 2 This is shown graphically as below : Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 Chapter 1 Continuous Time Signals Page 57 The function xo (t) is unit signum function. SOL 1.2.12 Option (B) is correct. Unit step signal is given as 1, t > 0 x (t) = * 0, t < 0 Odd part is given by xo (t) = 1 [x (t) x ( t)] 2 This is shown graphically as below : SOL 1.2.13 Option (D) is correct. Shift x (t) 3/4 units to the left and 3/4 units to the right and then adding both together, we get g (t) as shown below : The signal g (t) is symmetrical with respect to vertical axis so odd part go (t) = 0 SOL 1.2.14 SOL 1.2.15 Option (D) is correct. For an odd signal xo ( t) = xo (t) xo (t) = xo ( t) xo (0) = xo ( 0) The only number with a = a is a = 0 so xo (0) = 0 For a signal we write x (t) = xe (t) + xo (t) For t = 0 , x (0) = xe (0) + xo (0) = xe (0) + 0 = xe (0) Since xo (0) = 0 Option (B) is correct. For any odd signal xo ( t) = xo (t). Thus the complete odd part is in option (B). Sample Chapter GATE CLOUD Signals & System by Kanodia Page 58 SOL 1.2.16 Continuous Time Signals Chapter 1 Option (D) is correct. For any signal x (t) = xe (t) + xo (t) or xe (t) = x (t) xo (t) Since we have x (t) and xo (t) for t $ 0 only, from above equation we can plot xe ^ t h for t $ 0 as shown below. Even part of any signal is symmetric about vertical axis that is xe ( t) = xe (t). Thus the complete even part is as shown above. SOL 1.2.17 Option (D) is correct. Given signal is shown below : By folding the signal with respect to vertical axis Odd part, xo (t) = 1 [x (t) x ( t)] 2 which is shown below Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 SOL 1.2.18 Continuous Time Signals Page 59 Option (B) is correct. For signal g1 (t) 3 Energy, 2 # E1 = g1 (t) dt = 2 3 Average Power, # 25dt = 100 2 P1 = lim 1 E1 = 0 T T"3 Since g1 (t) has finite energy, it is an energy signal. For signal g2 (t) 3 Energy, # E2 = g2 (t) 2 dt = 3 3 Average power, P2 = 1 8 =1 8 4 # g2 (t) 2 dt 4 2 # 25dt = 1 # 100 = 12.5 8 2 The signal g2 (t) has finite power, so it is a power signal. Alternate Method : We know that most periodic signals are usually power signals and most non-periodic signals are considered to be energy signals. g1 (t) is non-periodic, so it is an energy signal. g2 (t) is periodic so it is a power signal. SOL 1.2.19 Option (B) is correct. 3 Energy, Eg = # 3 g (t) 2 dt = 3 Average Power, SOL 1.2.20 # 25dt = 150 3 Pg = lim 1 Eg = 0 T"3T Option (D) is correct. 3 Energy, Ex = # x (t) 2 dt = 3 3 Sample Chapter GATE CLOUD Signals & System by Kanodia Page 60 Continuous Time Signals Average Power, Px = 1 8 =1 8 Chapter 1 4 # 4 x (t) 2 dt 2 # 25dt = 100 = 12.5 8 2 SOL 1.2.21 Option (D) is correct. The signal is unbounded, therefore it is not an energy signal. SOL 1.2.22 Option (C) is correct. x (t) = 20 cos (5t) cos (10t) V = 10 [cos 15t + cos 5t] = 10 cos 15t + 10 cos 5t (10) 2 (10) 2 Power + = 100 W Px = 2 2 rms value Xrms = 100 = 10 volt SOL 1.2.23 2cosA cos B = cos (A B) + cos (A + B) Option (A) is correct. Here x (t) = e j (2t + /4) = 1 3 Ex = Energy of the signal # 3 3 P = lim 1 T " 32T The power of signal, = lim 1 T " 3 2T # 1 dt = 3 x (t) dt = 2 T T # 3 x (t) 2 dt T # 1 dt = lim 21 (2T) = 1 T T"3 T Since x (t) has finite power and infinite energy, therefore it is a power signal. SOL 1.2.24 Option (B) is correct. T # Px = 1 T Power, 0 =1 7 x (t) 2 dt , 7 # 0 T " Period x (t) 2 dt 2 5 7 Px = 1 > # (0) 2 dt + # (4) 2 dt + # (2) 2 dtH 7 2 0 5 1 0 + (16 3) + (4 2) =6 # #@ 7 = 8 unit SOL 1.2.25 Option (A) is correct. Energy Ex of signal x (t) is given as 3 Ex = # x (t) 2 dt = 12 units 3 Energy of 2x (t), E1 = (2) 2 # Ex = 4 # 12 = 48 unit Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 Continuous Time Signals Let, x2 (t) = x (3t) So, x2 (t) is defined over the range 1 # t # 1 1 Energy E2 = 1 # x2 (t) dt = 2 1 # x (3t) 2 dt 1 Let 3t = $ dt = 1 d 3 So E2 = 1 3 3 # 3 x ( ) 2 d = 1 # Ex = 4 unit 3 Energy of x (t 4) is same as x (t). Energy of 2x (2t) E 4 = (2) 2 # 1 Ex = 24 unit 2 SOL 1.2.26 Option (B) is correct. x (t) = e t , x ( t) = e t = e t = x (t) Since x (t) = x ( t), it is an even signal. Signal x (t) is bounded, so it is has some finite energy. SOL 1.2.27 Option (A) is correct. y (t) is multiplication of x1 (t) and x2 (t). For interval 0 # t # 1, x1 (t) = t, x2 (t) = 1 so, y (t) = x1 (t) x2 (t) = t For 1 # t # 2 , x1 (t) = 1, x2 (t) = 0.5 y (t) = x1 (t) x2 (t) = 0.5 For 2 # t # 3 , x1 (t) = 0.5, x2 (t) = 1.5 y (t) = x1 (t) x2 (t) = 0.75 Sample Chapter GATE CLOUD Signals & System by Kanodia Page 61 Page 62 SOL 1.2.28 Continuous Time Signals Option (C) is correct. Shift g (t) to the right by one time unit to obtain g (t 1) as shown below : For So, For So, For So, For So, SOL 1.2.29 Chapter 1 1 # t # 0, 0 # t # 1, 1 # t # 2, 2#t#3 f (t) x (t) f (t) x (t) f (t) x (t) f (t) = t 1, g (t 1) = 1 = t 1 = t, g (t 1) = 1 = t = 1, g (t 1) = t 2 = t 2 = t + 3, g (t 1) = 1 x (t) = t + 3 Option (D) is correct. Put t = 2 , 2 + 1, g (2 ) = *1, 0, 1 # 2 # 0 0 # 2 < 2 else where Changing the variable( " t ) 1 2t + 1 , 2 # t # 0 0#t<1 g (2t) = *1, 0, else where SOL 1.2.30 Option (C) is correct. The waveform for signal g (t) and g (t/2) are drawn as below. Signal g (t/2) is obtained by expanding the g (t) by a factor of 2 in the time domain. Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 SOL 1.2.31 Continuous Time Signals Page 63 Option (C) is correct. The signal g (t) and its expanded signal by factor of 2 and 3 is shown below : By adding all three, we get f (t) = g (t) + g (t/2) + g (t/3) SOL 1.2.32 Option (B) is correct. 3 ^2t/3h is obtained by expanding (t) with a factor of 3/2 and scaling amplitude by a factor of 3. Similarly, to get 3 ^t/3h, expand (t) by a factor of 3 and amplitude scale by 3 Sample Chapter GATE CLOUD Signals & System by Kanodia Page 64 Continuous Time Signals Chapter 1 Now adding both signal we get SOL 1.2.33 Option (A) is correct. Energy of a signal x (t), Ex = # 3 3 x (t) 2 t Now let the signal is time compressed by a factor of a y (t) = x (at) Energy of y (t) Ey = # 3 3 x (at) 2 dt at = & dt = 1 d a Ey = 1 a # 3 3 x ( ) 2 d = 1 Ex a So due to time compression energy reduces. SOL 1.2.34 Option (B) is correct. To get g (t + 2) shift g (t) to the left by 2 time units. The signal is advanced by 2 time units. SOL 1.2.35 Option (D) is correct. The signal y (t) is the time delayed version of x (t) i.e y (t) = x (t 2) SOL 1.2.36 Option (A) is correct. The delayed version of x (t), y (t) = x (t 3) can be obtained directly by shifting x (t) to the right by 3 sec. SOL 1.2.37 Option (C) is correct. The time delayed signal g (t 2) can be obtained by shifting g (t) to the right by 2 time units. SOL 1.2.38 Option (C) is correct. First time reverse the signal g (t) to get g ( t) and then shift g ( t), toward right to get g ( t + 1) as shown in figure Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 SOL 1.2.39 Continuous Time Signals Page 65 Option (A) is correct. 3 # Ex = We have x (t) 2 dt 3 Energy of signal x (at b), 3 # E2 = x (at b) 2 dt 3 Put at b = $ dt = 1 d a E2 = 1 a So SOL 1.2.40 3 # 3 x ( ) 2 d = 1 Ex a Option (C) is correct. The sequence of transformation is f (t) t " t time revenal f ( t) t " t 4 time shift f (4 t) t " 2t time scaling f (4 2t) This can be performed in following steps Alternate Method : As given in methodology of section 1.4, we can also follow the other sequence of operation which is given as f (t) t " t+4 time shift f (t + 4) t " 2t time scaling f (2t + 4) t " t time revenal Sample Chapter GATE CLOUD Signals & System by Kanodia f ( 2t + 4) Page 66 Continuous Time Signals Chapter 1 SOL 1.2.41 Option (C) is correct. First we obtain time reversal signal f ( t) by taking mirror image of f (t) along the vertical axis. Then by shifting f ( t) to the left by 3 units we get f ( t 3). SOL 1.2.42 Option (C) is correct. We can see that y (2) so 2a + b Similarly y (8/3) 8a+ b So 3 From eq (i) and (ii) a SOL 1.2.43 = x (0) [origin is shifted at 2] =0 = x (2) ...(i) =2 ...(ii) = 3 , b = 6 Option (C) is correct. From the graph we can write x2 (t) = x1 (3t 6) = x1 [3 (t 2)]. So x2 (t), can be obtained by compressing x1 (t) by a factor of 3 and then delaying by 2 time units. Alternate Method : As given in methodology of section 1.4, x2 (t) can be obtained by shifting x1 (t) by 6 time units to the right and then by scaling(compressing) it with a factor of 3. This is not given in any of the four options. SOL 1.2.44 Option (B) is correct. x1 (t) = x [0.5 (t 2)] or x1 (t) = x (0.5t 1) First shift x (t) to right by one unit to get x (t 1). Then, expand x (t 1) by a factor of 2 to get x a t 1k or x (0.5t 1) 2 If we change sequence of transformation by first doing scaling then shifting we get x (t) t " 0.5t time scaling x (0.5t) t " t 1 time shifting x [0.5 (t 1)] ! x [0.5t 1] Hence (B) is correct option. Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 SOL 1.2.45 Continuous Time Signals Page 67 Option (C) is correct. x2 (t) = x ( 0.5t 1) First shift x (t) to the right by 1 unit, we get x (t 1). Then, expand x (t 1) by a factor of 2 to get x (t/2 1) Now fold signal x (0.5t 1) about the vertical axis to get x ( 0.5t 1) If we change the order of transformation we get x (t ) t " 0.5t Time scaling x (0.5t) t " t 1 Time shifting x [0.5 (t 1)] t " t Time reversal x [ 0.5t 0.5] ! x [ 0.5t 1] Time scaling and time reversal are commutative, so we may change their order. SOL 1.2.46 Option (B) is correct. In multiple transformation, we first do shifting then time scaling. From y (t), we can see that x (t) is shifted to right by 4 time units to get x (t 4). Then it is time expanded by a factor of 2 to get x (2t 4) SOL 1.2.47 Option (C) is correct. First fold x (t), with respect to vertical axis. Then shift x ( t) toward right by 2 time units, to get x ( t + 2) Sample Chapter GATE CLOUD Signals & System by Kanodia Page 68 SOL 1.2.48 SOL 1.2.49 Continuous Time Signals Chapter 1 Option (C) is correct. From the graphs, we can see that signal has no time shift (because origin is not shifted), so t 0 = 0 . Signal x (t) is magnitude scaled by a factor of 2 . Since, y (t) has half duration of x (t), so it is time compressed by a factor of 2. W =1 2 y (t) = 2x c t m = 2x (2t) 1 2 Option (B) is correct. The sequence of transformation x (t) t " t/2 time scaling xa t k 2 t " t 1 time shifting xb t 1 l 2 6 amplitude scaling 6x b t 1 l 2 If we change the order of transformation. t " t/2 t " t 1 x (t 1) x a t 1k ! x b t 1 l x (t) 2 2 Graphically SOL 1.2.50 Option (C) is correct. We can perform following sequence of transformation. t " 2t t " t t " t 1 t x (1 2 ) time compression x (1 t) folding x (t + 1) time shifting Sample Chapter GATE CLOUD Signals & System by Kanodia x (t) Chapter 1 Continuous Time Signals Page 69 Graphically it is obtained as SOL 1.2.51 Option (C) is correct. 3 x (t) = #e 3 t (2t 2) dt = 3 3 #e 3 = # e 1 (t 1) dt = 1 2 2 t 3 = 1 e t 2 =1 2e t [2 (t 1)] dt [2 (t 1)] = 1 (t 1) 2 3 #e t (t 1) dt 3 3 # f (t) (t t ) = f (t ) 0 at t = 1 0 3 SOL 1.2.52 Option (C) is correct. From the scaling property of impulse function we can see that [a (t t 0)] = 1 (t t 0) a SOL 1.2.53 Option (C) is correct. g (t) = 6 (3t + 9) = 6 [3 (t + 3)] [a (t + b)] = 1 (t + b) = 6 (t + 3) a 3 = 2 (t + 3) So, g (t) is an impulse with magnitude of 2 unit at t = 3 . SOL 1.2.54 Option (B) is correct. Here we can apply the shifting property of impulse function as below 3 # f (t) (t t ) dt = f (t ) 0 0 3 Sample Chapter GATE CLOUD Signals & System by Kanodia Page 70 Continuous Time Signals Chapter 1 3 Thus x (t) = # (t + 5) cos ( t) dt 3 = cos ( t) t = 5 = cos ( 5 ) = 1 SOL 1.2.55 Option (C) is correct. First, fold the signal about t = 0 to get r ( t) and then shift r ( t) toward right to get r ( t + 2) as shown below SOL 1.2.56 Option (B) is correct. The signal x1 (t) is shown below 3 # E1 = 1 x (t) dt = 2 3 # 1dt = 1 unit 0 The signal x2 (t) is shown below 3 # E2 = x2 (t) 2 dt 3 2 = # t2 dt + 0 # E3 = x 3 (t) 2 dt = 3 3 # (1 + e = # 4dt = 3 x2 is unbounded 2 3 3 Energy of x 3 (t) 3 # (1 + e 6t 2 ) dt 0 12t + 2e 6t) dt = 3 ( x 3 is unbounded) 0 So, only x1 (t) has finite energy. SOL 1.2.57 Option (B) is correct. x (t) = u (t + 2) 2u (t) + u (t 2) To draw x (t), we observe change in amplitude at different instants. 1. First at t = 2 , x (t) steps up with amplitude 1. 2. At t = 0 , another step is added with amplitude of 2 . So, the net amplitude Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 Continuous Time Signals Page 71 becomes [1 + ( 2)] = 1. 3. SOL 1.2.58 Similarly at t = 2 , a step with amplitude 1 is added which causes net amplitude ( 1 + 1) = 0 . Option (C) is correct. To sketch x (t), we observe change in amplitude of step signals at different instants of time. 1. At t = 3 , a step with magnitude 1 is added. 2. 3. At t = 1, a step of magnitude 2 is added so net magnitude becomes (1 2) = 1. 4. SOL 1.2.59 At t = 1, another step of magnitude + 2 is added which causes net magnitude (2 1) = 1. At t = 3 , a step with magnitude 1 is added, Now magnitude is ( 1 + 1) = 0 . Option (B) is correct. x (t) = r (t + 2) r (t + 1) r (t 1) + r (t 2) To sketch x (t), we observe change in slope at different instants of time. 1. At t = 2 , a ramp with slope of 1 is added. 2. At t = 1, a ramp with slope of 1 is added, so net slope becomes ( 1 + 1) = 0 3. Similarly, at t = 1, a ramp of slope 1 is added with causes net slope ( 1 + 0) = 1 4. Again, at t = 2 a ramp of slope 1 is added and the net slope becomes zero. The correct sketch is ************ Sample Chapter GATE CLOUD Signals & System by Kanodia SOLUTIONS 1.3 SOL 1.3.1 Option (A) is correct. T = 2 = 2 = sec 60 30 Period of x (t), SOL 1.3.2 Option (C) is correct. Period of sin 5t , T1 = 2 5 Period of cos 7t , T2 = 2 7 Period of x (t), T = LCM b 2 , 2 l = 2 5 7 SOL 1.3.3 Option (D) is correct. Signal x (t) is not periodic because of the term 5t which is aperiodic in nature. SOL 1.3.4 Option (D) is correct. Not periodic because least common multiple of periods of sin 3t and sin t is infinite. SOL 1.3.5 Option (A) is correct. Even part of x (t), xe (t) = 1 [x (t) + x ( t)] 2 This can be obtained graphically in following steps : Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 SOL 1.3.6 Continuous Time Signals Option (C) is correct. xe (t) = 1 [x (t) + x ( t)] 2 Odd part of x (t), This can be obtained graphically in following steps : SOL 1.3.7 Option (A) is correct. Even part of x (t), xe (t) = 1 [x (t) + x ( t)] 2 Signal xe (t) is obtained as follows : Sample Chapter GATE CLOUD Signals & System by Kanodia Page 73 Page 74 SOL 1.3.8 Continuous Time Signals Option (C) is correct. This is energy signal because E3 = SOL 1.3.9 Chapter 1 # 3 3 x (t) dt < 3 = # 3 4t 3 e u (t) dt = # 0 3 4t e dt = 1 4 Option (A) is correct. Ex = # 3 = Energy of signal x (t), # 3 3 x (t) 2 dt (1) dt = 3 Since x (t) = 1 3 Energy of x (t) is infinite, therefore this is a power signal not an energy signal. T 2 Power of x (t), Px = lim 1 # x (t) dt = 1 T " 3 2T T SOL 1.3.10 Option (A) is correct. # 3 =2 4 Energy of signal x (t), # Ex = 3 =1 2 x (t) 2 dt = / 0 # 0 / # / / 1 (cos t + 1) 2 dt 4 (cos2 t + 2 cos t + 1) dt 1 1 b 2 cos 2 t + 2 + 2 cos t + 1l dt = 1 b 3 la k = 3 22 4 SOL 1.3.11 Option (B) is correct. First we shift x (t) and y (t) to the right by 1 unit, to get x (t 1) and y (t 1) respectively. Now by adding x (t 1) and y (t 1), we get f 1 (t) as shown below SOL 1.3.12 Option (A) is correct. First we shift x (t) to the right by 1 unit to get x (t 1) and y (t) to the left by 1 unit to get y (t + 1). Now, adding x (t 1) and y (t + 1) we will get f 2 (t) as shown below Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 Continuous Time Signals Page 75 SOL 1.3.13 Option (A) is correct. First we shift x (t) to the right by 0.5 unit, and y (t) to the left by 0.5 unit to get x (t 0.5) and y (t + 0.5) respectively. Now, adding x (t 0.5) and y (t + 0.5) we will get f 3 (t) as shown below SOL 1.3.14 Option (D) is correct. f4 (t) can be obtained by performing multiple operation on x (t). First delay x (t) by 1 unit, we get x (t 1). Now, time expand x (t 1) by a factor of 2, we get x ^t/2 1h or x (0.5t 1). In last step, f4 (t) can is obtained by multiplying x (0.5t 1) with a constant 1.5. Graphically, these steps are performed as shown below : SOL 1.3.15 Option (C) is correct. y (t) = x (10t 5) The sequence of transformation is t " t 5 t " 10t x (t) time shift x (t 4) time scaling x (10t 5) This can be performed in following steps SOL 1.3.16 Option (D) is correct. Multiplication of independent variable t by 5 will bring compression on time scale. It may be checked by x (5 # 0.8) = x (4). Sample Chapter GATE CLOUD Signals & System by Kanodia Page 76 SOL 1.3.17 SOL 1.3.18 Continuous Time Signals Option (A) is correct. Division of independent variable t by 5 will bring expansion on time scale. It may be checked by y (20) = x b 20 l = x (4) 5 Option (C) is correct. Mathematically, the function x (t) can be defined as Zt + 5, for 5 < t < 4 ] x (t) = [ t + 5, for 4<t<5 ]1, for 4 < t < 4 \ Z1, for 5 < t < 4 dx (t) ] y (t) = = [ 1, for 4<t<5 dt ]0, for 4 < t < 4 \ Energy of y (t) is calculated as Ey = SOL 1.3.19 Chapter 1 # 3 3 y (t) 2 dt = # 4 5 (1) 2 dt + # 4 5 ( 1) 2 dt = 2 Option (D) is correct. E= # 3 3 5 x (t) 2 dt = 2 # x 2 (t) dt 4 0 5 = 2 # (1) 1 dt + 2 # (5 t) 2 dt = 8 + 2 = 36 3 3 0 4 SOL 1.3.20 Option (C) is correct. The transformation of x (t) to y (t) is shown as below Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 SOL 1.3.21 Continuous Time Signals Page 77 Option (A) is correct. For an impulse function we have # 3 (t a) dt = 1, for t = a otherwise 0. 3 so, # 8 1 [ (t + 3) 2 (4t)] t = # 8 1 8 (t + 3) dt 2 # (4t) dt 1 Sample Chapter GATE CLOUD Signals & System by Kanodia Page 78 Continuous Time Signals 8 = 0 2 # (4t) 1 = 2 4 # 8 # 3 (t a) dt = 1, for t = a 3 8 since (at) = 1 (t) a (t) = 1 2 1 (t + 3) dt = 0 because t = 3 does not exist in the given interval ( 1 < t < 8 ). 1 SOL 1.3.22 # Chapter 1 Option (C) is correct. x (t) = 2 (2t) + 6 [3 (t 2)] = 2 (t) + 6 (t 2) 2 3 since a (t t 0) = 1 (t t 0) a = (t) + 2 (t 2) SOL 1.3.23 Option (A) is correct. From the shifting property of impulse function, we know that # 3 x (t) (t t 0) dt = x (t 0) 3 y ( ) = # 3 = So, # 3 3 3 x ( ) [ ( 2) + ( + 2)] d x ( ) [ ( 2)] d + # 3 x ( ) [ ( + 2)] d 3 = x (2) + x ( 2) SOL 1.3.24 Option (D) is correct. Substituting at = u & dt = 1 du, we get a # 3 3 (at b) sin2 (t 4) dt = # 3 3 =1 a # (u b) sin2 a u 4 kdu a a (u b) sin2 a u 4 k du a 3 3 sin2 b b 4 l a = a SOL 1.3.25 Option (C) is correct. x (t) is obtained in following steps : Sample Chapter GATE CLOUD Signals & System by Kanodia since # 3 3 x (t) (t t 0) dt = x (t 0) Chapter 1 SOL 1.3.26 Continuous Time Signals Option (B) is correct. All signal are as shown below SOL 1.3.27 Option (C) is correct. x (t) = 4 tri (t) = 4 (1 t ) 0 <| t |< 1 x b 1 l = 4 c1 1 m = 2 2 2 Sample Chapter GATE CLOUD Signals & System by Kanodia Page 79 Page 80 Continuous Time Signals Chapter 1 SOL 1.3.28 Option (D) is correct. Figure is as shown below SOL 1.3.29 Option (B) is correct. 1 x This is triangle with the same height as ( ) tri ( ), but 1/4 times the base width. Therefore, its area is 1/4 times as that of area of (x) or 1/4. SOL 1.3.30 Option (B) is correct. This is a triangle with the same height as (x) but 1/6 times the base width. The fact that the factor is 6 instead of 6, just, means that the triangle is reversed in time which does not change its shape or area. Thus its area is 1/6 times as that of (x) or 1/6. The area of function (bx) = lim 1 tri b bx l, a > 0 is 1 a a"0 a b SOL 1.3.31 Option (C) is correct. x (t) = 2tri [2 (t 1)] + 6rect a t k 4 x b 3 l = 2tri ;2 b 3 1lE + 6rect b 3 l 2 2 8 = 2tri (1) + 6rect b 3 l = 2 [1 (1)] + 6 = 6 8 SOL 1.3.32 Option (A) is correct. The figure is as shown below : Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 SOL 1.3.33 Continuous Time Signals Page 81 Option (D) is correct. v (t) is sum of 3 unit step signal starting from 1, 2, and 3, all signal ends at 4. SOL 1.3.34 Option (B) is correct. Unit step function u (t) and its folded version u ( t) are shown in the figures below Now, by shifting u ( t) to the right by a units and b units, we get u (a t) and u (b t) respectively. Sample Chapter GATE CLOUD Signals & System by Kanodia Page 82 Continuous Time Signals Chapter 1 Similarly, by shifting u (t) to the the right by a units and b units, we get u (t a) and u (t b). From the above graphs, we can see that v (t) = u (t a) u (t b) and, v (t) = u (b t) # u (t a) SOL 1.3.35 Option (B) is correct. The ramp function is shown as Signal r (t 4) and r (t 6) are obtained by shifting r (t) towards right by 4 units and 6 units respectively. Now we subtract r (t 6) from r (t 4) to get x (t). x (t) = r (t 4) r (t 6) Alternate Method : t 4, We have r (t 4) = * 0, t>4 t<4 t 6, r (t 6) = ) 0, t>6 t<6 and Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 SOL 1.3.36 Continuous Time Signals Page 83 Zt 4, 4<t<6 ] Now r (t 4) r (t 6) = [t 4 t + 6, t>6 ]0, t<4 \ Zt 4, 4<t<6 ] = [2, t>6 ]0, t<4 \ Option (C) is correct. To obtain the expression for x (t), we note the change in amplitude and slope at different instants of time and write expression for each change. The steps are as follows : 1. At t = 0 , the function steps from 0 to 3, for a change in amplitude of 3. Also the slope of function changes from 0 to 3 , for a change in slope of 3 ; so we write 2. 3. x1 (t) = (3 0) u (t 0) + ( 3 0) (t 0) u (t 0) = 3u (t) 3tu (t) = 3 (1 t) u (t) At t = 1, the function steps from 0 to 1.5, for a change in amplitude of 1.5. Also the slope of function changes from 3 to 1.5 , for a change in slope of 1.5; so we write x2 (t) = 1.5u (t 1) + 1.5 (t 1) u (t 1) = 1.5u (t 1) + 1.5tu (t 1) 1.5u (t 1) = 1.5tu (t 1) At t = 3 , the function steps up from 1.5 to 0, for a change in amplitude of 1.5. Also the slope of function changes from 1.5 to 0, for a change in slope of 1.5; so we write x3 (t) = 1.5u (t 3) + 1.5 (t 3) u (t 3) = 1.5u (t 3) + 1.5tu (t 3) 4.5u (t 3) = 1.5tu (t 3) 3u (t 3) = 1.5 (t 2) u (t 3) Hence the equation for x (t) is x (t) = x1 (t) + x2 (t) + x3 (t) = 3 (1 t) u (t) + 1.5tu (t 1) + 1.5 (t 2) u (t 3) SOL 1.3.37 Option (A) is correct. To obtain the waveform for x (t), we observe change in magnitude of unit step signals at different instants of time. 1. At t = 1, a step with magnitude 1 is added, so magnitude at t = 1 is 1. 2. At t = 1, another step of magnitude 2 is added, so net amplitude becomes (1 2) = 1 3. At t = 3 , a step of magnitude 1 is added which causes net magnitude ( 1 + 1) = 0 Sample Chapter GATE CLOUD Signals & System by Kanodia Page 84 Continuous Time Signals Chapter 1 Alternate Method : From the expression we get For 1 < t < 1, x (t) = 1 For 1 < t < 3 , x (t) = 1 For t > 3 , x (t) = 0 SOL 1.3.38 Option (D) is correct. Rearranging the given expression x (t) = 2u (t + 2) + u (t + 1) + u (t) The sketch of x (t) is obtained using following steps : 1. At t = 2 , a step of magnitude 2 is added, so magnitude at t = 2 is 2 2. At t = 1, another step of magnitude 1 is added which causes net magnitude to become ( 2 + 1) = 1 3. At t = 0 , another step of magnitude 1 is added, the net amplitude now becomes ( 1 + 1) = 0 . Alternate Method: For 2 < t < 1, x (t) = 2 For 1 < t < 0 , x (t) = 1 For 0 < t , x (t) = 0 SOL 1.3.39 Option (B) is correct. By observing both the change in amplitude and change in slope, we get x (t) as following : 1. At t = 1, a ramp of slope 2 is added, so the net slope of function becomes (0 + 2) = 2 2. At t = 2 , a ramp of slope 2 is added which causes net slope to becomes (2 2) = 0 3. At t = 3 , another ramp of slope 2 is added, now net slope of function becomes (0 + 2) = 2 Alternate Method : For 1 < t < 2 , x (t) = 2 (t 1) For 2 < t < 3 , x (t) = 2 For 3 < t , x (t) = 2t 2 SOL 1.3.40 Option (D) is correct. Rewriting the x (t) as below x (t) = tu (t) + (t 1) u (t 1) + 2u (t 1) u (t 2) 1. At t = 0 , a ramp of slope 1 is added. 2. At t = 1, another ramp of slope 1 is added, so net slope at this instant becomes ( 1 + 1) = 0 3. At t = 1, a step of amplitude 2 is added, so amplitude of x (t) becomes Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 Continuous Time Signals Page 85 ( 1 + 2) = 1 4. SOL 1.3.41 At t = 2 another step of amplitude 1 is added which causes net amplitude to become (1 1) = 0 Option (A) is correct. We may represent y (t) as the superposition of 4 rectangular pulses as follows y (t) = y1 (t) + y2 (t) + y 3 (t) + y 4 (t) y1 (t), y2 (t), y 3 (t) and y 4 (t) are the time shifted and time scaled version of function x (t) with different factors. In general yi (t) = x (ai t bi) i = 1, 2, 3, 4 y1 (t) = x (a1 t b) For t = 0 , y1 (0) = x (a1 # 0 b1) = x ( 1) & a1 # 0 b1 = 1 b1 = 1 For t = 4 , y1 (4) = x (a1 # 4 b1) = x (1) & a1 # 4 b 1 = 1 4a1 = 1 + b1 & a1 = 1/2 y1 (t) = x b 1 t 1l 2 y2 (t) = x (a2 t b2) For t = 1, y2 (1) = x (a2 # 1 b2) = x ( 1) ...(i) & a2 b2 = 1 For t = 4 , y2 (4) = x (a2 # 4 b2) = x (1) ...(ii) & 4a2 b2 = 1 Solving equation (i) and (ii), we get a = 2/3 and b = 5/3 Thus, y2 (t) = x b 2 t 5 l 3 3 Similarly, we can obtain y 3 (t) and y 4 (t) also y 3 (t) = x (t 3) y 4 (t) = x (2t 7) Accordingly, we may express the staircase signal y (t) in terms of the rectangular pulses x (t) as follows: Sample Chapter GATE CLOUD Signals & System by Kanodia Page 86 SOL 1.3.42 SOL 1.3.43 Continuous Time Signals Chapter 1 y (t) = x b 1 t 1l + x b 2 t 5 l + x (t 3) + x (2t 7) 2 3 3 Option (B) is correct. x1 (t) can be obtained using following methodology 1. At t = 0 , slope changes from 0 to 2, so we write x1l(t) = 2tu (t) 2. At t = 1, slope change from 2 to 2 for a change of 4 in slope; so we write x1m (t) = 4 (t 1) u (t 1) 3. At t = 2 , slope changes from 2 to 0 for a change of 2 in slope; so we write x1 n (t) = 2 (t 2) u (t 2) Thus, x (t) = x1l(t) + x1m (t) + x1 n (t) = 2tu (t) 4 (t 1) u (t 1) + 2 (t 2) u (t 2) Option (B) is correct. The expression for periodic waveform is x (t) = 3 / x (t kT ) 1 0 k = 3 Here, T0 = 2 , therefore x (t) = 3 / x (t 2k) 1 k = 3 *********** Sample Chapter GATE CLOUD Signals & System by Kanodia SOLUTIONS 1.4 SOL 1.4.1 Option (D) is correct. A signal conveys information on the nature of physical phenomenon. SOL 1.4.2 Option (A) is correct. x (t) = cos (1.2 t) + cos (2 t) + cos (2.8 t) Frequency of cos (1.2 t), f1 = 0.6 Hz Frequency of cos (2 t), f2 = 1 Hz Frequency of cos (2.8 t), f3 = 1.4 Hz Fundamental Frequency of x (t) will be greatest common divisor of f1, f = GCD (f1, f2, f3) = 0.2 Hz 2 f1 = 1.2 2 f2 = 1.2 2 f3 = 2.8 t f2, f3 SOL 1.4.3 Option (A) is correct. We have x (t) = cos (200 t) + 0.5 cos (40 t) cos (200 t) = cos (200 t) + 1 cos 240 t + 1 cos (360 t) 4 4 Fundamental frequency of (cos 200 t), f1 = 100 Hz 2 f1 = 200 Fundamental frequency of (cos 240 t), f2 = 120 Hz 2 f2 = 240 Fundamental frequency of (cos 360 t), f3 = 180 Hz 2 f3 = 360 t Fundamental frequency of x (t) is greatest common devisor of f1, f2 and f3 , i. e. f = GCD (f1, f2, f3) = 20 Hz SOL 1.4.4 Option (C) is correct. x (t) = 2 sin (2 t) + 3 sin (3 t) Period of sin (2 t), T1 = 2 = 1 sec 2 Period of sin (3 t), T2 = 2 = 2 sec 3 3 T1 = m = 1 = 3 Ratio n T2 ^2/3h 2 Period of x (t), T = LCM d1, 2 n = 2 3 SOL 1.4.5 Option (B) is correct. We have Period of f (t), f (t) = cos 9 (t 1)C 4 T = 2 = 2 = 8 sec ( /4) Sample Chapter GATE CLOUD Signals & System by Kanodia Page 88 SOL 1.4.6 Continuous Time Signals Chapter 1 Option (A) is correct. x1 (t) = 2 sin t + cos 4 t Period of sin t , T11 = 2 = 2 Period of cos 4 t , T12 = 2 = 1 4 2 T11 = 2 = 4 (rational) T12 (1/2) Since ratio of T11 and T12 is rational, x1 (t) is periodic. x2 (t) = sin 5 t + 3 sin 13 t Period of sin 5 t , T21 = 2 = 2 5 5 Period of sin 13 t , T22 = 2 = 2 13 13 T21 = (2/5) = 13 (rational) 5 T22 (2/13) Since ratio of T21 and T21 is rational, x2 (t) is also periodic. SOL 1.4.7 Option (B) is correct. The sum of two sinusoids is periodic if ratio of their periods is rational. SOL 1.4.8 Option (A) is correct. A signal is said to be periodic if it repeats at regular interval. If x (t) is periodic with period T0 it must satisfies. x (t + T0) = x (t) SOL 1.4.9 Option (B) is correct. We have x1 (t) = e j20t Period of x1 (t), T1 = 2 = 20 10 x2 (t) = e (2 + j) t 2 is not rational, so x (t) is not periodic. 2 (2 + j) Option (A) is correct. (A) x1 (t) = sin (10 t) + sin (20 t) Period of sin (10 t), T11 = 2 = 1 10 5 Period of sin (20 t), T12 = 2 = 1 20 10 11 = 1/5 = 2 (rational) Ratio T12 1/10 Since ration of T11 and T12 is rational, x1 (t) is periodic. Since, SOL 1.4.10 (B) Period of sin (10t), x2 (t) = sin (10t) + sin (20 t) T21 = 2 = 10 5 Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 Continuous Time Signals Page 89 T22 = 2 = 1 20 10 T21 = /5 = 2 Ratio, (not rational) T22 1/10 Since T21 /T22 is not rational, x2 (t) is not periodic. Similarly, we can check for option (C) and (D) also. Both are aperiodic. Period of sin (20 t), SOL 1.4.11 Option (D) is correct. T = 2 = 2 = 2.5 sec 0.8 Period of x (t), SOL 1.4.12 Option (D) is correct. x (t) = x1 (t) + jx2 (t) A complex valued signal always possess conjugate symmetry. SOL 1.4.13 Option (A) is correct. (t) = f (t) + f ( t) ( t) = f ( t) + f (t) Since (t) = ( t) SOL 1.4.14 SOL 1.4.15 SOL 1.4.16 SOL 1.4.17 SOL 1.4.18 Thus (t) is an even function. Option (B) is correct. We have x (t) = A cos ( t + ) We know that most of the periodic signals are power signal. x (t) is also a periodic signal and has finite power. 2 px = A 2 Option (D) is correct. Average power of signal is given by T/2 P = lim 1 # x (t) 2 dt T " 3 T T/2 Note : If x (t) is periodic, then T has finite value and above expression becomes as T/2 T P=1# x (t) 2 dt = 1 # x (t) 2 dt T0 T T/2 Option (C) is correct. Energy of a power signal is infinite while the power of an energy signal is zero. Option (A) is correct. s (t) = 8 cos ` 20 t j + 4 sin 15 t 2 = 8 sin 20 t + 4 sin 15 t Here A1 = 8 and A2 = 4 . Thus power is 2 2 2 2 P = A1 + A2 = 8 + 4 = 40 2 2 2 2 Option (B) is correct. A bounded signal always possesses some finite energy. E= t0 #-t g (t) 2 dt < 3 0 Sample Chapter GATE CLOUD Signals & System by Kanodia Page 90 SOL 1.4.19 Continuous Time Signals Chapter 1 Option (B) is correct. Let E be the energy of f (t) and E1 be the energy of f (2t), then E= and # 3 E1 = # 3 3 3 [f (t)] 2 dt [f (2t)] 2 dt Substituting 2t = p we get E1 = SOL 1.4.20 # 3 3 [f (p)] 2 dp 1 = 2 2 # 3 3 [f (p)] 2 dp = E 2 Option (C) is correct. If a function f (t) is shifted to right side by t 0 units, then the shifted function is expressed as f (t t 0) u (t t 0). Let, f (t) = t + 2 x (t) = f (t 1) u (t 1) If we write, x (t) = f (t) u (t 1) For t = 0 x (0) = f (0) = 2 But, x (0) = 0 (In the graph) So f (t) u (t t 0) is not correct expression for shifted signal. SOL 1.4.21 Option (A) is correct. The plot of given signal x (t) is shown below First reflect the signal about the vertical axis to obtain x ( t). Then shift x ( t) towards right by 1 unit to get x ( t + 1). Both operation is shown below Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1 SOL 1.4.22 Continuous Time Signals Page 91 Option (A) is correct. v [n] and v [ n] is drawn as y [n] = v [n] + v [ n] = 0 , for all n SOL 1.4.23 Option (B) is correct. Product property of impulse function f (t) (t t 0) = f (t 0) (t t 0) For t 0 = 0 , f (t) (t) = f (0) (t) Shifting property of impulse function 3 # 3 f (t) (t t0) dt = f (t 0) Area under Impulse function is unity. 3 # 3 (t) dt =1 For detailed discussion on properties of unit impulse function, refer page 32 of the book GATE GUIDE Signals & Systems by the same authors SOL 1.4.24 Option (D) is correct. Dirac delta function (t) is defined at t = 0 and it has infinite value a t = 0 . The area of dirac delta function is unity. SOL 1.4.25 Option (A) is correct. We know that (t) x (t) = x (0) (t) and Let x (t) = 3 cos ( 2 # (t) = 1 3 3 t), then x (0) = 1 Sample Chapter GATE CLOUD Signals & System by Kanodia Page 92 Continuous Time Signals Now SOL 1.4.26 # (t) x (t) = # x (0) (t) dt 3 3 3 3 = Chapter 1 # (t) dt = 1 3 3 Option (B) is correct. We know that 3 # 3 x (t) (t t0) = x (t0) so 3 # 3 at k6 sin (t) dt 6 Here x (t) = 6 sin t , t 0 = 6 = 6sin (t) t = /6 = 6 sin a k 6 = 6#1 = 3 2 SOL 1.4.27 Option (A) is correct. 1 2 x (t) = 3 # 3 t2 e t /2 (1 2t) dt 2 1 2t = & t = b + 1 l and dt = 1 d 2 2 2 1 +1 3 x (t) = 1 # b + 1 l e 2 b 2 l ( ) b 1 d l 2 2 2 3 2 1 +1 3 3 = 1 # b + 1 l e 2 b 2 l ( ) d # 3 f (t) (t) dt = f (0) 2 2 2 3 + 1 2 e 1 b + 1 l 1 = b2l 22 2 2 =0 1 1 2 e 1 b 1 l = 1 e 1 22 8 = bl 2 2 2 8 2 Let, 2 Now 2 2 2 SOL 1.4.28 Option (B) is correct. t #0 #0 SOL 1.4.29 t u (t) dt = #0 t 2 tu (t) dt = t , 2 Option (A) is correct. x (t) + x ( t) 2 x (t) x ( t) xo (t) = 2 g (t) = u (t) xe (t) = and Here u (t) + u ( t) =1 2 2 u (t) u ( t) x (t) = xo (t) = 2 2 Option (D) is correct. At t = 1, signal steps up from 0 " 1, so v1 (t) = (1 0) u (t 1) = u (t 1) At t = 2 , signal steps up from 1 " 2 , so v2 (t) = (2 1) u (t 2) = u (t 2) At t = 3 , signal steps up from 2 " 3 , so Thus SOL 1.4.30 xe (t) = Sample Chapter GATE CLOUD Signals & System by Kanodia (Parabola) Chapter 1 Continuous Time Signals Page 93 v 3 (t) = (3 2) u (t 3) = u (t 3) At t = 4 , signal steps down from 3 " 0 , so v 4 (t) = (0 3) u (t 4) = 3u (t 4) v (t) = v1 (t) + v2 (t) + v 3 (t) + v 4 (t) = u (t 1) + u (t 2) + u (t 3) 3u (t 4) For detailed discussion please refer to methodology of section 1.6 of the book GATE GUIDE Signals & Systems by same authors. SOL 1.4.31 Option (A) is correct. We know that ramp function is obtained by double differentiation of impulse function. differentiation differentiation (t) r (t) u (t) (Ramp) tu (t) (Step) differentiation u (t) (Impulse) differentiation (t) Given Function is f (t) = (t 1) (t 2) + (t 3) + (t 4) (t 5) + 2 (t 6) (t 7) In-terms of ramp function f (t) = tu (t 1) tu (t 2) + tu (t 3) + tu (t 4) tu (t 5) + 2tu (t 6) tu (t 7) SOL 1.4.32 Option (B) is correct. (A) v (t) = u (t 1) u (t 3) (B) v (t) = lim (t 1) a"0 (C) v (t) = u (t + 1) (D) v (t) = u (t) 2u (t 1) + 2u (t 2) 2u (t 3) + ... (A " 3) (B " 4) (C " 1) (D " 2) SOL 1.4.33 Option (C) is correct. At t = 0 , f (t) step up from 0 " 1, so we write f1 (t) = (1 0) u (t 0) = u (t) At t = 1, f (t) steps up from 1 " 2 , so we write f2 (t) = (2 1) u (t 1) = u (t 1) At t = 2 slope changes from 0 " 1 so we write f3 (t) = (1 0) (t 2) u (t 2) Now, f (t) = f1 (t) + f2 (t) + f3 (t) = u (t) + u (t 1) + (t 2) u (t 2) For detailed discussion please refer to methodology of section 1.6 on page 37, given in the book GATE GUIDE Signals & Systems by the same authors. SOL 1.4.34 Option (B) is correct. f1 (t) = tu (t) " option (3) Sample Chapter GATE CLOUD Signals & System by Kanodia Page 94 Continuous Time Signals f2 (t) = shift f1 (t) by 1 unit = (t 1) u (t 1) " option (6) f3 (t) = tu (t) + u (t) = (t + 1) u (t) " Option (4) f4 (t) = tu (t + 1) " option (1) SOL 1.4.35 Option (D) is correct. At t = 3 slope changes from 0 " 2 , so we write v1 (t) = (2 0) (t 3) u (t 3) = (2t 6) u (t 3) at t = 4 , v (t) becomes zero, so v (t) = (2t 6) [u (t 3) u (t 4)] *********** Sample Chapter GATE CLOUD Signals & System by Kanodia Chapter 1

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