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Consider a cascaded block of an LTI systemThe impulse response of S1. is h1. t = δ t -2 The impulse response of 52 is h2 t = t.e-ru tThe input x(t) is wide sense stationary random process with Rxx 0 = 3 and variance = 1Q. If w(t) is input to system 52, then mean of process Y(t) will bea)Y(t) will not be, wide sense stationary thus mean is not a constant value b)c)4d)2Correct answer is option 'B'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about Consider a cascaded block of an LTI systemThe impulse response of S1. is h1. t = δ t -2 The impulse response of 52 is h2 t = t.e-ru tThe input x(t) is wide sense stationary random process with Rxx 0 = 3 and variance = 1Q. If w(t) is input to system 52, then mean of process Y(t) will bea)Y(t) will not be, wide sense stationary thus mean is not a constant value b)c)4d)2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a cascaded block of an LTI systemThe impulse response of S1. is h1. t = δ t -2 The impulse response of 52 is h2 t = t.e-ru tThe input x(t) is wide sense stationary random process with Rxx 0 = 3 and variance = 1Q. If w(t) is input to system 52, then mean of process Y(t) will bea)Y(t) will not be, wide sense stationary thus mean is not a constant value b)c)4d)2Correct answer is option 'B'. Can you explain this answer?.

Solutions for Consider a cascaded block of an LTI systemThe impulse response of S1. is h1. t = δ t -2 The impulse response of 52 is h2 t = t.e-ru tThe input x(t) is wide sense stationary random process with Rxx 0 = 3 and variance = 1Q. If w(t) is input to system 52, then mean of process Y(t) will bea)Y(t) will not be, wide sense stationary thus mean is not a constant value b)c)4d)2Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electronics and Communication Engineering (ECE). Download more important topics, notes, lectures and mock test series for Electronics and Communication Engineering (ECE) Exam by signing up for free.

Here you can find the meaning of Consider a cascaded block of an LTI systemThe impulse response of S1. is h1. t = δ t -2 The impulse response of 52 is h2 t = t.e-ru tThe input x(t) is wide sense stationary random process with Rxx 0 = 3 and variance = 1Q. If w(t) is input to system 52, then mean of process Y(t) will bea)Y(t) will not be, wide sense stationary thus mean is not a constant value b)c)4d)2Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Consider a cascaded block of an LTI systemThe impulse response of S1. is h1. t = δ t -2 The impulse response of 52 is h2 t = t.e-ru tThe input x(t) is wide sense stationary random process with Rxx 0 = 3 and variance = 1Q. If w(t) is input to system 52, then mean of process Y(t) will bea)Y(t) will not be, wide sense stationary thus mean is not a constant value b)c)4d)2Correct answer is option 'B'. Can you explain this answer?, a detailed solution for Consider a cascaded block of an LTI systemThe impulse response of S1. is h1. t = δ t -2 The impulse response of 52 is h2 t = t.e-ru tThe input x(t) is wide sense stationary random process with Rxx 0 = 3 and variance = 1Q. If w(t) is input to system 52, then mean of process Y(t) will bea)Y(t) will not be, wide sense stationary thus mean is not a constant value b)c)4d)2Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of Consider a cascaded block of an LTI systemThe impulse response of S1. is h1. t = δ t -2 The impulse response of 52 is h2 t = t.e-ru tThe input x(t) is wide sense stationary random process with Rxx 0 = 3 and variance = 1Q. If w(t) is input to system 52, then mean of process Y(t) will bea)Y(t) will not be, wide sense stationary thus mean is not a constant value b)c)4d)2Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Consider a cascaded block of an LTI systemThe impulse response of S1. is h1. t = δ t -2 The impulse response of 52 is h2 t = t.e-ru tThe input x(t) is wide sense stationary random process with Rxx 0 = 3 and variance = 1Q. If w(t) is input to system 52, then mean of process Y(t) will bea)Y(t) will not be, wide sense stationary thus mean is not a constant value b)c)4d)2Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice Electronics and Communication Engineering (ECE) tests.