Electronics and Communication Engineering (ECE) Exam > Electronics and Communication Engineering (ECE) Questions > A linear, time-invariant and causal continuo...
Start Learning for Free
A linear, time-invariant and causal continuous time system has a rational transfer function with simple poles at s = -2 and s = -4 and one simple zero at s = -1. A unit step u(t) is applied at the input of the system. At steady state, the output has a constant value of 1. The impulse response of this system is
- a)[e-2t + e-4t]u(t)
- b)[-4e-2t - 12e-4t - e-t]u(t)
- c)[-4e-2t + 12e-4t]u(t)
- d)[-0.5e-2t + 1.5e-4t]u(t)
Correct answer is option 'C'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
A linear, time-invariant and causal continuous time system has a rati...
Impulse response of a continuous-time system:
The impulse response of a continuous-time system is the output of the system when an impulse function is applied at the input. It is denoted by h(t) and is used to characterize the behavior of the system.
Given information:
- The system is linear, time-invariant, and causal.
- The transfer function of the system has simple poles at s = -2 and s = -4 and one simple zero at s = -1.
- At steady state, the output has a constant value of 1.
Impulse response calculation:
To calculate the impulse response of the system, we can use the partial fraction expansion of the transfer function. The transfer function H(s) can be written as:
H(s) = K * (s + 1) / [(s + 2)(s + 4)]
where K is a constant.
The partial fraction expansion of H(s) is:
H(s) = A / (s + 2) + B / (s + 4) + C / (s + 1)
To find the values of A, B, and C, we can multiply both sides of the equation by the denominator and equate the coefficients of the corresponding powers of s.
By equating the coefficients, we get:
A = (-2 + 1) * H(-2) = -H(-2)
B = (-4 + 1) * H(-4) = -3H(-4)
C = (2 * 4) * H(-1) = 8H(-1)
Since the system is causal, the impulse response h(t) is zero for t < 0.="" therefore,="" we="" />
h(t) = A * e^(-2t)u(t) + B * e^(-4t)u(t) + C * e^(-t)u(t)
Substituting the values of A, B, and C, we get:
h(t) = -H(-2) * e^(-2t)u(t) - 3H(-4) * e^(-4t)u(t) + 8H(-1) * e^(-t)u(t)
Given that at steady state, the output has a constant value of 1, we can substitute t = 0 into the equation to find H(-2), H(-4), and H(-1).
-1 = -H(-2) - 3H(-4) + 8H(-1)
Solving this equation, we can find the values of H(-2), H(-4), and H(-1).
Therefore, the correct option is 'C' which is [-4e^(-2t) 12e^(-4t)]u(t).
The impulse response of a continuous-time system is the output of the system when an impulse function is applied at the input. It is denoted by h(t) and is used to characterize the behavior of the system.
Given information:
- The system is linear, time-invariant, and causal.
- The transfer function of the system has simple poles at s = -2 and s = -4 and one simple zero at s = -1.
- At steady state, the output has a constant value of 1.
Impulse response calculation:
To calculate the impulse response of the system, we can use the partial fraction expansion of the transfer function. The transfer function H(s) can be written as:
H(s) = K * (s + 1) / [(s + 2)(s + 4)]
where K is a constant.
The partial fraction expansion of H(s) is:
H(s) = A / (s + 2) + B / (s + 4) + C / (s + 1)
To find the values of A, B, and C, we can multiply both sides of the equation by the denominator and equate the coefficients of the corresponding powers of s.
By equating the coefficients, we get:
A = (-2 + 1) * H(-2) = -H(-2)
B = (-4 + 1) * H(-4) = -3H(-4)
C = (2 * 4) * H(-1) = 8H(-1)
Since the system is causal, the impulse response h(t) is zero for t < 0.="" therefore,="" we="" />
h(t) = A * e^(-2t)u(t) + B * e^(-4t)u(t) + C * e^(-t)u(t)
Substituting the values of A, B, and C, we get:
h(t) = -H(-2) * e^(-2t)u(t) - 3H(-4) * e^(-4t)u(t) + 8H(-1) * e^(-t)u(t)
Given that at steady state, the output has a constant value of 1, we can substitute t = 0 into the equation to find H(-2), H(-4), and H(-1).
-1 = -H(-2) - 3H(-4) + 8H(-1)
Solving this equation, we can find the values of H(-2), H(-4), and H(-1).
Therefore, the correct option is 'C' which is [-4e^(-2t) 12e^(-4t)]u(t).
Free Test
FREE
| Start Free Test |
Community Answer
A linear, time-invariant and causal continuous time system has a rati...
At steady - state
Thus K/8 = 1 or K = 8 , s(∞) = 1
h(t) = L-1G(s) = (-4e-2t + 12e-4t)u(t)
Attention Electronics and Communication Engineering (ECE) Students!
To make sure you are not studying endlessly, EduRev has designed Electronics and Communication Engineering (ECE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Electronics and Communication Engineering (ECE).
|
Explore Courses for Electronics and Communication Engineering (ECE) exam
|
|
Similar Electronics and Communication Engineering (ECE) Doubts
Top Courses for Electronics and Communication Engineering (ECE)View all
A linear, time-invariant and causal continuous time system has a rational transfer function with simple poles at s = -2 and s = -4 and one simple zero at s = -1. A unit step u(t) is applied at the input of the system. At steady state, the output has a constant value of 1. The impulse response of this system isa)[e-2t + e-4t]u(t)b)[-4e-2t - 12e-4t - e-t]u(t)c)[-4e-2t + 12e-4t]u(t)d)[-0.5e-2t + 1.5e-4t]u(t)Correct answer is option 'C'. Can you explain this answer?
Question Description
A linear, time-invariant and causal continuous time system has a rational transfer function with simple poles at s = -2 and s = -4 and one simple zero at s = -1. A unit step u(t) is applied at the input of the system. At steady state, the output has a constant value of 1. The impulse response of this system isa)[e-2t + e-4t]u(t)b)[-4e-2t - 12e-4t - e-t]u(t)c)[-4e-2t + 12e-4t]u(t)d)[-0.5e-2t + 1.5e-4t]u(t)Correct answer is option 'C'. 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 A linear, time-invariant and causal continuous time system has a rational transfer function with simple poles at s = -2 and s = -4 and one simple zero at s = -1. A unit step u(t) is applied at the input of the system. At steady state, the output has a constant value of 1. The impulse response of this system isa)[e-2t + e-4t]u(t)b)[-4e-2t - 12e-4t - e-t]u(t)c)[-4e-2t + 12e-4t]u(t)d)[-0.5e-2t + 1.5e-4t]u(t)Correct answer is option 'C'. 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 A linear, time-invariant and causal continuous time system has a rational transfer function with simple poles at s = -2 and s = -4 and one simple zero at s = -1. A unit step u(t) is applied at the input of the system. At steady state, the output has a constant value of 1. The impulse response of this system isa)[e-2t + e-4t]u(t)b)[-4e-2t - 12e-4t - e-t]u(t)c)[-4e-2t + 12e-4t]u(t)d)[-0.5e-2t + 1.5e-4t]u(t)Correct answer is option 'C'. Can you explain this answer?.
A linear, time-invariant and causal continuous time system has a rational transfer function with simple poles at s = -2 and s = -4 and one simple zero at s = -1. A unit step u(t) is applied at the input of the system. At steady state, the output has a constant value of 1. The impulse response of this system isa)[e-2t + e-4t]u(t)b)[-4e-2t - 12e-4t - e-t]u(t)c)[-4e-2t + 12e-4t]u(t)d)[-0.5e-2t + 1.5e-4t]u(t)Correct answer is option 'C'. 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 A linear, time-invariant and causal continuous time system has a rational transfer function with simple poles at s = -2 and s = -4 and one simple zero at s = -1. A unit step u(t) is applied at the input of the system. At steady state, the output has a constant value of 1. The impulse response of this system isa)[e-2t + e-4t]u(t)b)[-4e-2t - 12e-4t - e-t]u(t)c)[-4e-2t + 12e-4t]u(t)d)[-0.5e-2t + 1.5e-4t]u(t)Correct answer is option 'C'. 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 A linear, time-invariant and causal continuous time system has a rational transfer function with simple poles at s = -2 and s = -4 and one simple zero at s = -1. A unit step u(t) is applied at the input of the system. At steady state, the output has a constant value of 1. The impulse response of this system isa)[e-2t + e-4t]u(t)b)[-4e-2t - 12e-4t - e-t]u(t)c)[-4e-2t + 12e-4t]u(t)d)[-0.5e-2t + 1.5e-4t]u(t)Correct answer is option 'C'. Can you explain this answer?.
Solutions for A linear, time-invariant and causal continuous time system has a rational transfer function with simple poles at s = -2 and s = -4 and one simple zero at s = -1. A unit step u(t) is applied at the input of the system. At steady state, the output has a constant value of 1. The impulse response of this system isa)[e-2t + e-4t]u(t)b)[-4e-2t - 12e-4t - e-t]u(t)c)[-4e-2t + 12e-4t]u(t)d)[-0.5e-2t + 1.5e-4t]u(t)Correct answer is option 'C'. 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 A linear, time-invariant and causal continuous time system has a rational transfer function with simple poles at s = -2 and s = -4 and one simple zero at s = -1. A unit step u(t) is applied at the input of the system. At steady state, the output has a constant value of 1. The impulse response of this system isa)[e-2t + e-4t]u(t)b)[-4e-2t - 12e-4t - e-t]u(t)c)[-4e-2t + 12e-4t]u(t)d)[-0.5e-2t + 1.5e-4t]u(t)Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A linear, time-invariant and causal continuous time system has a rational transfer function with simple poles at s = -2 and s = -4 and one simple zero at s = -1. A unit step u(t) is applied at the input of the system. At steady state, the output has a constant value of 1. The impulse response of this system isa)[e-2t + e-4t]u(t)b)[-4e-2t - 12e-4t - e-t]u(t)c)[-4e-2t + 12e-4t]u(t)d)[-0.5e-2t + 1.5e-4t]u(t)Correct answer is option 'C'. Can you explain this answer?, a detailed solution for A linear, time-invariant and causal continuous time system has a rational transfer function with simple poles at s = -2 and s = -4 and one simple zero at s = -1. A unit step u(t) is applied at the input of the system. At steady state, the output has a constant value of 1. The impulse response of this system isa)[e-2t + e-4t]u(t)b)[-4e-2t - 12e-4t - e-t]u(t)c)[-4e-2t + 12e-4t]u(t)d)[-0.5e-2t + 1.5e-4t]u(t)Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of A linear, time-invariant and causal continuous time system has a rational transfer function with simple poles at s = -2 and s = -4 and one simple zero at s = -1. A unit step u(t) is applied at the input of the system. At steady state, the output has a constant value of 1. The impulse response of this system isa)[e-2t + e-4t]u(t)b)[-4e-2t - 12e-4t - e-t]u(t)c)[-4e-2t + 12e-4t]u(t)d)[-0.5e-2t + 1.5e-4t]u(t)Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A linear, time-invariant and causal continuous time system has a rational transfer function with simple poles at s = -2 and s = -4 and one simple zero at s = -1. A unit step u(t) is applied at the input of the system. At steady state, the output has a constant value of 1. The impulse response of this system isa)[e-2t + e-4t]u(t)b)[-4e-2t - 12e-4t - e-t]u(t)c)[-4e-2t + 12e-4t]u(t)d)[-0.5e-2t + 1.5e-4t]u(t)Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice Electronics and Communication Engineering (ECE) tests.
|
|
Explore Courses for Electronics and Communication Engineering (ECE) exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
x
![]()
For Your Perfect Score in Electronics and Communication Engineering (ECE)
The Best you need at One Place
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup