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The impulse response h(t) of a continuous time, linear time invariant system is described by h(t) = eat u(t) + ebt u(-t) where, u(t) denots the unit step function and a and b are real constants. This system is stable if
- a)both a and b are positive
- b)a is positive and b is negative
- c)a is negative and b is positive
- d)both a and b are negative
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The impulse response h(t) of a continuous time, linear time invariant ...
Concept:
An LTI system is stable if and only if its impulse response is absolutely integrable.
Calculation:
Given,
h(t) = eat u(t) + ebt u(-t)
h(t) = h1(t) + h2(t)
∵ τ > 0 therefore, for absolute integrability a < 0
∵ τ < 0 therefore, for absolute integrability b > 0
∴ h(t) should be in the form shown in the figure below:
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Community Answer
The impulse response h(t) of a continuous time, linear time invariant ...
Understanding System Stability
In continuous-time linear time-invariant (LTI) systems, stability is determined by the behavior of the impulse response h(t) as time approaches infinity. The given impulse response is:
h(t) = e^(at)u(t) + e^(bt)u(-t)
Where u(t) is the unit step function.
Analyzing Each Term
- First Term: e^(at)u(t)
- Active for t ≥ 0.
- For the system to be stable, this term must decay to zero as t approaches infinity.
- This occurs when a < />.
- Second Term: e^(bt)u(-t)
- Active for t < />
- For stability, this term must also decay to zero as t approaches negative infinity.
- This occurs when b < /> (i.e., b must be negative).
Conclusion on Stability
For the system to be stable:
- a must be negative (ensuring decay in the positive time region),
- b must also be negative (ensuring decay in the negative time region).
Thus, the correct condition for stability of the system is when a is negative and b is positive, confirming that the system will not exhibit unbounded output for bounded input.
Final Answer
Hence, the correct answer is option C: a is negative and b is positive.
In continuous-time linear time-invariant (LTI) systems, stability is determined by the behavior of the impulse response h(t) as time approaches infinity. The given impulse response is:
h(t) = e^(at)u(t) + e^(bt)u(-t)
Where u(t) is the unit step function.
Analyzing Each Term
- First Term: e^(at)u(t)
- Active for t ≥ 0.
- For the system to be stable, this term must decay to zero as t approaches infinity.
- This occurs when a < />.
- Second Term: e^(bt)u(-t)
- Active for t < />
- For stability, this term must also decay to zero as t approaches negative infinity.
- This occurs when b < /> (i.e., b must be negative).
Conclusion on Stability
For the system to be stable:
- a must be negative (ensuring decay in the positive time region),
- b must also be negative (ensuring decay in the negative time region).
Thus, the correct condition for stability of the system is when a is negative and b is positive, confirming that the system will not exhibit unbounded output for bounded input.
Final Answer
Hence, the correct answer is option C: a is negative and b is positive.
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The impulse response h(t) of a continuous time, linear time invariant system is described by h(t) = eatu(t) + ebtu(-t) where, u(t) denots the unit step function and a and b are real constants. This system is stable ifa)both a and b are positiveb)a is positive and b is negativec)a is negative and b is positived)both a and b are negativeCorrect answer is option 'C'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2025 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 The impulse response h(t) of a continuous time, linear time invariant system is described by h(t) = eatu(t) + ebtu(-t) where, u(t) denots the unit step function and a and b are real constants. This system is stable ifa)both a and b are positiveb)a is positive and b is negativec)a is negative and b is positived)both a and b are negativeCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The impulse response h(t) of a continuous time, linear time invariant system is described by h(t) = eatu(t) + ebtu(-t) where, u(t) denots the unit step function and a and b are real constants. This system is stable ifa)both a and b are positiveb)a is positive and b is negativec)a is negative and b is positived)both a and b are negativeCorrect answer is option 'C'. Can you explain this answer?.
The impulse response h(t) of a continuous time, linear time invariant system is described by h(t) = eatu(t) + ebtu(-t) where, u(t) denots the unit step function and a and b are real constants. This system is stable ifa)both a and b are positiveb)a is positive and b is negativec)a is negative and b is positived)both a and b are negativeCorrect answer is option 'C'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2025 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 The impulse response h(t) of a continuous time, linear time invariant system is described by h(t) = eatu(t) + ebtu(-t) where, u(t) denots the unit step function and a and b are real constants. This system is stable ifa)both a and b are positiveb)a is positive and b is negativec)a is negative and b is positived)both a and b are negativeCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The impulse response h(t) of a continuous time, linear time invariant system is described by h(t) = eatu(t) + ebtu(-t) where, u(t) denots the unit step function and a and b are real constants. This system is stable ifa)both a and b are positiveb)a is positive and b is negativec)a is negative and b is positived)both a and b are negativeCorrect answer is option 'C'. Can you explain this answer?.
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