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If a system has an open loop transfer function 1-s / 1+s, then the gain of the system at frequency of 1 rad/s will be
- a)1
- b)1/2
- c)Zero
- d)-1
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
If a system has an open loop transfer function 1-s / 1+s, then the gai...
Open Loop Transfer Function
The open loop transfer function of a system represents the relationship between the input and output of a system without any feedback. It is denoted by G(s) and is expressed as the ratio of the Laplace transform of the output to the Laplace transform of the input.
G(s) = Y(s) / X(s)
Where:
G(s) - open loop transfer function
Y(s) - Laplace transform of the output
X(s) - Laplace transform of the input
Given Transfer Function
In this question, the given open loop transfer function is 1 - s / 1 + s.
Frequency Response of a System
To determine the gain of a system at a specific frequency, we substitute jω for s in the transfer function, where ω represents the angular frequency in radians per second.
G(jω) = Y(jω) / X(jω)
To find the gain at a specific frequency, we substitute ω = 1 rad/s in the transfer function and evaluate the expression.
Gain at Frequency ω = 1 rad/s
Substituting ω = 1 rad/s in the transfer function:
G(j1) = Y(j1) / X(j1)
Calculating the Gain
G(j1) = (1 - j) / (1 + j)
To calculate the gain, we need to find the modulus of this complex number.
|G(j1)| = sqrt((1 - j)^2) / sqrt((1 + j)^2)
Simplifying the expression:
|G(j1)| = sqrt(1^2 + (-1)^2) / sqrt(1^2 + 1^2)
|G(j1)| = sqrt(1 + 1) / sqrt(1 + 1)
|G(j1)| = sqrt(2) / sqrt(2)
|G(j1)| = 1
Hence, the gain of the system at a frequency of 1 rad/s is 1.
Conclusion
The correct answer is option 'A' - 1. The gain of the system at a frequency of 1 rad/s is 1.
The open loop transfer function of a system represents the relationship between the input and output of a system without any feedback. It is denoted by G(s) and is expressed as the ratio of the Laplace transform of the output to the Laplace transform of the input.
G(s) = Y(s) / X(s)
Where:
G(s) - open loop transfer function
Y(s) - Laplace transform of the output
X(s) - Laplace transform of the input
Given Transfer Function
In this question, the given open loop transfer function is 1 - s / 1 + s.
Frequency Response of a System
To determine the gain of a system at a specific frequency, we substitute jω for s in the transfer function, where ω represents the angular frequency in radians per second.
G(jω) = Y(jω) / X(jω)
To find the gain at a specific frequency, we substitute ω = 1 rad/s in the transfer function and evaluate the expression.
Gain at Frequency ω = 1 rad/s
Substituting ω = 1 rad/s in the transfer function:
G(j1) = Y(j1) / X(j1)
Calculating the Gain
G(j1) = (1 - j) / (1 + j)
To calculate the gain, we need to find the modulus of this complex number.
|G(j1)| = sqrt((1 - j)^2) / sqrt((1 + j)^2)
Simplifying the expression:
|G(j1)| = sqrt(1^2 + (-1)^2) / sqrt(1^2 + 1^2)
|G(j1)| = sqrt(1 + 1) / sqrt(1 + 1)
|G(j1)| = sqrt(2) / sqrt(2)
|G(j1)| = 1
Hence, the gain of the system at a frequency of 1 rad/s is 1.
Conclusion
The correct answer is option 'A' - 1. The gain of the system at a frequency of 1 rad/s is 1.
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The system is all pass system and the gain of the system at frequency of 1 rad/sec.
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If a system has an open loop transfer function 1-s / 1+s, then the gain of the system at frequency of 1 rad/s will bea)1b)1/2c)Zerod)-1Correct answer is option 'D'. Can you explain this answer? for Electrical Engineering (EE) 2025 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about If a system has an open loop transfer function 1-s / 1+s, then the gain of the system at frequency of 1 rad/s will bea)1b)1/2c)Zerod)-1Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a system has an open loop transfer function 1-s / 1+s, then the gain of the system at frequency of 1 rad/s will bea)1b)1/2c)Zerod)-1Correct answer is option 'D'. Can you explain this answer?.
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