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The output in response to a unit step input for a particular continuous control system is c(t)= 1-e-t. What is the delay time Td?
- a)0.36
- b)0.18
- c)0.693
- d)0.289
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The output in response to a unit step input for a particular continuou...
Answer: c
Explanation: The output is given as a function of time. The final value of the output is limn->∞c(t)=1; . Hence Td (at 50% of the final value) is the solution of 0.5=1-e-Td, and is equal to ln 2 or 0.693 sec.
Explanation: The output is given as a function of time. The final value of the output is limn->∞c(t)=1; . Hence Td (at 50% of the final value) is the solution of 0.5=1-e-Td, and is equal to ln 2 or 0.693 sec.
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The output in response to a unit step input for a particular continuou...
Introduction:
The given continuous control system has an output c(t) that is expressed as c(t) = 1 - e^(-t). The unit step input is applied to the system, and we need to determine the delay time Td.
Explanation:
The unit step input, also known as the Heaviside step function, is a function that is zero for negative values of t and one for positive values of t. It represents an instantaneous change from zero to one at t = 0.
The given output c(t) = 1 - e^(-t) represents the response of the system to the unit step input. Let's analyze this function to determine the delay time Td.
Determining the Delay Time:
The delay time Td is the time it takes for the output to reach 50% of its final value. In other words, we need to find the value of t for which c(t) = 0.5.
1. Set c(t) = 0.5:
1 - e^(-t) = 0.5
2. Solve for t:
e^(-t) = 0.5
Taking the natural logarithm (ln) of both sides:
-t = ln(0.5)
t = -ln(0.5)
t ≈ 0.693
Therefore, the delay time Td is approximately 0.693.
Conclusion:
The delay time Td for the given continuous control system, with the output c(t) = 1 - e^(-t) in response to a unit step input, is approximately 0.693.
The given continuous control system has an output c(t) that is expressed as c(t) = 1 - e^(-t). The unit step input is applied to the system, and we need to determine the delay time Td.
Explanation:
The unit step input, also known as the Heaviside step function, is a function that is zero for negative values of t and one for positive values of t. It represents an instantaneous change from zero to one at t = 0.
The given output c(t) = 1 - e^(-t) represents the response of the system to the unit step input. Let's analyze this function to determine the delay time Td.
Determining the Delay Time:
The delay time Td is the time it takes for the output to reach 50% of its final value. In other words, we need to find the value of t for which c(t) = 0.5.
1. Set c(t) = 0.5:
1 - e^(-t) = 0.5
2. Solve for t:
e^(-t) = 0.5
Taking the natural logarithm (ln) of both sides:
-t = ln(0.5)
t = -ln(0.5)
t ≈ 0.693
Therefore, the delay time Td is approximately 0.693.
Conclusion:
The delay time Td for the given continuous control system, with the output c(t) = 1 - e^(-t) in response to a unit step input, is approximately 0.693.
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