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The response of a first-order measurement system to a unit step input is 1 - e-0.5t, where t is in seconds. A ramp of 0.1 units per second is given as the input to this system. The error in the measured value after transients have died down is (2009)
- a)0.02 units
- b)0.1 units
- c)0.2 units
- d)1 unit
Correct answer is option 'B'. Can you explain this answer?
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The response of a first-order measurement system to a unit step input...
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The response of a first-order measurement system to a unit step input...
Error in the measured value after transients have died down can be calculated by finding the difference between the actual output and the desired output. In this case, the desired output is the ramp input of 0.1 units per second.
Given that the response of the first-order measurement system to a unit step input is 1 - e^(-0.5t), we can calculate the response of the system to a ramp input using integration.
Let's integrate the response of the system to a unit step input to find the response of the system to a ramp input.
Integration of 1 - e^(-0.5t) dt = t + 2e^(-0.5t) + C
Since we are interested in the response after transients have died down, we can ignore the constant term (C) and simplify the equation to:
t + 2e^(-0.5t)
Now, let's substitute t = 0.1 units per second into the equation to find the output of the system to the ramp input:
0.1 + 2e^(-0.5 * 0.1) = 0.1 + 2e^(-0.05) ≈ 0.1 + 2(0.951) ≈ 0.1 + 1.902 ≈ 2.002 units
The actual output of the system to the ramp input is 2.002 units.
The desired output is 0.1 units per second.
Therefore, the error in the measured value after transients have died down is:
Error = Actual output - Desired output = 2.002 - 0.1 = 1.902 units
The correct answer is option 'B' - 0.1 units.
Given that the response of the first-order measurement system to a unit step input is 1 - e^(-0.5t), we can calculate the response of the system to a ramp input using integration.
Let's integrate the response of the system to a unit step input to find the response of the system to a ramp input.
Integration of 1 - e^(-0.5t) dt = t + 2e^(-0.5t) + C
Since we are interested in the response after transients have died down, we can ignore the constant term (C) and simplify the equation to:
t + 2e^(-0.5t)
Now, let's substitute t = 0.1 units per second into the equation to find the output of the system to the ramp input:
0.1 + 2e^(-0.5 * 0.1) = 0.1 + 2e^(-0.05) ≈ 0.1 + 2(0.951) ≈ 0.1 + 1.902 ≈ 2.002 units
The actual output of the system to the ramp input is 2.002 units.
The desired output is 0.1 units per second.
Therefore, the error in the measured value after transients have died down is:
Error = Actual output - Desired output = 2.002 - 0.1 = 1.902 units
The correct answer is option 'B' - 0.1 units.
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The response of a first-order measurement system to a unit step input is 1 - e-0.5t, where t is in seconds. A ramp of 0.1 units per second is given as the input to this system. The error in the measured value after transients have died down is (2009)a)0.02 unitsb)0.1 unitsc)0.2 unitsd)1 unitCorrect answer is option 'B'. Can you explain this answer?
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