The steady-state error due to a unit-ramp command and due to a unit-ramp disturbance of I controller with an internal feedback loop of first order plant for the given cases.

Question

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Answer 1

Question

Chapter 10, Problem 10.28P

To determine

The steady-state error due to a unit-ramp command and due to a unit-ramp disturbance of I controller with an internal feedback loop of first order plant for the given cases.

Expert Solution & Answer

Answer to Problem 10.28P

The steady-state error values for all the cases are:

Case 1. For $ζ=0.707$ ,

Error due to unit-ramp command: $ess=0.2$

Error due to unit-ramp disturbance: $ess=0.005$

Case 2. For $ζ=1$ ,

Error due to unit-ramp command: $ess=0.4$

Error due to unit-ramp disturbance: $ess=0.01$

Case 3. For a root separation factor of 10,

Error due to unit-ramp command: $ess=0.22$

Error due to unit-ramp disturbance: $ess=0.001$ .

Explanation of Solution

Given:

The I controller with an internal feedback loop of first order plant is as shown below:

System Dynamics, Chapter 10, Problem 10.28P

Where, the parameter values are as given:

$I=c=4$

Also, the performance specifications require the time constant of the system to be $τ=0.2$ .

The values of gains for the I controller are as follows:

Case 1. For $ζ=0.707$ , $KI=200$ and $K2=36$ .

Case 2. For $ζ=1$ , $KI=100$ and $K2=36$ .

Case 3. For a root separation factor of 10, $KI=1000$ and $K2=216$ .

Concept Used:

The transfer functions for the block diagram are as shown below:

$Ω(s)Ωr(s)=KIIs2+s(c+K2)+KIΩ(s)Td(s)=−sIs2+s(c+K2)+KI$

The steady-state error of a system using final value theorem is:

$ess=lims→0sE(s)$ .

Calculation:

From the block diagram as shown, the transfer functions are as:

$Ω(s)Ωr(s)=KIIs2+s(c+K2)+KIΩ(s)Td(s)=−sIs2+s(c+K2)+KI$

Therefore, the response $Ω(s)$ for the system is:

$Ω(s)=KIIs2+s(c+K2)+KIΩr(s)−sIs2+s(c+K2)+KITd(s)$

And from the block diagram shown in figure, we have

$E(s)=(Ωr(s)−Ω(s))⇒E(s)=(1−KIIs2+s(c+K2)+KI)Ωr(s)+sIs2+s(c+K2)+KITd(s)∵Ω(s)=KIIs2+s(c+K2)+KIΩr(s)−sIs2+s(c+K2)+KITd(s)$

On keeping the values of the parameters such that $I=c=4$

$E(s)=(1−KIIs2+s(c+K2)+KI)Ωr(s)+sIs2+s(c+K2)+KITd(s)⇒E(s)=(s(Is+c+K2)Is2+s(c+K2)+KI)Ωr(s)+sIs2+s(c+K2)+KITd(s)⇒E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)$

Case 1. When $ζ=0.707$ , the gain values are $KI=200$ and $K2=36$ :

Since,

$E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)∵KI=200 and K2=36⇒E(s)=(s(4s+4+36)4s2+s(4+36)+200Ωr(s)+s4s2+s(4+36)+200Td(s))⇒E(s)=(s(4s+40)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))$

Therefore, for unit-ramp command response $Ωr(s)$ and zero disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))⇒E(s)=(s(4s+40)4s2+40s+200⋅1s2+s4s2+40s+200⋅0)⇒E(s)=(4s+40)(4s2+40s+200)⋅1s⇒E(s)=(s+10)(s2+10s+50)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s(s+10)(s2+10s+50)⋅1s ∵E(s)=(s+10)(s2+10s+50)⋅1s⇒ess=lims→0(s+10)(s2+10s+50)=1050⇒ess=0.2$

Therefore, for zero command response $Ωr(s)$ and unit-ramp disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))⇒E(s)=(s(4s+40)4s2+40s+200⋅0+s4s2+40s+200⋅1s2)⇒E(s)=1(4s2+40s+200)⋅1s⇒E(s)=14(s2+10s+50)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s14(s2+10s+50)⋅1s ∵E(s)=14(s2+10s+50)⋅1s⇒ess=lims→014(s2+10s+50)=1200⇒ess=0.005$

Case 2. When $ζ=1$ , the gain values are $KI=100$ and $K2=36$ :

Since,

$E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)∵KI=100 and K2=36⇒E(s)=(s(4s+4+36)4s2+s(4+36)+100Ωr(s)+s4s2+s(4+36)+100Td(s))⇒E(s)=(s(4s+40)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))$

Therefore, for unit-ramp command response $Ωr(s)$ and zero disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))⇒E(s)=(s(4s+40)4s2+40s+100⋅1s2+s4s2+40s+100⋅0)⇒E(s)=(4s+40)(4s2+40s+100)⋅1s⇒E(s)=(s+10)(s2+10s+25)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s(s+10)(s2+10s+25)⋅1s ∵E(s)=(s+10)(s2+10s+25)⋅1s⇒ess=lims→0(s+10)(s2+10s+25)=1025⇒ess=0.4$

Therefore, for zero command response $Ωr(s)$ and unit-ramp disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))⇒E(s)=(s(4s+40)4s2+40s+100⋅0+s4s2+40s+100⋅1s2)⇒E(s)=1(4s2+40s+100)⋅1s⇒E(s)=14(s2+10s+25)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s14(s2+10s+25)⋅1s ∵E(s)=14(s2+10s+25)⋅1s⇒ess=lims→014(s2+10s+25)=1100⇒ess=0.01$

Case 3. For a root separation factor of 10, $KI=1000$ and $K2=216$ .

Since,

$E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)∵KI=1000 and K2=216⇒E(s)=(s(4s+4+216)4s2+s(4+216)+1000Ωr(s)+s4s2+s(4+216)+1000Td(s))⇒E(s)=(s(4s+220)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))$

Therefore, for unit-ramp command response $Ωr(s)$ and zero disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+220)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))⇒E(s)=(s(4s+220)4s2+220s+1000⋅1s2+s4s2+220s+1000⋅0)⇒E(s)=(4s+220)(4s2+220s+1000)⋅1s⇒E(s)=(s+55)(s2+55s+250)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s(s+55)(s2+55s+250)⋅1s ∵E(s)=(s+55)(s2+55s+250)⋅1s⇒ess=lims→0(s+55)(s2+55s+250)=55250⇒ess=0.22$

Therefore, for zero command response $Ωr(s)$ and unit-ramp disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+220)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))⇒E(s)=(s(4s+220)4s2+220s+1000⋅0+s4s2+220s+1000⋅1s2)⇒E(s)=1(4s2+220s+1000)⋅1s⇒E(s)=14(s2+55s+250)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s14(s2+55s+250)⋅1s ∵E(s)=14(s2+55s+250)⋅1s⇒ess=lims→0(s+1)4(s2+55s+250)=11000⇒ess=0.001$ .

Conclusion:

The steady-state error values for all the cases are:

Case 1. For $ζ=0.707$ ,

Error due to unit-ramp command: $ess=0.2$

Error due to unit-ramp disturbance: $ess=0.005$

Case 2. For $ζ=1$ , $Tm≈35 N⋅m$ .

Error due to unit-ramp command: $ess=0.4$

Error due to unit-ramp disturbance: $ess=0.01$

Case 3. For a root separation factor of 10, $Tm≈215 N⋅m$ .

Error due to unit-ramp command: $ess=0.22$

Error due to unit-ramp disturbance: $ess=0.001$ .

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Answer 2

Question

Chapter 10, Problem 10.28P

To determine

The steady-state error due to a unit-ramp command and due to a unit-ramp disturbance of I controller with an internal feedback loop of first order plant for the given cases.

Expert Solution & Answer

Answer to Problem 10.28P

The steady-state error values for all the cases are:

Case 1. For $ζ=0.707$ ,

Error due to unit-ramp command: $ess=0.2$

Error due to unit-ramp disturbance: $ess=0.005$

Case 2. For $ζ=1$ ,

Error due to unit-ramp command: $ess=0.4$

Error due to unit-ramp disturbance: $ess=0.01$

Case 3. For a root separation factor of 10,

Error due to unit-ramp command: $ess=0.22$

Error due to unit-ramp disturbance: $ess=0.001$ .

Explanation of Solution

Given:

The I controller with an internal feedback loop of first order plant is as shown below:

System Dynamics, Chapter 10, Problem 10.28P

Where, the parameter values are as given:

$I=c=4$

Also, the performance specifications require the time constant of the system to be $τ=0.2$ .

The values of gains for the I controller are as follows:

Case 1. For $ζ=0.707$ , $KI=200$ and $K2=36$ .

Case 2. For $ζ=1$ , $KI=100$ and $K2=36$ .

Case 3. For a root separation factor of 10, $KI=1000$ and $K2=216$ .

Concept Used:

The transfer functions for the block diagram are as shown below:

$Ω(s)Ωr(s)=KIIs2+s(c+K2)+KIΩ(s)Td(s)=−sIs2+s(c+K2)+KI$

The steady-state error of a system using final value theorem is:

$ess=lims→0sE(s)$ .

Calculation:

From the block diagram as shown, the transfer functions are as:

$Ω(s)Ωr(s)=KIIs2+s(c+K2)+KIΩ(s)Td(s)=−sIs2+s(c+K2)+KI$

Therefore, the response $Ω(s)$ for the system is:

$Ω(s)=KIIs2+s(c+K2)+KIΩr(s)−sIs2+s(c+K2)+KITd(s)$

And from the block diagram shown in figure, we have

$E(s)=(Ωr(s)−Ω(s))⇒E(s)=(1−KIIs2+s(c+K2)+KI)Ωr(s)+sIs2+s(c+K2)+KITd(s)∵Ω(s)=KIIs2+s(c+K2)+KIΩr(s)−sIs2+s(c+K2)+KITd(s)$

On keeping the values of the parameters such that $I=c=4$

$E(s)=(1−KIIs2+s(c+K2)+KI)Ωr(s)+sIs2+s(c+K2)+KITd(s)⇒E(s)=(s(Is+c+K2)Is2+s(c+K2)+KI)Ωr(s)+sIs2+s(c+K2)+KITd(s)⇒E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)$

Case 1. When $ζ=0.707$ , the gain values are $KI=200$ and $K2=36$ :

Since,

$E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)∵KI=200 and K2=36⇒E(s)=(s(4s+4+36)4s2+s(4+36)+200Ωr(s)+s4s2+s(4+36)+200Td(s))⇒E(s)=(s(4s+40)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))$

Therefore, for unit-ramp command response $Ωr(s)$ and zero disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))⇒E(s)=(s(4s+40)4s2+40s+200⋅1s2+s4s2+40s+200⋅0)⇒E(s)=(4s+40)(4s2+40s+200)⋅1s⇒E(s)=(s+10)(s2+10s+50)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s(s+10)(s2+10s+50)⋅1s ∵E(s)=(s+10)(s2+10s+50)⋅1s⇒ess=lims→0(s+10)(s2+10s+50)=1050⇒ess=0.2$

Therefore, for zero command response $Ωr(s)$ and unit-ramp disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))⇒E(s)=(s(4s+40)4s2+40s+200⋅0+s4s2+40s+200⋅1s2)⇒E(s)=1(4s2+40s+200)⋅1s⇒E(s)=14(s2+10s+50)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s14(s2+10s+50)⋅1s ∵E(s)=14(s2+10s+50)⋅1s⇒ess=lims→014(s2+10s+50)=1200⇒ess=0.005$

Case 2. When $ζ=1$ , the gain values are $KI=100$ and $K2=36$ :

Since,

$E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)∵KI=100 and K2=36⇒E(s)=(s(4s+4+36)4s2+s(4+36)+100Ωr(s)+s4s2+s(4+36)+100Td(s))⇒E(s)=(s(4s+40)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))$

Therefore, for unit-ramp command response $Ωr(s)$ and zero disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))⇒E(s)=(s(4s+40)4s2+40s+100⋅1s2+s4s2+40s+100⋅0)⇒E(s)=(4s+40)(4s2+40s+100)⋅1s⇒E(s)=(s+10)(s2+10s+25)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s(s+10)(s2+10s+25)⋅1s ∵E(s)=(s+10)(s2+10s+25)⋅1s⇒ess=lims→0(s+10)(s2+10s+25)=1025⇒ess=0.4$

Therefore, for zero command response $Ωr(s)$ and unit-ramp disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))⇒E(s)=(s(4s+40)4s2+40s+100⋅0+s4s2+40s+100⋅1s2)⇒E(s)=1(4s2+40s+100)⋅1s⇒E(s)=14(s2+10s+25)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s14(s2+10s+25)⋅1s ∵E(s)=14(s2+10s+25)⋅1s⇒ess=lims→014(s2+10s+25)=1100⇒ess=0.01$

Case 3. For a root separation factor of 10, $KI=1000$ and $K2=216$ .

Since,

$E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)∵KI=1000 and K2=216⇒E(s)=(s(4s+4+216)4s2+s(4+216)+1000Ωr(s)+s4s2+s(4+216)+1000Td(s))⇒E(s)=(s(4s+220)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))$

Therefore, for unit-ramp command response $Ωr(s)$ and zero disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+220)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))⇒E(s)=(s(4s+220)4s2+220s+1000⋅1s2+s4s2+220s+1000⋅0)⇒E(s)=(4s+220)(4s2+220s+1000)⋅1s⇒E(s)=(s+55)(s2+55s+250)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s(s+55)(s2+55s+250)⋅1s ∵E(s)=(s+55)(s2+55s+250)⋅1s⇒ess=lims→0(s+55)(s2+55s+250)=55250⇒ess=0.22$

Therefore, for zero command response $Ωr(s)$ and unit-ramp disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+220)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))⇒E(s)=(s(4s+220)4s2+220s+1000⋅0+s4s2+220s+1000⋅1s2)⇒E(s)=1(4s2+220s+1000)⋅1s⇒E(s)=14(s2+55s+250)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s14(s2+55s+250)⋅1s ∵E(s)=14(s2+55s+250)⋅1s⇒ess=lims→0(s+1)4(s2+55s+250)=11000⇒ess=0.001$ .

Conclusion:

The steady-state error values for all the cases are:

Case 1. For $ζ=0.707$ ,

Error due to unit-ramp command: $ess=0.2$

Error due to unit-ramp disturbance: $ess=0.005$

Case 2. For $ζ=1$ , $Tm≈35 N⋅m$ .

Error due to unit-ramp command: $ess=0.4$

Error due to unit-ramp disturbance: $ess=0.01$

Case 3. For a root separation factor of 10, $Tm≈215 N⋅m$ .

Error due to unit-ramp command: $ess=0.22$

Error due to unit-ramp disturbance: $ess=0.001$ .

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Answer 3

Question

Chapter 10, Problem 10.28P

To determine

The steady-state error due to a unit-ramp command and due to a unit-ramp disturbance of I controller with an internal feedback loop of first order plant for the given cases.

Expert Solution & Answer

Answer to Problem 10.28P

The steady-state error values for all the cases are:

Case 1. For $ζ=0.707$ ,

Error due to unit-ramp command: $ess=0.2$

Error due to unit-ramp disturbance: $ess=0.005$

Case 2. For $ζ=1$ ,

Error due to unit-ramp command: $ess=0.4$

Error due to unit-ramp disturbance: $ess=0.01$

Case 3. For a root separation factor of 10,

Error due to unit-ramp command: $ess=0.22$

Error due to unit-ramp disturbance: $ess=0.001$ .

Explanation of Solution

Given:

The I controller with an internal feedback loop of first order plant is as shown below:

System Dynamics, Chapter 10, Problem 10.28P

Where, the parameter values are as given:

$I=c=4$

Also, the performance specifications require the time constant of the system to be $τ=0.2$ .

The values of gains for the I controller are as follows:

Case 1. For $ζ=0.707$ , $KI=200$ and $K2=36$ .

Case 2. For $ζ=1$ , $KI=100$ and $K2=36$ .

Case 3. For a root separation factor of 10, $KI=1000$ and $K2=216$ .

Concept Used:

The transfer functions for the block diagram are as shown below:

$Ω(s)Ωr(s)=KIIs2+s(c+K2)+KIΩ(s)Td(s)=−sIs2+s(c+K2)+KI$

The steady-state error of a system using final value theorem is:

$ess=lims→0sE(s)$ .

Calculation:

From the block diagram as shown, the transfer functions are as:

$Ω(s)Ωr(s)=KIIs2+s(c+K2)+KIΩ(s)Td(s)=−sIs2+s(c+K2)+KI$

Therefore, the response $Ω(s)$ for the system is:

$Ω(s)=KIIs2+s(c+K2)+KIΩr(s)−sIs2+s(c+K2)+KITd(s)$

And from the block diagram shown in figure, we have

$E(s)=(Ωr(s)−Ω(s))⇒E(s)=(1−KIIs2+s(c+K2)+KI)Ωr(s)+sIs2+s(c+K2)+KITd(s)∵Ω(s)=KIIs2+s(c+K2)+KIΩr(s)−sIs2+s(c+K2)+KITd(s)$

On keeping the values of the parameters such that $I=c=4$

$E(s)=(1−KIIs2+s(c+K2)+KI)Ωr(s)+sIs2+s(c+K2)+KITd(s)⇒E(s)=(s(Is+c+K2)Is2+s(c+K2)+KI)Ωr(s)+sIs2+s(c+K2)+KITd(s)⇒E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)$

Case 1. When $ζ=0.707$ , the gain values are $KI=200$ and $K2=36$ :

Since,

$E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)∵KI=200 and K2=36⇒E(s)=(s(4s+4+36)4s2+s(4+36)+200Ωr(s)+s4s2+s(4+36)+200Td(s))⇒E(s)=(s(4s+40)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))$

Therefore, for unit-ramp command response $Ωr(s)$ and zero disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))⇒E(s)=(s(4s+40)4s2+40s+200⋅1s2+s4s2+40s+200⋅0)⇒E(s)=(4s+40)(4s2+40s+200)⋅1s⇒E(s)=(s+10)(s2+10s+50)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s(s+10)(s2+10s+50)⋅1s ∵E(s)=(s+10)(s2+10s+50)⋅1s⇒ess=lims→0(s+10)(s2+10s+50)=1050⇒ess=0.2$

Therefore, for zero command response $Ωr(s)$ and unit-ramp disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))⇒E(s)=(s(4s+40)4s2+40s+200⋅0+s4s2+40s+200⋅1s2)⇒E(s)=1(4s2+40s+200)⋅1s⇒E(s)=14(s2+10s+50)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s14(s2+10s+50)⋅1s ∵E(s)=14(s2+10s+50)⋅1s⇒ess=lims→014(s2+10s+50)=1200⇒ess=0.005$

Case 2. When $ζ=1$ , the gain values are $KI=100$ and $K2=36$ :

Since,

$E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)∵KI=100 and K2=36⇒E(s)=(s(4s+4+36)4s2+s(4+36)+100Ωr(s)+s4s2+s(4+36)+100Td(s))⇒E(s)=(s(4s+40)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))$

Therefore, for unit-ramp command response $Ωr(s)$ and zero disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))⇒E(s)=(s(4s+40)4s2+40s+100⋅1s2+s4s2+40s+100⋅0)⇒E(s)=(4s+40)(4s2+40s+100)⋅1s⇒E(s)=(s+10)(s2+10s+25)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s(s+10)(s2+10s+25)⋅1s ∵E(s)=(s+10)(s2+10s+25)⋅1s⇒ess=lims→0(s+10)(s2+10s+25)=1025⇒ess=0.4$

Therefore, for zero command response $Ωr(s)$ and unit-ramp disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))⇒E(s)=(s(4s+40)4s2+40s+100⋅0+s4s2+40s+100⋅1s2)⇒E(s)=1(4s2+40s+100)⋅1s⇒E(s)=14(s2+10s+25)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s14(s2+10s+25)⋅1s ∵E(s)=14(s2+10s+25)⋅1s⇒ess=lims→014(s2+10s+25)=1100⇒ess=0.01$

Case 3. For a root separation factor of 10, $KI=1000$ and $K2=216$ .

Since,

$E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)∵KI=1000 and K2=216⇒E(s)=(s(4s+4+216)4s2+s(4+216)+1000Ωr(s)+s4s2+s(4+216)+1000Td(s))⇒E(s)=(s(4s+220)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))$

Therefore, for unit-ramp command response $Ωr(s)$ and zero disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+220)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))⇒E(s)=(s(4s+220)4s2+220s+1000⋅1s2+s4s2+220s+1000⋅0)⇒E(s)=(4s+220)(4s2+220s+1000)⋅1s⇒E(s)=(s+55)(s2+55s+250)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s(s+55)(s2+55s+250)⋅1s ∵E(s)=(s+55)(s2+55s+250)⋅1s⇒ess=lims→0(s+55)(s2+55s+250)=55250⇒ess=0.22$

Therefore, for zero command response $Ωr(s)$ and unit-ramp disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+220)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))⇒E(s)=(s(4s+220)4s2+220s+1000⋅0+s4s2+220s+1000⋅1s2)⇒E(s)=1(4s2+220s+1000)⋅1s⇒E(s)=14(s2+55s+250)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s14(s2+55s+250)⋅1s ∵E(s)=14(s2+55s+250)⋅1s⇒ess=lims→0(s+1)4(s2+55s+250)=11000⇒ess=0.001$ .

Conclusion:

The steady-state error values for all the cases are:

Case 1. For $ζ=0.707$ ,

Error due to unit-ramp command: $ess=0.2$

Error due to unit-ramp disturbance: $ess=0.005$

Case 2. For $ζ=1$ , $Tm≈35 N⋅m$ .

Error due to unit-ramp command: $ess=0.4$

Error due to unit-ramp disturbance: $ess=0.01$

Case 3. For a root separation factor of 10, $Tm≈215 N⋅m$ .

Error due to unit-ramp command: $ess=0.22$

Error due to unit-ramp disturbance: $ess=0.001$ .

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Answer 4

Question

Chapter 10, Problem 10.28P

To determine

The steady-state error due to a unit-ramp command and due to a unit-ramp disturbance of I controller with an internal feedback loop of first order plant for the given cases.

Expert Solution & Answer

Answer to Problem 10.28P

The steady-state error values for all the cases are:

Case 1. For $ζ=0.707$ ,

Error due to unit-ramp command: $ess=0.2$

Error due to unit-ramp disturbance: $ess=0.005$

Case 2. For $ζ=1$ ,

Error due to unit-ramp command: $ess=0.4$

Error due to unit-ramp disturbance: $ess=0.01$

Case 3. For a root separation factor of 10,

Error due to unit-ramp command: $ess=0.22$

Error due to unit-ramp disturbance: $ess=0.001$ .

Explanation of Solution

Given:

The I controller with an internal feedback loop of first order plant is as shown below:

System Dynamics, Chapter 10, Problem 10.28P

Where, the parameter values are as given:

$I=c=4$

Also, the performance specifications require the time constant of the system to be $τ=0.2$ .

The values of gains for the I controller are as follows:

Case 1. For $ζ=0.707$ , $KI=200$ and $K2=36$ .

Case 2. For $ζ=1$ , $KI=100$ and $K2=36$ .

Case 3. For a root separation factor of 10, $KI=1000$ and $K2=216$ .

Concept Used:

The transfer functions for the block diagram are as shown below:

$Ω(s)Ωr(s)=KIIs2+s(c+K2)+KIΩ(s)Td(s)=−sIs2+s(c+K2)+KI$

The steady-state error of a system using final value theorem is:

$ess=lims→0sE(s)$ .

Calculation:

From the block diagram as shown, the transfer functions are as:

$Ω(s)Ωr(s)=KIIs2+s(c+K2)+KIΩ(s)Td(s)=−sIs2+s(c+K2)+KI$

Therefore, the response $Ω(s)$ for the system is:

$Ω(s)=KIIs2+s(c+K2)+KIΩr(s)−sIs2+s(c+K2)+KITd(s)$

And from the block diagram shown in figure, we have

$E(s)=(Ωr(s)−Ω(s))⇒E(s)=(1−KIIs2+s(c+K2)+KI)Ωr(s)+sIs2+s(c+K2)+KITd(s)∵Ω(s)=KIIs2+s(c+K2)+KIΩr(s)−sIs2+s(c+K2)+KITd(s)$

On keeping the values of the parameters such that $I=c=4$

$E(s)=(1−KIIs2+s(c+K2)+KI)Ωr(s)+sIs2+s(c+K2)+KITd(s)⇒E(s)=(s(Is+c+K2)Is2+s(c+K2)+KI)Ωr(s)+sIs2+s(c+K2)+KITd(s)⇒E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)$

Case 1. When $ζ=0.707$ , the gain values are $KI=200$ and $K2=36$ :

Since,

$E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)∵KI=200 and K2=36⇒E(s)=(s(4s+4+36)4s2+s(4+36)+200Ωr(s)+s4s2+s(4+36)+200Td(s))⇒E(s)=(s(4s+40)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))$

Therefore, for unit-ramp command response $Ωr(s)$ and zero disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))⇒E(s)=(s(4s+40)4s2+40s+200⋅1s2+s4s2+40s+200⋅0)⇒E(s)=(4s+40)(4s2+40s+200)⋅1s⇒E(s)=(s+10)(s2+10s+50)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s(s+10)(s2+10s+50)⋅1s ∵E(s)=(s+10)(s2+10s+50)⋅1s⇒ess=lims→0(s+10)(s2+10s+50)=1050⇒ess=0.2$

Therefore, for zero command response $Ωr(s)$ and unit-ramp disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))⇒E(s)=(s(4s+40)4s2+40s+200⋅0+s4s2+40s+200⋅1s2)⇒E(s)=1(4s2+40s+200)⋅1s⇒E(s)=14(s2+10s+50)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s14(s2+10s+50)⋅1s ∵E(s)=14(s2+10s+50)⋅1s⇒ess=lims→014(s2+10s+50)=1200⇒ess=0.005$

Case 2. When $ζ=1$ , the gain values are $KI=100$ and $K2=36$ :

Since,

$E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)∵KI=100 and K2=36⇒E(s)=(s(4s+4+36)4s2+s(4+36)+100Ωr(s)+s4s2+s(4+36)+100Td(s))⇒E(s)=(s(4s+40)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))$

Therefore, for unit-ramp command response $Ωr(s)$ and zero disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))⇒E(s)=(s(4s+40)4s2+40s+100⋅1s2+s4s2+40s+100⋅0)⇒E(s)=(4s+40)(4s2+40s+100)⋅1s⇒E(s)=(s+10)(s2+10s+25)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s(s+10)(s2+10s+25)⋅1s ∵E(s)=(s+10)(s2+10s+25)⋅1s⇒ess=lims→0(s+10)(s2+10s+25)=1025⇒ess=0.4$

Therefore, for zero command response $Ωr(s)$ and unit-ramp disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))⇒E(s)=(s(4s+40)4s2+40s+100⋅0+s4s2+40s+100⋅1s2)⇒E(s)=1(4s2+40s+100)⋅1s⇒E(s)=14(s2+10s+25)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s14(s2+10s+25)⋅1s ∵E(s)=14(s2+10s+25)⋅1s⇒ess=lims→014(s2+10s+25)=1100⇒ess=0.01$

Case 3. For a root separation factor of 10, $KI=1000$ and $K2=216$ .

Since,

$E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)∵KI=1000 and K2=216⇒E(s)=(s(4s+4+216)4s2+s(4+216)+1000Ωr(s)+s4s2+s(4+216)+1000Td(s))⇒E(s)=(s(4s+220)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))$

Therefore, for unit-ramp command response $Ωr(s)$ and zero disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+220)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))⇒E(s)=(s(4s+220)4s2+220s+1000⋅1s2+s4s2+220s+1000⋅0)⇒E(s)=(4s+220)(4s2+220s+1000)⋅1s⇒E(s)=(s+55)(s2+55s+250)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s(s+55)(s2+55s+250)⋅1s ∵E(s)=(s+55)(s2+55s+250)⋅1s⇒ess=lims→0(s+55)(s2+55s+250)=55250⇒ess=0.22$

Therefore, for zero command response $Ωr(s)$ and unit-ramp disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+220)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))⇒E(s)=(s(4s+220)4s2+220s+1000⋅0+s4s2+220s+1000⋅1s2)⇒E(s)=1(4s2+220s+1000)⋅1s⇒E(s)=14(s2+55s+250)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s14(s2+55s+250)⋅1s ∵E(s)=14(s2+55s+250)⋅1s⇒ess=lims→0(s+1)4(s2+55s+250)=11000⇒ess=0.001$ .

Conclusion:

The steady-state error values for all the cases are:

Case 1. For $ζ=0.707$ ,

Error due to unit-ramp command: $ess=0.2$

Error due to unit-ramp disturbance: $ess=0.005$

Case 2. For $ζ=1$ , $Tm≈35 N⋅m$ .

Error due to unit-ramp command: $ess=0.4$

Error due to unit-ramp disturbance: $ess=0.01$

Case 3. For a root separation factor of 10, $Tm≈215 N⋅m$ .

Error due to unit-ramp command: $ess=0.22$

Error due to unit-ramp disturbance: $ess=0.001$ .

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Answer 5

Chapter 10, Problem 10.28P

To determine

The steady-state error due to a unit-ramp command and due to a unit-ramp disturbance of I controller with an internal feedback loop of first order plant for the given cases.

Expert Solution & Answer

Answer to Problem 10.28P

The steady-state error values for all the cases are:

Case 1. For $ζ=0.707$ ,

Error due to unit-ramp command: $ess=0.2$

Error due to unit-ramp disturbance: $ess=0.005$

Case 2. For $ζ=1$ ,

Error due to unit-ramp command: $ess=0.4$

Error due to unit-ramp disturbance: $ess=0.01$

Case 3. For a root separation factor of 10,

Error due to unit-ramp command: $ess=0.22$

Error due to unit-ramp disturbance: $ess=0.001$ .

Explanation of Solution

Given:

The I controller with an internal feedback loop of first order plant is as shown below:

System Dynamics, Chapter 10, Problem 10.28P

Where, the parameter values are as given:

$I=c=4$

Also, the performance specifications require the time constant of the system to be $τ=0.2$ .

The values of gains for the I controller are as follows:

Case 1. For $ζ=0.707$ , $KI=200$ and $K2=36$ .

Case 2. For $ζ=1$ , $KI=100$ and $K2=36$ .

Case 3. For a root separation factor of 10, $KI=1000$ and $K2=216$ .

Concept Used:

The transfer functions for the block diagram are as shown below:

$Ω(s)Ωr(s)=KIIs2+s(c+K2)+KIΩ(s)Td(s)=−sIs2+s(c+K2)+KI$

The steady-state error of a system using final value theorem is:

$ess=lims→0sE(s)$ .

Calculation:

From the block diagram as shown, the transfer functions are as:

$Ω(s)Ωr(s)=KIIs2+s(c+K2)+KIΩ(s)Td(s)=−sIs2+s(c+K2)+KI$

Therefore, the response $Ω(s)$ for the system is:

$Ω(s)=KIIs2+s(c+K2)+KIΩr(s)−sIs2+s(c+K2)+KITd(s)$

And from the block diagram shown in figure, we have

$E(s)=(Ωr(s)−Ω(s))⇒E(s)=(1−KIIs2+s(c+K2)+KI)Ωr(s)+sIs2+s(c+K2)+KITd(s)∵Ω(s)=KIIs2+s(c+K2)+KIΩr(s)−sIs2+s(c+K2)+KITd(s)$

On keeping the values of the parameters such that $I=c=4$

$E(s)=(1−KIIs2+s(c+K2)+KI)Ωr(s)+sIs2+s(c+K2)+KITd(s)⇒E(s)=(s(Is+c+K2)Is2+s(c+K2)+KI)Ωr(s)+sIs2+s(c+K2)+KITd(s)⇒E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)$

Case 1. When $ζ=0.707$ , the gain values are $KI=200$ and $K2=36$ :

Since,

$E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)∵KI=200 and K2=36⇒E(s)=(s(4s+4+36)4s2+s(4+36)+200Ωr(s)+s4s2+s(4+36)+200Td(s))⇒E(s)=(s(4s+40)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))$

Therefore, for unit-ramp command response $Ωr(s)$ and zero disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))⇒E(s)=(s(4s+40)4s2+40s+200⋅1s2+s4s2+40s+200⋅0)⇒E(s)=(4s+40)(4s2+40s+200)⋅1s⇒E(s)=(s+10)(s2+10s+50)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s(s+10)(s2+10s+50)⋅1s ∵E(s)=(s+10)(s2+10s+50)⋅1s⇒ess=lims→0(s+10)(s2+10s+50)=1050⇒ess=0.2$

Therefore, for zero command response $Ωr(s)$ and unit-ramp disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))⇒E(s)=(s(4s+40)4s2+40s+200⋅0+s4s2+40s+200⋅1s2)⇒E(s)=1(4s2+40s+200)⋅1s⇒E(s)=14(s2+10s+50)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s14(s2+10s+50)⋅1s ∵E(s)=14(s2+10s+50)⋅1s⇒ess=lims→014(s2+10s+50)=1200⇒ess=0.005$

Case 2. When $ζ=1$ , the gain values are $KI=100$ and $K2=36$ :

Since,

$E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)∵KI=100 and K2=36⇒E(s)=(s(4s+4+36)4s2+s(4+36)+100Ωr(s)+s4s2+s(4+36)+100Td(s))⇒E(s)=(s(4s+40)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))$

Therefore, for unit-ramp command response $Ωr(s)$ and zero disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))⇒E(s)=(s(4s+40)4s2+40s+100⋅1s2+s4s2+40s+100⋅0)⇒E(s)=(4s+40)(4s2+40s+100)⋅1s⇒E(s)=(s+10)(s2+10s+25)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s(s+10)(s2+10s+25)⋅1s ∵E(s)=(s+10)(s2+10s+25)⋅1s⇒ess=lims→0(s+10)(s2+10s+25)=1025⇒ess=0.4$

Therefore, for zero command response $Ωr(s)$ and unit-ramp disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))⇒E(s)=(s(4s+40)4s2+40s+100⋅0+s4s2+40s+100⋅1s2)⇒E(s)=1(4s2+40s+100)⋅1s⇒E(s)=14(s2+10s+25)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s14(s2+10s+25)⋅1s ∵E(s)=14(s2+10s+25)⋅1s⇒ess=lims→014(s2+10s+25)=1100⇒ess=0.01$

Case 3. For a root separation factor of 10, $KI=1000$ and $K2=216$ .

Since,

$E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)∵KI=1000 and K2=216⇒E(s)=(s(4s+4+216)4s2+s(4+216)+1000Ωr(s)+s4s2+s(4+216)+1000Td(s))⇒E(s)=(s(4s+220)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))$

Therefore, for unit-ramp command response $Ωr(s)$ and zero disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+220)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))⇒E(s)=(s(4s+220)4s2+220s+1000⋅1s2+s4s2+220s+1000⋅0)⇒E(s)=(4s+220)(4s2+220s+1000)⋅1s⇒E(s)=(s+55)(s2+55s+250)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s(s+55)(s2+55s+250)⋅1s ∵E(s)=(s+55)(s2+55s+250)⋅1s⇒ess=lims→0(s+55)(s2+55s+250)=55250⇒ess=0.22$

Therefore, for zero command response $Ωr(s)$ and unit-ramp disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+220)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))⇒E(s)=(s(4s+220)4s2+220s+1000⋅0+s4s2+220s+1000⋅1s2)⇒E(s)=1(4s2+220s+1000)⋅1s⇒E(s)=14(s2+55s+250)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s14(s2+55s+250)⋅1s ∵E(s)=14(s2+55s+250)⋅1s⇒ess=lims→0(s+1)4(s2+55s+250)=11000⇒ess=0.001$ .

Conclusion:

The steady-state error values for all the cases are:

Case 1. For $ζ=0.707$ ,

Error due to unit-ramp command: $ess=0.2$

Error due to unit-ramp disturbance: $ess=0.005$

Case 2. For $ζ=1$ , $Tm≈35 N⋅m$ .

Error due to unit-ramp command: $ess=0.4$

Error due to unit-ramp disturbance: $ess=0.01$

Case 3. For a root separation factor of 10, $Tm≈215 N⋅m$ .

Error due to unit-ramp command: $ess=0.22$

Error due to unit-ramp disturbance: $ess=0.001$ .

Answer 6

Chapter 10, Problem 10.28P

To determine

The steady-state error due to a unit-ramp command and due to a unit-ramp disturbance of I controller with an internal feedback loop of first order plant for the given cases.

Expert Solution & Answer

Answer to Problem 10.28P

The steady-state error values for all the cases are:

Case 1. For $ζ=0.707$ ,

Error due to unit-ramp command: $ess=0.2$

Error due to unit-ramp disturbance: $ess=0.005$

Case 2. For $ζ=1$ ,

Error due to unit-ramp command: $ess=0.4$

Error due to unit-ramp disturbance: $ess=0.01$

Case 3. For a root separation factor of 10,

Error due to unit-ramp command: $ess=0.22$

Error due to unit-ramp disturbance: $ess=0.001$ .

Explanation of Solution

Given:

The I controller with an internal feedback loop of first order plant is as shown below:

System Dynamics, Chapter 10, Problem 10.28P

Where, the parameter values are as given:

$I=c=4$

Also, the performance specifications require the time constant of the system to be $τ=0.2$ .

The values of gains for the I controller are as follows:

Case 1. For $ζ=0.707$ , $KI=200$ and $K2=36$ .

Case 2. For $ζ=1$ , $KI=100$ and $K2=36$ .

Case 3. For a root separation factor of 10, $KI=1000$ and $K2=216$ .

Concept Used:

The transfer functions for the block diagram are as shown below:

$Ω(s)Ωr(s)=KIIs2+s(c+K2)+KIΩ(s)Td(s)=−sIs2+s(c+K2)+KI$

The steady-state error of a system using final value theorem is:

$ess=lims→0sE(s)$ .

Calculation:

From the block diagram as shown, the transfer functions are as:

$Ω(s)Ωr(s)=KIIs2+s(c+K2)+KIΩ(s)Td(s)=−sIs2+s(c+K2)+KI$

Therefore, the response $Ω(s)$ for the system is:

$Ω(s)=KIIs2+s(c+K2)+KIΩr(s)−sIs2+s(c+K2)+KITd(s)$

And from the block diagram shown in figure, we have

$E(s)=(Ωr(s)−Ω(s))⇒E(s)=(1−KIIs2+s(c+K2)+KI)Ωr(s)+sIs2+s(c+K2)+KITd(s)∵Ω(s)=KIIs2+s(c+K2)+KIΩr(s)−sIs2+s(c+K2)+KITd(s)$

On keeping the values of the parameters such that $I=c=4$

$E(s)=(1−KIIs2+s(c+K2)+KI)Ωr(s)+sIs2+s(c+K2)+KITd(s)⇒E(s)=(s(Is+c+K2)Is2+s(c+K2)+KI)Ωr(s)+sIs2+s(c+K2)+KITd(s)⇒E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)$

Case 1. When $ζ=0.707$ , the gain values are $KI=200$ and $K2=36$ :

Since,

$E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)∵KI=200 and K2=36⇒E(s)=(s(4s+4+36)4s2+s(4+36)+200Ωr(s)+s4s2+s(4+36)+200Td(s))⇒E(s)=(s(4s+40)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))$

Therefore, for unit-ramp command response $Ωr(s)$ and zero disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))⇒E(s)=(s(4s+40)4s2+40s+200⋅1s2+s4s2+40s+200⋅0)⇒E(s)=(4s+40)(4s2+40s+200)⋅1s⇒E(s)=(s+10)(s2+10s+50)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s(s+10)(s2+10s+50)⋅1s ∵E(s)=(s+10)(s2+10s+50)⋅1s⇒ess=lims→0(s+10)(s2+10s+50)=1050⇒ess=0.2$

Therefore, for zero command response $Ωr(s)$ and unit-ramp disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))⇒E(s)=(s(4s+40)4s2+40s+200⋅0+s4s2+40s+200⋅1s2)⇒E(s)=1(4s2+40s+200)⋅1s⇒E(s)=14(s2+10s+50)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s14(s2+10s+50)⋅1s ∵E(s)=14(s2+10s+50)⋅1s⇒ess=lims→014(s2+10s+50)=1200⇒ess=0.005$

Case 2. When $ζ=1$ , the gain values are $KI=100$ and $K2=36$ :

Since,

$E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)∵KI=100 and K2=36⇒E(s)=(s(4s+4+36)4s2+s(4+36)+100Ωr(s)+s4s2+s(4+36)+100Td(s))⇒E(s)=(s(4s+40)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))$

Therefore, for unit-ramp command response $Ωr(s)$ and zero disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))⇒E(s)=(s(4s+40)4s2+40s+100⋅1s2+s4s2+40s+100⋅0)⇒E(s)=(4s+40)(4s2+40s+100)⋅1s⇒E(s)=(s+10)(s2+10s+25)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s(s+10)(s2+10s+25)⋅1s ∵E(s)=(s+10)(s2+10s+25)⋅1s⇒ess=lims→0(s+10)(s2+10s+25)=1025⇒ess=0.4$

Therefore, for zero command response $Ωr(s)$ and unit-ramp disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))⇒E(s)=(s(4s+40)4s2+40s+100⋅0+s4s2+40s+100⋅1s2)⇒E(s)=1(4s2+40s+100)⋅1s⇒E(s)=14(s2+10s+25)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s14(s2+10s+25)⋅1s ∵E(s)=14(s2+10s+25)⋅1s⇒ess=lims→014(s2+10s+25)=1100⇒ess=0.01$

Case 3. For a root separation factor of 10, $KI=1000$ and $K2=216$ .

Since,

$E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)∵KI=1000 and K2=216⇒E(s)=(s(4s+4+216)4s2+s(4+216)+1000Ωr(s)+s4s2+s(4+216)+1000Td(s))⇒E(s)=(s(4s+220)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))$

Therefore, for unit-ramp command response $Ωr(s)$ and zero disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+220)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))⇒E(s)=(s(4s+220)4s2+220s+1000⋅1s2+s4s2+220s+1000⋅0)⇒E(s)=(4s+220)(4s2+220s+1000)⋅1s⇒E(s)=(s+55)(s2+55s+250)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s(s+55)(s2+55s+250)⋅1s ∵E(s)=(s+55)(s2+55s+250)⋅1s⇒ess=lims→0(s+55)(s2+55s+250)=55250⇒ess=0.22$

Therefore, for zero command response $Ωr(s)$ and unit-ramp disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+220)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))⇒E(s)=(s(4s+220)4s2+220s+1000⋅0+s4s2+220s+1000⋅1s2)⇒E(s)=1(4s2+220s+1000)⋅1s⇒E(s)=14(s2+55s+250)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s14(s2+55s+250)⋅1s ∵E(s)=14(s2+55s+250)⋅1s⇒ess=lims→0(s+1)4(s2+55s+250)=11000⇒ess=0.001$ .

Conclusion:

The steady-state error values for all the cases are:

Case 1. For $ζ=0.707$ ,

Error due to unit-ramp command: $ess=0.2$

Error due to unit-ramp disturbance: $ess=0.005$

Case 2. For $ζ=1$ , $Tm≈35 N⋅m$ .

Error due to unit-ramp command: $ess=0.4$

Error due to unit-ramp disturbance: $ess=0.01$

Case 3. For a root separation factor of 10, $Tm≈215 N⋅m$ .

Error due to unit-ramp command: $ess=0.22$

Error due to unit-ramp disturbance: $ess=0.001$ .

Answer 7

Chapter 10, Problem 10.28P

To determine

The steady-state error due to a unit-ramp command and due to a unit-ramp disturbance of I controller with an internal feedback loop of first order plant for the given cases.

Answer 8

Expert Solution & Answer

Answer to Problem 10.28P

The steady-state error values for all the cases are:

Case 1. For $ζ=0.707$ ,

Error due to unit-ramp command: $ess=0.2$

Error due to unit-ramp disturbance: $ess=0.005$

Case 2. For $ζ=1$ ,

Error due to unit-ramp command: $ess=0.4$

Error due to unit-ramp disturbance: $ess=0.01$

Case 3. For a root separation factor of 10,

Error due to unit-ramp command: $ess=0.22$

Error due to unit-ramp disturbance: $ess=0.001$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

The I controller with an internal feedback loop of first order plant is as shown below:

System Dynamics, Chapter 10, Problem 10.28P

Where, the parameter values are as given:

$I=c=4$

Also, the performance specifications require the time constant of the system to be $τ=0.2$ .

The values of gains for the I controller are as follows:

Case 1. For $ζ=0.707$ , $KI=200$ and $K2=36$ .

Case 2. For $ζ=1$ , $KI=100$ and $K2=36$ .

Case 3. For a root separation factor of 10, $KI=1000$ and $K2=216$ .

Concept Used:

The transfer functions for the block diagram are as shown below:

$Ω(s)Ωr(s)=KIIs2+s(c+K2)+KIΩ(s)Td(s)=−sIs2+s(c+K2)+KI$

The steady-state error of a system using final value theorem is:

$ess=lims→0sE(s)$ .

Calculation:

From the block diagram as shown, the transfer functions are as:

$Ω(s)Ωr(s)=KIIs2+s(c+K2)+KIΩ(s)Td(s)=−sIs2+s(c+K2)+KI$

Therefore, the response $Ω(s)$ for the system is:

$Ω(s)=KIIs2+s(c+K2)+KIΩr(s)−sIs2+s(c+K2)+KITd(s)$

And from the block diagram shown in figure, we have

$E(s)=(Ωr(s)−Ω(s))⇒E(s)=(1−KIIs2+s(c+K2)+KI)Ωr(s)+sIs2+s(c+K2)+KITd(s)∵Ω(s)=KIIs2+s(c+K2)+KIΩr(s)−sIs2+s(c+K2)+KITd(s)$

On keeping the values of the parameters such that $I=c=4$

$E(s)=(1−KIIs2+s(c+K2)+KI)Ωr(s)+sIs2+s(c+K2)+KITd(s)⇒E(s)=(s(Is+c+K2)Is2+s(c+K2)+KI)Ωr(s)+sIs2+s(c+K2)+KITd(s)⇒E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)$

Case 1. When $ζ=0.707$ , the gain values are $KI=200$ and $K2=36$ :

Since,

$E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)∵KI=200 and K2=36⇒E(s)=(s(4s+4+36)4s2+s(4+36)+200Ωr(s)+s4s2+s(4+36)+200Td(s))⇒E(s)=(s(4s+40)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))$

Therefore, for unit-ramp command response $Ωr(s)$ and zero disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))⇒E(s)=(s(4s+40)4s2+40s+200⋅1s2+s4s2+40s+200⋅0)⇒E(s)=(4s+40)(4s2+40s+200)⋅1s⇒E(s)=(s+10)(s2+10s+50)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s(s+10)(s2+10s+50)⋅1s ∵E(s)=(s+10)(s2+10s+50)⋅1s⇒ess=lims→0(s+10)(s2+10s+50)=1050⇒ess=0.2$

Therefore, for zero command response $Ωr(s)$ and unit-ramp disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))⇒E(s)=(s(4s+40)4s2+40s+200⋅0+s4s2+40s+200⋅1s2)⇒E(s)=1(4s2+40s+200)⋅1s⇒E(s)=14(s2+10s+50)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s14(s2+10s+50)⋅1s ∵E(s)=14(s2+10s+50)⋅1s⇒ess=lims→014(s2+10s+50)=1200⇒ess=0.005$

Case 2. When $ζ=1$ , the gain values are $KI=100$ and $K2=36$ :

Since,

$E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)∵KI=100 and K2=36⇒E(s)=(s(4s+4+36)4s2+s(4+36)+100Ωr(s)+s4s2+s(4+36)+100Td(s))⇒E(s)=(s(4s+40)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))$

Therefore, for unit-ramp command response $Ωr(s)$ and zero disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))⇒E(s)=(s(4s+40)4s2+40s+100⋅1s2+s4s2+40s+100⋅0)⇒E(s)=(4s+40)(4s2+40s+100)⋅1s⇒E(s)=(s+10)(s2+10s+25)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s(s+10)(s2+10s+25)⋅1s ∵E(s)=(s+10)(s2+10s+25)⋅1s⇒ess=lims→0(s+10)(s2+10s+25)=1025⇒ess=0.4$

Therefore, for zero command response $Ωr(s)$ and unit-ramp disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+40)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))⇒E(s)=(s(4s+40)4s2+40s+100⋅0+s4s2+40s+100⋅1s2)⇒E(s)=1(4s2+40s+100)⋅1s⇒E(s)=14(s2+10s+25)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s14(s2+10s+25)⋅1s ∵E(s)=14(s2+10s+25)⋅1s⇒ess=lims→014(s2+10s+25)=1100⇒ess=0.01$

Case 3. For a root separation factor of 10, $KI=1000$ and $K2=216$ .

Since,

$E(s)=(s(4s+4+K2)4s2+s(4+K2)+KI)Ωr(s)+s4s2+s(4+K2)+KITd(s)∵KI=1000 and K2=216⇒E(s)=(s(4s+4+216)4s2+s(4+216)+1000Ωr(s)+s4s2+s(4+216)+1000Td(s))⇒E(s)=(s(4s+220)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))$

Therefore, for unit-ramp command response $Ωr(s)$ and zero disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+220)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))⇒E(s)=(s(4s+220)4s2+220s+1000⋅1s2+s4s2+220s+1000⋅0)⇒E(s)=(4s+220)(4s2+220s+1000)⋅1s⇒E(s)=(s+55)(s2+55s+250)⋅1s$

Thus, the steady-state error for this design is:

$ess=lims→0sE(s)⇒ess=lims→0s(s+55)(s2+55s+250)⋅1s ∵E(s)=(s+55)(s2+55s+250)⋅1s⇒ess=lims→0(s+55)(s2+55s+250)=55250⇒ess=0.22$

Therefore, for zero command response $Ωr(s)$ and unit-ramp disturbance torque $Td(s)$ , we have

$E(s)=(s(4s+220)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))⇒E(s)=(s(4s+220)4s2+220s+1000⋅0+s4s2+220s+1000⋅1s2)⇒E(s)=1(4s2+220s+1000)⋅1s⇒E(s)=14(s2+55s+250)⋅1s$