System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 8, Problem 8.4P
To determine

(a)

The steady-state response xss and the corresponding time taken by the difference between x(0) and xss to get reduced by 98%for the model equation 2x˙+x=10f(t).

Expert Solution
Check Mark

Answer to Problem 8.4P

The steady-state response of the model equation 2x˙+x=10f(t) is xss=10 and the time taken for the difference between x(0) and xss to get reduced by 98% is t=8sec.

Explanation of Solution

Given information:

The given model equation is as:

2x˙+x=10f(t)

With initial conditions as follows:

x(0)=0 and f(t)=us(t).

Concept Used:

  1. Laplace transform is used for obtaining the response of the model equation.
  2. The steady-state response of model is obtained by using the final value theorem:
  3. xss=limtx(t)=lims0sX(s).

  4. For the response of the form T(t)=AeBt, the time constant τ=1B such that B>0

Calculation:

Equation to be solved is as:

2x˙+x=10f(t)

By taking the Laplace of the above equation,

2x˙+x=10f(t)2x˙+x=10us(t) f(t)=us(t)2(sX(s)x(0))+X(s)=10s2(sX(s)0)+X(s)=10s x(0)=0X(s).(2s+1)=10sX(s)=5s(s+12)

Now, on taking the inverse Laplace of the above transfer function, we get

X(s)=5s(s+12)=10(1s1(s+12))x(t)=10(1e12t)

Thus, this is the response for the given model equation. Now, for steady-stateusing final value theorem

xss=limtx(t)=lims0sX(s)xss=lims0s5s(s+12)=10

Therefore, steady-state value for the model equation is xss=10.

Also, from the obtained response x(t)=10(1e12t), the time constant of the model is

τ=2seconds.

Thus, at t=4τ

e12t=e4<0.02

Or say, at t=4τ

x(t)9.8

Thus, the time taken by the response to reach 98% of its steady state value or the time taken for the difference between x(0) and xss to get reduced by 98% is

t=4τ=4×2=8seconds.

Conclusion:

The steady-state response of the model equation 2x˙+x=10f(t) is xss=10 and the time taken for the difference between x(0) and xss to get reduced by 98% is t=8seconds.

To determine

(b)

The steady-state response xss and the corresponding time taken by the difference between x(0) and xss to get reduced by 98% for the model equation 2x˙+x=10f(t).

Expert Solution
Check Mark

Answer to Problem 8.4P

The steady-state response of the model equation 2x˙+x=10f(t) is xss=10 and the time taken for the difference between x(0) and xss to get reduced by 98% is t=7.824seconds.

Explanation of Solution

Given information:

The given model equation is as:

2x˙+x=10f(t)

With initial conditions as follows:

x(0)=5 and f(t)=us(t).

Concept Used:

  1. Laplace transform is used for obtaining the response of the model equation.
  2. The steady-state response of model is obtained by using the final value theorem:
  3. xss=limtx(t)=lims0sX(s).

  4. For the response of the form T(t)=AeBt, the time constant τ=1B such that B>0.

Calculation:

Equation to be solved is as:

2x˙+x=10f(t)

By taking the Laplace of this equation that is,

2x˙+x=10f(t)2x˙+x=10us(t) f(t)=us(t)2(sX(s)x(0))+X(s)=10s2(sX(s)5)+X(s)=10s x(0)=5X(s).(2s+1)=10(s+1)sX(s)=10(s+1)s(2s+1)

Now, on taking the inverse Laplace of the above transfer function, we get

X(s)=10(s+1)s(2s+1)=10(1s12(s+12))x(t)=10(112e12t)

Thus, this is the response for the given model equation. Now, for steady-state using final value theorem

xss=limtx(t)=lims0sX(s)xss=lims0s10(s+1)s(2s+1)=10

Therefore, steady-state value for the model equation is xss=10.

The difference between x(0) and xss is:

|x(0)xss|=|510|=5

x(0)

System Dynamics, Chapter 8, Problem 8.4P And the 98% of this difference is 4.9 units.

x(t) Now, at the time difference between and reduces by 98% or becomes 0.1 units, the response reaches its value equal to 99% of steady-state value or 9.9 units.

Therefore,

x(t)=10(112e12t)9.9=10(112e12t)12e12t=10.9912e12t=0.01e12t=0.02

Taking log on both sides, we get

12t=ln0.0212t=3.912t=7.824seconds

Thus, the time taken for the difference between xss and x(0) to get reduced by 98% is

t=7.824seconds.

Conclusion:

The steady-state response of the model equation 2x˙+x=10f(t) is xss=10 and the time taken for the difference between x(0) and xss to get reduced by 98% is t=7.824seconds.

To determine

(c)

The steady-state response xss and the corresponding time taken by the difference between x(0) and xss to get reduced by 98% for the model equation 2x˙+x=10f(t).

Expert Solution
Check Mark

Answer to Problem 8.4P

The steady-state response of the model equation 2x˙+x=10f(t) is xss=200 and the time taken for the difference between x(0) and xss to get reduced by 98% is t=8seconds.

Explanation of Solution

Given information:

The given model equation is as:

2x˙+x=10f(t)

With initial conditions as follows:

x(0)=0 and f(t)=20us(t).

Concept Used:

  1. Laplace transform is used for obtaining the response of the model equation.
  2. The steady-state response of model is obtained by using the final value theorem:
  3. xss=limtx(t)=lims0sX(s).

  4. For the response of the form T(t)=AeBt, the time constant τ=1B such that B>0.

Calculation:

Equation to be solved is as:

2x˙+x=10f(t)

By taking the Laplace of this equation that is,

2x˙+x=10f(t)2x˙+x=200us(t) f(t)=us(t)2(sX(s)x(0))+X(s)=200s2(sX(s)0)+X(s)=200s x(0)=0X(s).(2s+1)=200sX(s)=100s(s+12)

Now, on taking the inverse Laplace of the above transfer function, we get

X(s)=100s(s+12)=200(1s1(s+12))

x(t)=200(1e12t)

Thus, this is the response for the given model equation. Now, for steady-state using final value theorem

xss=limtx(t)=lims0sX(s)xss=lims0s100s(s+12)=200

Therefore, steady-state value for the model equation is.

x(t)=200(1e12t)

Also, from the obtained response, the time constant of the model is

τ=2seconds.

Thus, at t=4τ

e12t=e4<0.02

Or say, at t=4τ

x(t)196

Thus, the time taken by the response to reach 98% of its steady state value or the time taken for the difference between x(0) and xss to get reduced by 98% is

t=4τ=4×2=8seconds.

Conclusion:

The steady-state response of the model equation 2x˙+x=10f(t) is xss=200 and the time taken for the difference between x(0) and xss to get reduced by 98% is t=8seconds.

Thus, as in part a. and in c, we see that the time taken for the difference between x(0) and xss remains unchanged regardless of the changes in input response f(t). However, the steady-state response gets affected with a change in input response f(t).

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