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The system is described by the following differential equation:

\n

\\[{\\var{a_2}}\\frac{d^2y}{dt^2} + ({\\var{a_1}})\\frac{dy}{dt}+(\\var{a_0})y+(\\var{b_2})\\frac{d^2u}{dt^2}+(\\var{b_1})\\frac{du}{dt}+(\\var{b_0})u=0\\]

\n

The transfer function is given by \\[\\frac{Y(s)}{U(s)} = \\frac{b_2s^2+b_1s+b_0}{a_2s^2+a_1s+a_0}\\]

\n

\\(b_2 = \\)[[3]] \\(b_1 = \\)[[4]] \\(b_0 = \\)[[5]]

\n

\\(a_2 = \\)[[0]] \\(a_1 = \\)[[1]] \\(a_0 = \\)[[2]]

\n

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Find the transfer function of the following differential equation by taking the Laplace transform.

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This question tests the students skill on inverse laplace transforms of simple functions in the s-domain. The answer should be a constant and a single exponential expression. Coefficients and constants of the function in s-domain are created randomly.

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Find the inverse Laplace transform of the following function in s-domain.

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The function is described by the following equation:

\n

\\[Y(s)=\\frac{\\var{constant}}{s} + \\frac{\\var{exp_constant}}{s+\\var{exp_exponent}}\\]

\n

\n

The function in time-domain is given by \\(y(t) = \\) \\((\\)[[0]]\\( + \\) [[1]]\\(e^{-} \\)[[2]]\\(^t)u(t)\\)
 

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An open-loop system is represented by the transfer function \\[G(s)=\\frac{K}{s(s+1)(s+2)}, K>0\\]

\n

Closing the loop with unity feedback, determine which of the following is true.

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For the following characteristic polynomials, decide whether the corresponding system is stable, unstable or if you can't tell by simply observing the signs of the coefficients. Assume a,b,c,d are all greater than zero. For each row you get

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Try to, by inspection, determine what the roots of the expression will be equal to. Sign of the roots will influence stability of the system.

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For a second-order system with the following poles, where \\(a>0\\), \\(b>0, a\\neq b\\) and \\(a\\) and \\(b\\) are real numbers, indicate the kind of system from the given categories.

\n

For each row you get

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Assuming that the system is first-order, estimate both parameters \\(a\\) and \\(b\\) for the system transfer function \\(G(s)=\\frac{b}{s+a}\\)

\n

Specify the answer with a precision of two decimals. All intermediate calculations should be full precision. Failing to do so will be interpreted as a wrong answer.

\n

\\(a =\\) [[0]]

\n

\\(b =\\) [[1]]

\n

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In the lab, you have excited the unknown system \\(G(s)\\) several times with step input signals and after some signal processing determined that the system can best be described by parameters below. The parameters were obtained after the system responded to a step input \\(r(t)=\\var{step_size}u(t)\\) where \\(u(t)\\) is the unit step input.

\n

From testing, you found that the final value of the step response is given by \\(y(\\infty) = \\var{y_inf}\\). Also, you have found that \\(y(\\var{time_constant})=0.63y(\\infty)\\) for \\(\\tau=\\var{time_constant}\\) seconds.

\n

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Determine the percent overshoot, give your answer with one decimal.

\n

\\(\\%OS = \\) [[0]]

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Assuming the system is second-order, estimate the parameters \\(\\zeta\\), \\(\\omega_n\\) and \\(K\\) for the system transfer function \\(G(s)=\\frac{K}{s^2+2\\zeta\\omega_n s+\\omega_n^2}\\)

\n

Specify the answer with a precision of two decimals. All intermediate calculations should be full precision. Failing to do so will be interpreted as a wrong answer.

\n

\\(\\zeta = \\) [[0]]

\n

\\(\\omega_n = \\) [[1]]

\n

\\(K = \\) [[2]]

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In the lab, you have excited the unknown system \\(G(s)\\) several times with step input signals and after some signal processing determined that the system can best be described by parameters below. The parameters were obtained after the system responded to a step input \\(r(t)=\\var{step_size}u(t)\\) where \\(u(t)\\) is the unit step input.

\n

From testing, you found that the final value of the step response is given by \\(y(\\infty) = \\var{y_inf}\\). Also, the maximum value of the response is given by \\(y_{max}=\\var{y_max}\\). Finally, the \\(\\pm2\\%\\) settling time is given by \\(T_s=\\var{settling_time}\\) seconds.

\n

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Find the transfer function \\(\\frac{V_L(s)}{V(s)}\\) for the electrical network.

\n

\\(\\frac{V_L(s)}{V(s)} = \\frac{b_0s}{a_2s^2+a_1s+a_0}\\)

\n

\\(b_0 = \\) [[0]]
\\(a_2 = \\) [[1]]
\\(a_1 = \\) [[2]]
\\(a_0 = \\) [[3]]

\n

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The constant in the denumerator

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The resistance of R1 in ohm.

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The resistance of R2 in ohm.

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The number of the numerator.

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The coefficient of \\(s\\) in the denumerator

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The coefficient of \\(s^2\\) in the denumerator

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The inductance of coil L1 in Henry

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The inductance of coil L2 in Henry

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An electrical network is presented below. The following variables describe the circuit.

\n

\\(L1 = \\var{L1}H\\)
\\(L2 = \\var{L2}H\\)
\\(R1 = \\var{R1}\\Omega\\)
\\(R2 = \\var{R2}\\Omega\\)

\n

\n

(Image adapted from Nise's Control Systems Engineering (2011))

\n

\n

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Suggested steps:

\n
    \n
  1. Fill in values for all circuit elements and replace them with their Laplace transform
  2. \n
  3. Perform mesh or nodal analysis and find the set of relationships that describe the circuit.
  4. \n
  5. Solve for the desired transfer function.
  6. \n
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Consider the Simulink system below with unit step signals \\(R(s)\\) (left input) and \\(D(s)\\) (top input) that can be switched on or off as indicated.
\\(R(s)\\) is a reference signal and \\(D(s)\\) is a disturbance signal. The output of the subsystem \\(\\frac{25}{s^2+25s}\\) is the system output, which is connected with unity feedback and also to a scope.

\n

\n

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Assuming \\(D(s)\\) is switched to zero, find the transfer function \\(T(s)=\\frac{Y(s)}{R(s)} = \\frac{b_2s^2+b_1s+b_0}{a_2s^2+a_1s+a_0}\\).

\n

(If any coefficient of the resulting two polynomials is missing, fill in a zero for that respective coefficient below.)

\n

\\(b_2 = \\) [[0]]
\\(b_1 = \\) [[1]]
\\(b_0 = \\) [[2]]

\n

\\(a_2 = \\) [[3]]
\\(a_1 = \\) [[4]]
\\(a_0 = \\) [[5]]

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Determine the steady-state error due to a unit step reference input only (\\( D(s) \\) switched off). Give your answer with two decimals.
\\(e_R(\\infty) = y(\\infty) - r(\\infty) = \\) [[0]]

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Determine the overall steady-state error when both inputs are switched on. Give your answer with two significant digits. 
\\(e_R(\\infty) = y(\\infty) - r(\\infty) = \\) [[0]]

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Steady state error \\(e_R(\\infty)\\)

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The student is shown two radio choices: \"Yes\" and \"No\". One of them is correct.

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What is the determinant of the controllability matrix?

\n

\\(det(C_{Mz}) = \\) [[0]]

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Is the system controllable?

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What is the desired third-order characteristic equation \\(P_d = d_3s^3+d_2s^2+d_1s+d_0\\) for the system?

\n

Specify the coefficients using two decimals.

\n

\\(d_3 = \\) [[0]]
\\(d_2 = \\) [[1]]
\\(d_1 = \\) [[2]]
\\(d_0 = \\) [[3]]

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The system in phase-variable form is given by

\n

\\(\\textbf{A} = \\begin{bmatrix} a_{11} & a_{12} & a_{13} \\\\a_{21} & a_{22} & a_{23} \\\\ a_{31} & a_{32} & a_{33} \\end{bmatrix}, \\textbf{B} = \\begin{bmatrix} b_{1} \\\\ b_{2} \\\\ b_{3} \\end{bmatrix}, \\textbf{C} = \\begin{bmatrix} c_{1} & c_{2} & c_{3} \\end{bmatrix}, \\textbf{D} = 0 \\)

\n

where

\n

\\(a_{11} = \\) [[0]], \\(a_{12} = \\) [[1]], \\(a_{13} = \\) [[2]]
\\(a_{21} = \\) [[3]], \\(a_{22} = \\) [[4]], \\(a_{23} = \\) [[5]]
\\(a_{31} = \\) [[6]], \\(a_{32} = \\) [[7]], \\(a_{33} = \\) [[8]]

\n


\\(b_{1} = \\) [[9]], \\(b_{2} = \\) [[10]] , \\(b_{3} = \\) [[11]]
\\(c_{1} = \\) [[12]], \\(c_{2} = \\) [[13]] , \\(c_{3} = \\) [[14]]

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With the state feedback for the system in \\(\\textit{phase-variable}\\) form, the control signal becomes \\(u=-Kx\\) where the feedback gains \\(K = \\begin{bmatrix} k_{1} & k_{2} & k_{3} \\end{bmatrix}\\). Enter the gains below with two decimals.

\n

\\(k_1 = \\) [[0]]
\\(k_2 = \\) [[1]]
\\(k_3 = \\) [[2]]

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We can transform these gains to a suitable representation for the original cascaded system through a transformation \\(K_Z = KP^{-1}\\)

\n

The transformation matrix \\(P\\) is given by

\n

\\(\\textbf{P} = \\begin{bmatrix} p_{11} & p_{12} & p_{13} \\\\p_{21} & p_{22} & p_{23} \\\\ p_{31} & p_{32} & p_{33} \\end{bmatrix} \\)

\n

where

\n

\\(p_{11} = \\) [[0]] \\(p_{12} = \\) [[1]] \\(p_{13} = \\) [[2]]
\\(p_{21} = \\) [[3]] \\(p_{22} = \\) [[4]] \\(p_{23} = \\) [[5]]
\\(p_{31} = \\) [[6]] \\(p_{32} = \\) [[7]] \\(p_{33} = \\) [[8]]

\n

Use at least two decimals if needed.

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Transforming back to the original cascade system, the state feedback gains \\(K_Z = \\begin{bmatrix} k_{1,z} & k_{2,z} & k_{3,z} \\end{bmatrix}\\) are given by

\n

\\(k_{1,z} = \\) [[0]]
\\(k_{2,z} = \\) [[1]]
\\(k_{3,z} = \\) [[2]]

\n

Use at least 2 decimals.

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Finally, simulate a unit step response for the original cascade system in closed-loop with the state feedback gains \\(K_Z\\) and check the following response characteristics using the \\(\\textbf{stepinfo}\\) function:

\n

%OS =  [[0]] (use 1 decimal, e.g. 0.1)
Settling time = [[1]] (use 2 decimals, e.g. 0.12)
Steady-state value of the output = [[2]] (use 3 decimals, e.g. 0.123)

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Has the %OS criterion been satisfied?

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Has the settling time criterion been satisfied?

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An open-loop plant is given in cascade form as

\n

\\(\\textbf{A} = \\begin{bmatrix} -7 & 1 & 0 \\\\ 0 & -8 & 1 \\\\ 0 & 0 & -9 \\end{bmatrix}, \\textbf{B} = \\begin{bmatrix} 0 \\\\ 0 \\\\ 1 \\end{bmatrix}, \\textbf{C} = \\begin{bmatrix} -1 & 1 & 0 \\end{bmatrix}, \\textbf{D} = 0 \\)

\n

We will design a linear state-feedback controller to yield a 20% overshoot and a settling time of 2 seconds for a unit step input. The third pole should be placed such that it cancels the zero. We will use the \\(\\textit{phase-variable}\\) form in our design.

\n

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