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  UNDER CONSTRUCTION

Control Systems:
Solved Problems




\(\bb{Pr1}\ \)[Circuit \(\rightarrow\) State space]: Find the state and output equations for the electrical network shown below if the output vector is \(\bn{y}=[v_{R_2}\ i_{R_2}]^T\).

[Sol]
\(\hspace{55px} \bb{1}\) Label all the branch currents on the network, as in the figure
\(\hspace{55px} \bb{2}\) Select the state variables by listing the \(IV\)-relations for all energy storage elements:
\(\ds L\frac{di_L}{dt}=v_L,\ \ C\frac{dv_C}{dt}=i_C\quad \bb{1^*}\)

\(\hspace{55px}\) Select the state variables to be the differentiated variables: \(i_L, v_C\).
\(\hspace{55px} \bb{3}\) Transform the RHS of \(\bb{1^*}\) into linear combinations of state variables and input source. Applying KVL & KCL,
\(\ds v_L=v_C+i_{R_2}R_2,\ i_{R_2}=i_C+4v_L\Rightarrow v_L=v_C+(i_C+4v_L)R_2\)
\(\Rightarrow\boxed{v_L=\frac{1}{1-4R_2}(v_C+i_CR_2)}\quad \bb{2^*}\)

\(\hspace{55px}\) To obtain \(v_L\) in terms of state variables, find \(i_C\) in terms of state variables. Applying KCL,
\(\ds \begin{align} i_C &= i(t)-i_{R_1}-i_L \\ &= i(t)-\frac{v_{R_1}}{R_1}-i_L \\ &= \boxed{i(t)-\frac{v_L}{R_1}-i_L=i_C}\quad \bb{3^*}\end{align}\)
\(\hspace{55px}\) Rewrite \(\bb{2^*}\) and \(\bb{3^*}\) to obtain two simultaneous equations yielding \(v_L\) and \(i_C\) as linear combinations of state variables \(i_L\) and \(v_C\):
\(\ds \begin{align} (1-4R_2)v_L-R_2i_C &= v_C \\ -\frac{1}{R_1}v_L-i_C &= i_L-i(t)\end{align}\)
\(\hspace{55px}\) Solving for \(v_L\) and \(i_C\) yields
\(\begin{matrix} \begin{align} v_L &= \frac{1}{\Delta}[R_2i_L-v_C-R_2i(t)],\ \ \bb{4^*}\\i_C&=\frac{1}{\Delta}[(1-4R_2)i_L+\frac{1}{R_1}v_C-(1-4R_2)i(t)],\ \ \bb{5^*}\end{align} \hspace{30px}\Delta = -\lrbra{(1-4R_2)+\frac{R_2}{R_1}}. \end{matrix}\)
\(\hspace{55px}\) Substituting \(\bb{4^*}\) and \(\bb{5^*}\) into \(\bb{1^*}\) and writing the result in vector-matrix form renders the state equation:
\(\ds \Bxred{\mtx{\dot{i}_L}{\dot{v}_C}=\mtxx{R_2/(L\Delta)}{-1/(L\Delta)}{(1-4R_2)/(C\Delta)}{1/(R_1C\Delta)}\mtx{i_L}{v_C}+\mtx{-R_2/(L\Delta)} {-(1-4R_2)/(C\Delta)}i(t)}\)
\(\hspace{55px}\bb{4}\) Derive the output equation. Since the output variables are \(v_{R_2}\) and \(i_{R_2}\), note that around the mesh containing \(C,L\) and \(R_2\),
\(\ds \begin{align}v_{R_2}&=-v_C+v_L,&&\bb{6^*}\\i_{R_2}&=i_C+4v_L && \bb{7^*}\end{align}\)

\(\hspace{55px}\) Substituting \(\bb{4^*}\) & \(\bb{5^*}\) into \(\bb{6^*}\) and \(\bb{7^*}\), \(v_{R_2}\) and \(i_{R_2}\) are obtained as linear combinations of the state variables, \(i_L\) and \(v_C\). In vector-matrix
\(\hspace{55px}\) form, the output equation is
\(\ds \Bxred{\mtx{v_{R_2}}{i_{R_2}}=\mtxx{R_2/\Delta}{-(1+1/\Delta)}{1/\Delta}{(1-4R_1)/(\Delta R_1)}\mtx{i_L}{v_C}+ \mtx{-R_2/\Delta}{-1/\Delta}i(t)}\)
\(\hspace{55px}\vplup\) Tip: Write KCL/KVL's so to include as many state variables and variables in \(\bb{1^*}\) as possible, to minimize # of eq's to be solved
\(\bb{Pr2}\ \)[Transfer function \(\rightarrow\) State space]: Find the state-space representation of the transfer function shown below:

[Sol]
\(\hspace{55px} \bb{1}\) Separate the system into two cascaded blocks, as shown below; the first block contains the denominator and the second the numerator.
\(\hspace{55px} \bb{2}\) Write the differential equation for the first block:
\(\ds \begin{align}\frac{X_1(s)}{R(s)} &= \frac{1}{s^3+9s^2+26s+24} \\ (s^3+9s^2+26s+24)X_1(s) &= R(s) \Rightarrow \\ \dddot{x}_1 + 9\ddot{x}_1 + 26\dot{x}_1 + 24x_1 &= r \ \ \bb{1^*}\end{align}\)
\(\hspace{55px} \bb{3}\) Set the state variables:
\(\ds \begin{align} x_1 &= x_1 \\ x_2 &= \dot{x}_1 \\ x_3 &= \ddot{x}_1 \end{align}\)
\(\hspace{55px}\) Differentiating both sides and applying \(\bb{1^*}\) to find \(\dddot{x}_1=\dot{x}_3\), we obtain the state equations:
\(\ds\nmttx{\dot{x}_1=}{\dot{x}_2=}{\dot{x}_3=}\)\(\mtxxx{\hspace{-10px}}{x_2\hspace{-10px}}{}{\hspace{-10px}} {\hspace{-10px}}{x_3}{-\ 24x_1\hspace{-10px}}{-\ 26x_2\hspace{-10px}}{-\ 9x_3}\)\(\nmttx{}{}{+\ r\ }\)
\(\hspace{55px} \) In vector-matrix form,
\(\ds\bxred{\lrbra{\amttx{\dot{x_1}}{\dot{x_2}}{\dot{x_3}}}=\lrbra{\amtxxx{\ph{-2}0}{\ph{-2}1}{\ph{-}0}{\ph{-2}0} {\ph{-2}0}{\ph{-}1}{-24}{-26}{-9}}\lrbra{\amttx{x_1}{x_2}{x_3}}+\lrbra{\amttx{0}{0}{1}}r}\)
\(\hspace{55px} \bb{4}\) Introduce the effect of the numerator block. Writing in input-output form and taking inverse Laplace transform w/ I.C.'s \(=0\), we get
\(\ds \begin{align}C(s)&=(s^2+7s+2)X_1(s)\Rightarrow\\c&=\ddot{x_1}+7\dot{x_1}+2x_1\end{align}\)
\(\hspace{55px}\) Rewriting in terms of state variables,
\(\ds y = c(t)=x_3+7x_2+2x_1\Rightarrow\)
\(\bxred{y = [\hspace{1px} 2\ \ 7\ \ 1\hspace{1px} ]\lrbra{\nmttx{x_1}{x_2}{x_3}}}\)
\(\hspace{55px}\) Thus, the last box "collects" the states and generates the output equation. The equivalent block diagram is hence

\(\bb{Pr3}\ \)[State space \(\rightarrow\) Transfer function]: Find the transfer function of the system defined by

\(\ds \bn{\dot{x}}=\lrbra{\amtxxx{\ph{-}0}{\ph{-}1}{\ph{-}0}{\ph{-}0}{\ph{-}0}{\ph{-}1}{-1}{-2}{-3}}\bn{x}+\lrbra{\amttx{10}{0}{0}}u\), \(\ds \quad y=\lrbra{\hspace{1px}1\ \ 0\ \ 0\hspace{1px}}\bn{x}\)

[Sol]
\(\hspace{55px}\) Since \(\bn{D}=0\), the solution revolves around finding the resolvent, \((s\bn{I}-\bn{A})^{-1}\), in \(T(s)=\bn{C}(s\bn{I}-\bn{A})^{-1}\bn{B}+\bn{D}\ \bb{1^*}\). Hence, first find
\(\hspace{55px}\ (s\bn{I}-\bn{A})\):
\(\ds (s\bn{I}-\bn{A}) = \lrbra{\amtxxx{s}{0}{0}{0}{s}{0}{0}{0}{s}}-\lrbra{\amtxxx{\ph{-}0}{\ph{-}1}{\ph{-}0}{\ph{-}0}{\ph{-}0}{\ph{-}1}{-1}{-2}{-3}} = \lrbra{\amtxxx{s}{-1}{0}{0}{s}{-1}{1}{2}{s+3}}\)
\(\hspace{55px}\) Now, determine \((s\bn{I}-\bn{A})^{-1}\):
\(\ds (s\bn{I}-\bn{A})^{-1} = \frac{\t{adj}(s\bn{I}-\bn{A})}{\t{det}(s\bn{I}-\bn{A})} = \frac{\lrbra{\amtxxx{(s^2+3s+2)}{(s+3)}{1} {-1}{s(s+3)}{s}{-s}{-(2s+1)}{s^2}}}{s^3+3s^2+2s+1}\)
\(\hspace{55px}\) Substituting into \(\bb{1^*}\), with \(\bn{B}=[\hspace{1px}10\ \ 0\ \ 0\hspace{1px}]^T\), and \(\bn{C}=[\hspace{1px}1\ \ 0\ \ 0\hspace{1px}]\), we obtain
\(\ds \bxred{T(s)=\frac{10(s^2+3s+2)}{s^3+3s^2+2s+1}}\)

\(\bb{Pr4}\ \)[Output \(\rightarrow\) Transfer function](First-order): Given the unit step response below, estimate the system transfer function.

[Sol] The output waveform exhibits first-order characteristics: no overshoot and nonzero initial slope. Hence, it's same to assume the system is
  first-order, with general unit step response given by
\(\ds C(s)=\frac{1}{s}\cdot\frac{K}{s+a}=\frac{K/a}{s}-\frac{K/a}{s+a}\quad\bb{1^*}\)
Begin by finding the time constant \(a\): since the output final value is \(\approx 0.72\), the time constant is evaluated where the curve reaches \(0.63\times 0.72 = 0.45\), or about \(0.13\ \t{sec}\). Hence, \(0.13^{-1} = \boxed{a = 7.7}\).
To find \(K\), note from \(\bb{1^*}\) that the forced response reaches a steady-state value of \(K/a=0.72\). Substituting \(a\), we find \(\boxed{K=5.54}\) - and thus the system transfer function is
\(\ds\bxred{G(s)=\frac{5.54}{s+7.7}}\)
Note that the exact system transfer function used in generating the figure was \(G(s)=5/(s+7)\).

\(\bb{Pr5}\ \)[State-variable feedback design]: For plant below, design phase-variable feedback gains to yield \(9.5\)% overshoot and a setting time of \(0.74\) sec.:
\(\ds G(s)=\frac{20(s+5)}{s(s+1)(s+4)}\quad \bn{[1^*]}\)

[Sol] The solution shall be executed as follows:

[1]: Represent the plant in phase-variable form
[2]: Feed back each phase variable to the input of the plant through a gain \(k_i\)
[3]: Find the char. eq. for the cl-sys. in [2]
[4]: Decide upon all cl-p locations and determine an equivalent char. eq.
[5]: Equate like coefficients of the char. eqs. from [3] and [4] and solve for \(k_i\).

Begin by obtaining the desired char. eq. Using the transient response requirements, the cl-p's are \(s_{1,2}=-5.4\pm j7.2\) (see Time Response).
\(\up\)Next, select another cl'p (since sys. is third-order). The cl-sys. will have a zero at \(-5\), the same as the ol-sys. We could select the third cl-p to cancel the cl-z; however, to demonstrate the effect of the third pole and the design process, including the need for simulation, let us choose \(-5.1\) instead.
\(\up\)Now draw the signal-flow diagram for the plant, (a):

phase-variable plant representation
Next feed back all state variables to the control, \(u\), through gains \(k_i\), as shown in (b).
From (b), the cl-sys. state equations are:
\(\ds\bn{\dot{x}}=\lrbra{\mtxxx{0}{1}{0}{0}{0}{1}{-k_1}{-(4+k_2)}{-(5+k_3)}}\bn{x}+\lrbra{\amttx{0}{0}{1}}\)\(\quad \bn{[2^*]}\)
\(\dsup y = \lrbra{100\ \ 20\ \ 0}\bn{x}\)

Comparing \(\bn{[2^*]}\) to \(\bn{[1^*]}\), we identify the cl-sys. matrix as\(\up\)
\(\ds \bn{A}-\bn{BK} = \lrbra{\mtxxx{0}{1}{0}{0}{0}{1}{-k_1}{-(4+k_2)}{-(5+k_3)}}\hspace{50px}\)

To find the cl-sys. char. eq., form
\(\ds \t{det}(s\bn{I}-(\bn{A}-\bn{BK}))=s^3 + (5+k_3)s^2 + (4+k_2)s + k_1 = 0\)

This equation must match the desired char. eq.,
\(\ds s^3 + 15.9s^2 +136.08s +413.1 = 0\)

- formed from the poles \(s=-5.4+j7.2,-5.4-j7.2,\) and \(-5.1\), as determined previously.
\(\up\)Equating the coefficients from prior to eqs, we obtain
\(\bxred{k_1 = 413.1,\ k_2 = 132.08,\ k_3=10.9}\)

Completing the design; the zero term of the cl transfer function is the same as that of the ol-sys, or \((s+5)\).
\(\up\)Using \(\bn{[2^*]}\), we obtain the following state-space representation of the cl-sys:
\(\ds \bn{\dot{x}} = \lrbra{\mtxxx{0}{1}{0}{0}{0}{1}{-413.1}{-136.08}{-15.9}}\bn{x}+\lrbra{\amttx{0}{0}{1}}r\)
\(\dsup y=[100\ 20\ 0]\bn{x}\)

- and the transfer function is thus
\(\ds \bxred{T(s)=\frac{20(s+5)}{s^3+15.9s^2+136.08s+413.1}}\)

A simulation of the cl-sys is shown below; the plot displays a \(11.5\)% overshoot and a setting time of \(0.8\) sec. A redesign with the third pole canceling the zero at \(-5\) will yield performance equal to the requirements.
\(\up\)Lastly, since the ss-response approaches \(0.24\) instead of unity, the ss-error is large - which can be reduced using integral control.
Closed-loop system output





Dragon Notes,   Est. 2018     About

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