Control Systems Test 1

The characteristic equation of the system is:

s⁴ + K₁s³ + s² + K₂ s + 1 = 0

For the characteristic equation, we form the Routh’s array as:

For the system having two imaginary poles, the required condition is:

\(K_1^2 - {K_1}{K_2} + K_2^2 = 0\)  

But the above equation has no real roots, so there is no relationship between K₁ and K₂ that will yield just two jω-poles. 

Input = √2 sin 2t

ω = 2 rad/sec

Given:

\(G\left( s \right)=\frac{1}{s+4}\)

Replacing s with jω, we can write:

\(G\left( jω \right)=\frac{1}{jω +4}\)

\(\left| G\left( jω \right) \right|=\frac{1}{\sqrt{{{ω }^{2}}+16}}=\frac{1}{\sqrt{20}}\)

\(\angle (G\left( jω \right)={{\tan }^{-1}}\left( \frac{ω }{4} \right)=26.5{}^\circ\)

Now, the output will be:

\(=\sqrt{2}\times \frac{1}{\sqrt{20}}\sin \left( ω t-26.5{}^\circ \right)\)

\(=0.316\sin \left( ω t-26.5{}^\circ \right)\)

The standard transfer function of a PI controller is

\(T\left( s \right) = {K_p} + \frac{{{K_I}}}{s} = {K_p}\;\left( {1 + \frac{1}{{{T_I}s}}} \right)\)

Where \({T_I} = \frac{{{K_p}}}{{{k_I}}}\) is known as integral or reset time.

Proportional band \(\left( {PB} \right) = \frac{{100}}{{{K_p}}}\)

The transfer function of PI controller is

\(T\left( s \right) = 2.5\;\left( {1 + \frac{1}{{0.25s}}} \right)\)

\(= 2.5\;\left( {1 + \frac{4}{s}} \right)\)

Concept

The steady-state value of the output can be calculated using final value theorem

\(\mathop {{\rm{lt}}}\limits_{s \to 0} s\left[ {y\left( s \right)} \right]\)

\(x\left( t \right)\mathop \to \limits^{L.T} s\;X\left( s \right)\)

\(u\left( t \right)\mathop \to \limits^{L.T} \frac{1}{s}\)

Applications:

\({s^2}\;Y\left( s \right) + \;16\;sY\left( s \right)\; + \;15\;Y\left( s \right)\; = 1 + \frac{5}{s}\)

\(\left( {{s^2} + 16s + 15} \right)Y\left( s \right) = 1 + \frac{5}{s}\)

\(Y\left( S \right) = \frac{{s + 5}}{{s\left( {{s^2}+16s + 15} \right)}}\)

The steady-state value

\(\mathop {\lim }\limits_{S \to 0} \frac{{s\left( {s + 5} \right)}}{{s\left( {{s^2} + 16s + 15} \right)}} = \frac{1}{3}\)

The state-space representation of the system is

\(\left[ {\begin{array}{*{20}{c}} {{{\dot x}_1}}\\ {{{\dot x}_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - \frac{1}{{{T_1}}}}&0\\ 0&{ - \frac{1}{{{T_2}}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {\frac{1}{{{T_1}}}}\\ {\frac{1}{{{T_2}}}} \end{array}} \right]\)

Comparing it to the standard state-space model, we have the matrices:

\(A = \left[ {\begin{array}{*{20}{c}} { - \frac{1}{{{T_1}}}}&0\\ 0&{ - \frac{1}{{{T_2}}}} \end{array}} \right]\;and\;B = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{{T_1}}}}\\ {\frac{1}{{{T_2}}}} \end{array}} \right]\)

\(AB = \left[ {\begin{array}{*{20}{c}} { - \frac{1}{{{T_1}}}}&0\\ 0&{ - \frac{1}{{{T_2}}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\frac{1}{{{T_1}}}}\\ {\frac{1}{{{T_2}}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - \frac{1}{{T_1^2}}}\\ { - \frac{1}{{T_2^2}}} \end{array}} \right]\)

So, the controllability matrix is:

\({C_M} = \left[ {B\;AB} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{{T_1}}}}&{ - \frac{1}{{T_1^2}}}\\ {\frac{1}{{{T_2}}}}&{ - \frac{1}{{T_2^2}}} \end{array}} \right]\)

For complete state controllability, the matrix C_M must be of rank r = 2, i.e.

|C_M| ≠ 0

\(\left| {\begin{array}{*{20}{c}} {\frac{1}{{{T_1}}}}&{ - \frac{1}{{T_1^2}}}\\ {\frac{1}{{{T_2}}}}&{ - \frac{1}{{T_2^2}}} \end{array}} \right| \ne 0\)

\(- \frac{1}{{{T_1}T_2^2}} + \frac{1}{{{T_2}T_1^2}} \ne 0\)

\(\frac{1}{{{T_1}{T_2}}}\left( {\frac{1}{{{T_1}}} - \frac{1}{{{T_2}}}} \right) \ne 0\)

\(\frac{1}{{{T_1}{T_2}}}\left( {\frac{{{T_2} - {T_1}}}{{{T_1}{T_2}}}} \right) \ne 0\)

\(\frac{{\left( {{T_2} - {T_1}} \right)}}{{{{\left( {{T_1}{T_2}} \right)}^2}}} \ne 0\)

(T₂ – T₁) ≠ 0 or T₁ ≠ T₂

Therefore, the system is completely state controllable for T₁ ≠ T₂.

The given system has characteristic equation given by: |sI - A| = 0

\(\Rightarrow \left| {\begin{array}{*{20}{c}} {s - 1}&{ - 2}\\ { - 2}&{s - 1} \end{array}} \right| = 0\)

⇒ (s - 1)² = 4

s = +3, -1

The system has poles at s = 3 and s = -1

For option (d)

The position of poles from characteristic equation:

\(\left| {\begin{array}{*{20}{c}} {s - 27}&{42}\\ { - 16}&{s + 25} \end{array}} \right| = 0\)

(s – 27) (s + 25) + (16) (42) = 0

s² – 2s – 675 + 672 = 0

s² – 2s – 3 = 0