Control Systems Test 1

Concept:

KP = position error constant = \(\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)

Kv = velocity error constant = \(\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)

Ka = acceleration error constant = \(\mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)\)

Steady state error for different inputs is given by

Input	Type -0	Type - 1	Type -2
Unit step	\(\frac{1}{{1 + {K_p}}}\)	0	0
Unit ramp	∞	\(\frac{1}{{{K_v}}}\)	0
Unit parabolic	∞	∞	\(\frac{1}{{{K_a}}}\)

 

From the above table, it is clear that for type – 1 system, a system shows zero steady-state error for step-input, finite steady-state error for Ramp-input and \(\infty \) steady-state error for parabolic-input.

Application:

\(CLTF = \frac{{10\left( {s + 1} \right)}}{{{s^2} + 20s + 10}}\)

\(OLTF = \frac{{10\left( {s + 1} \right)}}{{{s^2} + 20s + 10 - 10\left( {s + 1} \right)}} = \frac{{10\left( {s + 1} \right)}}{{s\left( {10 + s} \right)}}\)

Thus, the given system is a type-1 system and the input is ramp input

\({k_v} = \mathop {\lim }\limits_{s \to 0} s\left( {\frac{{10\left( {s + 1} \right)}}{{s\left( {s + 10} \right)}}} \right) = 1\)

\({e_{ss}} = \frac{1}{1} = 1\)

Common Mistake: Steady state error formulas are applicable only to open loop transfer function. Therefore if closed loop transfer function is given, we must first calculate the open loop transfer function to apply the formula.

\(G\left( s \right)H\left( s \right) = \frac{1}{{s\left( {3s + 1} \right)}}{e^{ - 2Ts}}\)

\(G\left( {j\omega } \right)H\left( {j\omega } \right) = \frac{1}{{j\omega \left( {3j\omega + 1} \right)}}{e^{ - 2Tj\omega }}\)

\(\angle G\left( {j\omega } \right)H\left( {j\omega } \right) = - 90 - {\tan ^{ - 1}}\left( {3\omega } \right) - 2T\omega \)

At ω = ω₀, the phase angle is zero.

\( \Rightarrow 0 = - 90 - {\tan ^{ - 1}}\left( {3{\omega _0}} \right) - 2T{\omega _0}\)

\( \Rightarrow {\tan ^{ - 1}}\left( {3{\omega _0}} \right) = - \left( {90 + 2T{\omega _0}} \right)\)

⇒ 3ω₀ = - tan(90 + 2Tω₀)

⇒ 3ω₀ = cot(2ω₀T)

Concept:

From Nyquist stability criterion, N = P – Z

where, N = number of encirclements of point (–1, 0)

Z = number of zeros of (1 + G(s)·H(s)) or the number of poles of closed-loop system in the right half-plane.

For stability, z = 0,

Calculation:

Open loop transfer function:

\(G\left( s \right)H\left( s \right) = \frac{{2\left( {s + 8} \right)}}{{{s^2} + 7s - 8}} = \frac{{2\left( {s + 8} \right)}}{{\left( {s + 8} \right)\left( {s - 1} \right)}}\)

Number of right side open loop poles (p) = 1

Closed loop transfer function

\(= \frac{{\frac{{2\left( {s + 8} \right)}}{{{s^2} + 7s - 8}}}}{{1 + \frac{{2\left( {s + 8} \right)}}{{{s^2} + 7s - 8}}}}\)

\(= \frac{{2\left( {s + 8} \right)}}{{{s^2} + 7s - 8 + 2s + 16}}\)

\(= \frac{{2\left( {s + 8} \right)}}{{{s^2} + 9s + 8}}\)

\(= \frac{{2\left( {s + 8} \right)}}{{\left( {s + 1} \right)\left( {s + 8} \right)}}\)

Number of closed loop poles (z) = 0

From Nyquist stability

N = P – Z = 1

Therefore Nyquist contour encircle the -1 point once in the counterclockwise direction.

\(G\left( s \right) = \frac{{{e^{ - \frac{s}{{500}}}}}}{{s + 500}}\)

ϕ = ∠G(jω) at ω = 100π

\( = - {\tan ^{ - 1}}\left( {\frac{\omega }{{500}}} \right) - \frac{\omega }{{500}}\)

\( = - {\tan ^{ - 1}}\left( {\frac{{100\pi }}{{500}}} \right) - \frac{{100\pi }}{{500}}\)

\( = - {\tan ^{ - 1}}\left( {\frac{\pi }{5}} \right) - \frac{\pi }{5}\)

= -68.12°

y(t) = A sin (100 πt - 68.12°)

s³ – s² + 3s + 1 = 0

put s = z + 1

⇒ (z + 1)³ – (z + 1)² + 3(z + 1) + 1 = 0

⇒ z³ + 1 + 3z² + 3z – z² – 2z – 1 + 3z + 3 + 1 = 0

⇒ z³ + 2z² + 4z + 4 = 0

Taking Routh criteria for the given characteristic equation

z3	1	4
z2	2	4
z1	2	0
z0	4

There are no sign changes in the first column of the Routh array.

So, the number roots lie left side of s = 1, are 3.

Given \(\rm G\left( s \right)H\left( s \right) = \frac{K}{{s\left( {s + 4} \right)\left( {{s^2} + 4s + 12} \right)}}\)

\(\rm 1 + G\left( s \right)H\left( s \right) = 0\)

\(\rm 1 + \frac{K}{{s\left( {s + 4} \right)\left( {{s^2} + 4s + 12} \right)}} = 0\)

\(\therefore K = - s\left( {s + 4} \right)\left( {{s^2} + 4s + 12} \right)\)

for break point calculation we should have \(\rm \frac{{dK}}{{ds}} = 0\)

\(\rm \therefore \frac{d}{{ds}}\left[ {\left( {{s^4} + 8{s^3} + 28{s^2} + 48s} \right)\left( { - 1} \right)} \right] = 0\)

\(\rm 4{s^3} + 24{s^2} + 56s + 48 = 0\)

\(\rm {s^3} + 6{s^2} + 14s + 12 = 0\)

Break away point are, \(\rm -2, -2+j√2, -2-j√2\)

∴ Number of complex Break away points are = 2

\(\dot x\left( s \right) = \left[ {\begin{array}{*{20}{c}} 1&1\\ 0&1 \end{array}} \right]x\left( t \right) + \left[ {\begin{array}{*{20}{c}} {{b_1}}\\ {{b_2}} \end{array}} \right]u\left( t \right)\)

\(C\left( t \right) = \left[ {\begin{array}{*{20}{c}} {{d_1}}&{{d_2}} \end{array}} \right]x\left( t \right)\)

\(A = \left[ {\begin{array}{*{20}{c}} 1&1\\ 0&0 \end{array}} \right]B = \left[ {\begin{array}{*{20}{c}} {{b_1}}\\ {{b_2}} \end{array}} \right]C = \left[ {\begin{array}{*{20}{c}} {{d_1}}&{{d_2}} \end{array}} \right]\)

Controlling matrix, \({Q_C} = \left[ {\begin{array}{*{20}{c}} B&{AB} \end{array}} \right]\)

\({Q_C} = \left[ {\begin{array}{*{20}{c}} {{b_1}}&{{b_1} + {b_2}}\\ {{b_2}}&{{b_2}} \end{array}} \right]\)

\(\left| {{Q_C}} \right| = {b_1}{b_2} - {b_2}\left( {{b_1} + {b_2}} \right) = - b_2^2 \ne 0\)

Given the system is controllable for all values of b₂ except b₂ = 0

Observability matrix, \({Q_A} = \left[ {\begin{array}{*{20}{c}} C\\ {CA} \end{array}} \right]\)

\({Q_A}\left[ {\begin{array}{*{20}{c}} {{d_1}}&{{d_2}}\\ {{d_1}}&{{d_1} + {d_2}} \end{array}} \right]\)

\(\left| {{Q_A}} \right| = {d_1}\left( {{d_1} + {d_2}} \right) - {d_1}{d_2} = d_1^2 \ne 0\)

Given the system is observable for all values of d₁ except d₁ = 0.

Given system is homogeneous whose solution is X(t) = e^At X(0)

Now, e^At = L^-1 {(sI - A)^-1}

\(\left[ {sI - A} \right] = s\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}0&1\\{ - 2}&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}s&{ - 1}\\2&s\end{array}} \right]\)

\({\left( {sI - A} \right)^{ - 1}} = \frac{{Adj\left( {sI - A} \right)}}{{\left| {sI - A} \right|}}\)

\(Adj\left( {sI - A} \right) = {\left[ {\begin{array}{*{20}{c}}{{C_{11}}}&{{C_{12}}}\\{{C_{21}}}&{{C_{22}}}\end{array}} \right]^T} = {\left[ {\begin{array}{*{20}{c}}s&{ - 2}\\1&s\end{array}} \right]^T} = \left[ {\begin{array}{*{20}{c}}s&1\\{ - 2}&s\end{array}} \right]\)

|sI - A| = s² + 2

\({\left( {sI - A} \right)^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{\frac{s}{{{s^2} + 2}}}&{\frac{1}{{{s^2} + 2}}}\\{\frac{{ - 2}}{{{s^2} + 2}}}&{\frac{s}{{{s^2} + 2}}}\end{array}} \right]\)

\({e^{At}} = {L^{ - 1}}\left[ {\begin{array}{*{20}{c}}{\frac{s}{{{s^2} + 2}}}&{\frac{1}{{{s^2} + 2}}}\\{\frac{{ - 2}}{{{s^2} + 2}}}&{\frac{s}{{{s^2} + 2}}}\end{array}} \right]\)

\({L^{ - 1}}\left\{ {\frac{s}{{{s^2} + 2}}} \right\} = {L^{ - 1}}\left\{ {\frac{s}{{{s^2} + {{\left( {\sqrt 2 } \right)}^2}}}} \right\} = \cos \sqrt {2t} \)

\({L^{ - 1}}\left\{ {\frac{1}{{{s^2} + 2}}} \right\} = {L^{ - 1}}\left\{ {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt 2 }}}&{\frac{{\sqrt 2 }}{{{s^2} + {{\left( {\sqrt 2 } \right)}^2}}}}\end{array}} \right\} = \frac{1}{{\sqrt 2 }}\sin \sqrt 2 t\)

\({L^{ - 1}}\left\{ {\frac{{ - 2}}{{{s^2} + 2}}} \right\} = {L^{ - 1}}\left\{ { - \begin{array}{*{20}{c}}{\sqrt 2 }&{\frac{{\sqrt 2 }}{{{s^2} + {{\left( {\sqrt 2 } \right)}^2}}}}\end{array}} \right\} = - \sqrt 2 \sin \sqrt 2 t\)

\(\therefore {e^{At}} = \left[ {\begin{array}{*{20}{c}}{\cos \sqrt {2\;t} }&{\frac{1}{{\sqrt 2 }}\sin \sqrt 2 \;t}\\{ - \sqrt 2 \sin \sqrt 2 \;t}&{\cos \sqrt 2 \;t}\end{array}} \right]\)

\(\therefore X\left( t \right) = {e^{At}}X\left( 0 \right) = {e^{At}}\left[ {\begin{array}{*{20}{c}}1\\1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\cos \sqrt 2 \;t + \frac{1}{{\sqrt 2 }}\sin \sqrt 2 \;t}\\{\cos \sqrt 2 \;t - \sqrt 2 \sin \sqrt 2 \;t}\end{array}} \right]\)

∴ Output response Y(t) = [1 - 1] X(t)

\(= \left[ {1 - 1} \right]\left[ {\begin{array}{*{20}{c}}{\cos \sqrt 2 \;t + \frac{1}{{\sqrt 2 }}\sin \sqrt 2 \;t}\\{\cos \sqrt 2 \;t - \sqrt 2 \sin \sqrt 2 \;t}\end{array}} \right]\)

\(= \frac{3}{{\sqrt 2 }}\sin \sqrt 2 \;t\)

\(Y\left( t \right) = \frac{3}{{\sqrt 2 }}\sin \sqrt 2 \;t\) is the required response