Control Systems Test 2

Concept:

The characteristic equation for a given open-loop transfer function G(s) is

1 + G(s) H(s) = 0

According to the Routh tabulation method,

The system is said to be stable if there are no sign changes in the first column of Routh array

The number of poles lies on the right half of s plane = number of sign changes

Calculation:

Given polynomial is,

s⁵ + 5s⁴ + 11s³ + 23s² + 28s + 12

By applying Routh Hurwitz criteria,

s¹ row is zero. So, there are two poles on jω axis.

3s² + 12 = 0

⇒ s = ± 2j

AE : 3s² + 12 = 0

By differentiating

6s = 0

Total number of poles = 5

Number of sign changes = zero

Number of right side poles = zero

Number of left side poles = 5 – 2 = 3

As there are poles on jω axis, the system is marginally stable.

\(G\left( s \right) = \frac{{1000}}{{s\left( {s + 5} \right)\left( {s + 20} \right)}}\) 

\(\angle G\left( {j\omega } \right)H\left( {j\omega } \right) = - \frac{\pi }{2} - {\tan ^{ - 1}}\left( {\frac{\omega }{5}} \right) - {\tan ^{ - 1}}\left( {\frac{\omega }{{20}}} \right)\;\) 

At phase cross over frequency,

\( - \frac{\pi }{2} - {\tan ^{ - 1}}\left( {\frac{\omega }{5}} \right) - {\tan ^{ - 1}}\left( {\frac{\omega }{{20}}} \right) = - \pi \) 

\( \Rightarrow {\tan ^{ - 1}}\left( {\frac{{\frac{\omega }{5} + \frac{\omega }{{20}}}}{{1 - \frac{\omega }{5} \times \frac{\omega }{{20}}}}} \right) = \frac{\pi }{2}\) 

⇒ ω² = 100

⇒ ω = 10 rad/sec

At ω = 10,

\(\left| {G\left( {j\omega } \right)H\left( {j\omega } \right)} \right| = \frac{{1000}}{{10\sqrt {{5^2} + {{10}^2}} \sqrt {{{20}^2} + {{10}^2}} }}\) 

a = 0.4

Nyquist plot of G(s) passes through the negative real axis at the point (-a, 0)

= (-0.4, 0)

Concept:

The sensitivity of the overall gain ‘M’ to the variation in ‘G’ is defined as

Sensitivity = % change in M/% change in G

\(S_G^M = \frac{{\frac{{\delta M}}{M}}}{{\frac{{\delta G}}{G}}} = \frac{{\delta M}}{{\delta G}} \times \frac{G}{M}\)

Where δM denotes the incremental change in M due to the incremental change in G.

Application:

\(G = \frac{A}{{s\left( {s + a} \right)}} = \frac{A}{{{s^2} + as}}\)

Closed-loop transfer function is

\(TF = \frac{G}{{1 + GH}} = \frac{{\frac{A}{{{s^2} + as}}}}{{1 + \frac{A}{{{s^2} + as}}}} = \frac{A}{{{s^2} + as + A}}\)

\(\frac{{\partial T}}{{\partial a}} = \frac{{ - As}}{{{{\left( {{s^2} + as + A} \right)}^2}}}\)

Sensitivity \(= \frac{{\partial T}}{{\partial a}} \times \frac{a}{T}\) 

\(= \frac{{ - As}}{{{{\left( {{s^2} + as + A} \right)}^2}}} \times \frac{a}{{\frac{a}{{\left( {{s^2} + as + A} \right)}}}}\)

\(= \frac{{ - as}}{{\left( {{s^2} + as + A} \right)}}\)

For the system to be stable, the poles should lie on left half of the s-plane.

The roots of the characteristic equation are the poles. The characteristic equation is given below.

|sI - A| = 0

⇒ (s - x) [(s - y)(s + 2) + 1] = 0

For the system to be stable, the roots of the above equation must be less than zero.

⇒ x < 0

s² – 2y – sy + 2s + 1 = 0

⇒ s² + (2 - y) s + 1 – 2y = 0

⇒ 2 – y > 0 and (1 – 2y) > 0

⇒ y < 2 and \(y < \frac{1}{2}\)

Concept:

Final value theorem:

• A final value theorem allows the time domain behavior to be directly calculated by taking a limit of a frequency domain expression
• Final value theorem states that the final value of a system can be calculated by

\(f\left( \infty \right) = \mathop {\lim }\limits_{s \to 0} sF\left( s \right)\)

 Where F(s) is the Laplace transform of the function.

• For the final value theorem to be applicable system should be stable in steady-state and for that real part of poles should lie on the left side of s plane.

Initial value theorem:

\(C\left( 0 \right) = \mathop {\lim }\limits_{t \to 0} c\left( t \right) = \mathop {\lim }\limits_{s \to \infty } sC\left( s \right)\)

It is applicable only when the number of poles of C(s) is more than the number of zeros of C(s).

Calculation:

\(\ddot y\left( t \right) + 6\dot y\left( t \right) + 5y\left( t \right) = u\left( t \right)\)

By applying Laplace transform

\({s^2}y\left( s \right) + 6sy\left( s \right) + 5 = u\left( s \right)\)

\(\Rightarrow \frac{{y\left( s \right)}}{{u\left( s \right)}} = \frac{1}{{{s^2} + 6s + 5}}\)

For step input, the output of the system is

\(y\left( s \right) = \frac{1}{{{s^2} + 6s + 5}} \cdot \frac{1}{s}\)

By using final value theorem, the steady state value of the output is

\( = \mathop {\lim }\limits_{s \to 0} s \cdot y\left( s \right)\)

\(= \mathop {\lim }\limits_{s \to 0} s \cdot \frac{1}{{{s^2} + 6s + 5}} \cdot 1/s = \frac{1}{5} = 0.2\)

t_max is nothing but the peak time.

\({t_p} = \frac{{n\pi }}{{{\omega _d}}}\) 

first maxima occur at n = 1

\({t_p} = \frac{\pi }{{{\omega _d}}}\) 

Given transfer function,

\(\frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{25}}{{{s^2} + 6s + 25}}\) 

ω²_n = 25 ⇒ ω_n 5

\(2\xi {\omega _n} = 6 \Rightarrow \xi = \frac{6}{{2 \times 5}} = 0.6\) 

\({t_p} = \frac{\pi }{{{{\rm{\omega }}_d}}} = \frac{\pi }{{{{\rm{\omega }}_n}\sqrt {\left( {1 - {\xi ^2}} \right)} }}\) 

\(= \frac{\pi }{{5\sqrt {1 - {{\left( {0.6} \right)}^2}} }} = \frac{\pi }{4}\) 

\(G\left( s \right) = \frac{{5\left( {s + 4} \right)}}{{s\left( {s + 0.25} \right)\left( {{s^2} + 10s + 25} \right)}}\)

\(= \frac{{5 \times 4\left( {1 + \frac{s}{4}} \right)}}{{0.25 \times 25 \times s \times \left( {1 + \frac{s}{{0.25}}} \right)\left( {1 + \frac{{10s}}{{25}} + \frac{{{s^2}}}{{25}}} \right)}}\)

\(= \frac{{3.2\left( {1 + \frac{s}{4}} \right)}}{{s\left( {1 + \frac{s}{{0.25}}} \right)\left( {1 + \frac{{10s}}{{25}} + \frac{{{s^2}}}{{25}}} \right)}}\)

Constant gain = 3.2

Corner frequencies = 0.25, 4, 5

Highest corner frequency = 5 rad/sec

\(G\left( s \right)H\left( s \right) = \frac{K}{{s\left( {{s^2} + 6s + 25} \right)}}\) 

Poles: s = 0,s = -3 + 4j, -3 – 4j

At s = -3 + 4j

\(G\left( s \right)H\left( s \right) = \frac{K}{{\left( { - 3 + 4j} \right)\left( {{{\left( { - 3 + 4j} \right)}^2} + 6\left( { - 3 + 4j} \right) + 25} \right)}}\) 

\( = \frac{K}{{\left( { - 3 + 4j} \right)\left( {9 - 16 - 12j - 18 + 24j + 25} \right)}}\) 

\(= \frac{K}{{\left( { - 3 + 4j} \right)\left( {12j} \right)}}\) 

\(= \frac{K}{{\left( { - 48 - j36} \right)}}\) 

\(\angle G\left( s \right)H\left( s \right) = - \left[ {180 + {{\tan }^{ - 1}}\left( {\frac{{36}}{{48}}} \right)} \right]\) 

= -216.86

Angle of departure = 180 + (-216.86)

= -36.86°

= 323.14°

\(y = \left[ {1\;\;1} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right]\)

⇒ y(t) = x₁(t) + x₂(t)

By differentiating the above equation

\(\frac{d}{{dt}}y\left( t \right) = \frac{d}{{dt}}{x_1}\left( t \right) + \frac{d}{{dt}}{x_2}\left( t \right)\)

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

At t = 0

Now, \(\frac{d}{{dt}}y\left( 0 \right) = \frac{d}{{dt}}{x_1}\left( 0 \right) + \frac{d}{{dt}}{x_2}\left( 0 \right)\)

= ẋ₁(0) + ẋ₂(0)

= 2 – 2 = 0 

Concept:

The transfer function of the phase controller is given by \(G\left( s \right)=\frac{1+aTs}{1+Ts}\)

Where, a > 1 for phase lead controller

a < 1 for phase lag controller

Maximum phase lead/lag frequency \({{\omega }_{m}}=\frac{1}{T\sqrt{a}}\)

Maximum phase lead/lag \({{\phi }_{m}}={{\tan }^{-1}}\left( \frac{a-1}{2\sqrt{a}} \right)={{\sin }^{-1}}\left( \frac{a-1}{a+1} \right)\)

Calculation:

Maximum phase lead = 40°

\( \Rightarrow {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right) = 40^\circ \)

\( \Rightarrow \frac{{a - 1}}{{a + 1}} = 0.643\)

⇒ a – 1 = 0.643a + 0.643

⇒ a = 4.6

The maximum phase lead angle should be provided at a frequency of 5 rad/sec.

\( \Rightarrow \frac{1}{{T\sqrt a }} = 5\)

\( \Rightarrow \frac{1}{{T\sqrt {4.6} }} = 5\)

⇒ T = 0.093

Now, the transfer function is \({G_c}\left( s \right) = \frac{{1 + aTs}}{{1 + Ts}}\)

\( = K\frac{{1 + 0.428s}}{{1 + 0.093s}}\)

\( = K'\frac{{s + 2.3}}{{s + 10.7}}\)

Now, the open-loop transfer function of the compensated system is,

\(G\left( s \right){G_c}\left( s \right) = \frac{2}{{s\left( {s + 4} \right)}} \times K'\frac{{s + 2.3}}{{s + 10.7}}\)

Velocity error coefficient, \({K_v} = \mathop {\lim }\limits_{s \to 0} s \times \frac{2}{{s\left( {s + 4} \right)}} \times K'\frac{{s + 2.3}}{{s + 10.7}} = K'\left( {\frac{2}{4}} \right)\left( {\frac{{2.3}}{{10.7}}} \right)\)

\( \Rightarrow 10 = K'\left( {\frac{2}{4}} \right)\left( {\frac{{2.3}}{{10.7}}} \right)\)

⇒ K’ = 93

Now, the transfer function of the lead compensator is

\({G_c}\left( s \right) = \frac{{93\left( {s + 2.3} \right)}}{{\left( {s + 10.7} \right)}}\)

Gain margin \(= 20\log \left( {\frac{1}{{G\left( s \right)H\left( s \right)}}} \right)\) dB at ω = ω_pc

Where ω_pc = phase cross over frequency.

At ω = ω_pc, ∠G(s) H(s) = -180°

\(G\left( s \right)H\left( s \right) = \frac{K}{{s\left( {s + 1} \right)\left( {s + 2} \right)}}\)

\(G\left( {j\omega } \right)H\left( {j{\rm{\omega }}} \right) = \frac{k}{{j{\rm{\omega }}\left( {1 + j{\rm{\omega }}} \right)\left( {2 + j{\rm{\omega }}} \right)}}\)

∠G(jω_pc) H(jω_pc) = -180°

\(\Rightarrow - 90 - {\tan ^{ - 1}}\left( {\rm{\omega }} \right) - {\tan ^{ - 1}}\left( {\frac{{\rm{\omega }}}{2}} \right) = - 180\)

\(\Rightarrow {\tan ^{ - 1}}\left( {\rm{\omega }} \right) + ta{n^{ - 1}}\left( {\frac{{\rm{\omega }}}{2}} \right) = 90\)

\(\Rightarrow {\tan ^{ - 1}}\left( {\frac{{{\rm{\omega }} + \frac{{\rm{\omega }}}{2}}}{{1 - {\rm{\omega }}.\frac{{\rm{\omega }}}{2}}}} \right) = 90^\circ\)

\(\Rightarrow 1 - \frac{{{{\rm{\omega }}^2}}}{2} = 0\)

\(\Rightarrow {{\rm{\omega }}_{PC}} = \sqrt 2 \;\;rad/sec\)

\(\left| {G\left( {j{{\rm{\omega }}_{pc}}} \right)H\left( {j{{\rm{\omega }}_{pc}}} \right)} \right| = \left| {\frac{K}{{\left( {j\sqrt 2 } \right)\left( {1 + j\sqrt 2 } \right)\left( {2 + j\sqrt 2 } \right)}}} \right|\;\)

\(= \frac{K}{{\sqrt 2 \sqrt 3 \sqrt 6 }} = \frac{K}{6}\)

\(GM = 20\log \left( {\frac{1}{{\frac{K}{6}}}} \right) = 20\log \left( {\frac{6}{K}} \right)\)

Given that GM is zero

\(\Rightarrow 20\log \left( {\frac{6}{K}} \right) = 0\)