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.

The non-minimum phase transfer function is defined as the transfer function which has zeros (or) poles in right side of s-plane.

Type of System: The Type of system denotes the no. of poles at the origin of the open loop transfer function G(s)H(s).

Hence, there is no open loop pole at origin. Hence, type of system = 0.

Order of a System: The order of a control system is determined by the highest power of 's' in the denominator of its transfer function.

Hence, highest power of s of the characteristic equation 1+ G(s) = 0 will be 2.

Hence, order of given System = 2.

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

Superposition theorem states that for two signals additivity and homogeneity property must be satisfied and that is applicable for the LTI systems.

From given pole-zero plot, transfer function is

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

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°

The T.F. of the given block diagram can be reduced to:

Its block diagram is shown below, where a closed loop feedback system is transformed into an unity feedback system and vice-versa,

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

⇒ 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 

Answer A is correct.

Label the output of the first summing junction (the signal entering the block G1) as an intermediate node. The second summing junction receives two contributions coming from the first summing-node path: one through G1 and the other via the top branch (which is the block H2) that directly adds to the input of the second summing junction. Thus the combined gain seen by the blocks G2 and G3 is (G1 + H2).

The forward path from the first summing junction to the output therefore equals G2 · G3 · (G1 + H2).

There is a single feedback from the output back to the first summing junction through H1. This feedback multiplies the entire forward gain. Using the standard closed-loop formula for a single-loop negative feedback system, the closed-loop transfer function is the forward path divided by 1 + (loop gain). Here the loop gain is H1 · G2 · G3 · (G1 + H2).

Hence

C(s)/R(s) = G2 G3 (G1 + H2) / [1 + H1 G2 G3 (G1 + H2)].

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

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

On shifting the take-off point beyond block G, we have the reduced block diagram as shown below:

On further reducing the above block diagram, we get the block diagram as shown below.

Shifting the take-off point after block G₃ in block diagram-A, the reduced block diagram will be as shown below.

Thus, the block diagrams A and B are equivalent to each other.

On further reducing the block diagram-B, we get the block diagram-C. Hence, all the three block diagrams A, B and C are equivalent to each other,

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

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