Control Systems Test 1

The characteristic equation of standard second order system is given by

s² + 2ξ ω_n s + ω²_n = 0

The system is said to be

a) undamped if ξ = 0

b) critically damped if ξ = 1

c) underdamped if ξ < 1

d) overdamped if ξ > 1

1) s² + 8s + 15 = 0

ω²_n = 15 ⇒ ω_n = √15

\(2\xi {{\rm{\omega }}_n} = 8 \Rightarrow 2\xi \left( {\sqrt {15} } \right) = 8 \Rightarrow \xi = \frac{4}{{\sqrt {15} }} > 1\) 

So, the system is overdamped.

2) s² + 24s + 225 = 0

ω²_n = 225 ⇒ ω_n = 15

\(2\xi {{\rm{\omega }}_n} = 24 \Rightarrow 2\xi \left( {15} \right) = 24 \Rightarrow \xi = \frac{{12}}{{15}} < 1\) 

So, the system is underdamped.

3) s² + 20.25 = 0

\(\omega _n^2 = 20.25 \Rightarrow {\omega _n} = \sqrt {20.25} \) 

2ξω_n = 0 ⇒ ξ = 0

So, the system is undamped.

4) s² + 20s + 100 = 0

ω²_n = 100 ⇒ ω_n = 10

2ξω_n = 20 ⇒ 2ξ (10) = 20 ⇒ ξ = 1

Given characteristic equation is: s⁶ + 2s⁵ + 3s⁴ + 4s³ + 3s² + 2s + 1 = 0

Routh array:

\(\begin{array}{*{20}{c}}{{s^6}}\\{{s^5}}\\{{s^4}}\\{{s^3}}\\{{s^2}}\\{{s^1}}\\{{s^0}}\end{array}\left| {\begin{array}{*{20}{c}}1&3&3&1\\2&4&2&0\\1&2&1&{}\\0&0&0&{}\\{}&{}&{}&{}\\{}&{}&{}&{}\\{}&{}&{}&{}\end{array}} \right.\)

\(\frac{d}{{ds}}\left( {{s^4} + 2{s^2} + 1} \right)\)

⇒ 4s³ + 4s = 0

\(\begin{array}{*{20}{c}}{{s^6}}\\{{s^5}}\\{{s^4}}\\{{s^3}}\\{{s^2}}\\{{s^1}}\\{{s^0}}\end{array}\left| {\begin{array}{*{20}{c}}1&3&3&1\\2&4&2&0\\1&2&1&{}\\{0\left( 4 \right)}&{0\left( 4 \right)}&0&{}\\1&1&{}&{}\\0&0&{}&{}\\{}&{}&{}&{}\end{array}} \right.\)

s² + 1 = 0

⇒ s = ±j

⇒ ω_n = 1 rad/sec

\(G\left( s \right) = \frac{{k\left( {s + b} \right)}}{{{s^2}\left( {s + 20} \right)}}\)

Characteristic equation: 1 + G(s) H(s) = 0

\( \Rightarrow 1 + \frac{{k\left( {s + b} \right)}}{{{s^2}\left( {s + 20} \right)}} = 0\)

\( \Rightarrow k = \frac{{ - \left( {{s^3} + 20{s^2}} \right)}}{{\left( {s + b} \right)}}\)

\(\frac{{dk}}{{ds}} = 0\)

\( \Rightarrow \frac{{\left( {s + b} \right)\left( {3{s^2} + 40s} \right) - \left( {{s^3} + 20{s^2}} \right)\left( 1 \right)}}{{{{\left( {s + b} \right)}^2}}} = 0\)

⇒ 3s³ + 3bs²+ 40s² + 40bs – s³ – 20s² = 0

⇒ 2s³ + (3b + 20) s² + 40 bs = 0

⇒ s (2s² + (3b + 20) s + 40b) = 0

⇒ s = 0 and 2s² + (3b + 20) s + 40b = 0

To have single point,

(3b + 20)² – 4(2) (40b) = 0

⇒ 9b² + 400 + 120b – 320b = 0

⇒ 9b² – 200b + 400 = 0

\( \Rightarrow b = 20,\frac{{20}}{9}\)

For b = 20, the pole and zero gets cancelled.

Concept:

Phase margin = 180° + ∠GH

Calculation:

Let the system have a DC gain of ‘K’

Then the OLTF

\(G\left( s \right) = \frac{k}{{{{\left( {s + 1} \right)}^3}}}\)

PM = 180° + ∠GH(s) [H(s) = 1]

\(= 180 + \angle \frac{k}{{{{\left( {s + 1} \right)}^3}}} = 45^\circ\) 

= 180 + -3 tan^-1(ω) = 45

-3 tan^-1 ω = 45 – 180

-3 tan^-1 ω = -135

tan^-1 ω = 45

ω = 1

Phase margin is calculated at gain cross over frequency. Hence the calculated value of ω is ω_gc

The gain cross over frequency is the where gain of the system = 1

\({\left| {\frac{k}{{{{\left( {s + 1} \right)}^3}}}} \right|_{\omega = {\omega _{gc}}}} = 1\) 

\({\left| {\frac{k}{{{{\left( {j\omega + 1} \right)}^3}}}} \right|_{\omega = {\omega _{gc}}}} = 1\) 

\({\left| {\frac{k}{{\sqrt {{{\left( {s + 1} \right)}^3}} }}} \right|_{\omega = {\omega _{gc}}}} = 1\) 

\(\Rightarrow \frac{k}{{\sqrt {{{\left( 2 \right)}^3}} }} = 1\) 

k = 2√2

k = 2.828

\(G\left( s \right) = \frac{{0.5 + s}}{{0.25 + s}}\)

\(= \frac{{0.5\left( {1 + 2s} \right)}}{{0.25\left( {1 + 4s} \right)}} = \frac{{2\left( {1 + 2s} \right)}}{{\left( {1 + 4s} \right)}}\)

Now, it is in the form of  \(\left( {\frac{{1 + aTs}}{{1 + Ts}}} \right)\)

aT = 2 and T = 4

⇒ a = ½

Maximum phase lag  \({\phi _m} = {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right)\)

\(= {\sin ^{ - 1}}\left( {\frac{{\frac{1}{2} - 1}}{{\frac{1}{2} + 1}}} \right)\)

\(= {\sin ^{ - 1}}\left( { - \frac{1}{3}} \right)\)

⇒ ϕ_m = -19.47°

Here -ve sign indicates lagging angle.