Control Systems Test 2

The relative stability of a control system is determined by the position of the dominant pole.

Since the setting time is inversely proportional to the real part of pole hence faster-settling response is more stable i.e. the dominant pole of the system must be farther away from the origin (more negative) for the stability to improve.

For option 1 poles: -1, 0.1

For option 2 poles: -2, -2

For option 3 poles: -5, -1

For option 4 poles: -3, -1

The loop gain of a closed-loop system is:

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

So, the characteristic equation of the system is

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

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

Or \({s^3} + 6{s^2} + 8s + K = 0\) 

Therefore, we form the Routh’s array for the characteristic equation as

For K = 48: s¹ row is completely zero and two poles lie on jω axis. So, the system is marginally stable.

For 0 < K < 48 : No sign change in first column. Hence, the system is stable.

For K > 48: Two sign changes in the first column. Hence, unstable system.

So, K = 48 is the value for which the system just becomes unstable.

\(\begin{array}{l}\dot x = \left[ {\begin{array}{*{20}{c}}0&1\\0&{ - 3}\end{array}} \right]x + \left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]u\\y = \left[ {\begin{array}{*{20}{c}}1&0\end{array}} \right]x\\A = \left[ {\begin{array}{*{20}{c}}0&1\\0&{ - 3}\end{array}} \right]B = \left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]C = \left[ {\begin{array}{*{20}{c}}1&0\end{array}} \right]\end{array}\)

\(\begin{array}{l}\frac{{y\left( s \right)}}{{u\left( s \right)}} = c{\left[ {sI - A} \right]^{ - 1}}B + D\\\left[ {sI - A} \right] = \left[ {\begin{array}{*{20}{c}}s&{ - 1}\\0&{s + 3}\end{array}} \right]\\{\left[ {sI - A} \right]^{ - 1}} = \frac{1}{{s\left( {s + 3} \right)}}\left[ {\begin{array}{*{20}{c}}{s + 3}&1\\0&s\end{array}} \right]\end{array}\)

\(\begin{array}{l}\frac{{y\left( s \right)}}{{u\left( s \right)}} = \left[ {\begin{array}{*{20}{c}}1&0\end{array}} \right]\frac{1}{{s\left( {s + 3} \right)}}\left[ {\begin{array}{*{20}{c}}{s + 3}&1\\0&s\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]\\ = \frac{1}{{s\left( {s + 3} \right)}}\left[ {\begin{array}{*{20}{c}}{s + 3}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]\\ = \frac{1}{{s\left( {s + 3} \right)}}\left[ {s + 3} \right]\\ = \frac{1}{s}\end{array}\)

G(s) = Ke^-Ts‑

The characteristic equation:

CE: 1 + G(s) = 0

⇒ 1 + K(e^-Ts) = 0

By using approximation, e^-Ts ≈ 1 – Ts

⇒ 1 + K(1 – Ts) = 0

⇒ 1 + K – KTs = 0

\(\Rightarrow s = \frac{{1 + T}}{{KT}} = \frac{1}{T}\left( {1 + \frac{1}{K}} \right)\)

Given that K and T are greater than zero.

So, for all values of K and T, S will be positive.

If the roots have negative real parts, then the response is bounded and eventually decreases to zero.

The DC gain is the ratio of magnitude of steady state step response to the magnitude of step input.

For stable systems it is evaluated using S = 0

For unstable systems it is not defined

In magnitude bode plot each pole adds -20dB/dec slope and each zero adds +20dB/dec.

At very high frequency slope of bode plot asymptote

\(= 10\left( { - 20\frac{{dB}}{{{dec}}}} \right) + 2\left( {20\frac{{dB}}{{{dec}}}} \right)\)

= -160 dB/dec

Concept:

From the principal of Argument theorem,

The number of encirclements about (-1, 0) is \({\rm{N}} = {\rm{P}}-{\rm{Z}}\) 

Where P = Number of open-loop poles on Right – Half of s – plane

Z = Number of Closed Loop Poles on Right Half of s – plane

Given that the closed-loop system is stable means \({\rm{z}} = {\rm{}}0 \Rightarrow {\rm{N}} = {\rm{P}}\)

∴ So the Nyquist encircles the s – plane point (-1 +j0) in the counter-clockwise direction as many times as the number of right half s – plane poles

Note:

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

D(s) gives the roots of characteristic equation i.e. closed-loop poles.

Nyquist stability criteria state that the number of unstable closed-loop poles is equal to the number of unstable open-loop poles plus the number of encirclements of the origin of the Nyquist plot of the complex function D(s).

It can be slightly simplified if instead of plotting the function D(s) = 1 + G(s)H(s), we plot only the function G(s)H(s) around the point and count encirclement of the Nyquist plot of around the point (-1, j0).

If the system is given with the unbounded input then nothing can be clarified for the stability of the system.

Them is two sign change from the s⁴ mw down to the s° row. So two roots are on RHS. Because of symmetry rest two roots must be in LHP. From s⁶ to s⁴ there is 1 sign change so 1 on RHP and 1 on LHP.

Total LHP 3 root, RHP 3 root.

An optimal controller for an LTI system can be designed provided the system is both controllable and observable

Eigen value are given by | sI - A | = 0, which is the location of poles.

Given that,

s₁ = -2 + j2

s₂ = -2 – j2

Characteristic equation:

(s – (-2 + j2)) (s – (-2 –j2)) = 0

⇒ (s + 2 – j2) (s + 2 + j2) = 0

⇒ (s+2)² – (j2)² = 0

⇒ s² + 4s + 8 = 0

By comparing the above equation with standard second order characteristic equation,

s² + 2ξω_n s + ω_n² = 0

\({\rm{\omega }}_n^2 = 8 \Rightarrow {{\rm{\omega }}_n} = 2\sqrt 2 \;rad/sec\)

2ξω_n = 4

\(\Rightarrow \xi = \frac{4}{{2 \times 2\sqrt 2 }} = \frac{1}{{\sqrt 2 }}\) 

Undamped resonant frequency, \({{\rm{\omega }}_r} = {{\rm{\omega }}_n}\sqrt {1 - 2{\xi ^2}}\)

\(= 2\sqrt 2 \;\sqrt {1 - 2{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}} = 0\)

In general, the lead and lag compensator is represented by the below transfer function

If a > b then that is lag compensator because pole comes first.
If a < b then that is the lead compensator since zero comes first.
Analysis:

Lead compensator:
1) When sinusoidal input applied to this it produces sinusoidal output with the phase lead input.
2) It speeds up the Transient response and increases the margin for stability.

A circuit diagram is as shown:

\(G\left( s \right)H\left( s \right) = \frac{K}{{{{\left( {s + 2} \right)}^2}\left( {s + 3} \right)}}\)

Position error constant \(\mathop {{\rm{lt}}}\limits_{s \to 0} G\left( s \right)H\left( s \right)\)

\(= \mathop {{\rm{It}}}\limits_{s \to 0} \frac{K}{{{{\left( {s + 2} \right)}^2}\left( {s + 3} \right)}}\)

\(= \frac{K}{{12}}\)

Given that, K_P ≥ 2

⇒ K ≥ 24

Gain margin \(= \frac{1}{{\left| {G\left( {j{\rm{\omega }}} \right)H\left( {j{\rm{\omega }}} \right)} \right|}}\;at\;{\rm{\omega }} = {{\rm{\omega }}_{pc}}\)

At ω_pc, ∠G(jω) H(jω) = -180°

\(\Rightarrow - 2{\tan ^{ - 1}}\left( {\frac{{\rm{\omega }}}{2}} \right) - {\tan ^{ - 1}}\left( {\frac{{\rm{\omega }}}{3}} \right) = \; - 180^\circ\)

\(\Rightarrow {\tan ^{ - 1}}\left( {\frac{{\rm{\omega }}}{2}} \right) + {\tan ^{ - 1}}\left( {\frac{{\rm{\omega }}}{2}} \right) + {\tan ^{ - 1}}\left( {\frac{{\rm{\omega }}}{3}} \right) = 180^\circ\)

\(\Rightarrow {\tan ^{ - 1}}\left( {\frac{{\rm{\omega }}}{{\frac{{1 - {{\rm{\omega }}^2}}}{4}}}} \right) + {\tan ^{ - 1}}\left( {\frac{{\rm{\omega }}}{3}} \right) = 180^\circ\)

\(\Rightarrow \left( {\frac{{\rm{\omega }}}{{\frac{{1 - {{\rm{\omega }}^2}}}{4}}}} \right) + \left( {\frac{{\rm{\omega }}}{3}} \right) = 0\)

\(\Rightarrow \frac{{{{\rm{\omega }}^2}}}{4} - 1 = 3\)

⇒ ω = 4 rad/sec

Gain margin \(= \frac{{\left( {{{\rm{\omega }}^2} + 4} \right)\left( {\sqrt {{{\rm{\omega }}^2} + 9} } \right)}}{K} = \frac{{\left( {20} \right)\left( 5 \right)}}{K} = \frac{{100}}{K}\)

Given that,

Gain margin ≥ 3

\(\Rightarrow \frac{{100}}{K} \ge 3 \Rightarrow \;K \le \frac{{100}}{3}\)

Range of K is: \(24 \le K \le \frac{{100}}{3}\)

Maximum value \(= \frac{{100}}{3}\)

Minimum value = 24

Difference \(= \frac{{100}}{3} - 24 = \frac{{28}}{3} = 9.33\)

x(t) = e^-3t u(t)

\(\Rightarrow X\left( s \right) = \frac{1}{{s + 3}}\) 

y(t) = (2t + 1) e^-2t u(t)

= (2t e^-2t + e^-2t) u(t)

= 2t e^-2t u(t) + e^-2t u(t)

\( \Rightarrow y\left( s \right) = \frac{2}{{{{\left( {s + 2} \right)}^2}}} + \frac{1}{{s + 2}} = \frac{{s + 4}}{{{{\left( {s + 2} \right)}^2}}}\) 

Transfer function,

\(T\left( s \right) = \frac{{Y\left( s \right)}}{{X\left( s \right)}} = \frac{{\left( {s + 4} \right)\left( {s + 3} \right)}}{{{{\left( {s + 2} \right)}^2}}}\)

Zeros of T(s) = -4 and -3