Control Systems Test 5

Given \(G\left( s \right) = \frac{1}{{{s^2} + s - 6}} = \frac{1}{{\left( {s + 3} \right)\left( {s - 2} \right)}}\)

To stabilize the plant the pole s = 2 must be removed

The compensator \(\frac{{2\left( {s - 2} \right)}}{{s + 4}}\) is appropriate since

\(G\left( s \right){G_c}\left( s \right) = \frac{1}{{\left( {s + 3} \right)\left( {s - 2} \right)}}.\frac{{2\left( {s - 2} \right)}}{{s + 4}}\)

A phase crossover frequency, ∠G(jω) = -180°

-tan^-1(ω) - tan^-1(ω/9) - 90° = - 180°

⇒ ω_pc = 3 rad/sec

\(|G{\rm{(}}j{\omega _{pc}}H\left( {j{\omega _{pc}}} \right){\rm{|}} = \frac{{900}}{{s\left( {s + 1} \right)\left( {s + 9} \right)}} = 10\)

Gain margin. GM = 20 log 10 = -20 dB

System is unstable.

ω_pc should become ω_gc

⇒ ω_gc = 3 rad/sec

The magnitude should be 0 dB.

To make the magnitude 0 dB at ω_gc = 3 rad/sec a lag compensator which gives an attenuation of 20 dB is used.

To obtain 45° phase margin at ω_pc = 3 rad/sec.

A lead compensator with a phase lead of 45° used

Given the transfer function of a lag compensator as:

\({G_C}\left( s \right) = \frac{{1 + \alpha sT}}{{1 + sT}}\)

\({G_C}\left( {j\omega } \right) = \frac{{1 + \alpha j\omega T}}{{1 + j\omega T}}\)

The phase angle of the function is written as:

ϕ = tan^-1 (αωT) – tan^-1 (ωT)

The phase will be maximum if:

\(\frac{{d\phi }}{{d\omega }} = 0\)

\(\frac{{\alpha T}}{{1 + {\alpha ^2}{\omega ^2}{T^2}}} - \frac{T}{{1 + {\omega ^2}{T^2}}} = 0\)

\(\frac{\alpha }{{1 + {\alpha ^2}{\omega ^2}{T^2}}} = \frac{1}{{1 + {\omega ^2}{T^2}}}\)

α(1 + ω²T²) = (1 + α²ω² T²)

ω²(α²T² – αT²) = α – 1

\({\omega ^2} = \frac{{\left( {\alpha - 1} \right)}}{{\alpha {T^2}\left( {\alpha - 1} \right)}} = \frac{1}{{\alpha {T^2}}}\)

\(\omega = \frac{1}{{T\sqrt \alpha }}\)

Hence, the phase angle will be maximum for the frequency,

\(\omega = \frac{1}{{T\sqrt \alpha }}\) rad/sec       

i) Reducing the gain of the amplifier in the control system may improve stability

ii) Phase advance circuit improves gain margin and phase margin by adding a zero to the transfer function

iii) Integral error compensation minimizes steady-state error but doesn't improve stability

Integral compensator adds a pole at origin which reduces the stability.