Control Systems Test 2

Comparing the given equation with standard 2^nd order response

\(c\left( t \right) = 1 - \frac{{{e^{ - {\rm{\zeta \;}}{{\rm{\omega }}_{nt}}}}}}{{\sqrt {1 - {\rm{\zeta }}{{\rm{\;}}^2}} }}\sin \left( {{\omega _{d}t} + {{\cos }^{ - 1}}{\rm{\zeta \;}}} \right)\)

\(y\left( t \right) = 1 - \frac{2}{{\sqrt 3 }}{e^{ - {\rm{ \;}}{{\rm{}}_{t}}}}\cos \left( {\sqrt 3 t - \frac{\pi }{3}} \right)\)

\(\sqrt {1 - {\rm{\zeta }}{{\rm{\;}}^2}} = \frac{{\sqrt 3 }}{2}\)

Squaring both sides

\(1 - {\rm{\zeta }}{{\rm{\;}}^2} = \frac{3}{4}\)

\(\frac{1}{4} = {\rm{\zeta }}{{\rm{\;}}^2}\)

\({\rm{\zeta \;}} = \frac{1}{2}\)

\({\rm{\zeta \;}}{{\rm{\omega }}_n} = + 1\)

\(\left( {\frac{1}{2}} \right){\omega _n} = 1\)

Roots of a second order system = -1 + j2, -1 – j2 characteristic equation is,

(s – (-1 + j2)) (s – (-1 – j2)) = 0

⇒ (s + 1 – j2) (s + 1 + j2) = 0

⇒ (s + 1)² – (j2)² = 0

⇒ s² + 2s + 5 = 0

\(\omega _n^2 = 5\)

\({\omega _n} = \sqrt 5 \;rad/sec\)

\(2\xi {\omega _n} = 2 \)

\( \xi = \frac{2}{{2 \times \sqrt 5 }} = 0.45\)

Maximum % over-shoot will be:

\(= {e^{ - \frac{{\xi \pi }}{{\sqrt {1 - {\xi ^2}} }}}} \times 100\)

\(= {e^{\frac{{ - \pi \left( {0.45} \right)}}{{\sqrt {1 - {{\left( {0.45} \right)}^2}} }}}} \times 100\)

Concept:

Standard second-order closed-loop transfer function is given by:

\(\frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{\omega _n^2}}{{{s^2} + 2\xi {\omega _n}s + \omega _n^2}}\) 

ξ = damping ratio

ω_n = undamped natural frequency

Analysis:

Given:

ξ = 1.25, ω_n = 200 rad/sec

DC Gain: K = 1

The closed loop transfer function will be:

\(\frac{{G\left( s \right)}}{{R\left( s \right)}} = \frac{{{{\left( {200} \right)}^2}}}{{{s^2} + 2 \times 1.25 \times 200s + {{\left( {200} \right)}^2}}}\) 

\(= \frac{{40000}}{{\left( {s + 100} \right)\left( {s + 400} \right)}}\)   

For a unit step input:

\(R\left( s \right) = \frac{1}{s}.\) 

∴ The response C(s) of the system for a unit step input will be:

\(C\left( s \right) = \frac{{40000}}{{s\left( {s + 100} \right)\left( {s + 400} \right)}}\) 

Using partial fraction, the above expression can be written as:

\(C\left( s \right) = \frac{1}{s} - \frac{4}{{3\left( {s + 100} \right)}} + \frac{1}{{3\left( {s + 400} \right)}}\) 

Taking the inverse Laplace transform on both the sides, we get:

\(c\left( t \right) = 1 - \frac{4}{3}{e^{ - 100t}} + \frac{1}{3}{e^{ - 400t}}\) 

Given transfer function,

\(\frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{1}{{\tau s + 1}}\)

R(s) = 1/s

\(\begin{array}{l} \Rightarrow c\left( s \right) = \frac{1}{{s\left( {\tau s + 1} \right)}}\\ \Rightarrow c\left( s \right) = \frac{1}{s} - \frac{\tau }{{\tau s + 1}} \end{array}\)  

⇒ c(t) = (1-e^-t/τ)

Given that, c(t) = 0.6

t = 15

⇒ 0.6 = 1 - e^-15/τ

⇒ e^-15/τ = 0.4

 ⇒ e^15/τ = 2.5

⇒ 15/τ = In (2.5) = 0.916