Control Systems Test 4

Bode plot transfer function is represented in standard time constant form as \(T\left( s \right) = \frac{{k\left( {\frac{s}{{{\omega _{{c_1}}}}} + 1} \right) \ldots }}{{\left( {\frac{s}{{{\omega _{{c_2}}}}} + 1} \right)\left( {\frac{s}{{{\omega _{{c_3}}}}} + 1} \right) \ldots }}\)

ω_c1, ω_c2, … are corner frequencies.

In a Bode magnitude plot,

• For a pole at the origin, the initial slope is -20 dB/decade
• For a zero at the origin, the initial slope is 20 dB/decade
• The slope of magnitude plot changes at each corner frequency
• The corner frequency associated with poles causes a slope of -20 dB/decade
• The corner frequency associated with poles causes a slope of -20 dB/decade
• The final slope of Bode magnitude plot = (Z – P) × 20 dB/decade

Where Z is the number zeros and P is the number of poles

Application:

As the final slope is -60 dB/decade, it indicates that the difference between the poles and zeros is 3.

Therefore, to get a final slope of -60 dB/decade, the condition required is, p-q = 3

\(G\left( s \right) = \frac{{\pi {e^{ - 0.25s}}}}{s}\)

In G(s) plane, the Nyquist plot of G(S) passes through the negative real axis at the point (-a, j0)

a = magnitude of G(s) at ω = ω_pc

ω_pc is phase cross over frequency.

at \({\rm{\omega}} = {{\rm{\omega }}_{{\rm{pc}}}},\angle G\left( {j\omega } \right) = - 180^\circ\)

⇒ -0.25 ω_pc – 90 = -180

⇒ ω_pc = 2π

\({\left| {G\left( {j\omega } \right)} \right|_{{\rm{\omega }} = {{\rm{\omega }}_{{\rm{pc}}}}}} = \frac{{\pi \left( 1 \right)}}{{2\pi }} = 0.5\)

The gain margin of the system is given by:

\(GM = 20{\log _{10}}\left( {\frac{1}{a}} \right)\)

Where ‘a’ is the intersection point of the Nyquist plot on the negative real axis, i.e.

\(20\;dB = 20{\log _{10}}\left( {\frac{1}{a}} \right)\)

\(\frac{1}{a} = 10\)

a = 0.1    ---(1)

The open-loop transfer function is:

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

At the negative real axis, the phase angle of the system is – 180°, i.e.

\(\angle G\left( {jω } \right)H\left( {jω } \right) = - 180^\circ \)

\( - 90^\circ - {\tan ^{ - 1}}\frac{ω }{2} - {\tan ^{ - 1}}\frac{ω }{4} = - 180^\circ \)

\({\tan ^{ - 1}}\left( {\frac{{\frac{ω }{2} + \frac{ω }{4}}}{{1 - \frac{{{ω ^2}}}{8}}}} \right) = 90^\circ \)

\(1 - \frac{{{ω ^2}}}{8} = 0\)

ω = √8 rad/sec

Replacing s with jω, the transfer function can be written as:

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

So, the Nyquist plot crosses the negative real axis for ω = √8 rad/sec. Hence, the intersection point ‘a’ will be:

\(|G\left( {jω } \right)H\left( {jω } \right){\left. | \right|_{ω = \sqrt 8 }} = a\)

\(a={\left. {\frac{K}{{ω \sqrt {{ω ^2} + 4 }\sqrt {{ω ^2} + 16}}}} \right|_{ω = \sqrt 8 }} \)

With a = 0.1, we can write:

\({\left. {\frac{K}{{ω \sqrt {{ω ^2} + 4 }\sqrt {{ω ^2} + 16}}}} \right|_{ω = \sqrt 8 }}=0.1 \)

\(\frac{K}{{48}} = 0.1\)

K = 4.8

The characteristic equation for the given open-loop transfer function is defined as:

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

∴ For the given open-loop transfer function, we will get the characteristic equation as:

\({s^2} + \frac{1}{T}s + \frac{K}{T} = 0\)     ---(1)

The standard form of the characteristic equation of the second-order system will be:

\({s^2} + 2\xi {\omega _n}s + \omega _n^2 = 0\)

Comparing this with Equation (1), we get:

\({\omega _n} = \sqrt {\frac{K}{T}}\)    ---(2)

Also \( \;2\xi {\omega _n} = \frac{1}{T}\)

Using Equation (2), we get:

\(\xi = \frac{1}{{2\sqrt {KT} }}\)      ---(3)

From the given problem, we have the maximum overshoot as:

M_p = 20%

Since the maximum overshoot is defined as:

\({M_p} = \frac{{ - \xi \pi }}{{e\sqrt {1 - {\xi ^2}} }} = \frac{{20}}{{100}}\)

Putting on the respective values:

\(\frac{{\xi \pi }}{{\sqrt {1 - {\xi ^2}} }} = 1.6\)

ξ = 0.45

Now, the resonant frequency is given by:

\({\omega _r} = {\omega _n}\sqrt {1 - 2{\xi ^2}} = 6\)

Therefore, the natural frequency will be:

\({\omega _n} = \frac{6}{{\sqrt {1 - 2{{\left( {0.45} \right)}^2}} }}\)

= 7.82 rad/sec

Substituting the values of ξ and ω_n in Equations (1) and (2), we get:

\({\left( {7.82} \right)^2} = \frac{K}{T}\;and\)

\(0.45 = \frac{1}{{2\sqrt {KT} }}\)

Solving the above equations, we obtain the values of K and T as:

K = 8.67 and T = 0.14