Control Systems Test 3

\(\left. {\begin{array}{*{20}{c}} {{s^3}}\\ {{s^2}}\\ {\begin{array}{*{20}{c}} {{s^1}}\\ {{s^0}} \end{array}} \end{array}} \right|\begin{array}{*{20}{c}} 1\\ 2\\ {\begin{array}{*{20}{c}} 0\\ {12} \end{array}} \end{array}\begin{array}{*{20}{c}} 6\\ {12}\\ {\begin{array}{*{20}{c}} 0\\ 0 \end{array}} \end{array}\)

Row of zero occurs

Poles on jω axis

System is marginally stable

Hence (δ = 0)

After adding a zero at s = -4, the open transfer function becomes:

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

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

So, we have the characteristic equation as:

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

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

Or s³ + 2s² + 2s + K = 0

So, we form the Routh’s array for the system as

For the stability of the system, we have the condition:

\(\frac{{4 - K}}{2} > 0\) 

K < 4

And K > 0

∴ The common region is:

0 < K < 4

Hence, for a positive real number K (i.e. K > 0) the system is stable if K < 4.

The root locus plot gives the location of the closed-loop poles for different values of parameter gain K. So, we have the characteristic equation as:

\(1 + \frac{{K\left( {s + 1} \right)}}{{{s^2}\left( {s + 9} \right)}} = 0\)

s³ + 9s² + Ks + K = 0       ---(1)

For all the roots to be equal and real, we require

(s + P)³ = s³ + 3Ps² + 3P² s + P³ = 0       ---(2)

On comparing equations (1) and (2), we can write:

3P = 9

P = 3

And K = P³ = (3)³

K = 27

Important Points:

1) Every branch of a root locus diagram starts at a pole (K = 0) and terminates at a zero (K = ∞) of the open-loop transfer function.

2) The root locus diagram is symmetrical with respect to the real axis.

3. The number of branches of the root locus diagram are:

N = P if P ≥ Z

= Z, if P ≤ Z

4) Number of asymptotes in a root locus diagram = |P – Z|

5) Centroid: It is the intersection of the asymptotes and always lies on the real axis. It is denoted by σ.

\(\sigma = \frac{{\sum {P_i} - \sum {Z_i}}}{{\left| {P - Z} \right|}}\)

ΣPi is the sum of real parts of finite poles of G(s)H(s)

ΣZi is the sum of real parts of finite zeros of G(s)H(s)

6) Angle of asymptotes: 

l = 0, 1, 2, … |P – Z| – 1

7) On the real axis to the right side of any section, if the sum of the total number of poles and zeros are odd, the root locus diagram exists in that section.

8) Break in/away points: These exist when there are multiple roots on the root locus diagram.

At the breakpoints gain K is either maximum and/or minimum.

So, the roots of \(\frac{{dK}}{{ds}}\) are the breakpoints.

For the system to oscillate at ω = 2 rad/sec.

s = jω = 2j

Continuing with the Routh-Array, the s² row will contain:

\(\frac{3a-12}{3}~~~~8\) 

a - 4      8

Solving the s² row we get:

\(\left( {a - 4} \right){s^2} + 8 = 0\)

\( \Rightarrow \left( {a - 4} \right){\left( {j\omega } \right)^2} + 8 = 0\)

\( \Rightarrow \left( {a - 4} \right){\left( {j2} \right)^2} + 8 = 0\)

\( \Rightarrow \left( {a - 4} \right)4 = 8\)

\( \Rightarrow a = 6\)

The characteristic equation of the closed-loop system is:

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

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

Or s² + (K - 1)s + 2K = 0     ---(1)

For the characteristic equation, we have the Routh’s array as:

The required condition for the system to be stable will be:

K – 1 > 0, i.e. K > 1

And 2K > 0, i.e. K > 0

The common intersection range for stability will be:

K > 1

On comparing the standard 2^nd order characteristic equation with equation (1), we get:

\({\omega _n} = \sqrt {2K}\)

Also:

\(2\xi {\omega _n} = K - 1\)

\(With\;\xi = 0.707 = \frac{1}{{\sqrt 2 }}\)

\(2\left( {\frac{1}{{\sqrt 2 }}} \right)\sqrt {2K} = K - 1\)

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

2 × 2K = K² – 2K + 1

K² – 6K + 1 = 0

K = 5.83 or K = 0.17

For stability, we have the condition K > 1.

Thus, from the above results, we have the desired value of gain K as K = 5.83

\(G\left( s \right)H\left( s \right) = \frac{K}{{s\left( {s + 4} \right)\left( {s + 5} \right)}}\)

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

\( \Rightarrow 1 + \frac{K}{{s\left( {s + 4} \right)\left( {s + 5} \right)}} = 0\)

⇒ s(s + 4) (s + 5) + K = 0

⇒ s³ + 9s² + 20 s + K = 0

\(\left. {\begin{array}{*{20}{c}} {{s^3}}\\ {{s^2}}\\ {{s^1}}\\ {{s^0}} \end{array}} \right|\begin{array}{*{20}{c}} 1\\ 9\\ {\frac{{180 - K}}{9}}\\ K \end{array}\;\begin{array}{*{20}{c}} {20}\\ K\\ 0\\ {} \end{array}\)

The system to be marginally stable, 180 – K = 0

⇒ K = 180

9s² + K = 0

⇒ 9s² +180 = 0

\(\Rightarrow s = \pm j\sqrt {20} = \pm j2\sqrt 5\)

The characteristic equation of a unity feedback system is given by:

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

For the given system, we can write:

\(1 + \frac{{K{e^{ - s}}}}{{s\left( {{s^2} + 5s + 9} \right)}} = 0\)

At low frequencies, the exponential term can be approximated as:

e^-s = (1 - s), i.e. the higher powers of s are neglected.

The characteristic equation can now be written as:

\(1 + \frac{{K{{(1 - s)}}}}{{s\left( {{s^2} + 5s + 9} \right)}} = 0\)

s(s² + 5s + 9) + K(1 - s) = 0

s³ + 5s² + (9 - K) s + K = 0

Therefore, we from the Routh’s array for the system as:

For the system to be stable, there should not be any sign changes at the first column of the Routh array, i.e.

K > 0, and 

\(\frac{{45 - 6K}}{5} > 0\)

\(K < 7.5\)

Hence, the system is stable for the intersecting region, i.e.

0 < K < 7.5

∴ We conclude that the system is conditionally stable.