Break Points MCQ Quiz in தமிழ் - Objective Question with Answer for Break Points - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Break-in/away points 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 break points.

Calculation:

The characteristic equation is obtained as:

1+ OLTF = 0, i.e.

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

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

⇒ K = -s(s2 + 3s + 2)

⇒ K = -s3 – 3s2 – 2s

\(\Rightarrow \frac{{dK}}{{ds}} = - 3{s^2} - 6s - 2 = 0\)

⇒ 3s2 + 6s + 2 = 0

\(s = \frac{{ - 6 \pm \sqrt {36 - 24} }}{6} \)

\(s= \frac{{ - 6 \pm \sqrt {12} }}{6} = \frac{{ - 6 + \sqrt {12} }}{6}\)

\(s= - 0.422\)

• 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.
• The root locus diagram is symmetrical with respect to the real axis.
• The number of branches of the root locus diagram is:

          N = P if P ≥ Z

              = Z, if P ≤ Z

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

• The angle of asymptotes: \({\theta _l} = \frac{{\left( {2l + 1} \right)\pi }}{{P - Z}}\)

          l = 0, 1, 2, … |P – Z| – 1

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

Concept:

• The common method of obtaining the breakaway and break in point is to maximize and minimize the gain K, using differentiation, i.e equating \(\frac{{dK}}{{ds}}\) to zero and solving for s, we get the possible break away and break-in point.
• Gain K will be maximum at breakaway point and minimum at break-in.
• For gain K, the point at which multiple roots are present is known as the breakpoint. These are obtained from:\( \Rightarrow \frac{{dK}}{{ds}} = 0\)

Calculation:

Given open-loop pole-zero plot of the system. From the given plot, we have the zeroes and poles as:

Zeroes: s = 1 + j and s = 1 – j

Poles: s = -2 and s = -3

So, the Transfer Function of the system is obtained as;

\(H\left( s \right)G\left( s \right) = \frac{{K\left\{ {\left( {s\left( {1 + j} \right)} \right)\left( {s - \left( {1 - j} \right)} \right)} \right\}}}{{\left( {s + 2} \right)\left( {s + 3} \right)}}\)

\( = \frac{{K\left( {s - 1 - j} \right)\left( {s - 1 + j} \right)}}{{\left( {s + 2} \right)\left( {s + 3} \right)}}\)

\( = \frac{{K\left[ {{{\left( {s - 1} \right)}^2} - {{\left( j \right)}^2}} \right]}}{{\left( {s + 2} \right)\left( {s + 3} \right)}}\)

\( \Rightarrow G\left( s \right)H\left( s \right) = \frac{{K\left( {{{\left( {s + 1} \right)}^2} + 1} \right)}}{{\left( {s + 2} \right)\left( {s + 3} \right)}} = \frac{{K\left( {{s^2} - 2 + 2} \right)}}{{\left( {s + 2} \right)\left( {s + 3} \right)}}\)

The characteristic equation is 1 + G(s).H(s) = 0

i.e. \(1 + \frac{{K\left( {{s^2} - 2s + 2} \right)}}{{\left( {s + 2} \right)\left( {s + 3} \right)}} = 0\)

\( \Rightarrow \frac{{K\left( {{s^2} - 2s + 2} \right)}}{{\left( {s + 2} \right)\left( {s + 3} \right)}} = - 1\)

\( \Rightarrow K = \frac{{ - \left( {s + 2} \right)\left( {s + 3} \right)}}{{{s^2} - 2s + 2}} = \frac{{ - \left( {{s^2} + 5s + 6} \right)}}{{{s^2} - 2s + 2}}\)

Solving for \(\frac{{dK}}{{ds}} = 0\), we get

\( \Rightarrow \frac{{ - \left[ {\left( {{s^2} - 2s + 2} \right)\left( {2s + 5} \right) - \left( {{s^2} + 5s + 6} \right)\left( {2s - 2} \right)} \right]}}{{{{\left( {{s^2} + 2s + 2} \right)}^2}}} = 0\)

\( \Rightarrow \frac{{2{s^3} + 5{s^2} - 4{s^2} - 10s + 4s + 10 - 2{s^3} + 2{s^2} - 10{s^2} + 10s - 12s + 12}}{{{{\left( {{s^2} - 2s + 2} \right)}^2}}} = 0\)

⇒ s² – 8s² – 10s + 4s – 2s + 22 = 0

⇒ -7s² – 12 s + 4 s + 22 = 0

⇒ 7s² + 8 s - 22 s = 0

⇒ s = +1.29 and s = -2.43

Since, the point s = -2.43 is Maxima for gain K,

So, s = -2.43 is the break-away point.

Characteristic equation 1 + G(s) = 0

\(1 + \frac{{k\left( {s + \alpha } \right)}}{{{s^2}\left( {s + 3} \right)}} = 0\)

s³ + 3s² + k(s + α) = 0

\(k = - \frac{{{s^3} + 3{s^2}}}{{\left( {s + \alpha } \right)}}\)

\(\frac{{dk}}{{ds}} = \frac{{\left( {s + \alpha } \right)\left( {3{s^2} + 6s} \right) - \left( {{s^3} + 3{s^2}} \right)}}{{{{\left( {s + \alpha } \right)}^2}}} = 0\)

3s³ + 6s² + 3αs² + 6sα – s³ – 3s² = 0

2s³ + (3 + 3aα)s² + 6sα = 0

2s² + 3(1 + α)s + 6α = 0

The roots of the breakaway-point equation are:

\(s = \frac{{ - 3\left( {1 + \alpha } \right)\;}}{4} \pm \frac{{\sqrt {9{{\left( {1 + \alpha } \right)}^2} - 48\alpha } }}{4}\)

For no breakaway point other that at s = 0

Set,

9 (1 + α)² – 48 α < 0

3(α)2 – 10 α + 3 < 0

(3α-1)(α-3) < 0

Or, 0.333 < α < 3

Characteristic equation = 1 + G(s).H(s)

= s²(s + 3) + k(s + α) = 0

\( \Rightarrow k = \frac{{ - {s^2}\left( {s + 3} \right)}}{{\left( {s + \alpha } \right)}}\)

For breakaway point

\(\frac{{dk}}{{ds}} = \frac{{\left[ {3{s^2} + 6s} \right]\left[ {s + \alpha } \right] - \left[ {{s^3} + 3{s^2}} \right]}}{{{{\left( {s + \alpha } \right)}^2}}} = 0\)

\(\frac{{dk}}{{ds}} = \left( {3{s^3} + \left( {3\alpha + 6} \right){s^2} + 6\alpha s} \right) - \left( {{s^3} + 3{s^2}} \right) = 0\)

= 2s³ + 3 (1 + α) s² + 6 α s = 0

= s[2s² + 3 (1 + α) s + 6α] = 0

s = 0,

2s² + 3(1 + α) s + 6α = 0

\(s = \frac{{ - 3\left( {1 + \alpha } \right) \pm \sqrt {9{{\left( {1 + \alpha } \right)}^2} -48\;\alpha } }}{4}\)

For no real breakaway points other than s = 0

9(1 + α)² – 48 α < 0

9[1 + α² + 2α] – 48α < 0

9α² + 18α + 9 – 48α < 0

9α² – 30 α + 9 < 0

3α² – 10α + 3 < 0

0.33 < α < 3

\(\begin{array}{l} G\left( s \right)H\left( s \right) = \frac{{k\left( {s + 6} \right)}}{{s\left( {s + 2} \right)}}\\ k = \frac{{ - s\left( {s + 2} \right)}}{{\left( {s + 6} \right)}} = \frac{{ - \left( {{s^2} + 2s} \right)}}{{\left( {s + 6} \right)}}\\ \frac{{dk}}{{ds}} = 0 \end{array}\)

\(\Rightarrow - \left[ {\frac{{\left( {s + 6} \right)\left( {2s + 2} \right) - \left( {{s^2} - 2s} \right)\left( 1 \right)}}{{{{\left( {s + 6} \right)}^2}}}} \right] = 0\) 

⇒ 2s² + 12s + 2s + 12 – s² – 2s = 0

⇒ s² + 12s + 12 = 0

⇒ s = -1.1, -10.9

s = -1.1 is the breakaway point

Given, \(G\left( s \right)H\left( s \right) = \frac{{k\left( {s + 2} \right)}}{{{s^2} + 2s + 2}}\)

For breakaway point

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

\(\frac{{dK}}{{ds}} = 0\)

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

2s² + 6s + 4 – s² – 2s – 2 = 0

s² + 4s + 2 = 0

\(s = \frac{{ - 4 \pm \sqrt {16 - 8} }}{2}\)

\( = - 2 \pm \sqrt 2 \)

Given that \(G\left( s \right) = \frac{{k\left( {s + 4} \right)}}{{s\left( {s + 1} \right)}},H\left( s \right) = 1\)

\(G\left( s \right)H\left( s \right) = \frac{{k\left( {s + 1} \right)}}{{s\left( {s + 1} \right)}}\)

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

\(\begin{array}{l} \Rightarrow 1 + \frac{{k\left( {s + 4} \right)}}{{s\left( {s + 1} \right)}} = 0\\ \Rightarrow k = \frac{{ - s\left( {s + 1} \right)}}{{\left( {s + 4} \right)}} \end{array}\)

At break point,

\(\begin{array}{l} \frac{{dk}}{{dx}} = 0\\ \Rightarrow \frac{d}{{ds}}\left( {\frac{{ - s\left( {s + 1} \right)}}{{\left( {s + 4} \right)}}} \right) = 0\\ \Rightarrow \frac{{ - \left( {s + 4} \right)\left( {2s + 1} \right) + \left( {{s^2} + s} \right)\left( 1 \right)}}{{{{\left( {s + 4} \right)}^2}}} \end{array}\)

⇒ -(s + 4) (2s + 1) + s² + s = 0

⇒ -2s² – 4 – 8s – s + s² + s = 0

⇒ s² + 8s + 4 = 0

⇒ s = -0.53 and s = -7.46

The given system is critically damped.

At s = -0.53

\(k = \left| {\frac{{ - s\left( {s + 1} \right)}}{{\left( {s + 4} \right)}}} \right| = \frac{{0.53\left( { - 0.53 + 1} \right)}}{{\left( { - 0.53 + 4} \right)}} = 0.07\)

At s = -7.46,

\(k = \left| {\frac{{ - s\left( {s + 1} \right)}}{{\left( {s + 4} \right)}}} \right| = \frac{{\left( {7.46} \right) - \left( {7.46 + 1} \right)}}{{\left( { - 7.46 + 4} \right)}} = 13.93\)

\(\begin{array}{l} \frac{1}{{\sigma + 6}} = \frac{1}{{\sigma + 3}} + \frac{1}{{\sigma + 5}}\\ \Rightarrow \sigma = - 4.27,\; - 7.73 \end{array}\)

From the root locus it’s clear that breakaway point is between -3 and -5 and break in point is between -6 and –∞. Hence option (D) is correct.

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

For break points,

\(\begin{array}{l} \frac{{dk}}{{ds}} = 0\\ \Rightarrow \frac{d}{{ds}}\left( {{s^3} + 9{s^2} + 20s} \right) = 0\\ \Rightarrow 3{s^2} + 18s + 20 = 0\\ \Rightarrow s = - 1.472,\; - 4.5275 \end{array}\)

\(1 + G\left( s \right)H\left( s \right) = 0\)

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

\(k = - \frac{{{s^2} + 2s + 2}}{{{s^2} - 2s + 2}}\)

For break away & break in point differentiating above w. r. t s

\(\frac{{dk}}{{ds}} = - \frac{{\left( {{s^2} - 2s + 2} \right)\left( {2s + 2} \right) - \left( {{s^2} + 2s + 2} \right)\left( {2s - 2} \right)}}{{{{\left( {{s^2} - 2s + 2} \right)}^2}}} = 0\)

\(2{s^3} - 2{s^2} + 4 - \left( {2{s^3} + 2{s^2} - 4} \right) = 0\)

\({s^2} = 2\)

\(s = \pm \sqrt 2\)

Angle of departure at pole P, then

\({Q_d} = + 180 + \left( {{\rm{\Sigma }}{\phi _p} - {\rm{\Sigma }}{\phi _z}} \right)\)

\({Q_d} = 180 + \left( {90 - \left( {180 + 135} \right)} \right)\)

= -45°