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  are the break points.

Calculation:

The characteristic equation is obtained as:

1+ OLTF = 0, i.e.

⇒ K = -s(s2 + 3s + 2)

⇒ K = -s3 – 3s2 – 2s

⇒ 3s2 + 6s + 2 = 0

• 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.

          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: 

          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  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:

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;

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

i.e. 

Solving for , we get

⇒ 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

s³ + 3s² + k(s + α) = 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:

For no breakaway point other that at s = 0

Set,

9 (1 + α)² – 48 α

3(α)2 – 10 α + 3

(3α-1)(α-3)

Or, 0.333

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

= s²(s + 3) + k(s + α) = 0

For breakaway point

= 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

For no real breakaway points other than s = 0

9(1 + α)² – 48 α

9[1 + α² + 2α] – 48α

9α² + 18α + 9 – 48α

9α² – 30 α + 9

3α² – 10α + 3

0.33

⇒ 2s² + 12s + 2s + 12 – s² – 2s = 0

⇒ s² + 12s + 12 = 0

⇒ s = -1.1, -10.9

s = -1.1 is the breakaway point

s = -10.9 is the break in point

Given,

For breakaway point

2s² + 6s + 4 – s² – 2s – 2 = 0

s² + 4s + 2 = 0

= -0.586 or -3.41

Given that 

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

At break point,

⇒ -(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

At s = -7.46,

The value of gain k at system is critically damped is, k = 0.07 and 13.93

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.

For break points,

-4.5275 is not on the root locus branch.

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

Angle of departure at pole P, then

= -45°