Characteristic Equation MCQ Quiz - Objective Question with Answer for Characteristic Equation - Download Free PDF

Concept:

The transfer function of a closed-loop system is given by:

\( {C(s) \over R(s)} = {G\over 1+GH}\)

where, C(s) = Output

R(s) = Input

G = Forward  path gain

H = Feedback path gain

Explanation:

• 1 + GH is the characteristic equation of the transfer function.
• The roots of the characteristic equation are actually the zeroes of the characteristic equation.
• For the system to be stable, the number of zeroes of the characteristic equation on the right half of the S-plane must be 0.​

Additional Information

• The zeroes of the characteristic equation or the poles of the transfer function are the same thing. 
• So for the system to be stable, the number of poles on the right half of the S-plane must be zero.
• This is the another condition of stability of the system.

Root Locus:

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. Root locus diagram is symmetrical with respect to the real axis.

3. 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: \({\theta _l} = \frac{{\left( {2l + 1} \right)\pi }}{{P - Z}}\)

l = 0, 1, 2, … |P – Z| – 1

7. On the real axis to the right side of any section, if the sum of total number of poles and zeros are odd, 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 break points.

Analysis:

Given open-loop transfer function is:

\(G(s)H(s) = \frac{K}{{({s^3} + 6{s^2} + 11s + 6)}}\)

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

s3 + 6s2 + 11s + 6 + k = 0

By RH table

For k value put s1 row = 0

\(\frac{{66 - (k + 6)}}{{6}}\) = 0

k = 60

For ω value put s2 row = 0

6s2 + (k+6) = 0

-6ω2 + 66 = 0

ω2 = 11

ω = √11 rad/sec

Calculation:

\(T\left( s \right) = \frac{1}{{{s^2} + 2\alpha s + 1}}\) 

\( = \frac{{\frac{1}{{{s^2} + 1}}}}{{1 + \frac{{2\alpha s}}{{{s^2} + 1}}}}\) 

\(\therefore G\left( s \right) = \frac{1}{{{s^2} + 1}}\;H\left( s \right) = 2\alpha s\) 

\(OLTF = \frac{{2\alpha s}}{{{s^2} + 1}}\) 

zero = 0, pole = -j, +j

When the closed loop poles are equal:

The characteristic equation is:

s² + 1 + 2αs = 0

for α = 1,

(s + 1)² = 0

roots = -1, -1

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.

Concept:

The transfer function of a closed-loop system is given by:

\( {C(s) \over R(s)} = {G\over 1+GH}\)

where, C(s) = Output

R(s) = Input

G = Forward  path gain

H = Feedback path gain

Explanation:

• 1 + GH is the characteristic equation of the transfer function.
• The roots of the characteristic equation are actually the zeroes of the characteristic equation.
• For the system to be stable, the number of zeroes of the characteristic equation on the right half of the S-plane must be 0.​

Additional Information

• The zeroes of the characteristic equation or the poles of the transfer function are the same thing. 
• So for the system to be stable, the number of poles on the right half of the S-plane must be zero.
• This is the another condition of stability of the system.

Concept:

If the Open loop transfer function = G(s)H(s), then any point on root locus must satisfy two conditions

• Magnitude condition |G(s)H(s)| = 1
• Phase condition ∠G(s).H(s) = ±(2ξ + 1)180°

Calculation:

Open loop transfer function of given network is

\(G\left( s \right)H\left( s \right) = \frac{{90.K\left( {s + 3} \right)}}{{\left( {s + 2} \right)}}\) 

In order for a point s= -2.75to lie on root locus:

|G(-2.75).H(-2.75)| = 1

\({\left| {G\left( s \right)H\left( s \right)} \right|_{s = - 2.75}} = \left| {\frac{{10K\left( { - 2.75 + 3} \right)}}{{\left( { - 2.75 + 2} \right)}}} \right| = 1\) 

\(= \left| {10k\left( {\frac{{0.25}}{{ - 0.75}}} \right)} \right| = 1\) 

⇒ k = 3/10 = 0.3

Analysis:

Given open-loop transfer function is:

\(G(s)H(s) = \frac{K}{{({s^3} + 6{s^2} + 11s + 6)}}\)

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

s3 + 6s2 + 11s + 6 + k = 0

By RH table

For k value put s1 row = 0

\(\frac{{66 - (k + 6)}}{{6}}\) = 0

k = 60

For ω value put s2 row = 0

6s2 + (k+6) = 0

-6ω2 + 66 = 0

ω2 = 11

ω = √11 rad/sec

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.

Calculation:

\(T\left( s \right) = \frac{1}{{{s^2} + 2\alpha s + 1}}\) 

\( = \frac{{\frac{1}{{{s^2} + 1}}}}{{1 + \frac{{2\alpha s}}{{{s^2} + 1}}}}\) 

\(\therefore G\left( s \right) = \frac{1}{{{s^2} + 1}}\;H\left( s \right) = 2\alpha s\) 

\(OLTF = \frac{{2\alpha s}}{{{s^2} + 1}}\) 

zero = 0, pole = -j, +j

When the closed loop poles are equal:

The characteristic equation is:

s² + 1 + 2αs = 0

for α = 1,

(s + 1)² = 0

roots = -1, -1

Magnitude criterion:

It is used to find the value of the K at any pole lies on the root locus.

The magnitude condition is

|G(s)H(s)| = 1 

Calculation:

The open-loop transfer function is

\(G\left( s \right) = \frac{K}{{s\left( {{s^2} + 7s + 12} \right)}}\)

\(G\left( s \right) = \frac{K}{{s}(s+3)(s+4)}\)

Feedback H(s) = 1

s₁ = -1 + j1 lies on the root locus

By using magnitude criterion by substituting s₁

G(s) at s₁ = -1 + j is

\(G(s)|_{s = s_1} = \frac{K}{(-1+j)(-1+j+3)(-1+j+4)}\)

⇒ \(G(s)|_{s = s_1} = \frac{K}{(-1+j)(j+2)(j+3)}\)

The magnitude of G(s) is

\(|G(s)| =\frac {K}{\sqrt2(\sqrt5)(\sqrt10)}=1\)

∴ K = 10

Concept:

The transfer function of a closed-loop system is given by:

\( {C(s) \over R(s)} = {G\over 1+GH}\)

where, C(s) = Output

R(s) = Input

G = Forward  path gain

H = Feedback path gain

Explanation:

• 1 + GH is the characteristic equation of the transfer function.
• The roots of the characteristic equation are actually the zeroes of the characteristic equation.
• For the system to be stable, the number of zeroes of the characteristic equation on the right half of the S-plane must be 0.​

Additional Information

• The zeroes of the characteristic equation or the poles of the transfer function are the same thing. 
• So for the system to be stable, the number of poles on the right half of the S-plane must be zero.
• This is the another condition of stability of the system.

Root Locus:

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. Root locus diagram is symmetrical with respect to the real axis.

3. 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: \({\theta _l} = \frac{{\left( {2l + 1} \right)\pi }}{{P - Z}}\)

l = 0, 1, 2, … |P – Z| – 1

7. On the real axis to the right side of any section, if the sum of total number of poles and zeros are odd, 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 break points.

\(\rm \begin{array}{l} {s^3} + 7{s^2} + \left( {k + 12} \right)s + 5k = 0\\ \rm s\left( {{s^2} + 7{s} + 12} \right) + k\left( {s + 5} \right) = 0\\ \rm 1 + \frac{{k\left( {s + 5} \right)}}{{s\left( {s + 4} \right)\left( {s + 3} \right)}} = 0 \end{array}\)

The coordinate points of asymptote is nothing but centroid. 

Asymptote  x – coordinate \(\rm = \frac{{ - 4 - 3 - \left( { - 5} \right)}}{{3 - 1}}\)

\(= \frac{{ - 7 +5}}{2} = \frac{{ - 2}}{2} = - 1\)

Concept:

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 is:

N = P if P ≥ Z

= Z, if P ≤ Z

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

Analysis:

Given the open-loop transfer function as:

\(G\left( s \right)H\left( s \right) = \frac{{K\left( {s + 6} \right)}}{{\left( {s + 2} \right)\left( {s + 4} \right)}}\)

So, we have the open-loop poles and zeros as:

Poles: s = -2 and s = -4

Zeros: s = -6

Therefore, the number of asymptotes will be:

P – Z = 2 – 1 = 1

So, the characteristics mentioned in (1) is correct for the given system.

Now, the characteristic equation for the system is obtained by solving:

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

(s + 2) (s + 4) + K (s + 6) = 0

s² + (6 + K) s + 8 + 6K = 0

For the characteristic equation, we form the Routh’s array a

Root locus is plotted for K = 0 to ∞, i.e. K > 0

Here, for K > 0, the root locus does not intersect the jω axis because s¹ row will not be zero.  Thus, the characteristic mentioned in (2) is incorrect.

For the given system, we have two poles and one zero. So, one imaginary zero lies in infinity. Therefore, the characteristic mentioned in (4) is incorrect.

Hence, Option (2) must be the correct option. But, we check further for the characteristic mention in (3) as follows. We sketch the root locus for the given system as: