IAT_solutions.pdf

Solution of IAT 1 (**Figures to the right indicate scheme of marking)

1. a) Analogous electrical elements in force–voltage analogy for the elements of mechanical

translational system: (2)

Mechanical System Electrical System

Force Voltage

Velocity Current

Mass Inductance

Compliance (Reciprocal of Stiffness) Capacitance

Damping Resistance

b) Transfer Function: It is defined as the ratio of Laplace transform of output of a system to

the Laplace transform of input given to that system when all initial conditions are assumed to

be zero. i.e.,

Transfer Function, G(s) = C(s)/R(s) where, C(s) is

the output and R(s) is input when all initial

conditions are zero. (1)

Type of a system: It is defined as the no. of poles located at the origin. E.g. let,

( ) ( ), as there are two poles at origin, so the system is of type-2. (2)

c) Comparing with general 2nd

order system characteristic equation, ,

we get the value of i.e., (1)

and, i.e.,

(2)

d) Rise Time: It is defined as the time required

for the system waveform to go from 10% to 90%

of its final value. It is denoted by Tr.

(

√

) (1)

Settling Time: it is the time required for the

system to settle within ±2% to ±5% of the final

value. It is denoted by Ts.

where, is undamped natural

frequency, is damped natural frequency and

is damping ratio. (2)

e) A system is classified in four groups depending on the value of the damping as,

i. Undamped system with

ii. Underdamped system with

iii. Critically damped system with

iv. Overdamped system with (2)

f) Four steps of block diagram reduction:

i. Cascade form: When multiple subsystems are connected in cascade form, then final

output TF of equivalent single system is the product of all TFs of the subsystems.

ii. Parallel form: When multiple subsystems are connected in parallel form, then final

output TF of equivalent single system is the sum of all TFs of the subsystems.

(1)

iii. Feedback form: When multiple subsystems are connected in feedback form, then

final output TF of equivalent single system will be as below,

iv. Moving blocks: The blocks can be moved either to the left (fig. a) or right (fig.b) to

the summing junction as follows,

(2)

2. a) Block diagram reduction: (5 Steps X 1 = 5)

Step 1:

Step 2:

Step 3:

Step 4:

[*** Each step carries 1 mark.]

Step 5:

b) Mathematical modelling of Mechanical system: To develop free body diagram for the

given system, we sum all the forces to zero.

There are four forces,

i. An external force, F

ii. A force from the spring. To determine the

direction consider that the position "x" is defined

positive to the right. If the mass moves in the positive "x" direction, the spring is compressed

and exerts a force on the mass. So there will be a force from the spring, k*x, to the left.

iii. A force from the dashpot. By an argument similar to that for the spring there will be a

force from the dashpot, b*v, to the left. (The velocity, v, is the derivative of x with respect

to time.)

iv. Finally, there is the internal force which is

defined to be opposed to the defined direction of

motion. This is represented by M*a to the

left. (The acceleration, a, is the second derivative of

x with respect to time.) Free Body Diagram

So, ( ) ( ) ( ) i.e.,

This is the differential equation of a mechanical system as provided. (3)

Force to Voltage Analogy: Analogous equations are described in the following table,

Electrical Equation Mechanical Analogy (Force to Voltage)

∫

∫

∫

∫

So, Force-Voltage analogy is, (5)

Mechanical System Component Electrical System Component

Force (F) Voltage (e)

Velocity (v) Current (i)

Displacement (x) Charge (Q)

Mass (M) Inductance (L)

Damping Coefficient (b) Resistance (R)

Compliances(=1/Stiffness) (1/K) Capacitance (C)

3. a) RH Criterion: F(s) = s6 + 2s

5 + 8s

4+12s

3+20s

2+16s +16 = 0

s6 1 8 20 16

s5 2 (1) 12 (6) 16 (8) 0

s4 2 (1) 12 (6) 16 (8)

s3 0 (4) (1) 0 (12) (3)

s2 3 8

s1 1/3

s0 8

Since, row of coefficient of s3 becomes zero,

so the auxiliary equation, A = s4+6s

2+8 = 0

So,

(3)

No sign change in first column. But, there is a row of zeros. So, there is a possibility of roots

lying on the imaginary axis.

Auxiliary polynomial, s4 + 6s

2 + 8 = 0

let x = s2. This gives, x

2 + 6x + 8 = 0

Solution of this 2nd

order equation is, x = 2, 4

So, s = ± 2 and ± 4 = +j2, j2, +j2, j2

Roots of auxiliary polynomial are also roots of characteristic equation. Hence 4 roots lie on

imaginary axis and remaining 2 roots lie on left half of s-plane. Hence, the system is

marginally stable. (5)

b) Steady State Error for different Inputs & system Types: Steady State Error is defined as

the difference between desired input and measured output when the system ultimately

reaches to the steady state after its transient nature. So, e(t) = r(t) – c(t)

Or, in s-domain, E(s) = R(s) C(s)

Steady-state error can be calculated from the open or closed-loop transfer function for unity

feedback systems. For example, let's consider the system given below.

Since, the feedback H(s) = 1, error E(s) actually implies E(s) = R(s) C(s)

But, C(s) = G(s). E(s) So, E(s) = R(s) G(s) E(s)

E(s) = ( )

( )

e(∞) = ( ) ( ) ( )

Now, let's plug in the Laplace transforms for some standard inputs and determine equations

to calculate steady-state error from the open-loop transfer function in each case.

Step Input (R(s) = A / s): e(∞) = (

)

( )

( )

Where, Kp = position error constant = ( )

Ramp Input (R(s) = A/s2): e(∞) =

(

)

( )

( )

Where, Kv = velocity error constant = ( )

Parabolic Input (R(s) = A/s3): e(∞) =

(

)

( )

( )

Where, Ka = acceleration error constant = ( ) (2)

The number of poles at the origin of the loop gain transfer function (i.e. the number of

integrators) defines the system’s type number. For the following system, n represents the

type of the system.

(3)

Step Input Ramp Input Parabolic

Input

Type Steady state

error formula 1/(1+Kp) 1/Kv 1/Ka

Type 0

system

Static error

constant Kp = constant Kv = 0 Ka = 0

Error 1/(1+Kp) Infinity Infinity

Type 1

system

Static error

constant Kp = infinity Kv = constant Ka = 0

Error 0 1/Kv Infinity

Type 2

system

Static error

constant Kp = infinity Kv = infinity Ka = constant

Error 0 0 1/Ka

Type 3 &

higher

system

Static error

constant Kp = infinity Kv = infinity Ka = infinity

Error 0 0 0

(5)

--THE END--

8

Solution of IAT 2 (**Figures to the right indicate scheme of marking)

1. a) State Controllability: A system is said to be controllable at time t0 if it is possible by

means of an unconstrained control vector u(t) to transfer the system from any initial state

x(t0) to any other state x(tf) in a finite interval of time (tf-t0) ≥ 0.

For an LTI system, ̇ ( ) ( ) ( ) ( ) ( ) to be completely

controllable, it is necessary and sufficient that the following matrix has a rank of n.

[ ] (2)

b) Phase Margin: It’s a measure of relative stability, defined as the additional phase lag

for the open loop system required at unity gain (i.e. at gain crossover frequency, ωc) to

make the closed loop system unstable. Mathematically,

PM = 180o + ∟G(jωc)H(jωc) (1)

Gain Margin: It’s a measure of relative stability, defined as the change in open loop gain,

expressed in Decibels (dB), required at 180o of phase shift (i.e. at phase crossover

frequency, ωp) to make the closed loop system unstable. Mathematically,

GM =

| ( ) ( )| (2)

c) Centroid: If there are n no. of poles and m no. of zeros, then (n-m) no. of root locus

branches tends to infinity along straight line asymptotes drawn from a single point on the

real axis. This point is called Centroid. Mathematically it can be calculated as,

Centroid = ( Σpoles – Σzeros )/(n-m) (1)

Angle of Departure: It is an angle made by a complex pole to find the root locus of a

system when open loop gain K is varied from 0 to ∞. Mathematically it is calculated as,

d) Nyquist Stability Criterion: If a contour A that encircles the entire RHS plane, is

mapped through G(s)H(s), then the no. of closed loop poles in the RHS plane (Z) which is

equal to the no. of open loop zeros, is always equal to the difference between no. of open

loop poles (P) in the RHS plane and the no. of counter clockwise revolutions (N) around

-1 point of the mapping. i.e. Z = P-N. So if there is no encirclement around the point

(-1+j0), it implies that the system is stable. (2)

e) Provided, ( ) ( )

( )( )

So, open loop poles are at . So there are 3 poles (i.e., n = 3) and no

zeros (i.e., m = 0). So, no. of asymptotes = n m = 3.

( )

k = 0, 1, 2

So, angle of asymptotes are 60o, 180

o and 300

o (-60

o) (2)

9

f) Principle of Optimality: The system design is said to be optimal if it minimizes a

specified performance index (PI and/or cost function) with no constraints on the

controller configuration. A design becomes optimal when the system parameters are so

adjusted so that PI attains an extremum, normally a minimum value. Such a design should

provide a system that is not merely stable but also satisfies a specific criterion based on

minimization of settling time, peak overshoot or steady state error and has the desired

bandwidth, hence the word OPTIMAL. (2)

2. a) Bode Plot: Transfer function, as given is converted into standard form as,

( ) (

)

( ) (

)

Next substitute s = jω in the standard form,

( ) (

)

( ) (

)

Phase angle, ( ) (

) (

) (

) (1)

Table 1: Table of Factors: (2)

Sl

no Factors Magnitude Curve Phase Curve

1. Gain, K= 0.75 Straight line at 20 log K = -2.5 dB Φ = 0

2. Pole at origin,

Straight line of slope -20 dB/dec

passing through ω = 1, 0 dB point Φ = -90°

3. 1st order pole,

(

)

Line slopes are,

a) 0 dB/dec for ω ≤ 2

b) -20 dB/dec for ω > 2

Φ = -tan-1

(ω/2)

4. 1st order pole,

(

)

Line slopes are,

a) 0 dB/dec for ω ≤ 10

b) -20 dB/dec for ω > 10

Φ = -tan-1

(ω/10)

5. 1st order zero, (

)

Line slopes are,

a) 0 dB/dec for ω ≤ 5

b) +20 dB/dec for ω > 5

Φ = tan-1

(ω/5)

Table 2: Magnitude Plot Table: (3)

Sl

no Factors Resultant slope Start point End point

1. K Straight line at -2.5 dB 0.1 ∞

2.

-20 dB/dec 0.1 2

3.

( ) -20+(-20) = -40 dB/dec 2 5

4.

( ) -40+20 = -20 dB/dec 5 10

10

5. (

) -20+(-20) = -40 dB/dec 10 ∞

Table 3: Phase angle table: (4)

ω

(

) (

) (

) ( )

0.1 -90 -2.86 -0.572 1.15 -92.3

1 -90 -26.5 -5.71 11.3 -110.9

2 -90 -45 -11.31 21.8 -124.5

5 -90 -68.2 -26.56 45 -139.76

10 -90 -78.7 -45 63.43 -150.26

50 -90 -87.7 -78.7 84.3 -172.1

100 -90 -88.85 -84.3 87.13 -176.02

1000 -90 -89.85 -89.42 89.7 -179.55

Figure below shows the Bode plot for the system. (5)

Fig: Bode Plot

b) Root Locus Plot: Provided,

( )

( )

i) Poles: s = 0, (-3+j4) and (-3-j4) [So, n=3]

There are two complex poles. So a root locus branch exists between origin and ∞.

ii) Zeros: There are no zeros. So, m=0

iii) No. of Asymptotes: (n-m) = 3

11

iv) Angle of Asymptotes: ( )

= 60

o, 180

o, 300

o

v) Centroid: C = ( Σpoles – Σzeros )/(n-m) = (0-3-3)/3 = -2 (1)

vi) Breakaway Point: Characteristic equation, 1+G(s)H(s) = 0

( ) gives ( )

So,

( ) yields and for this value of s, the

value of

But to be an actual breakaway point, corresponding K value must be real and positive.

Hence neither of these two points are breakaway points. (2)

vii) Imaginary Axis Crossing: Substitute s=jω in the characteristic equation,

=> ( ) ( ) ( )

After simplification, ( ) ( )

Which yields ω = ±5 and K = 150. (3)

viii) Angle of Departure: OA = (-3+j4-0) = 5∟126.87o

AB = (-3+j4-(-3-j4)) = j8 = 8∟90o

So, ( ) ( )

(4)

Figure below shows the root-locus plot for the system.

Fig: Root Locus Plot (5)

3. a) Nyquist Plot: For the system as given,

( ) ( )

( )( )

poles are, s = 0,-2 and -10. Since there are no poles on the RHS plane, hence P = 0. So for

stability, no. of encirclement about the point (-1+j0) in the CCW direction should be zero

i.e. N = 0. (1)

Modified Nyquist contour is shown in fig (a) as there is a pole at origin.

Nyquist Paths are,

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1. Path a-b: s = jω

2. Path b-c-d: s = ; 90° ≤ θ ≤ -90°

3. Path d-e: s = -jω

4. Path e-f-a: s = ; -90° ≤ θ ≤ 90°

Path a-b: Here s = jω gives the polar plot ( ) ( )

( )( )

| ( ) ( )|

√ √

( ) ( ) (

) (

)

At ω = 0; ( ) ( ) and at ω = ∞; ( ) ( )

Polar plot is shown in fig (b).

Fig (a) Fig (b) (4)

Path b-c-d: Here s = ; ( ) ( )

( )( ) [As, R∞].

So this path maps to the centre.

Path d-e: Here s = -jω. This is a mirror

image of path a-b.

Path e-f-a: Here s = ;

( ) ( )

( )( )

Since there is a pole at origin, there is a

semi-circle in CW direction from e to a.

The entire Nyquist Plot is shown in fig

(c).

Fig (c) (5)

13

b) Controllability & Observability: Provided,

[

] [

] [ ]

Controllability matrix, Qc = [B AB A2B], since A is of order n = 3. (1)

AB = [

] [ ] [

] and A

2B = A*AB = [

] [ ] [

]

So, |Qc| = |

|

Qc is a non-singular matrix and hence, rank of Qc = order of A = 3.

So the system is controllable. (3)

Similarly, Observability matrix, Qo = [

]

CA = [ ] [

] [ ]

CA2 = CA*A = [ ] [

] [ ]

So, |Qo| = |

|

So, Qo is a singular matrix and hence, rank of Qo ≠ order of A.

So, the system is unobservable. (5)

--THE END--