Describing functions of some entries of Table 10.2 have been derived in Section 10.4 and Section 1

10.2 Revisiting Example 10.1 (Fig. 10.19) will be helpful.

G(jw) = 4

1

1 jw( + jw)2 -N E

; ( ) = - p E M 4 – 90∞ – 2 tan^–1w₁ = – 180∞

This givesw₁ = 1 rad/sec

- 1

N E

a f

1 ^{= |G(jw1)| = 2}

This gives E₁ = 8M/p

y(t) = – e(t) = – 8M p sin t

10.3 Revisiting Example 10.1 (Fig. 10.22) will be helpful.

G(jw) = 5

1 0 1 ²

jw( + j . w) ; N(E) = 4

1 0 1 ² p E -

FH IK

E^. – 90∞ – 2 tan^–1 0.1w₁ = – 180∞

This gives w₁ = 10 rad/sec.

For a stable limit cycle, 2D < E < •;D = 0.1

– 1

N E

a f

1 ^{= |G(jw1)| = 0.25}

This gives E₁ = 0.3

Amplitude of limit cycle is 0.3 and frequency is 10 rad/sec.

10.4 Revisiting Example 10.1 (Fig. 10.22) will be helpful.

For a limit cycle to exist, 2D < E < •;D = 0.2 –G(jw₁) = – 180∞ for w₁ = 10.95. |G(jw₁)| = 0.206

|G(jw₁)|π - 1

N E( ) for any value of E

The intersection of the two curves does not occur; therefore no limit cycle exists.

Intersection of the two curves occurs for

pD

2 < 0.206 or D < 0.131

10.5 G(jw) = 10

1+ 0 4 1 2 ( j . w) +( j w)

N (E) = 4 1

2

p E

H

E j H

-

FH IK

^- E

L NMM O

QPP

^{; H = 0.2}

The following figure shows G(jw)-plot and -( )

1

N E locus.

E H= Im

Re

G j( )w

E increasing -1

pH

N E( )

4

At the point of intersection of G(jw)-plot with -( )

1

N E locus, measure the angle of G(jw). This comes out to be –115.7∞.

–G(jw1) = – tan^–1 0.1w₁ – tan^–1 2w₁ = – 115.7∞ This gives w₁ = 5.9 rad/sec

|G(jw₁)| = 0.33

- 1

N E

a f

1 ^{= p E41 = 0.33}

This gives E₁ = 0.42

10.6 G( jw) = K jw(1+jw)²

N(E) = 2 2

1 1

1 1

1 2

p

L

p - - -

FH IK

NMM O

-

QPP

sin E E E

=1 2 1 1

1 1

1

-

L

+ - 2

NM O

-

QP

p sin

E E E = 1 – N_c(E)

The function N_c(E) is listed in Table 10.3.

For any K > 0, one of the following can happen.

(i) G(jw)-plot does not intersect -( )

1 N E locus (ii) G(jw)-plot intersects the

-( ) 1

N E locus; but the point of intersection corresponds to an unstable limit cycle.

10.7 Revisiting Review Example 10.3 (Fig. 10.40) will be helpful.

G(jw) = 10

1 1 0 5

jw

a

+ jw

fa

+ j ^. w

f

N(E) = 2 1 1

1 1

1

p

L

sin^- + - 2

NM O

QP

E E E = N_c(E) The function N_c(E) is listed in Table 10.3.

–G(jw₁) = – 180° gives w₁ = 2

|G(jw₁)| = - 1

N E

a f

1 ^{gives E1 = 4.25}

The limit cycle with amplitude 4.25 and frequency 2 rad/sec is a stable limit cycle.

10.8 Revisiting Example 10.2 (Fig. 10.24) will be helpful.

G(jw) = K

jw(1+ jw) +(1 j0 5. w) For K = 1, G(jw)-plot does not intersect

-( ) 1

N E locus. For K = 2, two intersection points are found. One corresponds to unstable limit cycle and the other corresponds to a stable limit cycle of amplitude 3.75 and frequency 1 rad/sec.

10.9 (a) Characteristic equation has two real, distinct roots in the left half of s-plane.

The origin in the (y, y) plane is a stable node.

(b) d y dt

d y dt

2 2

1 5 1

-( ) + ( - )

+ 6 (y – 1) = 0

The singular point is located at (1, 0) in the (y, y) plane. The charac-teristic equation is

s² + 5s + 6 = 0 = (s + 3) (s + 2) The singular point is a stable node.

(c) d y dt

d y dt

2 2

2 8 2

- -

-a f a f

+ 17(y – 2) = 0

The singular point is located at (2, 0) in the (y, y) plane. The charac-teristic equation is

s² – 8s + 17 = 0 with complex roots in right half s-plane. The singular point is an unstable focus.

10.10 e = f(q_R – q) = sin (q_R – q) = sin (– q)

q + aq + K sin q = 0

With x₁ = q and x₂ = q, we have

x₁ = x₂

x₂ = – K sin x₁ – ax₂

Singular points are given by the solution of the equations x₂ = 0

– K sin x₁ – a x₂ = 0

This gives x₁ = kp; k in an integer. We therefore have multiple singular points.

Linearized equation around singular point (0, 0):

q + aq + Kq = 0 The characteristic equation is

s² + as + K = 0

The singular point is stable and is either a node or a focus depending upon the magnitudes of a and K.

Linearization about the singular point (p, 0) gives

q + aq – Kq = 0 The characteristic equation is

s² + as – K = 0

The singular point is a saddle point.

10.11 d y dt

d y dt

2 2

1 2 1

-( ) + z ( - ) + y – 1 = 0

(i) z = 0; singularity (1, 0) on (y, y) plane is a centre

(ii) z = 0.15; singularity (1, 0) on (y, y) plane is a stable focus.

10.12 Revisiting Example 10.4 (Fig. 10.34) will be helpful.

Jq + T_c sgn

c h

q ^{= KA K1e; e = qR – q} Therefore

 sgn 

e K

J e T J^c e

+ + ^{( )} = 0; K = K_AK₁

 sgn 

e e T

K e

n

+^w2

LNM

+ c ^{( )}

OQP

^{= 0; wn = K J/}

With x₁ = e and x₂ = e/w_n, we have

x₁ =w_n x₂

x2 = – w_n x T

K^c _nx

1+ 2

LNM

^sgn

a f

^w

OQP

For T_c = 0, we have

x₁ =w_nx₂; x₂ = – w_nx₁ d x

d x

2 1

= – x x

1 2

; this gives x²₁ + x²₂ = x²₁(0)

The effect of the Coulomb friction is to shift the centre of the circles in phase plane to + T_c/K for x₂ < 0 and to – T_c/K for x₂ > 0. The steady-state error found from phase trajectory is – 0.2 rad. From the phase trajectory, we see that

(a) the system is obviously stable for all initial conditions and inputs; and (b) maximum steady-state error = ± 0.3 rad.

10.13 With x₁ = y and x₂ = y, we get

x₁ = x₂

x2 =

- - <

- - - >

- - + <

-R S|

T|

x x

x x x

x x x

2

2 1 1

2 1

1

2 1 1

2 1 1

;

;

;

Region I 1 <

Region II Region III

1

1

a f

a f a f

a f a f

From the phase portrait shown in the figure below, it is seen that the system is stable in the equilibrium zone.

10.14 Revisiting Review Example 10.6 (Fig. 10.42) will be helpful. The system equation in the linear range is

e + e + e = 0 The characteristic equation

s² + s + 1 = 0

has complex roots in left half of s-plane. Therefore, on the (e, e) plane, the origin is a stable focus.

The system equation in saturation region is

e + e = sgn (e)

The trajectories are asymptotic to the ordinates – 1 and + 1 depending on whether the error is +ve or – ve.

From the phase trajectories plotted from the given initial conditions (with and without saturation) we find that velocity is always greater for trajecto-ries without saturation; thus the error saturation has a slowing down effect on the transient.

10.15 With x₁ = e, x₂ = e and u = f (e), we have

x₁ = x₂

x₂ = – x₂ – f (x₁)

Without hysteresis, the system behaviour is oscillatory with decreasing (tending towards zero) period and amplitude of oscillations (refer Fig.

10.32). With hysteresis, the system enters into a limit cycle, as is shown in the figure below.

10.16 (a) e = – u = – f(e); x₁ = e, x₂ = e

The trajectories corresponding to x₁ > 0 are given by (refer Eqns (10.24))

x₁(t) = – 1 2 ²

x2(t) + x₁(0) + 1 2 ²

x2(0) and trajectories for x₁ < 0 are given by x₁(t) = 1

2 ²

x2(t) + x₁(0) – 1 2 ²

x2(0)

The system response is a limit cycle as shown in the figure.

(b) e = – u = – f (e + K_De); x₁ = e, x₂ = e

The trajectories corresponding to (x₁ + K_Dx₂) > 0 are given by x₁(t) = – 1

2 ²

x2(t) + x₁(0) + 1 2 ²

x2(0)

and the trajectories corresponding to (x₁ + K_Dx₂) < 0 are given by x₁(t) = 1

2 ²

x2(t) + x₁(0) – 1 2 ²

x2(0)

With derivative feedback, the limit cycle gets eliminated as shown in the figure.

(c) Constructing a trajectory for the case of large K_D proves the point. An illustrative trajectory is shown in the figure.

Slope = - 1 KD

Trajectory

Trajectory Trajectory

Switching line

Switching line

x2

x2

x2

x1

x1

x1

10.17 q + 0.5q = 2 sgn(e + 0.5q) With x₁ = e and x₂ = e, we have

x1 = x₂

x₂ = – 0.5x₂ – 2 sgn(x₁ + 0.5 x₂)

As seen from the phase trajectory in the figure, the system has good damp-ing, no oscillations but exhibits chattering behaviour. Steady-state error is zero.

+ e +

-

-q_R= const

x1

x2

q q

1 1

s + 0.5 0.5

s 2

-2

1

2

x1+ 0.5x2= 0

10.18 (a) With x₁ = e, x₂ = e and u = f (e), we have

x1 = x₂

x₂ = – x₂ – f(x₁)

The system oscillates with everdecreasing amplitude and everin-creasing frequency (refer Fig. 10.32).

(b) A rough sketch of the phase trajectory is shown in the figure.

(i) Deadzone provides damping; oscillations get reduced.

(ii) Deadzone introduces steady-state error; maximum error = ± 0.2.

(c) x1 = x₂

x₂ = – x₂ – f x₁ 1x₂ +3

FH IK

A rough sketch of the phase trajectory is shown in the figure. By derivative control action (i) settling time is reduced, but (ii) chattering effect appears.

10.19 Section 10.10 provides solution to this problem. Optimum switching curve is given by Eqns (10.48). Figure 10.37 shows a few trajectories.

10.20 e + e = – u; x₁ = e, x₂ = e For u = +1 (refer Eqns (10.26)) x₁ – x₁(0) = – (x₂ – x₂(0)) + ln 1

1 0

2 2

+

F

+

HG I

x

KJ

x ( )

The trajectories are asymptotic to the ordinate – 1.

For u = – 1

x₁ – x₁(0) = – (x₂ – x₂(0)) – ln 1

1 0

2 2

-F

+

HG I

x

KJ

x ( )

The trajectories are asymptotic to the ordinate + 1.

For x₁(0) = x₂(0) = 0,

x₁ = – x₂ + ln (1 + x₂); u = + 1 x₁ = – x₂ – ln (1 – x₂); u = – 1 Switch curve is given by

x₁ = – x₂ + x x

2

| 2| ln 1 ²

2

2

F

+

HG I

x

KJ

x

| |

The figure given below shows the switching curve and a few typical mini-mum-time trajectories.

In document Solution Manual Digital Control and State Variable Methods (Page 112-123)