CHNG3802 Tutorial 7 Solutions

Feedback Control Systems 1. Consider the following heat exchanger. The flow rate of the process fluid (Fi)

is acting as a disturbance.

(a) Draw a block diagram of the temperature feedback control loop. (b) Draw a simplified block diagram of the feedback control loop.

TI

TC

Steam

Tset

To Process Fluid

Condensate

Gc Gf

Gm

Gp

To(s)Tset(s)

Gd

C(s) m(s)

d(s)

εεεε(s)

ym(s)

Gc Gp

To Tset

Gdd(s)

C(s) εεεε(s)

(c) Develop the transfer function between the outlet temperature (To) and the Process flow (Fi).

( ) ( ) ( )smGsdGsT pdo += …(1)

( ) )(sTGsy omm = …(2)

( ) ( ) ( )sysTs mset −=ε …(3)

( ) ( )sGsC cε= …(4)

( ) ( )sCGsm f= …(5)

Next we assume that the set point does not vary – that is, its deviation variable is zero;

0)( =sTset

Eliminate all the intermediate variables, ε(s), C(s), m(s) and ym(s), by combining equations 1 through 5. The result is

( ) ( )[ ] ( )sdGsTGGGGsT domcfpo +−=

( )( ) mcfp

do

GGGG

G

sd

sT

+=

1 #

This is the closed-loop transfer function between the process flow and the outlet temperature.

(d) Develop the transfer function between the outlet temperature (To) and the

set point (Tset). Then write down the overall closed-loop transfer function.

Similar to part (c), if we let d(s) = 0 and combine equations 1 through 5, the closed loop transfer function between the set point and the outlet temperature results;

( )( ) mcfp

cfp

set

o

GGGG

GGG

sT

sT

+=

1 #

The overall closed loop transfer function can now be written as

( ) ( ) ( )sdGGGG

GsT

GGGG

GGGsT

mcfp

dset

mcfp

cfpo +

++

=11

#

(e) Write down the characteristic equation of the loop.

The characteristic equation is the denominator of the closed-loop transfer functions:

mcfp GGGG+1

(f) Qualitatively explain what is observed if the poles of the transfer function are complex conjugates.

Complex conjugate poles, such as p1 and p2. Complex poles always appear in conjugate pairs, p1 = a + bi and p2 = a - bi. These roots give rise to terms such as eat sin(βt+φ). sin(βt+φ) is a periodic, oscillating function, while the behavior of eat depends on the value of the real part a. Thus,

• If a > 0, then eat � ∞ as t� ∞ , and eatsin(βt+φ) grows to infinity in an oscillating manner (Fig. 1a).

• If a < 0, then eat � 0 as t � ∞ , and eatsin(βt+φ) decays to zero in an

oscillating manner with decreasing amplitude (Fig. 1b).

• If a = 0, then eat = 1 for all times, and eatsin(βt+φ) = sin(βt+φ), the

response oscillates continuously with constant amplitude (Fig. 1c).

Figure 1a Figure 1b.

Figure 1c.

Output

Time

Output

Time

Output

Time

2. Each of the following two systems is controlled with a proportional feedback controller. In each case find the range of values of the proportional gain Kc that will produce stable closed-loop responses. Also identify the characteristic equation, in each case. Assume that Gm = Gf =1.

(a) Firstly we write the characteristic equation of the system: 01 =+ mpfc GGGG

011.0

2111 =

+⋅⋅⋅+

sK c

Rewriting the characteristic equation as ( ) 0211.0 =++ cKs

Solving, we get

( )

01.0

21=

+−= cK

s

For stable closed-loop responses the root, s, must be negative i.e.

1.0

12 −− cK ≤ 0

Kc < 2

1− is the range of values for Kc for which the system response is stable.

(b) Firstly we write the characteristic equation of the system: 01 =+ mpfc GGGG

05210

1111

23=

−++⋅⋅⋅+

sssK c

Rewriting this as the characteristic equation: ( ) 05210 23 =−+++ cKsss

Using the Routh-Hurwitz array as follows,

S3 10 1 0 S2 2 (Kc –5) 0 S1 ( )

2

5102 −− cK

0

S0 (Kc –5) 0

All values in the first column must be positive for a stable closed-loop response; therefore there is must be no sign change and so

05 ≥−cK and ( )

02

5102≥

−− cK

5≥cK and 2.5≤cK

which gives the following range of stability 2.55 ≤≤ cK

3. For the control system below, find the closed-loop transfer function between

the output C and setpoint R. Then,

The closed-loop transfer function between C and R is:

( )

( ) 1

1

1

11

1

1

13

3

+++

+=+

=

ssK

sK

GGGG

GGG

R

C

mcfp

cfp

So the closed-loop transfer function will be:

( )( )

( )( )Kssss

sK

Ks

sK

R

C

++++++=

+++=

1464

1

1

12344

Therefore the characteristic equation of the system will be;

( ) 01464 234 =+++++ Kssss

( )31

1

+s

( )1

1

+s

+

-

R C K

(a) Determine the value of K above which the system is unstable.

To find the amount of K which makes the system unstable, use the Routh-Hurwitz array as follows:

S4 1 6 1 + K S3 4 4 0 S2 5 1 + K S1

5

416 K−

0

S0 1 + K To have a stable system, all the elements in the first column must be positive, therefore,

05

416 ≥− K � 4≤K

and,

01 ≥+ K � 1−≥K At K = 4, the system is on the verge of instability. (b) Determine the value of K for which two of the roots are on the imaginary

axis, and determine the values of these imaginary roots and the remaining two roots.

At K = 4, the characteristic equation will be,

05464 234 =++++ ssss or, ( )( ) 0541 22 =+++ sss Therefore the poles of the system which are located on the imaginary axis are the

roots of (s2 + 1), which are s = ± j. The other two poles are the roots of (s2 + 4s + 5), which are s = -2 ± j.