1. Block Diagram Manipulation
SE1CA Week 3
Block Diagram Manipulation
Combining blocks in series:
Changing the direction of a block:
Control Systems Engineering Dr Will Browne

2. Block Diagram Manipulation
SE1CA Week 3
Block Diagram Manipulation
Move a block to before a summing junction:
Move a block to after a summing junction:
Source Nise 2004
Control Systems Engineering Dr Will Browne

3. Block Diagram Manipulation
SE1CA Week 3
Block Diagram Manipulation
Move a block to before a take-off point:
Move a block to after a take-off point :
Control Systems Engineering Dr Will Browne

4. Negative in the Feedback Loop
SE1CA Week 3
Negative in the Feedback Loop
Can be expressed as functions in respect to the changing variable:
We can form three equations: 1) 2) 3) + _
Control Systems Engineering Dr Will Browne

5. SE1CA Week 3
We need to form a relationship between input & output by removing the intermediate variables: 4)
Combine 1 & 3
2 & 4
Multiply out bracket
Collect output terms
Rearrange
Forward
Loop =
Control Systems Engineering Dr Will Browne

6. Negative in the Feedback Loop
SE1CA Week 3
Negative in the Feedback Loop
Common in useful systems:
Can be expressed as functions in respect to the changing variable:
Input
Error
Output + _
Plant
Feedback + _
Control Systems Engineering Dr Will Browne

7. Negative in the Feedback Loop
SE1CA Week 3
Negative in the Feedback Loop
Useful parts:
Create transfer function from the
variables (input and output) and
constants (bits inside the boxes).
Output = Forward
Input Loop
Forward
Input
Output + _
Plant
Loop
Feedback
Control Systems Engineering Dr Will Browne

8. Dynamic Systems F I L As an introduction, consider dynamic systems:
these change with time
As an example consider water system with two tanks
Water will flow from first tank to second I F L
Stops when levels equal.
Plot time response of system….?

9. Dynamic Systems Pressure F Pressure
Water flows because of pressure difference
(Ignore atmospheric pressure - approx equal at both ends of pipe)
If have water at one end - what is its pressure?
(Tanks with constant cross sectional area A)
Pressure is force per unit area, = Force / A
Force is mass of water * g
Mass of water is volume of water * density
M = V * 
Volume is height of water, h, times its area A:
V = h * A
Combining:
pressure is

10. Dynamic Systems For first tank, pressure is I *  * g
SE1CA Week 4
Dynamic Systems
For first tank, pressure is I *  * g
For second tank, pressure is L *  * g
Thus flow depends on (I-L) *  * g as well as on the pipe (its restrictance, R)
Flow =
Flow changes volume of tanks:
Volume change = A * change in height (L) = Flow
Thus change in L with time =
A tank has a capacitance, C =
Thus change in L with time is
Flow stops, and there is no change in height,
when I = L
Control Systems Engineering Dr Will Browne

11. Dynamic Systems For first tank, pressure is I *  * g
SE1CA Week 4
Dynamic Systems
For first tank, pressure is I *  * g
For second tank, pressure is L *  * g
Thus flow depends on (I-L) *  * g as well as on the pipe (its restrictance, R)
Flow changes volume of tanks:
Volume change = A * change in height (L) = Flow
Thus change in L with time =
A tank has a capacitance, C =
Thus change in height L is
Flow stops, and there is no change in height,
when I = L
Control Systems Engineering Dr Will Browne

12. Water system To justify the block diagram: Flow F = (I - L) * 1/R
SE1CA Week 5
Water system
To justify the block diagram:
Flow F = (I - L) * 1/R
Transfer function of Pipe = 1/R
L = 1/C * Integral of F
Transfer function of tank = 1/C * 1/s = 1/sC
Control Systems Engineering Dr Will Browne

13. Water system L + sCR L = I Dynamic change in height : Rearrange
SE1CA Week 5
Water system
Dynamic change in height :
Rearrange
The transfer function:
Multiplying both sides by I * (1+sCR) gives: or
that is
Slightly rearranged, this gives
L + sCR L = I
Control Systems Engineering Dr Will Browne

14. Dynamic Flow Level change – not instantaneous Initially:
SE1CA Week 4
Dynamic Flow
Level change – not instantaneous
Initially:
Large height difference  Large flow  L up a lot
Then:
Height difference less  Less flow  L increases, but by less
Later:
Height difference ‘lesser’  Less flow  L up, but by less, etc
Graphically we can thus argue the variation of level L and flow F is: L F t T I t T
Control Systems Engineering Dr Will Browne

15. Time Response We have dynamic equation:
SE1CA Week 4
Time Response
We have dynamic equation:
Put into ‘s’ domain and rearrange:
Integrate a step input with respect to time:
Variation of L is
As t gets larger, exponential term disappears,
L tends to I.
At t = T, L is 63% of final value, I
At t = 5T, L is within 1% of final value.
Control Systems Engineering Dr Will Browne

16. Time Response The variation of F is inverse exponential,
SE1CA Week 4
Time Response
The variation of F is inverse exponential,
T is the time constant of the system = C * R.
At t = T, F down to 37% of initial value, I/R
At t =5T, F is less than 1% of initial value.
As t gets larger, so F tends to 0.
Variation of L is
As t gets larger, exponential term disappears, and L thus tends to I.
At t = T, L is 63% of final value, I
At t = 5T, L is within 1% of final value.
Control Systems Engineering Dr Will Browne

17. Time Response of System
SE1CA Week 4
Time Response of System
Any system of the form:
Has a time response (depending on input):
Output
Exponential
Time
Control Systems Engineering Dr Will Browne

18. Unit Step Response of System
SE1CA Week 4
Unit Step Response of System
Has a time response to a unit step input:
Output
Input
Output when K = 1
Time
Output
Input
Output when K = 1.6
Control Systems Engineering Dr Will Browne

19. Unit Step Response of System
SE1CA Week 4
Unit Step Response of System
Has a time response to a unit step input:
Output
Input
Output when K = 1, T= 0.01
Time
Output
Input
Output when K = 1, T= 0.02
Control Systems Engineering Dr Will Browne

20. Determine the block diagram for each component?
ECEN Week 2
Tutorial question
Determine the block diagram for each component?
Source Nise 2004
Control Systems Engineering Dr Will Browne

21. Mass, Spring & Damper System
SE1CA Week 3
Mass, Spring & Damper System
Mathematical solution:
Much harder to see how system responds!
Control Systems Engineering Dr Will Browne

22. Mass, Spring & Damper System
SE1CA Week 3
Mass, Spring & Damper System
Determine transfer function:
Think how the system responds!
Larger force results in larger distance,
Larger mass, spring stiffness or damping
results in smaller distance.
Control Systems Engineering Dr Will Browne

23. Mass, Spring & Damper System
SE1CA Week 3
Mass, Spring & Damper System
Combine components k _ + _ k _ + _
Control Systems Engineering Dr Will Browne

24. Mass, Spring & Damper System
SE1CA Week 3
Mass, Spring & Damper System
Reduce block diagram: k 1. 2. 3. _ + _ + + _
Control Systems Engineering Dr Will Browne

25. Differentiation Alternative method:
SE1CA Week 4
Differentiation
Alternative method:
Level, L, changes because of the flow of liquid
Mathematically, change of L with time is
In fact, the change in L is proportional to flow, F
Flow related to difference in levels
Thus
C*R is the time constant, T.
Note, the above is a differential equation
It has the differential of L being a function including L.
Control Systems Engineering Dr Will Browne

26. Differentiation: Slopes
SE1CA Week 4
Differentiation: Slopes
Consider the graph of L
At any instant of time we can see value of L
The change in L is the slope of the graph, which varies with time.
Initially steep (high value), then less, then less
But the flow is initially high, then less, then less
Thus slope of L is like F, but slope is change of L.
In fact F is proportional to
Control Systems Engineering Dr Will Browne

27. Integration: Area The reverse process is integration
SE1CA Week 4
Integration: Area
The reverse process is integration
Graphical interpretation: area under a graph.
Consider the flow graph:
the area at different times is shown.
After a short time, area is as shown. Later, area has grown, but by less, etc.
Consider the height of water in the tank:
Thus L like Area under F:
L is prop to integral of F with time.
Control Systems Engineering Dr Will Browne

28. Integration: Area In fact, for this system we have and
SE1CA Week 4
Integration: Area
In fact, for this system we have
and
F is differential of L and L is integral of F.
Differentiation and integration are opposites.
Note, here they are used to model a water systems.
Can also model electronic circuits, mechanical systems, motors, etc.
In fact, the differential equation has the same form and hence the same exponential response as that for many systems.
Note also there are analogies between water systems and electronics:
pipe like a resistor, tank like a capacitor
Also, for thermals, walls have thermal resistance, rooms have capacity.
Control Systems Engineering Dr Will Browne

29. Exercises Exercises Two tanks are connected by pipe.
SE1CA Week 4
Exercises
Exercises
Two tanks are connected by pipe.
Heights of liquid are 0.1m and 0.02m
Liquid density,  = 2 kg/m3, g = 9.8 ms-2
What is the pressure difference?
If restrictance is 0.1 N s / m5, what is flow?
Write down differential equation for system if the area of the second tank is 0.5 m2
Write down equations showing variation of F and L.
(In each case, use equations given in notes, and put in appropriate component values)
Control Systems Engineering Dr Will Browne

30. Notes http://ecs.victoria.ac.nz/Courses/ECEN315_2010T1/CourseOutline
First order control - UAV
X is a method of modeling complex dynamic systems as a set of first order differential equations. Control design in the state space ...
Mass-spring system