Block Diagram Manipulation SE1CA5 Week 3 Block Diagram Manipulation Combining blocks in series: Changing the direction of a block: Control Systems Engineering Dr Will Browne

Block Diagram Manipulation SE1CA5 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

Block Diagram Manipulation SE1CA5 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

Negative in the Feedback Loop SE1CA5 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

SE1CA5 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 1 - Loop = Control Systems Engineering Dr Will Browne

Negative in the Feedback Loop SE1CA5 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

Negative in the Feedback Loop SE1CA5 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 1 - Loop Forward Input Output + _ Plant Loop Feedback Control Systems Engineering Dr Will Browne

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….?

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

Dynamic Systems For first tank, pressure is I *  * g SE1CA5 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

Dynamic Systems For first tank, pressure is I *  * g SE1CA5 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

Water system To justify the block diagram: Flow F = (I - L) * 1/R SE1CA5 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

Water system L + sCR L = I Dynamic change in height : Rearrange SE1CA5 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

Dynamic Flow Level change – not instantaneous Initially: SE1CA5 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

Time Response We have dynamic equation: SE1CA5 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

Time Response The variation of F is inverse exponential, SE1CA5 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

Time Response of System SE1CA5 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

Unit Step Response of System SE1CA5 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

Unit Step Response of System SE1CA5 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

Determine the block diagram for each component? ECEN315 Week 2 Tutorial question Determine the block diagram for each component? Source Nise 2004 Control Systems Engineering Dr Will Browne

Mass, Spring & Damper System SE1CA5 Week 3 Mass, Spring & Damper System Mathematical solution: Much harder to see how system responds! Control Systems Engineering Dr Will Browne

Mass, Spring & Damper System SE1CA5 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

Mass, Spring & Damper System SE1CA5 Week 3 Mass, Spring & Damper System Combine components k _ + _ k _ + _ Control Systems Engineering Dr Will Browne

Mass, Spring & Damper System SE1CA5 Week 3 Mass, Spring & Damper System Reduce block diagram: k 1. 2. 3. _ + _ + + _ Control Systems Engineering Dr Will Browne

Differentiation Alternative method: SE1CA5 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

Differentiation: Slopes SE1CA5 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

Integration: Area The reverse process is integration SE1CA5 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

Integration: Area In fact, for this system we have and SE1CA5 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

Exercises Exercises Two tanks are connected by pipe. SE1CA5 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

Notes http://ecs.victoria.ac.nz/Courses/ECEN315_2010T1/CourseOutline First order control - UAV http://www.youtube.com/watch?v=YlbEbQ6TJMc X is a method of modeling complex dynamic systems as a set of first order differential equations. Control design in the state space ... http://www.youtube.com/watch?v=puOLD3-abwU Mass-spring system http://www.youtube.com/watch?v=aZNnwQ8HJHU