Control output i.e. Output = Input, despite disturbances This is achieved by feedback. Open loop systems - i.e. without feedback Process with transfer function P perturbed by a disturbance D. Suppose P is 10 and disturbance D is 0. If the output O is to be 1, just make input I = 0.1. But, if P changes by 10% to 11 then O changes by 10% to 1.1. If disturbance D is 0.1, then O will also change by 0.1. Feedback control

Closed loop system: feedback added! Now consider the ‘closed-loop’ system below: P represents the device being controlled; C is the controller. Ignoring disturbances (D = 0), by forward over 1 minus loop rule Let P = 10, as before, and C = 10; If I is 1, then O is 0.99 i.e. it is within 1% of being 1 If P changes to 11; O = I * 110/111; if I is 1, O is still about Feedback control

To see the effect of disturbances, assume I is 0. Then If C = 10, P = 10 and D = 0.1 Negative Feedback reduces effects on output of disturbances reduces effects on output of parameter changes if |closed loop gain| < |open loop gain| Disturbances control

If CP large: O ~ I + 0 = I So Feedback makes output almost same as input, minimises effects of disturbances and reduces effect of change in device. This is true because the ‘loop gain’, C * P, is high. Note, can’t just keep increasing the ‘gain’ of C. Also, need to consider the dynamics of the blocks Also, there can be a block in the feedback path which we must consider Principle of Superposition

Assume armature inductance is negligible. Armature resistor: Back emf of motor: Torque proportional to armature current: Torque is opposed by the inertia torque: Hint: apply Kirchhoff’s voltage law to the armature circuit We need to form a relationship between input voltage and output velocity: R2-D2 Motor System

Combine components Components of Motor System R + _

Reduce block diagram: Block Diagram of Motor System + _ /R + _

Output linked to input: Can be expressed much more simply!: Where: Time Constant: Gain: Transfer Function of System

Specify a motor and resistor for R2-D2? Assume we walk at 2 m/s & R2-D2’s wheel diameter is 4cm. Therefore, required angular velocity is: 16 revs/s -> 100 radians/s Input voltage is 4 D cells giving 6V input. Ke = 6/100 = 0.06 V/rad/s Weight of R2-D2 gives an inertia torque (J): J = 0.05 kgm 2 Assume current is 1 amp -> R = 1  Want T = 0.2s K T = 25 Nm/A We need to form a relationship between input voltage and output velocity: R2-D2 Motor System

Has a time response to a unit step input: Unit Step Response of System Time OutputOutput Output when K = 1, T= 0.01 Input OutputOutput Output when K = 1.6, T= 0.02 Input

Now include armature inductance: Armature resistor: Back emf of motor: Torque proportional to armature current: Torque is opposed by the inertia torque: Hint: apply Kirchhoff’s voltage law to the armature circuit We need to form a relationship between input voltage and output velocity: R2-D2 Motor System

Include new component: Components of Motor System R + _ LasLas

Reduce block diagram: + _ _ + _ 1/R

Output linked to input: Previously (inductor = 0) Can be expressed much more simply!: Previously (inductor = 0) Transfer Function of System

Any system of the form: Has a time response (depending on input): Time Response of System Time O ut p ut Varies!

Has a time response to a unit step input: Over Damping Time OutputOutput Output is over damped Input OutputOutput Output changes with k, T1 & T2 Input

Has a time response to a unit step input: Critical Damping Time OutputOutput Output is critically damped Input OutputOutput Output from unique T1, T2 & k Input

Has a time response to a unit step input: Under Damping Time OutputOutput Output is under damped Input OutputOutput Output changes with k, T1 & T2 Input

d0 = tf(1.0,[1 0 1]) %undamped d1 = tf(1.0,[1 2 1]) %critically damped d2 = tf(4.0,[1 2 4]) %under damped d3 = tf(0.5,[ ]) %over damped T= [0: 0.01: 20];%set up the time increments [y0,t]=step(d0,T);%step response over one second [y1,t]=step(d1,T);%step response over one second [y2,t]=step(d2,T);%step response over one second [y3,t]=step(d3,T);%step response over one second stept = 1 + 0*t; %graph to show step response clf; %clear all graphs hold on % put each graph on top of each other plot(t,y0, 'r'); plot(t,y1,'k'); plot(t,y2,'g'); plot(t,y3,'b'); plot(t,stept,'m'); Matlab code

1) Write down the transfer function for the RC circuit when R= 2 kΩ and C = 5 mF. Sketch the response of the system to a unit step input, marking the time constant's position on the time axis and the final value on the other axis. 2)Find the transfer function of the thermal system for which R = 4 KW -1 and C = 2 JK -1. Sketch the response of the output if the input is a step change of 2 W 3) For each of the following work out O/I and sketch response if I is a step. Exercises

Notes EN315_2010T1/CourseOutlinehttp://ecs.victoria.ac.nz/Courses/EC EN315_2010T1/CourseOutline Second order control - UAV vCxey19O4http:// vCxey19O4 X is a method of modeling complex dynamic systems as a set of first order differential equations. Control design in the state space... uOLD3-abwUhttp:// uOLD3-abwU Mass-spring system ZNnwQ8HJHUhttp:// ZNnwQ8HJHU