3.0 Time Response OBJECTIVE Influence of poles toward the time response Routh-Hurwitz Criteria used for determining the stability of a system Transient response for a system Parameter used for steady state response for a system

First Order System Standard form for a first order transfer function is Eqn. (3.1) Where K is the dc gain and is the time constant For a unit step of Its response are constant. For and then and For a=1, » num=[1];den=[1 1];[y,x,t]=step(num,den,0:0.01:10);plot(t,y);grid » xlabel('t (saat)');ylabel('Amplitud');title('Sambutan Unit Langkah bagi Tertib Pertama');

(i) Time constant,  when the final value =0.6321. (ii) Rise time, tie range for response of 0.1 to 0.9 from final value (iii) Settling time, time range when the response is 0.98-1.02 of the final value.

(cm3.s-1). The out-flow rate, (cm3.s-1) is related to the Example A process control for controlling the height of a water tank, h (cm) is done through controlling the in-flow rate, (The out-flow rate, cm3.s-1). (cm3.s-1) is related to the A process control for controlling the height of a water tank, h (cm) is done through controlling the in-flow rate, (cm3.s-1). The out-flow rate, (cm3.s-1) is related to the (a) Obtain the system block diagram (b) If s.cm-2 and cm2, determine the open-loop transfer function (c) Determine the system closed-loop transfer function, dc gain and time constant.

(cm3.s-1) 0.5 t

Solution (a) Taking Laplace transform of Rearrange By converting the Laplace transform of

+ - (b) This gives the open loop transfer function as Using dan

(c) While the closed loop transfer function is given as Comparing with standard first order transfer function of Eqn. (3.1) Thus, the dc gain And time constant s

Second Order System For impulse response where Standard form Eqn. (3.2) Where K is the dc gain,  is the damping ratio is the undamped natural frequency Roots of the denominator Eqn. (3.3) will give the poles of the second order system Eqn. (3.4)

(a) Standard form of second order transfer function is Example For the transfer function below, obtain the damping ratio, undamped natural frequency and its dc gain for a unit step input. (a) (b) Penyelesaian (a) Standard form of second order transfer function is Normalized the transfer function Compare with the standard form Giving the natural undamped frequency of rad.s-1

(b) Comparing with standard form of second order system of Eqn (3.2) Damping ratio of For a step input, the dc gain is given by (b) Comparing with standard form of second order system of Eqn (3.2) Undamped natural frequency rad.s-1 Damping ratio of For a step input the dc gain is given by

Under damped     Its transfer function is. When where Pole position    

Underdamped step response of a second order system

Peak time, The time taken for the first peak of second order response Percentage of overshoot, Percentage different of the maximum peak from the final value

Settling time, Time taken for the final value to be within 2% of the final value Example: Consider a second order RLC circuit with current, i that varies in time, t is subjected to a step voltage, v and is represented as a differntial equation below. Assume the initial value of the current is zero. (a) Obtain the transfer function of the system. (b) Determine damping ration, undamped and damped natural frequencies for a unit step input.

Rearrange the equation give the transfer function as solution: Using the diffrential property of Laplace transform Rearrange the equation give the transfer function as or

(b) Comparing with standard form of Nutral undamped natural frequency rad.s-1 Dc gain of And the damping ratio is obtained from Which gives And the damped natural frequency rad.s-1

»wn=2;zeta=0. 5;num=[wn. wn];den=[1,2. zeta. wn,wn »wn=2;zeta=0.5;num=[wn*wn];den=[1,2*zeta*wn,wn*wn];[y,x,t]=step(num,den,0:0.01:10);» plot(t,y);grid; » xlabel('t (saat)');ylabel ('Amplitud');title('Sambutan Unit Langkah bagi Tertib Kedua')

Over damped transfer function »wn=2;zeta=5;num=[wn*wn];den=[1,2*zeta*wn,wn*wn];[y,x,t]=step(num,den,0:0.01:50); » plot(t,y);grid » xlabel('t (saat)');ylabel('Amplitud');title('Sambutan Unit Langkah bagi Tertib Kedua');

Critical damped , to produce transfer function of »wn=2;zeta=1;num=[wn*wn];den=[1,2*zeta*wn,wn*wn];[y,x,t]=step(num,den); grid; xlabel('t (saat)');ylabel('Amplitud');title('Sambutan Unit Langkah bagi Tertib Kedua');

Undamped , transfer function becomes >>wn=2;zeta=0;num=[wn*wn];den=[1,2*zeta*wn,wn*wn];step(num,den) grid;xlabel('t (saat)');ylabel('Amplitud');title('Sambutan Unit Langkah bagi Tertib Kedua');

Pole position of a second order The movement of the poles will determine the time response behaviour of the system. We will study these behaviours (1) Real poles moving horizontally Satah s -3 -1

(2) Complex poles on the jw-axis moving vertically s-plane 5j j -j -5j

(3) Complex poles moving vertically 10j s-plane 5j 3j -3j -5j -10j

(4) Complex poles moving vertically s-plane -1 -3

Dominant pole Poles that dominate the response over the others Consider Example Conider the response of two transfer function of and We can consider the second order of is dominated by the pole at -1, hence the second order model can be resperented with a first order model of Wherelse the model of can’t be approximated by a first order model

Stability Routh-Hurwitz Stability Criteria If a polynomial is given by where are constants and necessary condition for stability are: (i) All the coefficients of the polynomial are of the same sign. If not, there are poles on the right hand side of the s-plane . (ii) All the coefficient should exist accept for the For the sufficient condition, we must formed a Routh-array,

Routh-Hurwitz Criteria states that the number of roots of characteristic equation is the same as the numberof sign changed of the first column.

Case 1: No zero on the first column After the Aray has been tabled, all the elements on the first column is not equal to zero. If there is no sign changed, all the poles are in the LHP. While the number of poles on the RHP is equal to number of sign change on the first column of the Routh’s array. Example: Consider a fourth order characteristic equation

Solution: Form theRouth’s array 2 12 1 8 There are two sign change on rows 2 and 3. Hence, there two poles on the RHP (Right-half f s-palne).

MATLAB solution >>roots([2 1 12 8 2]) ans = 0.0885 + 2.4380i 0.0885 - 2.4380i -0.3385 + 0.2311i -0.3385 - 0.2311i Case 2: Coeffiecient of the first column is zero but not the others. Change the zero element by a small number positive number, . The number of pole on the RHP will depend on the number of sign change. Example: Consider a fifth order characteristic equation

,  and also at row 4 and 5 Solution: Form the Routh’s array 1 3 5 2 6 ,  is a small positive number there are two sign change at row 3 and 4 and also at row 4 and 5 . Hence, there two poles on the RHP.

MATLAB Solution » roots([1 2 3 6 5 3]) ans = 0.3429 + 1.5083i 0.3429 - 1.5083i -1.6681 -0.5088 + 0.7020i -0.5088 - 0.7020I Case 3: All the coefficients on a row are zeros. Form an auxillary equation from the row above it and replace the coefficient of the row with the differentiated coefficient of the auxillary equation. For this case, ifthere is no sign change, the characteristic equation has a pair of poles with opposite sign of real component or/and a pair of conjugate poles on the imaginary axis. Example: Consider this fifth order characteristic equation

Formed Routh array 1 6 8 7 42 56 21 9.3 28 84

Form the auxillary equation on the second row: Differentiate the equation: As there is no sign change, there is a pair of conjugate poles on the axis and/or a pair of poles with opposite sign of real component. To be sure we can use MATLAB » roots([1 7 6 42 8 56]) ans = -7.0000 0.0000 + 2.0000i 0.0000 - 2.0000i 0.0000 + 1.4142i 0.0000 - 1.4142i

Use of Routh Hurwitz Criteria Main use is to determine the position of the poles, which in turns can determine the stability of the response. j TAK STABIL STABIL  Example A closed-loop transfer function is given by Determine the range for K for the system to be always stable and its oscillating frequency before it becomes unstable.

1 3 K 2 Solution: Charactristic equation is Expand the equation Form the Routh’s array 1 3 K 2

For nosign chage on the first column: Refering to row 4 which gives and row 5 Hence its range Oscillating frequency rad.s-1.

Steady state R(s) E(s) Y(s) + - B(s) From the diagram Consider and Use the final value theorem and define steady state error, that is given by

Unit step Steady state error, Unit step input, From We define step error coefficient, Thus, the steady state error is By knowing the type of open-loop transfer function, we can know step error coefficient and thus the steady state error

For open-loop transfer function of type 0: , For open-loop transfer function of type 2: ,

Example: A first order plant with time constant of 9 sec and dc gain of 5 is negatively feedback with unity gain, determine the steady state error for a unit step input and the final value of the output. Solution: The block diagram of the system is . R(s) Y(s) + - As we are looking for a stedy state error for a step input, we need to know Knowing the open-loop transfer function, then And steady state error of Its final value is

Unit Ramp , while its Laplace form is As in the above section, we know that Thus, its steady state error is Define ramp error coefficient, Which the steady state error as Just like for the unit step input we can conclude the steady state error for a unit ramp through the type of the open-loop transfer function of the system. For open-loop transfer function of type 0: For open-loop transfer function of type 1: For open-loop transfer function of type 2:

- + Example: A missile positioning system is shown. (i) Find its closed-loop transfer function (ii) Determine its undamped natural frequency and its damping ratio if (iii) Determine the steady state error, if the input is a unit ramp. (iv) Cadangkan satu kaedah bagi menghapuskan ralat keadaan mantap untuk (iii). Compensatot DC motor + -

(a) By Mason rule, the closed-loop transfer function is Solution: (a) By Mason rule, the closed-loop transfer function is , (b) If Comparing with a standard second order transfer function

Comparing Thus undamped natural frequency rad.s-1 and damping ratio of (c) To determine the ramp error coefficient, we must obtain its open-loop transfer function As it is a type 1, the system will have a finite ramp error coefficient, putting Hence steady state error of

Unit Parabola Its time function ,while its Laplace ,thus its steady state error is Define parabolic error coefficient, Similarly we can determine its steady state error by knowing the type of the open-loop transfer function For open-loop transfer function of type 0: For open-loop transfer function of type 1: For open-loop transfer function of type 2:

 Unit step Unit ramp Unit parabolic Type 0 Type 1 Type 2 In summary we can make a table of the staedy state error for the above input Unit step Unit ramp Unit parabolic Type 0  Type 1 Type 2