Elizabeth Jose Asst. Prof. EEE Dept. AJCE NON - LINEAR SYSTEMS COMMON PHYSICAL NONLINEARITIES Incidental NonlinearitiesIntentional Nonlinearities Which are inherently present in the system Which are deliberately inserted in the system to modify system characteristics Eg: Saturation, Dead-zone, Coulomb friction, backlash etc Eg: Relay SATURATION In this type of nonlinearity the o/p is proportional to i/p for a limited range of input signal. When the i/p exceeds this range, the o/p tends to become linearly constant DEADZONE Deadzone is the region in which the o/p is zero for a given i/p. When the i/p is increased beyond this deadzone value, the o/p will be linear. FRICTION This type of nonlinearity exist when there is a relative motion between contacting surfaces. Viscous frictionCoulomb frictionStiction Is directly proportional to the relative velocity of sliding surfaces. Coulomb friction offers a constant retarding force (Always opposing the relative motion) It is the force required to initiate the motion Is linear in nature. Nonlinear in nature Nonlinear in nature Viscous friction Ideal Stiction and coulombs friction Actual Stiction and coulombs frictionStiction, viscous and coulombs friction BACKLASH This type of nonlinearity is the hysteresis in mechanical system, such as gear trains and linkages. It is basically the play between the teeth of the drive gear and those of the driven gear RELAY A relay is a nonlinear power amplifier which can provide large power amplification inexpensively and is therefore deliberately introduced in control systems. DESCRIBING FUNCTION Describing function is a function which is used to replace the nonlinear element in the system for analytical purpose. If the nonlinearity is replaced by a describing function then all linear theory can be used for the analysis of the system. Describing function of a nonlinear element if finding out by applying a sinusoidal signalto the element. t X x e sin =Then generating the Fourier series expansion, from the behavior of the nonlinearity. Then Describing function of a nonlinearity is ( )11, | e Z =XYX KNFourier series expansion after applying the sinusoidal signal, after neglecting the higher order harmonics is, 1 1 1 1cos sin | e e Z = + = Y t B t A yWhere 2121 1B A Y + =1111tanAB= |and We Know ) ( sin201t d t yPeriodAPeriode e}=) ( cos201t d t yPeriodBPeriode e}=and DESCRIBING FUNCTION OF SATURATION NONLINEARITY Input Output relationship of saturation nonlinearity is The response of the nonlinearity when the i/p is sinusoidal is t X x e sin =The i/p isWhere X is the maximum value of xFrom above fig, When S x t = = , | e1 Thus equ. (1) becomes | sin X S =2 The output y of the nonlinearity can be divided in to three regions as( )( ) t e | t| t e || es s s ss st Kxt KSt Kx;;0 ;= yThe output y has half wave and quarter symmetry, so 01 = B) ( sin22201t d t y A e ett}= and( ) ( ) t d t y t d t y A e ete ett||} }+ =201sin4sin4( ) ( ) t d t KS t d t Kx e ete ett||} }+ =20sin4sin4Substitutet X x e sin =Then ( ) ( ) t d tKSt d t t XKA e ete e ett||} }+ =201sin4sin sin4Solving |t||tcos422 sin 21KS KXA +((

=Substitute value of S from equ. (2), then, | | | | |tcos sin21+ =KXAWe have 2121 1B A Y + =1111tanAB= |and | | | | |tcos sin21+ = KXYand01 = |Thus the describing function is ( ) | |0110 cos sin2, Z + = Z = | | |t| eKXYX KNDESCRIBING FUNCTION OF DEAD-ZONE NONLINEARITY Input Output relationship of dead-zone nonlinearity is The response of the nonlinearity when the i/p is sinusoidal is t X x e sin =The i/p is 3 From above fig, When2,Dx t = =o eThus equ. (3) becomeso sin2XD=4 The output y of the nonlinearity can be divided in to three regions asThe output y has half wave and quarter symmetry, so 01 = B ) ( sin22201t d t y A e ett}=andSubstitutet X x e sin =Then Solving Substitute value of D from equ. (4), then, So, and01 = |Thus the describing function is Also, from equ (4), we can generate the triangle shown and from it Depending on the maximum value of input X, describing function can be written as 1 21|.|

\|XD|.|

\|XDoDESCRIBING FUNCTION OF DEAD-ZONE AND SATURATION NONLINEARITY Input Output relationship of dead-zone and saturation nonlinearity is The response of the nonlinearity when the i/p is sinusoidal is t X x e sin =The i/p is 5 From above fig, When2,Dx t = =o eThus equ. (5) becomeso sin2XD= 6 The output y of the nonlinearity can be divided in to three regions asThe output y has half wave and quarter symmetry, so 01 = B ) ( sin22201t d t y A e ett}=and| sin X S = 7 S x t = = , | eSubstitutet X x e sin =Then Solving Substitute value of D and S from from equ. (6) and (7), then, So, and01 = |Thus the describing function is Depending on the maximum value of input X, describing function can be written as DESCRIBING FUNCTION OF RELAY WITH DEAD-ZONE AND HYSTERESIS Input Output relationship of the nonlinearity is The response of the nonlinearity when the i/p is sinusoidal is HDx t = =2), ( | t et X x e sin =The i/p is 8 From above fig, When2,Dx t = =o eThus equ. (8) becomeso sin2XD= 9 The output y of the nonlinearity can be divided in to three regions as) ( sin201t d t y A e et t}=| sin2X HD= 10 ThenFrom triangle From triangle 1 21 |.|

\|XD|.|

\|XHXD2|1 221 |.|

\|XD|.|

\|XD2o Ideal Relay D = H = 0 i.e., Relay with Dead zone H = 0 Relay with Hysteresis H = D Elizabeth Jose Asst. Prof EEE Dept. AJCE PHASE PLANE ANALYSIS Construction of Phase Trajectories by Isocline method A locus passing through the points of same slope in phase plane is called isocline.Let S be the slope at any point in phase plane ( )( )2 1 12 1 212,,x x fx x fdxdxS = =andS1tan= o Construction of Phase Trajectories by Isocline method 1. Find equation forand in terms of x1 and x2 -1x-2x2. Generate equation (1), i.e,( )( )2 1 12 1 212,,x x fx x fdxdxS = =3. Find the equation for x2 in terms of S and x1 4. Take different vales for S and put two values for x1 at each value of S, then calculate x2 for each value of x1 5.Tabulate the value of S, o, x1 and x2 6.Draw the isoclines corresponding to each vales of S with x1 and x2

7. Take point A corresponding to the initial condition given. 8. Draw two lines from point A, one at a slope of S1(i.e., at angle o1=tan-1(S1) ) and the other at a slope of S2 (i.e., at angle o2=tan-1(S2) ). 9.Let these two lines meet isocline-2 at p and q. 10.Mark point B on isocline-2, at midway between p and q.The phase trajectory will leave point A at a slope S1. When the trajectory reaches isocline-2, the slopechanges to S2. 11.Repeat the above steps 8-10 with point B to get a point C on isocline-3. Join A, B, C, D, .. with a smooth curve. Example: sec / 1 radn = eA linear second order system is, where,,0 22= + +- - -e e en ne ,e 15 . 0 = , , and . Determine the singular point and plot phase trajectory. 5 . 1 ) 0 ( = e0 ) 0 ( =-eSolution: Let, x1 = and x2 =-eeThen21x x =-and2 1222 x x e xn n,e e = =- - -Singular points ( ) are obtained by substitutingand0201 , x x01 =-x02 =-xThen singular points will be (0, 0), i.e., origin Slope of trajectory1212--= =xxdxdxS( )22 122xx xSn n,e e + =i.e.,,Given,15 . 0 = , sec / 1 radn = e and3 . 021=xxSThusSxx =3 . 012or0oS -2.0-1.0-0.500.51.02.0 -63-45-270274563 x1x2x1x2x1x2x1x2x1x2x1x2x1x2 10.611.40.251.250.25-0.81-1.251-0.771-0.43 21.21.52.10.52.50.75-2.52-2.52-1.542-0.86 Choose some values for S, let it be -2, -1, -0.5, 0, 0.5, 1 and 2 For each value of S, take two values for x1 and find values for x2 Construction of Phase Trajectories by Delta method Delta method is using to plot phase trajectory if the system equation is in the following form 0 , , =|.|

\|+- - -t x x f xThen equation can be converted to the following form 0 , ,2=((

|.|

\|+ +- - -t x x x xno eIn above equation, the changes in the variablesare negligible for short intervals. |.|

\|-t x x , ,Step: 1 Rewrite the equation in the following form,( ) 02= + +- -o e x xnChoose state variables asx x =1 andnxxe-=2Differentiating, we get, the state equations asnx x x e2 1= = - -( ) o e + =-1 2x xn( )211212xxdxdxxx o + = =--

FindStep: 2 Step: 3 Step: 4 Step: 5 2 Take a point A on phase plane with coordinates (B1, B2) as shown, Normally the point corresponds to initial condition Step: 6 Findthevalueofobysubstitutingthevalueofcoordinatesof point A in equ.(2), o may be either positive or negative. Step: 7 If o is positive, mark point C1 on the negative side of x1 axis such that length OC1 = o.Step: 8 If o is negative, mark point C1 on the positive side of x1 axis such that length OC1 = o.Through point A, draw two perpendicular lines C1A and PQ.Step: 9 Draw a small circular arc AB with C1 as centre and C1A as radius.Step: 10 Then repeat above steps with point B starting with calculation of o.Step: 11 Example: Construct a phase trajectory by the delta method for a nonlinear system represented by the differential equation,0 4 4 = + +- - - -x x x xChoose the initial conditions as 1 ) 0 ( = xand 0 ) 0 ( =-x. Answer: Convert the given equation in to the form( ) 02= + +- -o e x xnHence 0 4 =((

+ +- - - -x x x xThus we get,- -= x x o sec / 2 radn = e and Choose state variables asx x =1 and22- -= =x xxnei.e.,22 x x =-2 2 2 24 2 2 x x x x = =oFirstmarkpointA,bytakingthegiveninitialconditions,x1(0)=1 and x2(0) = 0 Calculate the value of o1 at point A by using the coordinates (1, 0) Since x2(0) = 0, o1 = 0Then C1 will be at origin.Draw the arc AB with C1 as centre. From graph, the value of x2 corresponding to B is - 0.1875.Repeat the above steps with value of B and so on. Elizabeth Jose Asst. Prof. EEE Dept. AJCE PHASE PLANE ANALYSIS Phase plane analysis is used to determine the stability and transient behavior of a second order system Phase PlaneThe coordinate plane with the state variables x1 and x2 as two axes is called the phase plane. The curve describing the state points (x1 , x2) in the phase plane with time as running parameter is called phase trajectory (i.e., locus of state point (x1 , x2) in phase plane is called phase trajectory). A trajectory can be constructed for each initial condition, so a family of trajectories can be constructed for a system and such a family of phase trajectory is called phase portrait Singular points A point in phase plane at which the derivatives of all state variables are zero is called singular point. It is also called equilibrium point If a system is placed at a singular point, it will continue to lie there if left undisturbed Singular points are classified in to Nodal point Saddle point Focus point Centre or Vortex point Depending on the value of eigen value of the systemEigen valuesLocation of Eigen values Type of singular point Phase portrait with singular point at origin Distinct, real and negative eigen vales Stable Node Distinct, real and positive eigen vales Unstable Node Distinct, real, one eigen value is positive and other is negative Saddle point Eigen valuesLocation of Eigen values Type of singular point Phase portrait with singular point at origin Complex conjugates with negative real parts Stable focus Complex conjugates with positive real parts Unstable focus Purely imaginary and conjugates Centre or Vortex point Stability analysis using Phase trajectory For a linear time invariant system,When the input is zero, the system is stable for arbitrary initial conditions if the resulting trajectory tends towards the equilibrium state When the system is excited by a bounded i/p, the system is stable if the system o/p is bounded. A linear autonomous system has only one equilibrium state. So the behaviour of system about the equilibrium state completely determines the qualitative behaviour of it in the entire state space In a nonlinear system there may be multiple equilibrium states. So in nonlinear systems, stability is discussed relative to the equilibrium state. The general stability of a system cannot be defined.Limit cycles in Phase portrait The limit cycles are oscillations of fixed amplitude and period. The existence of limit cycles can be predicted from closed trajectories in the phase portrait. Linear System Nonlinear System The limit cycle in the nonlinear system is stable, if the paths in its neighbourhood converge towards the limit cycle. The limit cycle in the nonlinear system is unstable, if the paths in its neighbourhood diverge away from the limit cycle. Construction of Phase Trajectories The state equations of a second order autonomous system are( )2 1 1 1, x x f x =-( )2 1 2 2, x x f x =-Where x1 and x2 are the state variables of systemThen,( )( )2 1 12 1 212,,x x fx x fxx=--i.e.,( )( )2 1 12 1 212,,x x fx x fdxdx=Phase trajectory can be constructed using the slope equation (1) either analytically or graphically1 Construction of Phase Trajectories by analytical method 1. Find equation forandin terms of x1 and x2 -1x-2x2. Generate equation (1), i.e,( )( )2 1 12 1 212,,x x fx x fdxdx=3. Integrate the equation (1) to get the solution of state equation, and solve for x2 4. For a given set of initial conditions, assume different vales for x1 and calculate x2 for each value of x1

5. Tabulate the value of x1 and x2 6. Plot the vales of x1 and x2 on a graph sheet by taking the axes as x1 and x2

7. Join all points by a smooth curveExample: Consider a system with ideal relay as shown. Determine the singular point. Construct phase trajectories, corresponding to initial conditions (i)and (ii) . Take r = 2 volts, and M = 1.2 volts. 1 ) 0 ( , 2 ) 0 ( = =-c c5 . 1 ) 0 ( , 2 ) 0 ( = =-c c+ recu - u e M -M 21sAnswer: ) ( ) ( t u t c =- -Lets takeand ) (1t c x =) (2t c x-= On differentiating 21) ( x t c x = = - -u t c x = =- - -) (2 and From fig. ,u = - M; when c > r u = M; when c < r So the state equations will be 21x x =-M x =-2To find singular point It can be found out by substituting and01 =-x02 =-xLet the coordinates of singular points be ()0201 , x x002 = x r x =01i.e., we will get Thus singular point is (r, 0) To construct phase trajectory 2 1212xMdxdxxx = =--We have 1 2 2dx M dx x =i.e., Integrating from t = 0 to t. } } = 1122) 0 (1) 0 (2 2xxxxdx M dx x| |1122) 0 ( 1) 0 (222xxxxx Mx =((

| | ) 0 ( ) 0 ( 222 1 122x x x M x + = When,x1 > r,) 0 ( ) 0 ( 2 222 1 122x x M x M x + + =) 0 ( ) 0 ( 2 222 1 122x x M x M x + =x1 < r, First set of initial condition, x1(0) = 2 and x2(0) = 1, and substituting values of r and M, then, 8 . 5 4 . 21 2+ = x x8 . 3 4 . 21 2 = x xFor x1 > r For x1 < r Put different values for x1 and find the corresponding values of x2

Second set of initial condition, x1(0) = 2 and x2(0) = 1.5, and substituting values of r and M, then, 05 . 7 4 . 21 2+ = x x55 . 2 4 . 21 2 = x xFor x1 > r For x1 < r Put different values for x1 and find the corresponding values of x2

x1 > rx1 < r x1x2x1x2 21-11.58300 2.10.87-0.871.60.2-0.2 2.20.72-0.721.70.53-0.53 2.30.53-0.531.80.72-0.72 2.40.2-0.21.90.87-0.87 2.4160021-1 x1 > rx1 < r x1x2x1x2 21.5-1.51.06200 2.11.42-1.421.30.75-0.75 2.31.24-1.241.51.02-1.02 2.51.02-1.021.71.24-1.24 2.70.75-0.751.91.42-1.42 2.9370021.5-1.5