Chapter 3 : Linear Differential Eqn. Chapter 3 : Linear Differential Eqn.

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1 1.0 Introduction Linear differential equations is all about to find the total solution y(t), where : y(t) = homogeneous solution [ y h (t) ] + particular solution y p (t) General form of differential equation : where C = constant coefficient 1 The order of differential equation is defined to be the order of the highest derivative appearing in the equation. 1.0 Introduction a) Homogenous solution The homogenous solution is obtained when x(t) is set to zero. 2 The homogenous solution consists of the form A e αt where : where A : arbitrary constant α : complex constant determined by differential equation Substituting y(t) = A e αt in 2 3 1

2 Chapter 2 : Linear Differential Eqn. 1.0 Introduction a) Homogenous solution If α is a root of 3, y h (t) = Ae αt will satisfy 2, 3 is called characteristic equation of the differential 1. α 1, α 2,., α n-1, α n are called the characteristics roots of DE 1.0 Introduction a) Homogenous solution When the characteristics of DE are all distinct, the homogeneous solution will be : 4 When the characteristics of DE are not all distinct, the homogeneous solution will be of slightly different form. Suppose that α 1 is a k-multiple root of the CE, then corresponding to α 1 will be k terms in the homogeneous solution 5 2

3 1.0 Introduction b) Particular solution Depends explicitly on x(t) and satisfies 1 Is called the forced response of the system. Can be obtained by inspection by using Table 2.1 below : x(t) Particular solution t n Ae αt - if α is to the characteristics root cos ωt sin ωt Ate αt - if α is = to the characteristics root A n-1 t n-1 e αt - if α is a (n-1) : multiple characteristics root A 1 cos ωt + A 2 sin ωt A 1 cos ωt + A 2 sin ωt 1.0 Introduction b) Particular solution To determine the particular solution : a) Guess the general form of the constant. b) Determine the constant e.g. comparing coefficient The total solution of y(t) : y(t) = y h (t) + y p (t) Where the constant A 1, A 2, A 3,..., A n-1, A n in the homogeneous solution can be determined by a set of boundary conditions by using the initial conditions. 3

4 1.1 Linear First Order Differential Equation There two methods in order to solve the 1 st order differential equation : a) Solving by multiplication factor b) Solving by undetermined coefficient (classical method) Solving by multiplication factor Hand - written 4

5 1.1.2 Solving by undetermined coefficient (classical method) Usually the form of 1 st order LDE with constant coefficients : 1 a) Homogeneus solution The homogeneous solution is a solution of the equation : 1 when x(t) = 0, where satisfies 2 Chapter 2 : Linear Differential Eqn Solving by undetermined coefficient (classical method) a) Homogeneus solution The homogenous solution consists of the form A e αt where : where A : arbitrary constant α : complex constant determined by differential equation] Substituting y(t) = A e αt in 2, we obtain : Aαe αt + p Ae αt = 0 Ae αt (α + p) = 0 3 If α is a root of 3, y h (t) = Ae αt will satisfy 2. 5

6 1.1.2 Solving by undetermined coefficient (classical method) b) Particular solution Depends on x(t) and satisfies 1 Can be obtained by using Table 2.1 as a guide. To determine the particular solution : - Guess the general form of the constant. - Determine the multiplying constant in the general form (Table 2.1) by matching coefficients Solving by undetermined coefficient (classical method) The total solution of y(t) : y(t) = y h (t) + y p (t) Where the constant A 1, A 2, A 3,..., A n-1, A n in the homogeneous solution can be determined by a set of boundary conditions by using the initial conditions. 6

7 Example The following systems is described by the 1 st order differential equations relating the input and the output. Solve the output using the classical method. a) y (t) y(t) = e -2t, y(0) = 2 b) y (t) 4y(t) = 2e -2t + 2 cos 3t, y(0) = 8 c) y (t) + y(t) = sin (t), y(0) = 7 d) y (t) - 3y(t) = t 2 e -3t, y(0) = 4 e) y (t) + 5y(t) = e -7t cos (6t), y(0) = Application to electrical systems Related equation Component Voltage Current v(t) = i(t) R i(t) = v(t) / R 7

8 Example Consider the circuit shown below. The 5 V source is connected to the capacitor via the switch that has been closed for a long time and opened at t = 0. The input voltage x(t) is assumed to be zero for t < 0. Derive the differential equation. R x(t) S 1 MΩ uf C y(t) 5 V - - Example For the following figure switch S is closed at t= 0. Using the classical method, find i(t) for t > 0. Assume i(0) = 0 t=0 R 12 V + - S 50Ω i(t) 0.1 H 8

9 Example Derive the differential equation for the system below. Hence, calculate v(t) if i(t) = 2 cos (4t) u(t). Switch S is closed at t = 0 and V(0) = 10 V S i c (t) i R (t) i(t) 1 mf 1kΩ v(t) Example For the circuit below, find the differential equation relating y(t) and x(t). If x(t) = 2t e -10t, solve for y(t) and assume y(0) = 0 x(t) F 10 Ω y(t) 9

10 1.3 Second Order Differential Equation A second order differential equation is linear if it has the form : 1 The system is linear time invariant if A(t) and B(t) are constants Solving by undetermined coefficient (classical method) a) Homogeneous solution Let x(t) = 0 for equation 1 2 Substitute y h (t) = Ae αt in equation 2 3 Characteristics equation 10

11 1.3.1 Solving by undetermined coefficient (classical method) a) Homogeneous solution Characteristics root There are 4 possibilities might exist for characteristic roots : i) Two distinct real values for α 1 and α 2 (when A 2 4B > 0) ii) Two equal real roots for α 1 and α 2 (when A 2 4B = 0) iii) Two complex conjugate roots for α 1 and α 2 (when A 2 4B < 0) iv) Two pure imaginary roots for α 1 and α 2 (when A = 0 and B is positive) The form of y h (t) depends on the different cases for α 1 and α Solving by undetermined coefficient (classical method) i) Two distinct real values for α 1 and α 2 (when A 2 4B > 0) 4 Homogeneous solution 11

12 1.3.1 Solving by undetermined coefficient (classical method) ii) Two equal real roots for α 1 and α 2 (when A 2 4B = 0) From equation 4 : α 1 and α 2 are equal. Homogeneous solution Solving by undetermined coefficient (classical method) iii) Two complex conjugate roots for α 1 and α 2 (when A 2 4B < 0) From equation 4 Homogeneous solution : α 1 = p + jq : α 2 = p - jq 12

13 1.3.1 Solving by undetermined coefficient (classical method) iv) Two pure imaginary roots for α 1 and α 2 (when A = 0 and B is positive) From equation Homogeneous solution 4 : α 1 = jq : α 2 = - jq Solving by undetermined coefficient (classical method) b) Particular solution Depends on x(t). Can be obtained by using Table 2.1 as a guide. To determine the particular solution : - Guess the general form of the constant. - Determine the multiplying constant in the general form (Table 2.1) by matching coefficients. 13

14 1.3.1 Solving by undetermined coefficient (classical method) The total solution of y(t) : y(t) = y h (t) + y p (t) Where the constant A 1, A 2, A 3,..., A n-1, A n in the homogeneous solution can be determined by a set of boundary conditions by using the initial conditions. Example The following systems is described by the 2 nd order differential equations relating the input and the output. Solve the output using the classical method. a) y (t) + 3y (t) - 4y(t) = e -4t, y (0) = 0, y(0) = 1 b) y (t) + 4y (t) + 4y(t) = 4 cos (10t), y (0) = 0, y(0) = 1 c) y (t) + 2y (t) + 6y(t) = 10, y (0) = 0, y(0) = 1 d) y (t) + y(t) = te 2t, y (0) = 0, y(0) = 1 e) y (t) + 7y (t) + 12y(t) = 5t e -t, y (0) = 1, y(0) = 0 f) y (t) + 2y (t) + 10y(t) = 3e 2t t 2, y (0) = 10, y(0) = 1 14

15 1.4 Zero Input Response And Zero Output Response Zero input response : the response of the system when the input is set to 0. : y zi (t) is determined by homogeneous solution y h (t) where the constant is determined by the initial conditions. y zi (t) = y h (t) = y (0) y zi (t) = y h (t) = y (0) Zero state response : the response of the system when the initial condition is set to zero. 1.4 Zero Input Response And Zero Output Response Zero state response : the response of the system when the initial condition is set to zero. : y zs (t) is determined by the additional of homogeneous solution y h (t) and particular solution y p (t) where the constant of y h (t) is then determined by the initial condition is set to zero for y zs (t). y zs (t) = y h (t) + y p (t) = y (0) y zs (t) = y h (t) + y p (t) = y (0) The total response is equal to : y(t) = zero input response [ y zi (t) ] + zero state response [ y zs (t) ] 15

16 Example A systems is described as follows. Determine the zero input response and zero state response. Hence, find the output y(t). a) y (t) + 3y (t) - 4y(t) = t 2, y (0) = 0, y(0) = 1 b) y (t) + 4y (t) + 5y(t) = te -2t, y (0) = 0, y(0) = 0 c) y (t) + 5y (t) + 6y(t) = t + e -3t, y (0) = 1, y(0) = Application to electrical systems Example R 1 i(t) e(t) + - i L (t) L i c (t) C v(t) i) Find the differential equation relating v(t) and e(t) for the above figure. ii) Given that R = 2/5 Ω, C = ½ F and L = 1/3 H. Determine v(t) if e(t) = cos (10t) 16

17 Application to electrical systems R1 Solution ic(t) il(t) + e(t) i(t) i1(t) C L v(t) - i) Find the differential equation relating v(t) and e(t) for the above figure. i1 (t) R1 + v(t) = e(t) 1 (il (t) + ic (t))r1 + v(t) = e(t) 2 Note that : Application to electrical systems Solution i) Find the differential equation relating v(t) and e(t) for the above figure. 3 Differentiate equation

18 . 1.5 Application to electrical systems Solution Substitute value of R 1, L and C in equation 4 5 ii) Given that R = 2/5 Ω, C = ½ F and L = 1/3 H. Determine v(t) if : e(t) = cos (10t) v(t) can be determined using undetermined coefficient method (classical method) 18