Lecture 2 : Mathematical Review

Transcription

1 UCSI University Kuala Lumpur, Malaysia Faculty of Engineering Department of Mechatronics Lecture Mathematical Review Mohd Sulhi bin Azman Lecturer Department of Mechatronics UCSI University 1 August 011 Page : 1 Contents Partial Fraction Expansion Matrices: Determinant Inverse Eigenvalues and Eigenvectors Cramer s Rule Differential Equations Laplace Transforms Transfer Function State Space Representation Page : 1

2 Partial Fraction Expansion A complete (single) fraction can be decomposed or expanded into its partial form. There are few methods used for partial fraction expansion, however, we will be using the following method: Heaviside Cover-up Method. Equating or comparing the coefficients method. Page : 3 Partial Fraction Expansion Case I : Linear Factors f ( x) k1 k = + x + a x + b x + a x + b ( )( ) To find k 1 and k, we use Heaviside Cover-up method: f ( x) f ( x) k = lim = 1 x a ( x + b) ( x + b) x b ( x + a) ( x + a) x a k f ( x) f ( x) = lim = x b Page : 4

3 Partial Fraction Expansion Case II : Repeated Factors f ( x) k k kn = n ( x + a)( x + a) ( x + a) x + a ( x + b) ( x + b) n To find k 1, k,, k n, we may use the comparing coefficients method. Alternatively, we may also use Heaviside cover-up method. Page : 5 Determinant A determinant is a special number associated with a square matrix. We may determine a determinant by expanding along a row or a column by using Laplace s formula for expanding the determinant: n j= 1 i ( ) det( A) = A = A 1 M i, j i, j Where M is a minor matrix obtained by deleting specific row and column. This formula works with a square matrix of the n x n order. + j Page : 6 3

4 x Cramer s Rule Most equations can be represented in matrix form. And therefore, equations in matrix form can be solved by using Cramer s Rule, and of course any method suitable and relevant method. For now, let us focus on Cramer s Rule as it is the easiest method of solving matrix equations of the form Ax=b. If: Then: 1 c b = = = a1 b1 x1 c1 a b x = c Dx c b c1b cb1 D a b a b a b a D b x a c 1 1 Dx a c a c a c D a b a b a b = = = a b Page : 7 Cramer s Rule We can also use Cramer s rule to solve for a 3x3 systems of linear equations. The relevant formula is given as: x=d x /D y=d y /D z=d z /D Page : 8 4

5 Differential Equations Differential equation describes the behaviour of a particular system. There are two types of differential equations: Ordinary differential equations Partial differential equations (not covered in this course) The order of a differential equation is the highest order of derivative contained within the expression. Page : 9 Differential Equations Definition (n th order): n n 1 n 3 d x d x d x d x d x dx n n n 1 n n n a + a + a + + a + a + a + a x = f ( t) 1 3 dt dt dt dt dt dt First order: Second order: dx a1 a0 x f ( t) dt + = d x dx 1 0 a + a + a x = f ( t) dt dt Page : 10 5

6 First Order ODE The highest derivative in the first order ODE is one (1). There are two general method for solving the first order ODE: Separation of variables (SOVA) Integrating factor (for linear first order ODE) Page : 11 First Order ODE : Integrating Method Given : dx a1 a0 x f ( t) dt + = Dividing by a 1 throughout: dx a0 f ( t) + x = dt a a 1 1 Define P(t) and Q(t) : dx P( t) x Q( t) dt + = Page : 1 6

7 First Order ODE : Integrating Method The solution is then given by: u( t) Q( t) dt + C x( t) = u( t) Where u(t) is the integrating factor, defined by: ( ) ( ) u( t) = exp P( t) dt = e P t dt Page : 13 Second Order ODE The highest order of derivative contain within the expression of a second order ODE is. General form: d x dx 1 0 a + a + a x = f ( t) dt dt Solving second order ODE: Method of undetermined coefficients; or Variation of parameters; or Series solution of differential equation; or Laplace transforms. There are two type of solutions for a second order ODE: x( t) = x + x = x ( t) + x ( t) homogeneous particular h p Page : 14 7

8 Solution of Second Order ODE The homogenous solution is found by setting f(t)=0 and letting: m dx dt = = d x Which will then lead to a quadratic equation: a m a m a m = 0 dt And the solution is given by: m 1, a ± a 4a a = a Page : 15 If: Solution of Second Order ODE m 1, = real and distinct roots i.e. m 1 and m, then: x ( t) = k e + k e h m t 1 1 m 1, = real and repeated roots i.e. m 1 and m 1, then: m 1, = complex conjugate i.e. α ± jβ, then: m t ( ) x ( t) = k e + k xe = k + k x e h m 1t m 1t m 1t 1 1 ( β β ) α x ( t) = e t cos t + j sin t h Page : 16 8

9 Solution of Second Order ODE The particular solution depends on the trial function: Page : 17 Laplace Transform The Laplace transform is a very powerful method that is used widely in control engineering to solve differential equations. It basically transforms a time-domain equation to a frequency domain equation. Mathematical definition: L { } st f ( t) F( s) f ( t) e dt = = 0 Page : 18 9

10 Some Important Theorems Page : 19 Some Important Theorems Page : 0 10

11 Some Important Theorems Page : 1 Transfer Functions The transfer function is the ratio between the output to the input. It is usually obtained by taking the forward Laplace transform of differential equation, assuming zero initial conditions. Definition: (Source : Wikipedia, 010) Page : 11

12 Transfer Functions A transfer function is basically written in the following format: Output C( s) G( s) = = Input R( s) Again, the transfer function is found by taking forward Laplace transform of differential equations, assuming zero initial condition. We will also come across the terms gain, zeros and poles when dealing with transfer function. Another alternative form of a transfer function is: ( + z) K s G( s) = s + p Page : 3 Transfer Functions Now, K stands for gain, z stands for zeros and p stand for poles. A zero is a number that causes the numerator of the transfer function to be zero. A pole is a number that causes the denominator of the transfer function to be zero. Another name for a pole is root. Another form of a transfer function is the factor form, where the numerator and the denominator of the transfer function is written as factors of linear function. The general format is: ( s zi ) ( s p ) Π + G( s) = K Π + i Page : 4 1

13 Transfer Functions We can plot the values of the pole(s) and zero(s) on a pole-zero map, also known as the pz-map. A pole is marked with the symbol cross (x) while a zero is marked with the symbol empty circle (o). Example: Plot the poles and zeros for the following: G( s) = s ( s + 1)( s + ) Page : 5 Transfer Functions Solution to the example in the previous slide: Im Re - -1 Page : 6 13

14 State Space Representation A state space representation is a modern method of representing differential equations in matrix form. We can then use MATLAB to perform a state space analysis of a control system. Definition: (source : Wikipedia, 010) Page : 7 Example : From DEs to State Space Observe the following conversion: d y dy 1 0 a + a + a y = f ( x) dx dx y = y yɺ = yɺ = y y = yɺ ɺɺ y = ɺɺ y 1 1 a ɺɺ y + a y + a y = f ( x) ɺɺ y = a y + a y f x ( ( )) a Page : 8 14

15 Example : From DEs to State Space And finally, in a state space matrix: yɺ y1 0 y = a + a 0 a 1 y 1 ɺ f ( x) Page : 9 Next Step Textbook reference : Chapter. Homework has been posted on the course website. Attempt them. You do not have to submit Homework as it will not be graded. Thank You. Page : 30 15

16 Wise Word "Failure will never overtake me if my determination to succeed is strong enough. Og Mandino Page : 31 16