ECE634S09_L8_DifferentialEquations

ECE634S09_L8_DifferentialEquations - ECE634 Signals and...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
ECE634 Signals and Systems II, Spring 2009 - Lecture 8, February 9 Daniel S. Brogan 1 4.3 Solution of Differential and Integro-Differential Equations: Example 4.10 (pp.371-372) (Linear differential equation with constant coefficients) Solve the second-order linear differential equation ( ) ( )( )( ) 2 56 1 DD y tDx t ++ =+ for the initial conditions () 02 y = and ( ) 01 y = ± and the input ( )( ) 4 t x te u t = . The equation is 2 2 d y dy dx y tx t dt dt dt = + If we let () () yt Ys , then using the time-differentiation property, () () () dy sY s y sY s dt ⇔−=− () () () () 2 22 2 00 2 1 dy sY s sy y s dt −− ⇔−− =− ± For the input 4 t x u t = , 1 4 Xs s = + and 44 dx s s sX s x dt s s ⇔− = = + + We could also take the derivative in the time domain and then take the Laplace transform… ( ) ( ) 4 4 tt t t de ut de dut dx t ut e e ut e t e ut t dt dt dt dt δδ ==+ = + = + 1 41 dx s dt s s + = Note that this method required more work. Using the Laplace transform on the differential equation 2 2 2 1 2 1 5 2 6 d y dy dx s xt sYs s dt dt dt s s ⎡⎤ = +⇔ + + = + ⎣⎦ Now for some algebra…
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
ECE634 Signals and Systems II, Spring 2009 - Lecture 8, February 9 Daniel S. Brogan 2 () ( ) 2 1 56 21 1 4 s Ys s s s s + ⎡⎤ ++− + = ⎣⎦ + () ( ) ( ) ( ) 14 2 1 1 23 4 sss s s +++ + ++ = + 22 1 2 19 44 2 20 45 234 2 3 4 ss s s s A B C Ys s + + + + === + + + 2 2 2 20 45 8 40 45 13 34 1 2 2 s A =− + == = ( ) 2 3 0 4 5 1 8 6 0 4 5 3 24 1 1 s B + = 2 4 0 4 5 3 2 8 0 4 53 2 1 2 s C + = 13/2 3 3/2 The inverse Laplace transform of this is… 4 13 3 3 tt t y te e e u t −− ⎢⎥ Zero-Input and Zero-State Components of the Response The zero-input response is the output of the system based on initial conditions only, i.e. no input. The zero-state response is the output of the system based on the input only, i.e. the initial conditions are zero. In the above example, these two can be separated as follows: N 2 initial condition terms input terms 1 1 4 s s s s + ++= + + + ±²³ ²´
Background image of page 2
ECE634 Signals and Systems II, Spring 2009 - Lecture 8, February 9 Daniel S. Brogan 3 () zero-input component zero-state component zero-input component zero-state component 21 1 1 2 3 2 3 4 23 234 ss A B C D E Ys sss s s s ++ =+ = + + + + +++ + + + ±²³²´ ±²²²³²²²´ ±²²³ ² ²´
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 7

ECE634S09_L8_DifferentialEquations - ECE634 Signals and...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online