102_1_Lecture16

102_1_Lecture16 - UCLA Spring 2009-2010 Systems and Signals...

Info iconThis preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
UCLA Spring 2009-2010 Systems and Signals Lecture 16: Applications of the Laplace Transform May 24, 2010 EE102: Systems and Signals; Spr 09-10, Lee 1
Background image of page 1

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

View Full DocumentRight Arrow Icon
Solutions of LCCODE’s The Laplace transforms provide an easy solution for linear di f erential equations with Constant coe F cients Initial conditions Input signals Solution procedure: Laplace transform converts di f erential equation, initial conditions, and input signals into an algebraic equation Solve for the Laplace transform of the output Inverse Laplace transform provides the solution EE102: Systems and Signals; Spr 09-10, Lee 2
Background image of page 2
Example Solve the LCCODE y ± ( t )+5 y ± ( t )+6 y ( t )= f ± ( t )+ f ( t ) where y ( t ) is the output signal, and initial conditions are y (0 - ) = 2 , and y ± (0) = 1 . Assume that the input signal is f ( t e - 4 t u ( t ) . Solution: Taking the Laplace transform of the left side of the equation is s 2 Y ( s ) - sy (0 - ) - y ± (0 - ) + 5( sY ( s ) - y (0 - )) + 6 Y ( s ) = Y ( s )( s 2 +5 s + 6) - 2 s - 1 - 10 = Y ( s )( s 2 s + 6) - (2 s + 11) EE102: Systems and Signals; Spr 09-10, Lee 3
Background image of page 3

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

View Full DocumentRight Arrow Icon
The Laplace transform of the right side is sF ( s ) - f (0 - )+ F ( s )= F ( s )( s + 1) = s +1 s +4 where we have used the fact that that L [ f ( t )] = L [ e - 4 t u ( t )] = 1 s +4 . The Laplace transform of the equation is then Y ( s )( s 2 +5 s + 6) - (2 s + 11) = s s . Solving for Y ( s ) , Y ( s 2 s + 11 s 2 s +6 ± ²³ ´ from init. cond. + s ( s + 4)( s 2 s + 6) ± ²³ ´ from input . The Frst part can be traced back to the initial conditions, and the second part is due to the input. We’ll return to this in a few pages. EE102: Systems and Signals; Spr 09-10, Lee 4
Background image of page 4
Combining these terms we get Y ( s )= (2 s + 11)( s + 4) + ( s + 1) ( s + 4)( s 2 +5 s + 6) = 2 s 2 + 20 s + 45 ( s + 4)( s + 3)( s + 2) To fnd the inverse Laplace transForm we frst fnd the partial expansion, 2 s 2 + 20 s + 45 ( s + 2)( s + 3)( s + 4) = r 1 s +2 + r 2 s +3 + r 3 s +4 . Using the cover up algorithm, r 1 = 2( - 2) 2 + 20( - 2) + 45 ( - 2 + 4)( - 2 + 3) = 8 - 40 + 45 (2)(1) = 13 2 r 2 = 2( - 3) 2 + 20( - 3) + 45 ( - 3 + 4)( - 3 + 2) = 18 - 60 + 45 (1)( - 1) = - 3 r 3 = 2( - 4) 2 + 20( - 4) + 45 ( - 4 + 3)( - 4 + 2) = 32 - 80 + 45 ( - 1)( - 2) = - 3 2 EE102: Systems and Signals; Spr 09-10, Lee 5
Background image of page 5

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

View Full DocumentRight Arrow Icon
Then Y ( s )= 13 / 2 s +2 - 3 s +3 - 3 / 2 s +4 . The solution y ( t ) is then found by taking the term-by-term inverse Laplace transform y ( t ± 13 2 e - 2 t - 3 e - 3 t - 3 2 e - 4 t ² u ( t ) Zero-State and Zero-Input Solutions: This solution combines the e f ect of the initial conditions, and the input. We can also keep them separate, and solve for the zero-input and zero-state signals, Y ( s 2 s + 11 s 2 +5 s +6 + s +1 ( s + 4)( s 2 s + 6) = 2 s + 11 ( s + 2)( s + 3) ³ ´µ zero input + s ( s + 2)( s + 3)( s + 4) ³ ´µ zero state . EE102: Systems and Signals; Spr 09-10, Lee 6
Background image of page 6
The zero input component is 2 s + 11 ( s + 2)( s + 3) = r 1 s +2 + r 2 s +3 = 7 s - 5 s . which corresponds to the signal y zi ( t )= ± 7 e - 2 t - 5 e - 3 t ² u ( t ) The zero state component is s +1 ( s + 2)( s + 3)( s + 4) = r 1 s + r 2 s + r 3 s +4 = - 1 / 2 s + 2 s - 3 / 2 s EE102: Systems and Signals; Spr 09-10, Lee 7
Background image of page 7

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

View Full DocumentRight Arrow Icon
which corresponds to the signal y zs ( t )= ± - 3 2 e - 4 t +2 e - 3 t - 1 2 e - 2 t ² u ( t ) The complete solution is then y ( t y zi ( t )+ y ( t ) = ³ 7 e - 2 t - 5 e - 3 t ´ u ( t ± - 1 2 e - 2 t e - 3 t - 3 2 e - 4 t ² u ( t ) = ± 13 2 e - 2 t - 3 e - 3 t - 3 2 e - 4 t ² u ( t ) EE102: Systems and Signals; Spr 09-10, Lee 8
Background image of page 8
Zero-State Response: Transfer Function of LTI Systems A linear time invariant system is complete characterized by its impulse response. If the impulse response is h ( t ) , the output y ( t
Background image of page 9

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

View Full DocumentRight Arrow Icon
Image of page 10
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/09/2010 for the course EE ee102 taught by Professor Levan during the Spring '09 term at UCLA.

Page1 / 42

102_1_Lecture16 - UCLA Spring 2009-2010 Systems and Signals...

This preview shows document pages 1 - 10. Sign up to view the full document.

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