Lecture 11

# Lecture 11 - ECE 3101 Signals and Systems Reading...

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Lecture 11, 3/2/11, Makeup Class ECE 3101, Copyright P. B. Luh 1 ECE 3101 Signals and Systems Reading Assignment: Sections 4.1, 4.2 and B.5 Problem Set 4: Due today Problem Set 5: 4.1-1 (b, d, f, and h), 4.1-3 (b, d, f and h), 4.2-1 (b, d, f, and h), 4.2-2, 4.2-3 (b, d, f, and h), 4.2-9. Due Tuesday, 3/15/11 Exam. 2, Tue. 3/22/11, end of Ch. 3. Old sample Exam 1 Last Time: The Laplace Transform 0 st dt e ) t ( f ) s ( F 0 0, (t) 1, e - t 1/(s+ ), U(t) 1/s Linearity: a 1 f 1 (t) + a 2 f 2 (t) a 1 F 1 (s) + a 2 F 2 (s) Derivative Theorem: f(t) F(s), the f (1) (t) sF(s) - f(0 - ) f (2) (t) s 2 F(s) - sf(0 - ) - f (1) (0 - )

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2 f (n) (t) s n F(s) - s n-1 f(0 - ) - s n-2 f (1) (0 - )-. .-f (n-1) (0 - ) Converting differentiation equations into algebraic equations Convolution: h(t) u(t) H(s) U(s) Additional Properties of the Laplace Transform Time shifting: f(t-t 0 )U(t-t 0 ) e -st 0 F(s) e s 0 t f(t) F(s-s 0 ) ~ Symmetry between s and t Today: Additional Properties of the Laplace Transform f (-1) (t) F(s)/s; t f(t) (-1)F (1) (s) f(at) (1/a)F(s/a) for a > 0 Initial Value Theorem and Final Value Theorem Inverse Laplace Transform Next Time: Sections 4.3 and 4.4 ; dt e ) t ( f ) s ( F 0 st j c j c st ds e ) s ( F j 2 1 ) t ( f
Lecture 11, 3/2/11, Makeup Class ECE 3101, Copyright P. B. Luh 3 Frequency Shifting Recall that f(t-t 0 )u(t-t 0 ) e -st 0 F(s) What is the Laplace transform of e s 0 t f(t) ? e s 0 t f(t) F(s-s 0 ) To prove this:     0 st t s t s dt e ) t ( f e ) t ( f e 0 0 L 0 t ) s s ( dt e ) t ( f 0 = F(s-s 0 ) Example. f(t)cos t ? f(t)cos t = f(t)(e j t + e -j t )/2 = (f(t)e j t + f(t)e -j t )/2 f(t)cos t [F(s-j ) + F(s+j )]/2 Similarly, f(t)sin t j[F(s+j ) - F(s-j )]/2

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4 Time Integration Recall that f(t) F(s), the f (1) (t) sF(s) - f(0 - ) Then ? d ) ( f t 0 s ) s ( F Proof t 0 d ) ( f ) t ( g Let g(t) G(s) g (1) (t) sG(s) - g(0 - ) = sG(s) f(t) F(s) = sG(s) G(s) = F(s)/s Example ? d sin t 0  ; s t sin 2 2   2 2 t 0 s s d sin   t cos 1 1 d sin t 0    2 2 2 2 s s s s s 1 1 Correct?
Lecture 11, 3/2/11, Makeup Class ECE 3101, Copyright P. B. Luh 5 Input h(t) Output u(t) y ZS (t) = h(t) u(t) U(s) Y ZS (s) =H(s)U(s) f(t-t 0 )U(t-t 0 ) U(s)e -st 0 H(s)U(s)e -st 0 = Y ZS (s)e -st 0 y(t-t 0 )U(t-t 0 ) Delayed input ~ Delayed output t 0 d ) ( u U(s)/s H(s)U(s)/s t 0 ZS d ) ( y Integrated input ~ Integrated output

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6 Frequency Differentiation
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## This note was uploaded on 03/29/2011 for the course ECE 3101 taught by Professor Luh during the Spring '11 term at UConn.

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Lecture 11 - ECE 3101 Signals and Systems Reading...

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