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Unformatted text preview: S. Boyd EE102 Lecture 3 The Laplace transform • definition & examples • properties & formulas – linearity – the inverse Laplace transform – time scaling – exponential scaling – time delay – derivative – integral – multiplication by t – convolution 3–1 Idea the Laplace transform converts integral and differential equations into algebraic equations this is like phasors, but • applies to general signals, not just sinusoids • handles nonsteadystate conditions allows us to analyze • LCCODEs • complicated circuits with sources, Ls, Rs, and Cs • complicated systems with integrators, differentiators, gains The Laplace transform 3–2 Complex numbers complex number in Cartesian form: z = x + jy • x = < z , the real part of z • y = = z , the imaginary part of z • j = √ 1 (engineering notation); i = √ 1 is polite term in mixed company complex number in polar form: z = re jφ • r is the modulus or magnitude of z • φ is the angle or phase of z • exp( jφ ) = cos φ + j sin φ complex exponential of z = x + jy : e z = e x + jy = e x e jy = e x (cos y + j sin y ) The Laplace transform 3–3 The Laplace transform we’ll be interested in signals defined for t ≥ the Laplace transform of a signal (function) f is the function F = L ( f ) defined by F ( s ) = Z ∞ f ( t ) e st dt for those s ∈ C for which the integral makes sense • F is a complexvalued function of complex numbers • s is called the (complex) frequency variable , with units sec 1 ; t is called the time variable (in sec); st is unitless • for now, we assume f contains no impulses at t = 0 common notation convention: lower case letter denotes signal; capital letter denotes its Laplace transform, e.g. , U denotes L ( u ) , V in denotes L ( v in ) , etc. The Laplace transform 3–4 Example let’s find Laplace transform of f ( t ) = e t : F ( s ) = Z ∞ e t e st dt = Z ∞ e (1 s ) t dt = 1 1 s e (1 s ) t fl fl fl fl ∞ = 1 s 1 provided we can say e (1 s ) t → as t → ∞ , which is true for < s > 1 : fl fl fl e (1 s ) t fl fl fl = fl fl fl e j ( = s ) t fl fl fl  {z } =1 fl fl fl e (1< s ) t fl fl fl = e (1< s ) t • the integral defining F makes sense for all s ∈ C with < s > 1 (the ‘region of convergence’ of F ) • but the resulting formula for F makes sense for all s ∈ C except s = 1 we’ll ignore these (sometimes important) details and just say that L ( e t ) = 1 s 1 The Laplace transform 3–5 More examples constant: (or unit step) f ( t ) = 1 (for t ≥ ) F ( s ) = Z ∞ e st dt = 1 s e st fl fl fl fl ∞ = 1 s provided we can say e st → as t → ∞ , which is true for < s > since fl fl e st fl fl = fl fl fl e j ( = s ) t fl fl fl  {z } =1 fl fl fl e ( < s ) t fl fl fl = e ( < s ) t • the integral defining F makes sense for all s with < s > • but the resulting formula for F makes sense for all s except s = 0 The Laplace transform 3–6 sinusoid: first express...
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This note was uploaded on 02/17/2011 for the course CHEM ENGG 101 taught by Professor Mister during the Spring '11 term at University of Engineering & Technology.
 Spring '11
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