102_1_Lecture14

# 102_1_Lecture14 - UCLA Spring 2009-2010 Systems and Signals...

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UCLA Spring 2009-2010 Systems and Signals Lecture 14: The Laplace Transform May 17, 2010 EE102: Systems and Signals; Spr 09-10, Lee 1

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Limitations of the Fourier Transform The Fourier transform allows us to analyze many signals and systems To be useful, Fourier transforms must exist, or be defined in a generalized sense. For many areas, this will be all you will need (communications, optics, image processing). For many signals and systems the Fourier transform is not enough: Signals that grow with time (your bank account, or the GDP of the US) Systems that are unstable (many mechanical or electrical systems). These are important problems. How can we analyze these? EE102: Systems and Signals; Spr 09-10, Lee 2
Consider the signal f ( t ) = e 2 t u ( t ) This is an increasing exponential, so it doesn’t have a Fourier transform. However, we can create a new function φ ( t ) = f ( t ) e - σ t . If σ > 2 , then this is a decreasing exponential. It does have a Fourier transform. The Fourier transform represents φ ( t ) in terms of spectral components e j ω t . We can express f ( t ) as f ( t ) = φ ( t ) e σ t Each spectral component is multiplied by e σ t , so f ( t ) can be represented by spectral components e σ t e j ω t = e ( σ + j ω ) t . EE102: Systems and Signals; Spr 09-10, Lee 3

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t t ! ( t ) = f ( t ) e - " t f ( t ) 1 1 t t e j ! t e ( ! + j " ) t EE102: Systems and Signals; Spr 09-10, Lee 4
How big should we choose σ ? For f ( t ) = e 2 t , any σ > 2 will produce a decaying, Fourier transformable signal, and a di ff erent Fourier transform. If σ 0 is the smallest value for which f ( t ) e - σ 0 t goes to zero, then any σ > σ 0 will do ! 0 ! j ! Region of Convergence This means the spectrum of f ( t ) is not unique. The part of the complex plane where the spectrum exists is the region of covergence . EE102: Systems and Signals; Spr 09-10, Lee 5

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Bilateral Laplace Transform The Fourier transform is: F ( j ω ) = -∞ f ( t ) e - j ω t dt f ( t ) = 1 2 π -∞ F ( j ω ) e j ω t d ω The Fourier transform of f ( t ) e - σ t is F f ( t ) e - σ t = -∞ f ( t ) e - σ t e - j ω t dt = -∞ f ( t ) e - ( σ + j ω ) t dt EE102: Systems and Signals; Spr 09-10, Lee 6
= -∞ f ( t ) e - st dt = F ( s ) were s the complex frequency s = σ + j ω . The complex frequency s includes: the oscillation component j ω that we’re used to, plus a decay/growth component σ . This is not what you intuitively think of as a frequency. EE102: Systems and Signals; Spr 09-10, Lee 7

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This is the bilateral Laplace transform , which is defined as F ( s ) = -∞ f ( t ) e - st dt. The inverse can be shown to be f ( t ) = 1 2 π j c + j c - j F ( s ) e st ds where c > σ 0 (but we will never use this!). The notation for the Laplace transform is F ( s ) = L [ f ( t )] and f ( t ) = L - 1 [ F ( s )] or simply f ( t ) F ( s ) . EE102: Systems and Signals; Spr 09-10, Lee 8
The Fourier transform is a special case of the Laplace transform F ( j ω ) = F ( s ) | s = j ω provided the j ω axis is in the region of convergence.

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• Spring '09
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