Lesson+30

# Lesson+30 - 1 Lesson 32 Lesson 30 No Challenge 29...

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Lesson 32 1

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2 Lesson 32 2 Lesson 30 No Challenge 29 CTFT (Fourier Transform) Ch17.1-2 Challenge 30
Lesson 30 Lesson 32 3 CTFT! So why do I need to learn another frequency transform trick? Isn't the CTFS enough? What is wrong with the CTFS? Works only with periodic signals and periodic signals are non-causal! The CTFS has difficulties working with “jump” discontinuities. 3

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Periodic discontinuous signals Lesson 32 4 Why the large error? Harmonics x(t) 0 T 2T Point of discontinuity 4 Ideal
Lesson 32 5 dt e t x t j ) ( ) X( What happens when the signal being studied is causal, therefore non-periodic? Introduce a new tool called the continuous-time (bilateral) Fourier transform (CTFT) Analysis equation d e t x t j ) X( 2 1 ) ( Synthesis equation Bilateral or 2-sided transform (CTFT) 5

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Lesson 32 6 dt t x dt t x 2 2 2 1 1 x(t) x(t) or x(t) x(t) ) ( ; ) ( ; Existence The CTFT of x(t) exists if: What about a unit step? 2 1 u(t) ; u(t) Does mean a CTFT of a unit step doesn't exist? Under these conditions, the inverse Fourier transform converges to x(t) where x(t) is continuous and to the mean of x(t) at points of discontinuity. 2 1 u(t) ; u(t) 6
Lesson 32 7 0 t OK then, what is (u(t)) if it does exist? How do engineers overcome such a mathematical incongruence? Cheat 7

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Lesson 32 8 What is ( u t )? j a a a j a a j a t u e X a a a a at 1 lim lim 1 lim ) ( ) ( 2 2 0 2 2 2 2 0 0 0 Trick 0 t 0 t 0 a at t u e t x ) ( ) ( Fourier Transforms in the limit. 8 ???
Lesson 32 9 a d a a 1 2 2 tan j X 1 ) ( 0 t 0 ) ( ) ( a at t u e t x Sidebar Area

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