Lesson+32 - 1 Lesson 32 Lesson 32 Challenge 31...

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Lesson 32 1
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Challenge 31 Continuous-Time Fourier Transform Chap. 17 Challenge 32 (No In-Class Problem) Lesson 32 2 Lesson 32
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Challenge 31 Lesson 32 3 A filter’s CTFT frequency response is shown below. - 0 0 0 |H( )| - 0 0 0 H( ) - t 0 Linear phase This is called an ideal or boxcar filter. What is h(t)? (impulse response) H( ) = |H( )| H( )=e -j t 0 1.0
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Lesson 32 4 )) ( ( ) ( ) ( ( ) ( ) ( ) ( ) ( 0 0 0 0 0 0 0 sinc sin 2 1 2 1 2 1 0 0 0 0 0 0 0 0 0 0 0 t t t t t t e t t j d e d e e t h t t j t t j t j t j ω π ω π ω π ω π ω π ω ω ω ω ω ω ω ω ω ω ω ω - 0 0 0 |H( )| - 0 0 0 H( ) - t 0 Inverse CTFT Could also have used the CTFT Properties tables For – 0 0 H( ) = |H( )| H( )=e -j t 0 1 0
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Challenge 31 Lesson 32 5 )) ( ( ) ( 0 0 0 sinc t t t h - 0 0 0 |H( )| - 0 0 0 H( ) - t 0 Interpretation t t 0 Group delay g = -d ( )/d = t 0 (constant delay for all ) Impulse response 1 0 Can you build this filter in hardware?
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Lesson 32 6 t t e t e t d je d je t h t j t j t j t j ) ( cos 1 2 1 2 1 2 1 2 1 ) ( 0 0 0 0 0 0 0 0 0 - 0 0 0 |H( )| - 0 0 0 H( ) Suppose /2  /2 Different phase spectrum – different outcome Non-linear phase (Hilbert filter) Same magnitude response Inverse CTFT
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CTFT Overview Lesson 32 7 T T j a t j a T t j at j a e dt e dt e e X 0 0 1 ) ( ) ( ) ( 0 T t x(t)=e -at , t [0,T] square integrable Causal input (1-sided Fourier transform)
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Lesson 32 8 T T j a t j a T t j at j a e dt
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