lecture 22

# lecture 22 - Summary of Lecture#22 Convolution algebra y t...

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1 1 Summary of Lecture #22 10/15/2010 Convolution algebra What are filters for? An example of imp. matching circuit LP, HP, BP and BR filters ϖ o and ϖ r for circuits with one L and one C Bandpass filters Peak frequency: ϖ m Half power frequencies: ϖ 1 and ϖ 2 Half power bandwidth: B ϖ = ϖ 2 - ϖ 1 Pass band: [ ϖ 1 , ϖ 2 ] Quality factor Q: Q = ϖ m / B ϖ - - τ τ - τ = τ τ τ - = = d ) t ( h ) ( f d ) ( h ) t ( f ) t ( h ) t ( f ) t ( y 2 t h (1) (t) 3 ) t t ( Af d ) t ( A ) t ( f d ) ( h ) t ( f ) t t ( A ) t ( f ) t t ( A ) t ( h 0 0 0 0 - = τ - τ δ τ - = τ τ τ - = - δ - δ = - - Convolution with a delta function f max Af max Magnified and shifted, if A > 1 Minimized and shifted if A < 1 4 Convolution algebra Define Then pp. 781-786 f * h = f ( - 1) *h (1) if f ( - 1) (t) exists and h( - ∞ ) = 0. f * h = f (1) *h ( - 1) if h ( - 1) (t) exists and f( - ∞ ) = 0. f * h = f ( - 2) *h (2) = f (2) *h ( - 2) if …. , " dt ) " t ( f ' dt f , ' dt ) ' t ( f f f f " f dt f d , f f ' f dt df ] [ t ' t ) 2 ( t ) 1 ( ) 2 ( 2 2 ) 1 ( - - - - - = = = = = = = = ɺ ɺ ɺ 5 Ex. 10. y(t)=f(t) * g(t) , by convolution algebra f(t) = 4[u(t) - u(t - 4)], g(t) = 4[u(t) - 2u(t - 2) + 2u(t - 4) - 2u(t - 6) + u(t - 8)] f(t) 4 0 4 t g(t) 4 - 4 t f ( - 1) 16 g (1) τ τ 4t f( -∞ ) = 0 g( -∞ ) = 0 4 - 8 8 By convolution algebra g (1) (t) = 4[ δ (t) - 2 δ (t - 2)+ 2 δ (t - 4) - 2 δ (t - 6)+ δ (t - 8)] f (-1) τ 2 4 6 8 - 128 - 128 128 64

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2 7 Ex. 11. Given f(t) = 2[u(t) - u(t - 1)+u(t - 2) - u(t - 3)] and g(t) = 4[u(t) - u(t - 1)+u(t
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lecture 22 - Summary of Lecture#22 Convolution algebra y t...

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