lecture 17

lecture 17 - Summary of Lecture #17 10/01/2010 Frequency...

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1 1 Summary of Lecture #17 10/01/2010 Transfer function H(s) Impulse response h(t) Step response s(t) Frequency response H(j ϖ ) Two examples Review of algebra of complex numbers Qualitative and graphical estimate of frequency response based on pole-zero plots Example s Final: 12/17/2010, Friday 1:00 to 3:00 PM SMTH 108 2 Frequency response Given a stable system with a transfer function H(s) and a sinusoidal input v in (t) = Acos( ϖ t + θ ), the steady state response is v out (t) = A |H(j ϖ )| cos[ ϖ t + θ + H(j ϖ ) ] ) j ( H | ) j ( H | ) p j )( p j ( ) p j )( p j ( ) z j )( z j ( ) z j )( z j ( K ) j ( H ) p s )( p s ( ) p s )( p s ( ) z s )( z s ( ) z s )( z s ( K ) p s )( p s ( ) s ( n ) s ( H 2 c 1 c 2 1 2 c 1 c 2 1 2 c 1 c 2 1 2 c 1 c 2 1 2 1 ϖ ϖ = - ϖ - ϖ - ϖ - ϖ - ϖ - ϖ - ϖ - ϖ = ϖ - - - - - - - - = - - = 3 For an input of A cos( ϖ t + θ ), the frequency response is ) ) j ( H t cos( | ) j ( H | A e A ˆ e A ˆ ) t ( v t p 2 t p 1 out 2 1 ϖ + θ + ϖ ϖ + + + = co | s H( ( A j ) | ) H t (j ) ϖ + θ ϖ + ϖ Which part is the ZIR ? Which part is the ZSR? 4 Alternate method: Use phasor method to calculate H(j ϖ ) Magnitude | H(j ϖ ) | Phase angle H(j ϖ ) Then for an input of A cos( ϖ t + θ ), the frequency response is Review Chapter 10
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lecture 17 - Summary of Lecture #17 10/01/2010 Frequency...

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