# SolSec7.5 - Problems and Solutions Section 7.5(7.20-7.24...

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Problems and Solutions Section 7.5 (7.20-7.24) 7.20 Using the definition of the mobility transfer function of Window 7.4, calculate the Re and Im parts of the frequency response function and hence verify equations (7.15) and (7.16). Solution: From Window 7.4: k cs ms s s F s sX + + = 2 ) ( ) ( c j m k j ω α + = ) ( ) ( 2 αω ωω () ( ) = −− [] −+ jk m j c km c 2 22 2 2 2 2 2 2 ) ( ) ( ) ( ) ( c m k m k j c + + = The previous expression can be separated into real and imaginary parts: Re ( ) = 2 2 c c Im 2 2 2 2 ) ( ) ( ) ( ) ( c m k m k + =

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7.21 Using equations (7.15) and (7.16), verify that the Nyquist plot of the mobility frequency response function does in fact form a circle. Solution: Define A c km c c c = −+ −= ω ωω α 2 () ( ) Re( ) 22 2 1 2 1 2 B c = = (- 2 ) ( ) Im( ) 2 Show that AB c 2 1 2 += which is a circle of radius c 2 1 with center at Re ( α ) = c 2 1 , Im ( α ) = 0. c c c c 2 2 2 2 2 2 1 2 + ( ) ( ) c c c c c 2 2 2 2 2 2 2 1 2 [] + + ( ) ( ) ( ) c c c c c 2 2 2 2 2 2 2 1 2 + ( ) ( ) ( ) ( ) c 2 1 2 Which is the equation of a circle.
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SolSec7.5 - Problems and Solutions Section 7.5(7.20-7.24...

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