Solution8

Solution8 - Solution8 Ch6_43,45,(47),50,57 6.43 Solution:...

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(a) The frequency response is obtained by letting s = j ω , D ( j ω )= K Tj ω +1 α Tj ω +1 The phase is given by, φ =tan 1 ( T ω ) tan 1 ( α T ω ) (b) Using the trigonometric relationship, tan( A B )= tan( A ) tan( B ) 1+tan( A )tan( B ) then tan( φ )= T ω α T ω 1 α T 2 ω 2 and since, sin 2 ( φ )= tan 2 ( φ ) 1+tan 2 ( φ ) then sin( φ )= s ω 2 T 2 (1 α ) 2 1+ α 2 ω 4 T 4 +(1+ α 2 ) ω 2 T 2 To determine the frequency at which the phase is a maximum, let us set the derivative with respect to ω equal to zero, d sin( φ ) d ω =0 which leads to 2 ω T 2 (1 α ) 2 (1 αω 4 T 4 )=0 The value ω = 0 gives the maximum of the function and setting the second part of the above equation to zero then, ω 4 = 1 α 2 T 4 or ω max = 1 α T The maximum phase contribution, that is, the peak of the D ( s ) curve corresponds to, sin φ max = 1 α 1+ α or α = 1 sin φ max 1+sin φ max tan φ max = ω max T αω max T 1+ ω 2 max T 2 = 1 α 2 α
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(c) The maxmum frequency occurs midway between the two break fre- quencies on a logarithmic scale, log ω max =l o g 1 T α T =l o g 1 T +log 1 α T = 1 2 log 1 T +log 1 α T as shown in Fig. 6.52. (d) Alternatively, we may state these results in terms of the pole-zero locations. Rewrite D ( s )as , D ( s )= K ( s + z ) ( s + p ) then D ( j ω )= K ( j ω + z ) ( j ω + p ) and φ =tan 1 ω | z | tan 1 ω | p | or tan φ = ω | z | ω | p | 1+ ω | z |
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This note was uploaded on 01/23/2012 for the course EE 4580 taught by Professor Gu during the Fall '10 term at LSU.

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Solution8 - Solution8 Ch6_43,45,(47),50,57 6.43 Solution:...

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