Solution to HW4 EE4580 Fall04

Solution to HW4 EE4580 Fall04 - Then you can easily sketch...

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Correction to 5-11 of HW4 EE4580 Fall04 1) The closed-loop transfer function is 1 1 1 10 2 ) ( ) 10 )( ( 10 ) ( ) ( ) ( K s K s s s s s K s R s Y s G c + + + + + + = = λ (1) You can use direct computation, Mason’s rule or block diagram simplification method to get ) ( s G c . Let 10 2 ) ( ) 10 )( ( ) ( 1 + + + + + + = s K s s s s s s T (2) The characteristic equation is 0 ) ( = s T (3) There are two parameters and 1 K in the characteristic equation, and they will determine the root loci. 2) part a): 0 10 2 ) ( ) 10 )( ( ) ( 1 = + + + + + + = s K s s s s s s T (4) 0 ) ( ) 10 )( ( 10 2 1 1 = + + + + + + s s s s s s K (5) Let 2 = , (5) is written as 0 ) 2 ( ) 10 )( 2 ( 10 2 1 1 = + + + + + + s s s s s s K (6) Then you can easily sketch the root locus of (6). It is shown in Figure 1. 3) part b): Let 5 = , (5) is written as 0 ) 5 ( ) 10 )( 5 ( 10 2 1 1 = + + + + + + s s s s s s K (7)
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Unformatted text preview: Then you can easily sketch the root locus of (7). It is shown in Figure 2. 4) part c): 10 2 ) ( ) 10 )( ( ) ( 1 = + + + + + + = s K s s s s s s T ⇔ 10 2 11 11 1 1 1 2 3 2 = + + + + + K s K s s s s (8) Let 2 1 = K , (8) is written as 20 4 11 11 1 2 3 2 = + + + + + s s s s s (9) Then you can easily sketch the root locus of (9). It is shown in Figure 3. Figure 2: Root Locus of 5-11b Figure 1: Root Locus of 5-11a Figure 3: Root Locus of 5-11c...
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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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Solution to HW4 EE4580 Fall04 - Then you can easily sketch...

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