MTE1_A_Solution

MTE1_A_Solution

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Unformatted text preview: part a, we have z=w*=a‐jb=1‐j2. All we have to do is now to get r and which is easy: r 5 arctan(2) 1.107rads For part b, we start by calculating zw and z2w. First note, z2=r2e2j. Then zw (a jb)(r cos( ) jr sin( )) z 2 w (a jb)(r 2 cos(2 ) jr 2 sin(2 )) We are told that zw is purely real and z2w is purely imaginary. We can then set the imaginary part of zw and real part of z2w equal to zero: r[a sin( ) b cos( )] 0 r 2 [a cos(2 ) b sin(2 )] 0 Now if r is zero (and thus z=0), all values will satisfy these equations, and thus be admissible. If r is nonzero, then we have two equations a sin( ) b cos( ) a cos(2 ) b sin(2 ) Using basic trigonometry formulae, we get a (cos 2 ( ) sin 2 ( )) a 0 Therefore, bcos()=0. Now, if b is also zero, then w=0 and all values of are admissible. If b is non‐zero, then cos() = 0 or 2 7. Suppose G (s) ( s 2 s 2) ( s 1)n a. Suppose n=3. Find the partial fraction expansion of G(s). [3 pts] b. Suppose n is a positive integer and n > 3. Find the partial fraction expansion of G(s). [3pts] Answer: Let us start with part a. The generalpartial fraction formula is G ( s) A B C 2 ( s 1) ( s 1) ( s 1)3 A( s 1) 2 B ( s 1) C ( s 1)3 Now we can compare coefficients to get three equations: A 1 2A B 1 A B C 2 This leads to A=1, B=‐1, and C=2. Part b is not so easy. Here is one of the easiest ways to solve it. The main idea is to write the numerator also as a polynomial in (s+1): (( s 1) 1) 2 (( s 1) 1) 2 G(s) ( s 1)3 ( s 1) 2 ( s 1) 2 ( s 1) n 1 1 2 n2 n 1 ( s 1) ( s 1) ( s 1) n This is the partial fraction expansion. A longer approach is to express G(s) as follows: G ( s...
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This note was uploaded on 12/27/2011 for the course EEL 3105 taught by Professor Boykins during the Fall '10 term at University of Florida.

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