MTE1_A_Solution - EEL 3105 ‐ Mid Term Exam 1A ‐...

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Unformatted text preview: EEL 3105 ‐ Mid Term Exam 1A ‐ Solution 1. Consider two vectors p and q in two dimensions. Suppose p and q are given in polar coordinates as p 3.5rad q 4.8rad a. Define z = p+q. Suppose z is expressed as z re j . Find r and . [2pts] b. Find the scalar product of p and q. [2pts] c. Find the cross product of p and q. [2pts] Answer: First, we convert p and q to Cartesian coordinates. p 3cos(0.5) 3sin(0.5) [2.63 1.44] q 4 cos(0.8) 4sin(0.8) 2.79 2.87 a. z=p+q= 5.42 4.31 . We need to express z in polar coordinates to get r and . We can calculate r and to get r 5.422 4.312 6.92 4.31 0.67rads 5.42 arctan b. Scalar product is easy to calculate since we already have the Cartesian representation of p and q. Thus p.q 2.69 * 2.79 1.44 * 2.87 11.47 Another way to get this answer is to use the polar representation. Angle between p and q is [0.8‐0.5]rads. Therefore, the scalar product is 12cos(0.3)=11.47. c. Cross product of p and q will be a vector perpendicular to the plane of p and q given by the right hand rule. Let’s represent p and q in the x,y plane via unit vectors along x and y directions, i, j. Thus, p 2.63i 1.44 j q 2.79i 2.87 j p q 2.63* 2.87 1.44* 2.79 k 3.54k where k is the unit vector along the z direction. Another way to get this answer is from the polar representation for p and q. Again, angle between p and q is 0.3rads. Therefore, the magnitude of pxq is 12sin(0.3)=3.54 and its direction is perpendicular to the plane of p and q. 2. Suppose p and q are two vectors in 3 dimensions. Magnitude of p is 3 and the magnitude of q is 5. Moreover, it is known that the magnitude of the scalar product of p and q is equal to the magnitude of the cross product of p and q. a. Find the value(s) of the scalar product of p and q. [2pts] b. Suppose we change p and q so that the magnitudes of p and q remain 3 and 5 respectively but the angle between p and q is doubled. What is the scalar product of p and q? [2pt] c. Suppose p and q are changed as described in part b above. What is the magnitude of the cross product of p and q? [2pt] Answer: Let denote the angle between p and q. Then the magnitude of the scalar product is |15cos()| while the magnitude of the cross product is |15sin()|. We are given that these are equal. Therefore, |cos()|=|sin()|. Using elementary trigonometry, it must be the case that |cos()|=|sin()| = 1/ 2 . Therefore, possible values of are /4, 3/4, ‐ /4, ‐3/4. a. Scalar product of p and q is given by 15cos(). Possible values are 10.61 . b. Now, we are asked to double the angle while keeping magnitudes the same. Therefore, possible values of the angle are /2, 3/2, ‐ /2, ‐3/2 or just two values /2, ‐ /2. Scalar product of p and q then becomes 15cos(/2)=0=15cos(/2). Thus, the answer is zero. c. Magnitude of the cross product is |15sin(/2)|=15 in both cases. 3. Consider vectors x and y in 6 dimensions given by x [3 0 4 1 2 3] y [1 2 0 1 3 1] a. Find the component of vector x along vector y. [3pts] b. Define new vectors z=x+y and w=x‐y. Find the scalar product of z and w. [2pt] Answer: a. The formula for component of a vector x along vector y is (x.y)y/||y||. So what we need to do is to calculate x.y and ||y||. We get x.y=‐13 and ||y||=4. Therefore, the component of x along y is 13 / 4 13 / 2 0 13 / 4 39 / 4 13 / 4 For part b, note that z.w=(x+y).(x‐y)=x.x‐y.y=39‐16=23. 4. Let x and y be two vectors. Define the vector z = x + y where is a real (scalar) number. Find the value of that minimizes the norm of z. [3pts] Answer: This was one of the level C problems. Recall that to calculate norm of z, we must first calculate z.z. So, we start from here. z.z ( x y ).( x y ) x.x 2 x. y 2 y Now, we must take square root of z.z and find that minimizes it. This looks complicated since square root is not going to be easy. Now an easy but key observation is that the that maximizes square root of z.z will also maximize z.z. So, we can try to minimize z.z. This is easy! It is a quadratic function of . We can take derivative with respect of and set it to zero: d ( z.z ) 2 x. y 2 y. y 0 d If y is zero, then z=x and all values of lead to the same value for norm of z, i.e., ||x||. On the other hand, if y is non‐zero, we can divide by y.y and solve for : x. y y. y It will be instructive and fun for you to understand this answer geometrically by drawing vectors x, y, and z in the plane and interpreting norm as the length of the vector. Notice the close connection to component of x along y idea. 5. Find the partial fraction expansion of G ( s) ( s 2) [4pts] s ( s 1)( s 3) Answer: We start by writing the general form of the partial fraction expansion: A B C s s 1 s 3 A sG ( s ) |s 0 2 / 3 G ( s) B ( s 1)G ( s ) |s 1 1/ 2 C ( s 3)G ( s ) |s 3 1/ 6 G ( s) 2 1 1 3s 2( s 1) 6( s 3) 6. Consider two complex numbers w a jb z re j a. Suppose a=1, b=2. It is known that z is complex conjugate of w. Find r and . [3pts] b. Now suppose a, b, r and are unknown real numbers. Suppose we are told that zw is purely imaginary and z2w is purely real. Find value(s) of . [3pts] Answer: For 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) A3 An 1 An A1 A2 ... 2 3 n 1 ( s 1) ( s 1) ( s 1) ( s 1) ( s 1) n A ( s 1) n 1 A2 ( s 1) n 2 A3 ( s 1) n 3 ... An 1 ( s 1) An 1 ( s 1) n Now notice that the numerator must equal (s2+s+2). Therefore, A1=0 since n‐1>2. Now if n‐2>2, then A2=0. We can repeat this argument and conclude that Ai = 0 if i>n‐2. Thus, the only coefficients that are possibly nonzero are An, An‐1 and An‐2. We can now get three equations in three unknowns and get the partial fraction expansion. Note that for n=3, we get back the answer in part a. There is yet another easy way to do this problem and that is by leveraging part a solution. We begin by factoring the denominator: G(s) s2 s 2 1 s2 s 2 . ( s 1) n 3 ( s 1)3 ( s 1) n 3 ( s 1)3 1 1 2 ( s 1) ( s 1) 2 ( s 1)3 1 1 2 n2 n 1 ( s 1) ( s 1) ( s 1) n 1 ( s 1) n 3 ...
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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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