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Unformatted text preview: EEL 3105 Solution Practice Problem Set 1. Consider the 6 dimensional vectors x [2 1 3 1 2 a ] y [b 4 2 2 1 6] (i) For this part, suppose a=2 and b=4. Calculate the scalar product of x and y. x. y x1. y1 x 2. y 2 x3. y 3 x 4. y 4 x5. y5 6 (ii) Suppose a=4. For what value of b will the scalar product be zero? x. y 2(b) 4 6 2 2 4(6) 0 b 5 (iii) Suppose the scalar product of x and y is 10. What can you say about values of a and b? x. y 2(b) 4 6 2 2 a(6) 10 2b 6a 4 b 3a 2 Thus, a and b are any real numbers satisfying the equation b+3a=2. Clearly, there are infinitely many such choices. They lie on the straight line given by b+3a=2. (iv) Now suppose the Euclidean norm of x is 5. Find value(s) of a and b. 2
2
2
2
2
x x12 x2 x3 x4 x5 x6 a 2 19 25 a 6
b 23 6 2. Consider the vectors (i) Find the angle between x and y. First, note that x.y = ‐3,  x  18, y  15 . x 2 1 3 2 y 3 1 2 1 x. y x y cos cos x. y
x. y
1
) cos 1 (
) 1.7544rad cos 1 (
xy
xy
30 Note that the angle must be more than /2 since cosine is negative. (ii) Define a new vector w y a[1 1 1 1] . For what value of a will the scalar product of x and w be zero? w [3 a 1 a 2 a 1 a] x.w x. y a[2 1 3 2] 3 a(4) 0 a (iii) 3 4 Now suppose we change x and y to new vectors p and q such that their magnitude remains the same, i.e., x=p and y=q but the angle between p and q is twice the angle between x and y. Find the scalar product of p and q. p.q  p  q  cos 2  x  y  cos 2 18 15 cos(3.5088) 15.33 3. Let p, q, r be three dimensional vectors. Suppose the vector product of p and q is r, i. e. pq r (i) Suppose p = [1 0 ‐1] and q = [1 1 1]. Find r. j k
i p q p1 p2 p3 i ( p 2 q 3 p 3q 2) j ( p1q 3 p 3q1) k ( p1q 2 p 2 q1) q q q 2
3
1 r p q [1 2 1] (ii) Suppose p = [1 0 ‐1] and r=[1 1 1]. Find q. j k
i p q 1 0 1 i (0 q 2) j (q 3 q1) k (q 2 0) q q q 2
3
1 q 2 1, q 3 q1 1
q3 and q1 are any real numbers such that q3 q1 1 (iii) Suppose p = [1 0 ‐1]. Suppose r = 0, i. e., r is the zero vector. What can you say about q? Since pxq = 0, the magnitude of pxq must be zero. But we know that the magnitude of pxq is pqsin() where is the angle between p and q. So, the angle must be zero or p and q must be parallel. Therefore, q = p where can be any real number. Another way to see this is to use the equation for pxq from part (ii) above. Using this, we can conclude that q2 0, q3 q1 0 Therefore, q =[q1 0 –q1] = q1 [1 0 ‐1], and q1 is completely free. With = q1, we get the same answer. 4. Let y and z be two complex numbers. Let w ye z (i) For this part only, suppose y = 1+j1 and z=1‐j1. Find and express w in Cartesian and exponential forms. w (1 j1)e1 j1 2e j /4 ee j e 2e j ( 0.215) e 2[cos(.215) j sin(.215) 3.756 j 0.819 (ii) Suppose z = jb. Find w/2y. ez
e jb 1
w 2y
2
2
2 (iii) Suppose we are told that w=y. Suppose z = a+jb. Find a and b. w ye a jb w y e a e a 1 a 0 b can be any real number 5. Find partial fraction expansion of G ( s) G ( s) 1 s ( s 1) 2
2 AB
C
D 2 ss
s 1 ( s 1)2 A( s 3 2s 2 1) B ( s 2 2s 1) C ( s 3 s 2 ) Ds 2 1 A C 0, 2 A B C D 0, A 2 B 0, B 1 A 2, B 1, C 2, D 1
G ( s) 2 1
2
1 2 ss
s 1 ( s 1) 2 6. We are given z 1 j1
w 1
3
j
2 2 Now note that the angle of z is /4 while angle of w is arc tangent of 3
= 0.886 rads. Next, we will 2 calculate the angle of y. The key fact is that the angle of a product is sum of angles. Therefore angle of y is given by angle( y ) angle( z m ) angle( wn ) m.angle( z ) n.angle( w) m .886n
4 (i)
(ii) If m=2, n=‐1, we get angle of y = .685 rads. I had a typographical error in defining w which makes parts (ii) and (iii) not so interesting. I meant to give z 1 j1
w 1
3
j
2
2 Now, we can repeat the problem. In this case, angle of w becomes /3. And angle( y ) angle( z m ) angle( wn ) m.angle( z ) n.angle( w) m n 4
3
(i)
(ii) If m=2, n=‐1, then the angle of y is (/2 – /3)=/6. We want m n 0
4
3 m
4 n
3
We can pick any integer values to satisfy the above equality, e.g., m = 4, n=‐3. But there are lots of other choices as well. We can get additional possibilities by noticing that 2k is also equal to zero as far as angle is concerned. Thus, m n 2 k 4
3 3m 4n 24k
(iii) Thus, for each k, we can solve for integer value of m and n to satisfy the above equation. First, notice that since m and n are integers, m/4 + n/3 must be an integer multiple of /12. By choosing n = 1 and m=‐1, we can get the angle of y to be /12. Thus, by selecting m and n, we can get any integer multiple of /12. Therefore, angle of y must be integer multiples of /12, i. e., q/12 where q is any positive or negative integer. ...
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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.
 Fall '10
 boykins

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