ee20-hw01-f10-sol

ee20-hw01-f10-sol - EECS 20N Structure and Interpretation...

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EECS 20N: Structure and Interpretation of Signals and Systems Department of Electrical Engineering and Computer Sciences U NIVERSITY OF C ALIFORNIA B ERKELEY Problem Set 1 SOLUTIONS HW 1.1 (a) z 1 = 1 + i 3 = 2 parenleftBig 1 2 + i 3 2 parenrightBig = 2 exp ( i π 3 ) z 2 = exp ( i 2 π 3 ) = cos ( 2 π 3 ) + i sin ( 2 π 3 ) = 1 2 + i 3 2 See Fig. 1 2 1 1/2 z1 z1* 1/z1* 1/z1 z2 = 1/z2* z2* = 1/z2 Real Axis Imaginary Axis -1/2 -1 -2 Figure 1: Problem 1.1a (b) z 1 + 2 z 2 = 1 1 + i ( 3 + 3 ) = i 2 3 z 2 1 + z 2 = ( 2 exp ( i π 3 )) 2 + exp ( i 2 π 3 ) = 4 exp ( i 2 π 3 ) + exp ( i 2 π 3 ) = 5 exp ( i 2 π 3 ) 1 2 z 1 + z * 2 = 1 2 ( 1 + i 3 ) + parenleftBig 1 2 + i 3 2 parenrightBig * = 1 2 + i 3 2 1 2 i 3 2 = 0 (c) | z 1 | = vextendsingle vextendsingle 2 exp ( i π 3 )vextendsingle vextendsingle = 2 | z 2 | = vextendsingle vextendsingle exp ( i 2 π 3 )vextendsingle vextendsingle = 1 | z 1 z 2 | = | z 1 | | z 2 | = 2 | z 1 z * 2 | = | z 1 | | z * 2 | = | z 1 | | z 2 | = 2 | z 1 /z 2 | = | z 1 | / | z 2 | = 2 | z 2 /z 1 | = | z 2 | | z 1 | = 1 / 2 (d) (i) z 2 1 = ( 2 exp ( i π 3 )) 2 = 4 exp ( i 2 π 3 ) (ii) z 3 1 = ( 2 exp ( i π 3 )) 3 = 8 exp ( i 3 π 3 ) = 8 1

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(iii) z 6 1 = ( 2 exp ( i π 3 )) 6 = 64 exp ( i 6 π 3 ) = 64 exp ( i 2 π ) = 64 (iv) z 4 2 = ( exp ( i 2 π 3 )) 4 = exp ( i 8 π 3 ) = exp ( i ( 2 π + 2 π 3 )) = exp ( i 2 π 3 ) = z 2 (e) z 1 / 4 2 = exp ( i ( 2 + 2 π 3 )) 1 / 4 n ∈ { 0 , 1 , 2 , 3 } z 1 / 4 2 = exp ( i ( 2 π 3 )) 1 / 4 , exp ( i ( 2 π + 2 π 3 )) 1 / 4 , exp ( i ( 4 π + 2 π 3 )) 1 / 4 , exp ( i ( 6 π + 2 π 3 )) 1 / 4 z 1 / 4 2 = exp ( i π 6 ) , exp ( i ( π/ 2 + π 6 )) , exp ( i ( π + π 6 )) , exp ( i ( 3 π/ 2 + π 6 )) z 1 / 4 2 = exp ( i π 6 ) , exp ( i ( 2 π 3 )) , exp ( i ( 7 π 6 )) , exp ( i ( 10 π 6 )) See Figure 2. The circles are the roots and the cross is z 2 . 1 Real Axis Imaginary Axis -1 0 Figure 2: Problem 1.1e HW 1.2 (a) Using the given hint we write e i 2 θ = ( e ) 2 , so cos (2 θ ) + i sin (2 θ ) = cos ( θ ) 2 sin ( θ ) 2 + i 2 cos ( θ ) sin ( θ ) The result follows by comparing the real and imaginary parts of this equality: cos (2 θ ) = cos ( θ ) 2 sin ( θ ) 2 and sin (2 θ ) = 2 cos ( θ ) sin ( θ ) . 2
(b) Using the same procedure as in part (a) we obtain cos (3 θ ) + i sin (3 θ ) = ( cos ( θ ) 2 sin ( θ ) 2 + i 2 cos ( θ ) sin ( θ ) ) · (cos ( θ ) + i sin ( θ )) and simplifying we obtain: cos (3 θ ) = cos ( θ ) 3 3 cos ( θ ) sin ( θ ) 2 and sin (3 θ ) = 3 cos ( θ ) 2 sin ( θ ) sin ( θ ) 3 . (c) First we establish the identity n summationdisplay k =1 A k cos ( ωt + φ k ) = A cos ( ωt + φ ) and then we determine A and φ in terms of the known amplitude and phase parameters A k and φ k , k = 1 , . . . , n . Consider n complex numbers written in polar form as follows: A 1 e 1 , . . . , A k e k , . . . , A n e n If we add these complex numbers, we get a complex number whose polar form representation we write as Ae . In other words A 1 e 1 + · · · + A k e k + · · · + A n e n = n summationdisplay k =1 A k e k = Ae . We can multiply both sides by e iωt (which we know is never zero) to get n summationdisplay k =1 A k e i ( ωt + φ k ) = Ae i ( ωt + φ ) . (1) Two complex numbers are equal if, and only if, their respective real and imag- inary parts are equal. So it must be true that n summationdisplay k =1 A k cos ( ωt + φ k ) = A cos ( ωt + φ ) and n summationdisplay k =1 A k sin ( ωt + φ k ) = A sin ( ωt + φ ) , which we’ve obtained by applying Euler’s formula to the complex-valued Equa- tion (1).

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