PS1-solutions

PS1-solutions - EE261 Raj Bhatnagar Summer 2010-2011 EE 261...

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Unformatted text preview: EE261 Raj Bhatnagar Summer 2010-2011 EE 261 The Fourier Transform and its Applications Problem Set 1 Due Wednesday, June 29 1. (10 points) Some practice with complex numbers (a) Express the following numbers in polar form: (i) 2 + i 3 (ii)- 2 + i 1 (iii)- 2- i 3 (iv) 1- i 3 (b) For z 1 = 2 e i/ 4 and z 2 = 8 e i/ 3 , nd (i) 2 z 1- z 2 in Cartesian form. (ii) 1 /z 1 in polar form. (iii) z 1 /z 2 2 in polar form. (iv) 3 z 2 in polar form. Hint: There are three solutions to this part. Solution: (a) (i) | z | = 2 2 + 3 2 = 13 . To nd the angle, we must be a little careful. Notice that in the complex plane, this number is in the rst quadrant. So, z = tan- 1 ( 3 2 ) = 0 . 983rad . Therefore, 2 + i 3 = 13 e i . 983 . (ii) | z | = p (- 2) 2 + 1 2 = 5 . By examining the signs of the real and imaginary parts, we see that this number is in the second quadrant. So, z = - tan- 1 ( 1 2 ) = 2 . 68rad . Therefore,- 2 + i 1 = 5 e i 2 . 68 . (iii) | z | = p (- 2) 2 + (- 3) 2 = 13 . This number is in the third quadrant. So, z =- ( - tan- 1 ( 3 2 ) ) =- 2 . 16rad . Therefore,- 2- i 3 = 13 e- i 2 . 16 . (iv) | z | = p (1 2 + (- 3) 2 = 10 . This number is in the fourth quadrant. So, z = tan- 1 (- 3 1 ) =- 1 . 25rad . Therefore, 1- i 3 = 10 e- i 1 . 25 . (b) (i) z 1 = 2 e i/ 4 = 2[cos( 4 ) + i sin( 4 )] = 2 + i 2 z 2 = 8 e i/ 3 = 8[cos( 3 ) + i sin( 3 )] = 4 + i 4 3 2 z 1- z 2 = 2( 2 + i 2)- (4 + i 4 3) =- 1 . 17- i 4 . 1 (ii) ( z 1 )- 1 = ( 2 e i/ 4 )- 1 = 1 2 e- i/ 4 (iii) z 1 z 2 2 = 2 e i/ 4 (8 e i/ 3 ) 2 = 2 e i/ 4 64 e i 2 / 3 = 1 32 e i ( / 4- 2 / 3) = 1 32 e- i 5 / 12 (iv) 3 z 2 = ( z 2 ) 1 3 = (8 e i/ 3 ) 1 3 = 2 e i/ 9 . If we represent the / 3 as 7 / 3 or 13 / 3 (by adding 2 or 4 respectively), then we nd two additional cube roots: 2 e i 7 / 9 and 2 e i 13 / 9 . 1 2. (10 points) Powers of a complex exponential (a) Let = e i 2 /n . Verify that 1 , , 2 , . . . , n- 1 are n distinct complex numbers whose n th power is 1. (These are the n th roots of unity. Among the many places they come up, we'll use them in de ning the discrete Fourier transform.) (b) Evaluate the sum 1 + + 2 + + n- 1 . How are the n th roots of unity located on the unit circle? Solution: (a) Recall that re i is a phasor. This is nothing more than a vector in the complex plane. The vector has length r and is oriented at an angle (measured counterclockwise from the real axis). Thus, w = e 2 i/n is a vector of length 1 and is oriented at an angle of 2 n . Similarly, w k = e 2 ik/n has length 1 and is oriented at an angle of 2 k n . Therefore, the roots are distinct and their n th power is 1 because e i 2 k = 1 for all k Z . They will be equally spaced 2 n radians apart around the unit circle....
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PS1-solutions - EE261 Raj Bhatnagar Summer 2010-2011 EE 261...

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