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Unformatted text preview: EE261 Raj Bhatnagar Summer 20102011 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 iπ/ 4 and z 2 = 8 e iπ/ 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 = π 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 = ( π 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 iπ/ 4 = 2[cos( π 4 ) + i sin( π 4 )] = √ 2 + i √ 2 z 2 = 8 e iπ/ 3 = 8[cos( π 3 ) + i sin( π 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 iπ/ 4 ) 1 = 1 2 e iπ/ 4 (iii) z 1 z 2 2 = 2 e iπ/ 4 (8 e iπ/ 3 ) 2 = 2 e iπ/ 4 64 e i 2 π/ 3 = 1 32 e i ( π/ 4 2 π/ 3) = 1 32 e i 5 π/ 12 (iv) 3 √ z 2 = ( z 2 ) 1 3 = (8 e iπ/ 3 ) 1 3 = 2 e iπ/ 9 . If we represent the π/ 3 as 7 π/ 3 or 13 π/ 3 (by adding 2 π or 4 π respectively), then we nd two additional cube roots: 2 e i 7 π/ 9 and 2 e i 13 π/ 9 . 1 2. (10 points) Powers of a complex exponential (a) Let ω = e i 2 π/n . Verify that 1 , ω , ω 2 , . . . , ω 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 + ω + ω 2 + ··· + ω n 1 . How are the n th roots of unity located on the unit circle? Solution: (a) Recall that re iθ 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 θ (measured counterclockwise from the real axis). Thus, w = e 2 πi/n is a vector of length 1 and is oriented at an angle of 2 π n . Similarly, w k = e 2 πik/n has length 1 and is oriented at an angle of 2 πk n . Therefore, the roots are distinct and their n th power is 1 because e i 2 πk = 1 for all k ∈ Z . They will be equally spaced 2 π n radians apart around the unit circle....
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This document was uploaded on 07/28/2011.
 Summer '09

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