HW01f09_soln - ECE-2025 Homework #1 Solutions Problem 1.1:...

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ECE-2025 Homework #1 Solutions Fall-2009 Problem 1.1: Approach: For each part, the rectangular-to-polar conversion formulas discussed in class were used to make the conversion. Details: (a) A negative real number has a phase (or angle) of π . The magnitude is obviously 2 π . (b, d, f) Convert to polar first, and then raise to the power by multiplying the angle and raising the angle to the power. (b) 33 j 33 = 33 e j 5 π / 4 33 e j 3 / 4 ( ) 2 = 33 2 (2) e j 3 / 2 = 2178 e j / 2 (d) j 8 ( ) 3 = 8 e j / 2 ( ) 3 = 8 3 e j 3 / 2 = 512 e j / 2 (f) Use the same strategy. 3 + j 4 ( ) 7 = 78125 e j 2.934 (c) 1 j ( ) = 2 e j / 3 (e) 3 j 4 ( ) = 5 e j 0.9273 MATLAB verification is given below: %- Problem 1.1 disp('Problem 1.1') zprint([-2*pi, (-33-33i).^2, 1-j*sqrt(3), (-8i).^3, 3-4i, (-3+4i).^7]) Problem 1.1 Z = X + jY Magnitude Phase Ph/pi Ph(deg) -6.283 0 6.283 3.142 1.000 180.00 0 2178 2178 1.571 0.500 90.00 1 -1.732 2 -1.047 -0.333 -60.00 0 512 512 1.571 0.500 90.00 3 -4 5 -0.927 -0.295 -53.13 -7.644e+004 1.612e+004 7.813e+004 2.934 0.934 168.09 Problem 1.2: Approach: For each part, the rectangular-to-polar conversion formulas discussed in class were used to make the conversion. Details: (a) The angle lies in the 3 rd quadrant, so both the real and imaginary parts will be negative. (b) The sum in the exponent can be expanded to a product of exponentials:
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(c) , so the angle is , which means the number has a real part of zero (i.e., is purely imaginary) and its imaginary part is negative. (d) , so the angle is , which means the number is a negative real number with no imaginary part. %- Problem 1.2 disp('Problem 1.2') zprint([8*exp(-5i*pi/6), exp(pi-5i*pi/4), pi*exp(71i*pi/2), (pi^exp(1))*exp(- 41i*pi)]) Problem 1.2 Z = X + jY Magnitude Phase Ph/pi Ph(deg) -6.928 -4 8 -2.618 -0.833 -150.00 -16.36 16.36 23.14 2.356 0.750 135.00 3.084e-015 -3.142 3.142 -1.571 -0.500 -90.00 -22.46 -3.521e-013 22.46 -3.142 -1.000 -180.00 Problem 1.3: Approach: For each part, the complex number properties discussed in class were used to find the answer. Details: (a) Conjugate means "change the sign of all j's", thus we get which lies in the 2 nd quadrant.
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This note was uploaded on 11/24/2010 for the course ECE 2025 taught by Professor Juang during the Spring '08 term at Georgia Institute of Technology.

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HW01f09_soln - ECE-2025 Homework #1 Solutions Problem 1.1:...

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