Solutions to ECE 2026 Problem Set #1. 2.5e j0.46

Transcription

1 Solutions to ECE 2026 Problem Set #1 PROBLEM 1.1.* Convert the following to polar form: (a) z 3 + 4j e jtan 1 ( 4/3) 5e j j 5e (b) z j j 2e j /6 2.5e j0.46 (c) z e j /3 + e j2 /3 cos( /3) + jsin( /3) + cos(2 /3) + jsin(2 /3) j j j 3 3e j0.5 (d) z (1 + j) 8 ( 2 e j0.25 ) 8 ( 2 ) 8 e j8(0.25 ) 16e j0 16. (e) z jcos( /6) sin( /6) j(cos( /6) + jsin( /6)) j(e j /6 ) (e j /2 )(e j /6 ) e j2 /3 e j0.667 (f) z (1 + j) 2 (1 j) 2 (1 + 2j + j 2 ) (1 2j + j 2 ) (1 + 2j 1) (1 2j 1) 2j ( 2j) 4j 4e j0.5

2 PROBLEM 1.2.* The figure below shows the locations in the complex plane of z re j when raised to the powers of k {0, 1, 2, 3, 4, 5,... 10}. In other words, it show the locations in the complex plane of z 0, z 1, z 2, z 3, z 4, z 5, z 6, z 7, z 8, z 9 and z 10 : z 1 re j z 3 z 5 z 7 z 9 z 10 z 8 z 0 1 z 6 z 4 z 2 r 2 e j2 Although the axes are not labeled, and you are not told which points correspond to which powers of k, you will still be able to determine the value of z re j : (a) Find r (an estimate is OK). Start with z 0 : We know that z 0 1. This is a real number, so it will fall on the real axis, to the right of the origin. The only point that falls on the real axis is the rightmost point above, so that must be z 0 1. Now the sequence of powers z 1, z 2, z 3 etc in polar form can be written as: z 1 re j, z 2 r 2 e j2, z 3 r 3 e j3, z 4 r 4 e j4, z 5 r 5 e j5 Note that the distances from the origin in this sequence are r, r 2, r 3, r 4, r 5. If r > 1, the points would grow further from the origin. But the figure shows that z 0 1 is the furthest from the origin; therefore, it must be that r < 1. Furthermore, the outermost point that isn t z 0 1 must correspond to z 1 re j. (This is the left-most point, highlighed in yellow.) To estimate r we can get out a ruler and measure the distance to the origin of the yellow point. Since it is about 80% of the distance to the origin of the point z 0 1, we conclude that r 0.8. (b) Find. Here we could proceed by estimating the angle of the yellow point, at a glance it looks to be more than 160 (about 0.9 ) and less than 180 ( ). To get the precise angle we can instead observe that the tenth power z 10 r 10 e j 10 (whose angle is 10 ) falls precisely on the imaginary axis, its angle appears to be precisely /2. solving 10 /2 2 m (for any integer m) for yields m {..., 0.05, 0.15, 0.35, 0.55, 0.75, 0.95, 1.15, 1.35,...}. Of these, the one that matches the at-a-glance-estimate of between 0.9 and is: Verify your answer in MATLAB: z 0.8*exp(0.95i*pi); plot(z.^(0:10),'o'); axis image

3 PROBLEM 1.3.* Find all solutions to the equation: z In other words, find the complex numbers z for which the above equation is true. (Hint: Since this is a 8-th order polynomial, there are 8 solutions.) These numbers are the so-called 8-th roots of unity. Plot them in the complex plane. Moving the 1 to the right-hand side, and using the fact that 1 e j2 m for any integer m, yields: z 8 e j2 m, for any integer m. Raising both sides to the power of 1/8 yields: z e j2 m/8, for any integer m. There are only 8 unique values, corresponding to m {0, 1, 2, 3, 4, 5, 6, 7}, namely: z {1, e j2 /8, e j4 /8, e j6 /8, e j8 /8, e j10 /8, e j12 /8, e j14 /8 }. These are eight equally spaced points on the unit circle in the complex plane, as shown below: Im{ z } e j4 /8 e j6 /8 e j2 /8 e j8 /8 1 Re{ z } e j10 /8 e j14 /8 e j12 /8

4 PROBLEM 1.4.* Plot two periods of the following sinusoids over the time-interval 0 t 2T, where T is the period: All of the plots will look the same, the only difference will be the scaling of the axes and the location of time zero. Given a sinusoid in standard form x( t ) Acos( t + ), there are three steps: Determine y-axis scale from A, since the sinusoid will range between ±A. Determine the scale of the x axis from the frequency 2 /T, where T is the period (distance between neighboring peaks). Determine the location of time zero from the phase. Since we know that a sinusoid with zero phase will achieve a peak at time 0, it follows that delaying such a sinusoid by t p will move the peak to time t p : cos( 0 (t t p )) cos( 0 t + ) t p / 0. When the phase is restricted to the standard range to, the corresponding delay t p will identify the location of the peak that is closest to the origin. (a) A sinusoid having a period of 3 secs, an amplitude of 14, and a phase of radians. TWO PERIODS t t p / / (b) x( t ) 1.2cos( -- t + -- ). 5 2 TWO PERIODS t t d / /

5 (c) x( t ) 2 cos(4 (t 0.05)). T 0.5, 4 (0.05) 0.2. TWO PERIODS t t d /

6 PROBLEM 1.5.* The waveform shown below can be represented by x( t ) Re{Xe j 0 t }: x( t ) Time t (seconds) t p (a) Find 0. By measuring the distance between peaks we see that the period is T 0.02 s. Therefore, the radian frequency is 0 2 /T 100 rad/s. Or 50 Hz. (b) Find the complex phasor X, expressed in polar form. The amplitude and phase of the phasor are the same as the amplitude and phase of the sinusoid. The amplitude can be read directly from the graph as A 6. The phase is 0 t p, where t p is the time of the peak nearest to time zero. From the plot the time delay of the peak closest to time zero is about t p , or equivalently 2.5 ms. Plugging this in, we find that 0.25 radians. 6e j0.25 Thus the phasor is X. (As a sanity check, this would imply that the sinusoid at time 0 would be x( 0 ) 6cos(0.25 ) 4.2, which appears to be pretty close to accurate.) (c) The waveform can also be written as x( t ) Acos( 0 t + ). Find the amplitude A and phase. The amplitude A and phase of the sinusoid are related to the complex amplitude X by: X Ae j A 6 and 0.25.