SOLUTIONS to ECE 2026 Summer 2017 Problem Set #2

Transcription

1 SOLUTIONS to ECE 06 Summer 07 Problem Set # PROBLEM..* Put each of the following signals into the standard form x( t ) = Acos( t + ). (Standard form means that A 0, 0, and < Use the phasor addition theorem Before proceeding to the solution, let s review: If we combine the amplitude A and phase of a sinusoid Acos( 0 t + ) into its complex amplitude Ae j, then we can write this sinusoid as the real part of something: Acos( 0 t + ) = Re{Ae j e j 0t }. Using this representation, adding two sinusoids with the same frequency yields: A cos( 0 t + ) + A cos( 0 t + ) = Re{A e j e j 0t } + Re{A e j e j 0t } = Re{(A e j + A e j )e j 0t } = Acos( t + ), where the amplitude A and phase of the sum satisfy Ae j = A e j + A e j. (This is the phasor addition theorem.) The approach for each of these problems is thus: () extract the complex amplitude from each sinusoid in the sum; () convert each to rectangular form; (3) add them; and (4) then convert back to polar form to get the amplitude and phase of the sum sinusoid. (a) x ( t ) = cos(3 t ) cos(3 t ). The complex amplitude Ae j of the sum is: e j0.3 e j0.4 =.09e j0.5 x ( t ) = Acos( t + ) =.09cos(3 t 0.5 ). (b) x ( t ) = cos(06 (t 0.000)) 3sin(06 (t 0.000)). = cos(06 t 0.06 ) 3sin(06 t ) = cos(06 t 0.06 ) 3cos(06 t /) = cos(06 t 0.06 ) 3cos(06 t ) = cos(06 t 0.06 ) + 3cos(06 t ) = cos(06 t 0.06 ) + 3cos(06 t ) The complex amplitude Ae j of the sum is: e j e j0.095 = 4.49e j0.0 x ( t ) = Acos( t + ) = 4.49cos(06 t 0.0 ). (c) x 3 ( t ) =.cos( t) +.cos( t + /3) +.cos( t /3) +.sin( t + /3) +.sin( t + /3). =.cos( t) +.cos( t + /3) +.cos( t /3) +.cos( t + /3 /) +.cos( t + /3 /)

2 =.cos( t) +.cos( t + /3) +.cos( t /3) +.cos( t /6) +.cos( t + /6). The complex amplitude Ae j of the sum is:. +.e j / 3 +.e j /3 +.e j /6 +.e j /6 =.( + e j / 3 + e j /3 + e j /6 + e j /6 ) =.( + cos( /3) + cos( /6)) =.( ) =.( + 3 ) = x 3 ( t ) = Acos( t + ) = 4.48cos( t).

3 PROBLEM..* Suppose that adding and subtracting a pair of sinusoids yields the following: (a) Find 0. A cos(5 t + ) + A cos(5 t + ) = cos( 0 t + /4), A cos(5 t + ) A cos(5 t + ) = cos( 0 t /4 ). where the real variables 0, A, A, and satisfy A > 0, A > 0, <, and <. Although these unknowns are not specified explicitly, they can be determined uniquely from the given equations. Adding or subtracting sinusoids at the same frequency will yield another sinusoid at that same frequency, so 0 = 5. (b) Find the amplitudes A and A and the phases and of the two sinusoids. According to the phasor addition theorem, the two given equations imply that the corresponding complex amplitudes satisfy the following two equations: A e j + A e j = e j /4 A e j A e j = e j /4 or equivalently, in terms of X = A e j and X = A e j, X + X = e j /4 X X = e j /4. We have here two equations and two complex-valued unknowns, so we can solve for the complex amplitudes X and X. In particular, adding the two equations yields: X =. (e j /4 + e j /4 ) =. cos( /4) = X = -- = A =, =. Similarly, subtracting the second equation from the first yields: X =. (e j /4 e j /4 ) =. jsin( /4) = j X = j = e j / A =, = 0.5.

4 e j0 = Ae j e j0.4 Ae j = + e j0.4 PROBLEM.3.* Find a positive amplitude A and phase so that the following equation is true: Acos(800 t + ) cos(800 t ) = cos(800 t). Use the phasor method. According to the phasor addition theorem, the complex amplitude of the sum is: =.68e j0. A =.68, = 0..

5 PROBLEM.4.* Consider the signal x( t ) = cos(06 t )cos(000 t). (a) Express x( t ) as the sum of four complex exponentials. (Hint: You can do this because each sinusoid can be expressed as the sum of two complex exponentials, according to Euler.) Decomposing each sinusoid according to Euler yields: x( t ) = cos(06 t )cos(000 t) = -- (e j06 t + e j06 t )-- (e j000 t + e j000 t ) = -- e j406 t + -- e j406 t + -- e 6 t + -- e j6 t (b) Plot the two-sided spectrum for x(t), carefully labeled the frequencies (in either Hz or rad/s) of each spectral line, and also carefully labeling the complex values associated with each spectral line. Each exponential in the answer to part (a) results in its own spectral line:

6 PROBLEM.5.* Consider the waveform x( t ) shown below: x( t ).5 T = t The above signal x( t ) consists of a DC component plus a sinusoidal signal. (The terminology DC component means a component that is constant versus time.) (a) What is the frequency of the DC component? Look at what happens when we take a sinusoid in standard form Acos( 0 t + ) and set its frequency to zero, 0 = 0: We get a constant signal Acos( ), a signal that does not change with time. For this reason, the frequency of a constant signal is zero. The terminology DC component refers to the constant component. So the frequency of the DC component is zero. (b) What is the frequency of the sinusoidal component? From the picture we can see that the period of the sinusoidal component is T = s. Therefore, its frequency is: f = -- = = 00 Hz or equivalently = f = 400 rad/s. T 0.005

7 (c) Write an equation for the signal x( t ). You can determine numerical values for all the amplitudes, frequencies, and phases in your equation by inspection of the above graph. The DC component is the average component, or. The signal ranges from 0.5 to.5, so the amplitude of the sinusoidal component is 0.5. There is a peak at time , so we can write the sinusoidal component as a sinusoid with zero phase that has been delayed by Putting it all together yields: x( t ) = + 0.5cos(400 (t )) = + 0.5cos(400 t.36 ) = + 0.5cos(400 t ) Any of these is correct, but the highlighted solution is in standard form. (d) Plot the two-sided spectrum of the signal x( t ). Show the complex amplitudes for both the positive and negative frequency components contained in x( t ). Using Euler we can rewrite the above equation as follows: x( t ) = + 0.5cos(400 t ) = + 0.5e j(400 t ) + 0.5e j(400 t )t = + 0.5e j0.64 e j400 t + 0.5e j0.64 e j400 t. Each term in this sum yields a line in the spectrum, so the spectrum is: 0.5e j e j

8 PROBLEM.6.* Consider a real signal x( t ) that has the following two-sided spectrum: e j e j 0.5e j / 0.5e j / (a) Write an equation for x( t ) as a sum of cosines. Each line in the spectrum leads to an exponential term, so that the equation is: x( t ) = + e j e j6 t + e j e j6 t + 0.5e j / e j8 t + 0.5e j / e j8 t = = + cos(6 t + ) + cos(8 t /) cos(6 t) + sin(8 t). (b) Plot the two-sided spectrum of the signal z( t ) = x( t ). Observe that delaying an arbitary sinuoid Acos( 0 t + ) by yields Acos( 0 (t ) + ) = Acos( 0 t + 0 ). In other words, delaying a sinusoid by results in a phase shift of 0. z( t )= x(t ) = + cos(6 (t ) + ) + cos(8 (t ) /) = + cos(6 t 6 + ) + cos(8 t 8 /) = + cos(6 t 5 ) + cos(8 t /) = + cos(6 t + ) + cos(8 t /) = + e j e j6 t + e j e j6 t + 0.5e j / e j8 t + 0.5e j / e j8 t. We have exactly the same spectrum as before: e j e j 0.5e j / 0.5e j /

9 (c) Find two distinct and nonzero values for the delay t 0 such that the difference signal x( t ) x(t t 0 ) is a single sinusoid, i.e., such that the difference signal can be written in the standard form x( t ) x(t t 0 ) = Acos( 0 t + ). The difference signal is: y( t ) = x( t ) x(t t 0 ) = 4 + 4cos(6 t + ) + cos(8 t + /) 4 4cos(6 (t t 0 ) + ) cos(8 (t t 0 ) + /) = A cos(6 t + ) + A cos(8 t + ), where we see that the difference operation cancels the constant term, and that (according to the phasor addition theorem) the difference between the two 3-Hz sinusoid must be another 3-Hz sinusoid, whose amplitude and phase we denote by A and, and where similarly the difference between the two 4-Hz sinusoids is another 4-Hz sinusoid, whose amplitude and phase we denote by A and. In order to get a single sinusoid, therefore, we need either A or A to be zero. But according to the phasor addition theorem: A = 4e j 4e j( 6 t ) = 4 e j6 t, and similarly A = 4 e j8 t. We will get A =0 when 6 t = for any, e.g. when t = /3 s. We will get A =0 when 8 t = for any, e.g. when t = /4 s. Intuition The first answer is easily justified in retrospect (no math!) by observing that /3 is simply the period of the 3-Hz sinusoid. Delaying the first sinusoid by one full period will result in the same sinusoid back, which leads to cancellation when subtracting. Similarly, the second answer of /4 is simply the period of the 4-Hz sinusoid, which leads to cancellation of the second sinusoid when cancelling.

10 (The remaining are not starred. They are not to be turned in. They are provided for extra practice.) PROBLEM.7. Find a positive amplitude A and phase so that the following equation is true: Acos(800 t ) cos(800 t + ) = cos(800 t). Use the phasor method. [Hint : Taking the magnitude squared of both sides of the corresponding phasor equation leads to a value for A. Once A is known, the original phasor equation can be solved for. Hint : Another approach is to think geometrically about the figure in the complex z plane given by the set {z : z = + e j, < }.] (An observation: This problem is almost exactly the same as Prob..3*, the only difference is that the positions of 0.4 and were swapped. This minor change changes dramatically the nature of the problem!) The phasor equation that corresponds to the given equation is: Ae j0.4 e j =, e j = Ae j0.4. The problem is to solve this boxed equation for the unknowns A and. There are lots of approaches. At a glance it might look difficult, because at a glance it looks like we must solve for two unknowns from only a single equation. But because the equation is complex, we can break it down into a pair of equations: either real and imaginary, or magnitude and phase. So really there are two real-valued equations, and two unknowns. Following Hint : Taking the magnitude squared of both sides of the boxed equation yields: e j = Ae j0.4 = A + Acos(0.4 ) A = Acos(0.4 ). There are two solutions to this last equation. One of the solutions is the trivial solution A = 0, but this is not allowed because the problem asks for a positive A. The other (nontrivial) solution can be found by dividing both sides by A, yielding: A = cos(0.4 ) = Plugging back into the boxed equation yields: e j = Ae j0.4 = cos(0.4 )e j0.4 = e j0.8 = 0.8

11 Following Hint : The solution can be also approached geometrically. Look at the left-hand side of the boxed equation: As the angle ranges from to, the term e j traces the unit circle, as illustrated below (shown in blue): = ± + Ae j0.4 This line is the range of values that the right-hand side of the boxed equation can take, depending on the value of A. = / This circle is the range of values that the left-hand side of the boxed equation can take, depending on. = = / The right-hand side of the boxed equation is Ae j0.4, which traces a line in the complex plane that starts at (when A = 0) and moves up as A increases, as illustrated by the black line in the figure above. The left-hand side of the boxed equation equals the right-hand side where the black line intersects the blue circle. They intersect in two places: At the point, which corresponds to A = 0 and ; this solution is not allowed because the problem statement requires that A be positive. At the highlighted point ( ). Looking closer at the triangle formed by the highlighted point, the origin, and : A Two of the sides have length (because the circle has radius ). This makes it an isosceles triangle. Moreover, because two angles in an isosceles triangle are equal, and because the sum of all three angles of any triangle add to, it follows that the remaining unknown angle, which is, must be 0.. Solving = 0. for yields = 0.8. Bisecting the triangle yields a right-angle triangle whose base is A/; setting this equal to cos(0.4 ) yields A = cos(0.4 ) = (Same answers as before!)

12 PROBLEM.8. Suppose that the following sum of n sinusoids is identically zero for all time t: P n k = k cos(06 t ) = 0. 8 How many terms are there in the sum? Answer the question by specifying a positive value for the integer n such that the sum is identically zero for all time t. Specify all possible answers, if the answer is not unique. The corresponding phasor equation is: P n k = e jk /8 = 0. Dividing both sides by yields: P n k = ejk /8 = 0. This is the sum of n consecutive of the 6-th roots of unity. There are 6 unique roots, equally spaced on the unit circle in the complex plane. The sum will be zero whenever n is a positive integer multiple of 6, namely n {6, 3, 48,...}. PROBLEM.9. Suppose a real-valued signal x( t ) that has the following two-sided spectrum: + jc Ae j /5 0.8e j Be j / The unknown variables A, B, c,,, and are all real, satisfying A > 0, B > 0, and <. Although these variables are not given explicitly, they can be determined uniquely from the spectrum plot. (a) Find and. = 00 = 300 (b) Find A and. A = 0.8, = /5 (c) Find B and c. B =, c = (d) Write an expression for x( t ) involving only real numbers and cosine functions. x( t ) = +.6cos(00 t /5) + 4cos(300 t /3).