Electrical Engineering 20N - Fall 1998 - Midterm 1

# Electrical Engineering 20N - Fall 1998 - Midterm 1 - EECS...

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Unformatted text preview: EECS 20. Solutions to Midterm 1. 2 October 1998 1. 15 points (a) Find  so that Re[(1 + i) exp i = 1: Answer Using exp(i) = cos() + i sin(), Re[(1 + i) exp i = Re[cos() sin() + i(cos() + sin()] = 1 so cos() sin() = 1; one solution of which is  = =2. Another solution is  =  . The general solution is =2  2n;   2n . (b) De ne x : Reals ! Reals 8t 2 Reals; x(t) = sin(!0t + 1=4): Find A 2 Comps so that 8t 2 Reals; x(t) = A exp(i!0t) + A exp( i!0t); where A is the complex conjugate of A. Answer Using sin() = 1=2i[exp(i) exp( i)], sin(!0t + 1=4 ) = 1=2i[exp(i(!0t + 1=4 )) exp( i(!0 t + 1=4 )) so A = 1=2i exp(i1=4 ) = 1=2[sin(=4) i cos(=4)]. 1 2. 15 points Draw the following sets (a) f(x; y ) 2 Reals2 j xy = 1g. (b) f(x; y ) 2 Reals2 j y x2  0g. (c) fz 2 Comps j z 5 = 1 + 0ig. Answer In drawing the third set, we use the fact that z5 = 1 = exp i(2n), so z = exp i(2n=5), n = 0; 1; 2; 3; 4. There are no more solutions, since exp i(2  5=5) = expi2 = 1 which we have already. Note A very important result of complex variables is that a polynomial of degree n in a complex variable z has exactly n roots. In the case here check that z5 1 = Y (z 4 n=0 exp i(2n=5)): y y = x2 (1,1) x (2,4) (-1,-1) 5 points with magnitude 1 and angle at multiples of 2/5 pi Figure 1: The three sets 2 3. 25 points (a) Evaluate the truth values of S = [P ^ (:Q)] _ R for the following values of P; Q; R. P Q R Answer True False False False True False True False True P Q R S S True False False True False True False False True False True True (b) The following sequence of statements is a complete context. Let x = 5; y = 6 Then, x 6= y Now let Z = fz 2 Reals j z  x + y g Then Let (1) (2) (3) x2Z (4) w = smallest non-negative number in Z (5) Answer the following: i. Are the two expressions in (1) both assignments or assertions? Ans Both are assignments ii. Is the expression (2) an assertion or a predicate? Ans It is a true assertion iii. Is the equality in (3) an assignment or an assertion? Ans It is an assignment in which Z is assigned the set on the right-hand side. iv. Is the expression \z  x + y " in (3) an assertion or a predicate? Ans It is a predicate which is true if and only if z  11 v. Is (4) an assertion or a predicate? Ans It is a false assertion vi. Is (5) an assignment or an assertion? Ans It is an assignment equivalent to the assignment w = 11. 3 4. 20 points A signal is a function. We have studied signals that are functions of time and space and functions that are data and event sequences. Mathematically, we model a signal as a function with some range and common domain. For example, Sound : Time ! Pressure. Propose mathematical models for the signals with the following intuitive descriptions. Give a very brief justi cation for your proposed models. (a) A gray-scale video with 256 gray-scale values . Ans A video is a sequence of images. Let Images = [HorSpace  V erSpace ! f0;    ; 255g] Then a video is represented by a function V ideo : Time ! Images where Time = f0; 1=30; 2=30;   g. (b) The position of a bird in ight. Ans The bird's position at time t in ight can be represented as a point in three-dimensional space, (x(t); y (t); z (t)), so Position : Time ! Reals3 where Time = [a; b] is the duration of the ight. (c) The buttons you press with your TV remote control. Ans Let Buttons = fpower; play; fwd; rew;   g be the buttons we can press. Then the sequence of button presses can be modeled by a function ButtonPress : Indices ! Buttons where Indices = f1; 2;   g 4 5. 25 points The function x : Reals ! Reals is given by its graph shown in Figure 2. Note that 8t 62 [0; 1]; x(t) = 0, and x(0:4) = 1. De ne y by 1 0 0.4 1 Figure 2: The graph of x 8t 2 Reals; y(t) = 1 X x(t k= 1 kp) where p 2 Reals. (a) Prove that y is periodic with period p, i.e. 8t 2 Reals; y(t) = y(t + p): Ans We must verify this using the de nition of y. Substituting t + p for t we get y(t + p) = = = 1 X x(t + p kp) k 1 1 X x(t + (1 k)p) k 1 1 X x(t mp); by taking m = 1 = = m= 1 k = y (t); by de nition of y (b) Plot y for p = 1. (c) Plot y for p = 2. (d) Plot y for p = 0:5. Ans See Figure 3. Note that the period in the top plot is 1.0, in the middle it is 2.0 and in the lower plot it is 0.5. (e) Suppose the function z is obtained by advancing x by 0.4, i.e. 8t; z(t) = x(t + 0:4): De ne w by 8t 2 Reals; w(t) = 5 1 X z (t k= 1 kp) w 0 y 0.4 1 p=1 p=2 p=0.5 Figure 3: The graphs of y; w What is the relation between w and y . Use this relation to plot w for p = 1. Ans We have w(t) = = 1 X z(t kp) k 1 1 X x(t + 0:4 = k= 1 kp) = y (t + 0:4) So the plot of w is obtained by moving the plot of y to the left by 0.4, as shown in the top panel of Figure 3 in bold. 6 ...
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