EECS20n, Solution to Mock Midterm 2, /7/00. 5 points Write the following in Cartesian coordinates (i.e. in the form x + jy) (a) point j 3 j 2 + j =0 (b) 2 points k=0 e jkπ/6 = ej2π/6 =0 e jπ/6 (c) 2 points( + j)/( j) = j (d) 2 points cos π/4+jsin π/4 = e jπ/4 = ±e jπ/8 = ±(cos π/8+jsin π/8). Write the following in polar coordinates (i.e. in the form re jθ ) (a) 2 points +j = 2e jπ/4 (b) 2 points ( + j)3 = 3 2e jπ/4 (c) 2 points [cos π/4+j sin π/4] /2 = ±.e jπ/8 (d) 2 points ( + j)/( j) =.e jπ/2
2 2. 5 points Which of the following discrete-time or continuous-time signals is periodic. Answer yes or no. If the signal is periodic, give its fundamental period and state the units. Suppose that for a discrete-time signal, n denotes seconds, and for a continuous-time signal, t denotes minutes. (a) 2 points n Ints, x(n) =e j 2n Periodic NO (b) 2 points t Reals, x(t) =e j 2t Periodic YES, Period = 2π/ 2 min. (c) 2 points n Ints, x(n) = cos3πn + sin(3πn + π/7) Periodic YES, Period = 2 sec. (d) 2 points t Reals, x(t) = cos3t + sin 3t Periodic YES, Period = 2π/3 min. (e) 2 points n Ints, x(n) = cos 3πn + sin(3πn + π/7) Periodic YES, Period = 2 sec. 5 points Find A, θ, in the following expression: A cos(t + θ) = cos(2π 0, 000t + π 4 )+sin(2π 0, 000t + π 4 ) So, A = 2, =20, 000π, θ =0. = 2cos(2π 0, 000t)
3 H () H () π/2 2/2 π/4 H 2 () H 2 () 2-/2 π/4 π/2 H 3 () H 3 () 2 π Figure : Plots for Problem 3 3. 5 points On Figure plot the amplitude and phase response of the following frequency responses. On your plots carefully mark the values for =0and for one other non-zero value of. (a) 4 points Reals, H () =+j. So, H () = [ + 2 ] /2, H () = tan. H (0) =, H () = 2, H (0) = 0, H () = π/4. (b) 4 points Reals, H 2 () = +j. So, H 2() =[+ 2 ] /2, H 2 () = tan. H 2 (0) =, H 2 () =/ 2, H 2 (0) = 0, H () = π/4. (c) 4 points Reals, H 3 () =+cos. So H 3 () =+cos, H 3 () =0. H 3 (0) = 2,H 3 (π) =0. 3 points Which of H,H 2,H 3 can be the frequency response of a discrete-time system? H 3. Since it is periodic with period 2π.
4 h 5 5 y 4 3 2 n -3-2 - 0 2 3-3 -2-0 2 3 n Figure 2: Impulse and step response for Problem 4 4. 5 points A discrete-time system H has impulse response h : Ints Reals given by {, n = 2,, 0,, 2 h(n) = 0, otherwise (a) 2 points Sketch h. (b) 5 points What is the step response of H, i.e. the output signal when the input signal is step, where step(n) =,n 0, and step(n) =0,n < 0? You can give your answer as a plot or as an expression. (c) 5 points What is the frequency response of H? (d) 3 points What is the output signal of H for the following input signals? i. n, x(n) =cosn ii. n, x(n) =cos(n + π/6) iii. n, x(n) = sin 00n (a,b)figure 2 shows h and the step response y. An alternative way to calculate y is n Ints, y(n) = (c) The frequency response is m= h(m)x(n m) = x(n +2)+x(n +)+x(n)+x(n ) + x(n 2) = 0,n 3;,n = 2; 2,n = ; 3,n =0;4,n =;5,n 2. Ĥ() = = h(n)e jn n= 2 e jn n= 2 = +2cos +2cos2 (d) Use the fact that the response to a signal x(n) =cos(n+ θ) is y(n) = Ĥ() cos(n+ θ + Ĥ()). i. n, y(n) =Ĥ()x(n)=[+2cos+2cos2]cosn ii. n, y(n) =Ĥ()x(n)=[+2cos+2cos2]cos(n + π/6) iii. n, y(n) =Ĥ(00)x(n) = [ + 2 cos 00 + 2 cos 200] sin 00n
5 5. 5 points (a) 5 points Find the frequency response for the LTI systems described by these differential equations (input is x, output is y) i. ẏ(t) 0.5y(t) =x(t) ii. ÿ(t) 0.5ẏ(t)+0.25y(t) =ẋ(t)+x(t) (b) 5 points What is the response of the second system above for the input t, x(t) = e j(00t+π/4)? (c) 5 points Find the frequency response for the LTI systems described by these difference equations (input is x, output is y) i. y(n) 0.5y(n ) = x(n) ii. y(n) y(n ) + 0.25y(n 2) = x(n)+x(n ) (a) Using y(t) =Ĥ()ejt is the response to x(t) =e jt, we get i. j 0.5 ii Ĥ() = j+ 2 0.5j+0.25 = +j 0.25 2 0.5j (b) The response is t, Ĥ(00)e j(00t+π/4). (c) Using y(n) =Ĥ()ejn is the response to x(n) =e jn, we get i. 0.5e j ii. +e j e j +0.25e 2j.
6 x u 0 0 0.25 t 0 n 0 2 3 9 y v 0 t 0 0.5 0.75 0 2 3 6 9 n Figure 3: Periodic signals for Problem 6 6. 5 points Figure 3 plots two continuous-time periodic signals x and y both with period second, and two discrete-time signals u and v both with period 0 samples. The plots are given only for one period. Suppose the exponential Fouriers Series representations of these signals are given as: t Reals, x(t) = = t Reals, y(t) = = n Ints, u(n) = n Ints, v(n) = k= k= 9 k=0 9 k=0 X k e jkxt Y k e jkyt U k e jkun V k e jkvn (a) 3 points Give the values of x =2πrad/sec, y =2πrad/sec, u = π/5 rad/sample, v = π/5 rad/sample. (b) 4 points Calculate the values of the coefficients X 0 =0.25, Y 0 =0.25, U 0 =0.4, V 0 =0.4. These are just the average values of the signal over one period. (c) 2 points Suppose the signals x is measured in Volts. What is the unit of X 0? Volts. (d) 4 points Calculate the values of the coefficients X = 0.25 t=0 e j2πt dt = j 2π Y = 0.75 t=0.5 e j2πt dt = +j U = 0 V = 0 2π 9n=0 x(n)e jπ/5n = 0 6n=3 e jπ/5n = e j3π/5 3n=0 e jπ/5n = e j4π/5 0( e jπ/5 ) e j4π/5 0( e jπ/5 ) (e) 3 points Express y as a delayed version of x and v as a delayed version of u. t, y(t) =x(t 0.5), n, v(n) =x(n 3).
(f) 4 points Express the FS coefficients {Y k } in terms of {X k } and {V k } in terms of {U k }. k, Y k = X k e jkπ, k, V k = U k e jk3π/5. 7
8 + H() H2() x u v y H3() H4() + + H() H2() x u v y H3() Figure 4: Feedback systems for Problem 7 7. 5 points Figure 4 shows two feedback systems. In these figures, H k (),k =, 2, 3 is the frequency response of the three LTI systems. (a) 9 points Calculate the closed-loop frequency response H() of the first feedback system in terms of the H k. Hint: Use the fact that the two systems are the same. By the hint, H 4 = H /( H ). So, H = H 4 H 2 H 4 H 2 H 3 = H H 2 H H H 2 H 3 () (b) 6 points Suppose H k () =/( + j2) for all k =, 2, 3. Calculate H(0), H() and lim H(). We have H k (0) =,H k () = /( + j2), lim H k () =0. Substitution in () gives, H(0) =,H() = (+2j) 2 +2j (+2j) 3 = +2j ( + 2j) 3 ( + 2j) 2 lim H() =0.
9 v 2 y h y v - 0 t - t Figure 5: Impulse response for Problem 8 8. 5 points A continuous-time LTI system has the impulse response {, t < t Reals, h(t) = 0, otherwise (a) 3 points Sketch the impulse response, and mark carefully the relevant points on your plot. This is shown in Figure 5 (b) 3 points Is this system causal? Answer yes or no. NOT CAUSAL. (c) 3 points What is the step response of this system, i.e. the response to step(t) =,t 0 and =0,t <0? The step response is the integral of the impulse response, t 0, t y(t) = h(s)ds = t +, t s= 2, t A sketch of y is in the figure. (d) 3 points What is the ramp response of this system, i.e. the response to ramp(t) =t, t 0, and =0,t <0? The ramp response v is the integral of the step response, t 0, t v(t) = y(s)ds = 2 s= (t +)2, t, 2+2(t ), t (e) 3 points What is the respone of this system to the input signal impulsetrain, where The response is t Reals, impulsetrain(t) = t, w(t) = h impulsetrain(t) = τ= k= δ(t 2k). h(t τ)impulsetrain(τ)dτ
0 = = k= k= τ= h(t 2k) =. h(t τ)δ(τ 2k)