Bộ đáp án sách Signals and Systems của Simon HayKin. Là tài liệu bổ ích cho môn Tín hiệu và Hệ Thống thuộc khoa Điện Điện Tử và Tự động hóa. Là một bộ khó cho các sinh viên trong hệ thống các trường Đại học và Cao Đẳng trong toàn quốc. Hy vọng tài liệu này sẽ giúp ích cho các bạn. Chúc các bạn qua môn thành công
CHAPTER 1.1 to 1.41 - part of text 1.42 (a) Periodic: Fundamental period = 0.5s (b) Nonperiodic (c) Periodic Fundamental period = 3s (d) Periodic Fundamental period = samples (e) Nonperiodic (f) Periodic: Fundamental period = 10 samples (g) Nonperiodic (h) Nonperiodic (i) Periodic: Fundamental period = sample l.43 π y ( t ) = cos 200t + - 6 π = cos 200t + - 6 π = - cos 400t + - 3 (a) DC component = π (b) Sinusoidal component = - cos 400t + - 3 Amplitude = 200 Fundamental frequency = - Hz π 1.44 The RMS value of sinusoidal x(t) is A ⁄ Hence, the average power of x(t) in a 1-ohm resistor is ( A ⁄ ) = A2/2 1.45 Let N denote the fundamental period of x[N] which is defined by 2π N = -Ω The average power of x[n] is therefore N -1 P = ∑ x [ n ] N = -N n=0 N -1 ∑A n=0 N -1 A = -N 1.46 2 2πn cos + φ N ∑ cos n=0 2πn + φ N The energy of the raised cosine pulse is E = π⁄ω ∫–π ⁄ ω 4- ( cos ( ωt ) + ) dt π⁄ω = - ∫ ( cos ( ωt ) + cos ( ωt ) + ) dt π ⁄ ω 1 - cos ( 2ωt ) + - + cos ( ωt ) + 1 dt = - ∫ 2 π = - - = 3π ⁄ 4ω 2 ω 1.47 The signal x(t) is even; its total energy is therefore E = ∫ x ( t ) dt = ∫ ( ) dt + ∫ ( – t ) dt 2 = [ t ] t=0 + – - ( – t ) t=4 26 = + - = -3 1.48 (a) The differentiator output is y ( t ) = –1 for – < t < – for < t < otherwise (b) The energy of y(t) is E = –4 ∫–5 ( ) dt + ∫ ( – ) dt 2 = 1+1 = 1.49 The output of the integrator is t y ( t ) = A ∫ τ dτ = At for ≤ t ≤ T Hence the energy of y(t) is E = 1.50 T ∫0 A T A t dt = 2 (a) x(5t) 1.0 -1 -0.8 (b) 0.8 t 25 t x(0.2t) 1.0 -25 -20 20 1.51 x(10t - 5) 1.0 1.52 0.1 0.5 0.9 1.0 t (a) x(t) -1 t -1 y(t - 1) -1 t -1 x(t)y(t - 1) 1 -1 t -1 1.52 (b) x(t + 1) x(t - 1) 1 -1 t -1 y(-t)y(-t) -2 -1 4 t -1 x(t - 1)y(-t) t -2 -1 -1 1.52 (c) -2 -1 3 4 t -1 -2 -1 t x(t + 1)y(t - 2) -2 -1 t 1.52 (d) x(t) -3 -2 -1 t -1 y(1/2t + 1) -5 -4 -3 -2 -1 t -1.0 x(t - 1)y(-t) t -3 1.52 -2 -1 -1 (e) x(t) -4 -3 -2 -1 t -1 y(2 - t) -4 -3 -2 t -1 x(t)y(2 - t) -1 -1 t 1.52 (f) x(t) -2 -1 t -1 y(t/2 + 1) 1.0 -5 -3 -2 -1 -6 1 t -1.0 x(2t)y(1/2t + 1) +1 -0.5 -1 t -1 1.52 (g) x(4 - t) -7 -6 -5 -4 -3 t -2 -1 y(t) -2 -1 t x(4 - t)y(t) = -3 -2 -1 t 1.53 We may represent x(t) as the superposition of rectangular pulses as follows: g1(t) 1 11 t g2(t) t g3(t) 1 t g4(t) 1 t To generate g1(t) from the prescribed g(t), we let g ( t ) = g ( at – b ) where a and b are to be determined The width of pulse g(t) is 2, whereas the width of pulse g1(t) is We therefore need to expand g(t) by a factor of 2, which, in turn, requires that we choose a = The mid-point of g(t) is at t = 0, whereas the mid-point of g1(t) is at t = Hence, we must choose b to satisfy the condition a(2) – b = or b = 2a = - = 2 Hence, g ( t ) = g - t – 1 2 Proceeding in a similar manner, we find that g ( t ) = g - t – - 3 g3 ( t ) = g ( t – ) g ( t ) = g ( 2t – ) Accordingly, we may express the staircase signal x(t) in terms of the rectangular pulse g(t) as follows: x ( t ) = g - t – 1 + g - t – - + g ( t – ) + g ( 2t – ) 3 2 1.54 (a) x(t) = u(t) - u(t - 2) t (b) x(t) = u(t + 1) - 2u(t) + u(t - 1) -2 -1 t -1 (c) x(t) = -u(t + 3) + 2u(t +1) -2u(t - 1) + u(t - 3) -3 t -1 (d) x(t) = r(t + 1) - r(t) + r(t - 2) -2 -1 t (e) x(t) = r(t + 2) - r(t + 1) - r(t - 1)+ r(t - 2) -3 -2 -1 t 1.55 We may generate x(t) as the superposition of rectangular pulses as follows: g1(t) -4 -2 4 t g2(t) -4 -2 t g3(t) -4 -2 t All three pulses, g1(t), g2(t), and g3(t), are symmetrically positioned around the origin: g1(t) is exactly the same as g(t) g2(t) is an expanded version of g(t) by a factor of 3 g3(t) is an expanded version of g(t) by a factor of Hence, it follows that g1 ( t ) = g ( t ) g ( t ) = g - t 3 g ( t ) = g - t 4 That is, 1 x ( t ) = g ( t ) + g - t + g - t 3 4 1.56 (a) x[2n] o o o -1 n (b) x[3n - 1] o o1 o -1 10 n figure(3); clf; nyquist(num,den) Root Locus Imag Axis −1 −2 −3 Figure −4 −7 −6 −5 −4 −3 −2 −1 Real Axis Bode Diagram Gm = 2.6725 dB (at 1.924 rad/sec), Pm = 15.132 deg (at 1.5257 rad/sec) 20 Magnitude (dB) −20 −40 −60 −80 −100 −135 Phase (deg) −180 −225 −270 −1 10 10 10 10 Figure Frequency (rad/sec) Nyquist Diagram 0.8 0.6 0.4 Imaginary Axis 0.2 −0.2 −0.4 −0.6 Figure −0.8 −2 −1.8 −1.6 −1.4 −1.2 −1 −0.8 −0.6 −0.4 Real Axis 76 −0.2 %Solution to Problem 9.63 figure(1); clf; K = 1; num1 = 0.5*K*[1 2]; den1 = [1 16 48 0]; subplot(3,1,1) rlocus(num1,den1) %Make figure 9.22 K = 1; num2 = K; den2 = [1 0]; subplot(3,1,2) rlocus(num2,den2) %Make figure 9.28 K = 1; num3 = 0.5*K*[1 2]; den3 = [1 -48 0]; subplot(3,1,3) rlocus(num3,den3) figure(2); clf; subplot(2,2,1) nyquist(num1,den1) subplot(2,2,3) nyquist(num2,den2) subplot(2,2,4) nyquist(num3,den3) Imag Axis Root Locus 10 −10 −12 −10 −8 −6 −4 −2 Real Axis Root Locus Imag Axis 0.5 −0.5 −1 −0.9 −0.8 −0.7 −0.6 −0.5 −0.4 −0.3 −0.2 −0.1 Real Axis Imag Axis Root Locus Figure 20 −20 −12 −10 −8 −6 −4 −2 Real Axis 77 Nyquist Diagram Imaginary Axis −1 −2 −1 −0.8 −0.6 −0.4 −0.2 Real Axis Nyquist Diagram Nyquist Diagram 20 Imaginary Axis Imaginary Axis 10 0 −2 −0.8 −0.6 −0.4 −0.2 −1 Real Axis 9.64 Figure −1 −10 −20 −1 −0.8 −0.6 −0.4 −0.2 Real Axis We are given a unity feedback system with loop transfer function K L ( s ) = s(s + 1) K = -2 s +s The closed-loop transfer function of the system is L(s) T ( s ) = -1 + L(s) K = s +s+K (a) With K = 1, T(s) takes the value T ( s ) = -2 s +s+1 In general, the transfer function of a second-order system is described as (1) (2) (3) ωn T ( s ) = 2 s + 2ζ n ω n s + ω n Comparing Eqs (3) and (4): ωn = (4) ζ n = 0.5 The closed-loop poles of the uncompensated system with K = are located at s = – - ± j - 2 78 Figure shows the root locus of the uncompensated system jω closed-loop pole for K=1 s-plane x −1 j√3/2 σ x0 − j√3/2 Figure (b) The feedback system is to be compensated so as to produce a dominant pair of closedloop poles with ωn = and ζ n = 0.5 That is, the damping factor is left unchanged but the natural frequency is doubled This set of specifications is equivalent to pole locations at s = – ± j , as indicated in Fig 2, shown below It is clear from the root locus of Fig that this requirement cannot be satisfied by a change in the gain K of the uncompensated loop transfer function L(s) Rather, we may have to use a phase-lead compensator as indicated in the problem statement Closed-loop pole jω j√3 s-plane x - τ θ2 o 120ο θ1 -1 - ατ x σ x − j√3 Figure We may restate the design requirement: • Design a phase-lead compensated system so that the resulting root locus has a dominant pair of closed-loop poles at s = – ± j as indicated in Fig For this to happen, the angle criterion of the root locus must be satisfied at s = – ± j With the open-loop poles of the uncompensated system at s = and s = -1, we readily see from Fig that the sum of contributions of these two poles to the angle criterion is – 120° – 90° = – 210° We therefore require a phase advance of 30o to satisfy the angle criterion 79 Figure also includes the pole-zero pattern of the phase-lead compensator, where the angles θ1 and θ2 are respectively defined by θ = tan - – 1 ατ – 1 (5) θ = tan -1 - – 1 τ where the parameters α and τ pertain to the compensator To realize a phase advance of 30, we require θ – θ = 30° or θ = 30° + θ – 1 (6) Taking the tangents of both sides of this equation: tan 30° + tan θ tan θ = – tan 30° tan θ - + tan θ = -1 – - tan θ (7) Using Eqs (5) and (6) in (7) and rearranging terms, we find that with α > the time constant τ is constrained by the equation τ – 0.95τ + 0.025 = Solving this equation for τ: τ = 0.923 or τ = 0.027 The solution τ = 0.923 corresponds to a compensator whose transfer function has a zero to the right of the desired dominant poles and a pole on their left On the other hand, for the solution τ = 0.027 both the pole and the zero of the compensator lie to the left of the dominant closed-loop poles, which conforms to the picture portrayed in Fig Se we choose τ = 0.027, as suggested in the problem statement (c) The loop transfer function of the compensated feedback system for α = 10 is L c ( s ) = G c ( s )L ( s ) ατs + K = τs + s ( s + ) 80 K ( 0.27s + ) = -s ( s + ) ( 0.027s + ) Figure 3(a) shows a complete root locus of L(s) An expanded version of the root locus around the origin is shown in Fig 3(b) Using the RLOCFIND command of MATLAB, the point – + j is located on the root locus The value of K corresponding to this closed-loop pole is 3.846 %Solution to Problem 9.64 figure(1); clf; K = 1; t = 0.027; num = K*[10*t 1]; den = [t 1+t 0]; rlocus(num,den) figure(2); clf; rlocus(num,den) axis([-5 -4 4]) K = rlocfind(num,den) Root Locus 40 30 20 Imag Axis 10 −10 Figure −20 −30 −40 −35 −30 −25 −20 −15 −10 −5 Real Axis Root Locus K=3.897 Imag Axis Figure −1 −2 −3 −4 −5 −4 −3 −2 −1 Real Axis 81 9.65 The loop transfer function of the compensated feedback system is L ( s ) = G ( s )H ( s ) 10K ( ατs + ) = - , α