Rutgers UniversityCSE 345ch06-9244-vectorist

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Solutions to Additional Problems 6.26. A signal x(t) has Laplace transform X(s) as given below. Plot the poles and zeros in the s-plane and determine the Fourier transform of x(t) without inverting ... X(s). (a) X(s) = s2s+5 2+1 s+6 X(s) = (s + j)(s − j) (s +3)(s +2) zeros at: ±j poles at: −3, −2 X(jω) = X(s)|s=jω = −ω2 + 1 −ω2 + 5jω + 6 Pole−Zero Map Real Axis Imag Axis −3.5 −3 −2.5 −2 −1.5 −1 −0.5 0 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 Figure P6.26. (a) Pole-Zero Plot of X(s) (b) X(s) = s2s+2−s+1 1 X(s) = (s +1)(s − 1) (s + 0.5 − j3 4)(s + 0.5 + j3 4) zeros at: ±1 1poles at: −1 ± j√3 2 X(jω) = X(s)|s=jω = −ω2 − 1 −ω2 + jω + 1 Pole−Zero Map Real Axis Imag Axis −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 Figure P6.26. (b) Pole-Zero Plot of X(s) (c) X(s) = s−14 + s−22 X(s) = 3(s − 10 3 ) (s − 4)(s − 2) zero at: 10 3 poles at: 4, 2 X(jω) = X(s)|s=jω = 1 jω − 4 + 2 jω − 2 2Pole−Zero Map Real Axis Imag Axis 0 0.5 1 1.5 2 2.5 3 3.5 4 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 Figure P6.26. (b) Pole-Zero Plot of X(s) 6.27. Determine the bilateral Laplace transform and ROC for the following signals: (a) x(t) = e−tu(t +2) X(s) = −∞ ∞ x(t)e−st dt = −∞ ∞ e−tu(t +2)e−st dt = −∞2 e−t(1+s) dt = e2(1+s) 1 + s ROC: Re(s) > -1 (b) x(t) = u(−t +3) X(s) = −∞ 3 e−st dt = −e−3s s ROC: Re(s) < 0 3(c) x(t) = δ(t +1) X(s) = −∞ ∞ δ(t +1)e−st dt = es ROC: all s (d) x(t) = sin(t)u(t) X(s) = 0∞ 21j ejt − e−jt e−st dt = 0∞ 21j et(j−s) dt − 0∞ 21j e−t(j+s) dt = 12j j−−1s − j +1 s = 1 (1 + s2) ROC: Re(s) > 0 6.28. Determine the unilateral Laplace transform of the following signals using the defining equation: (a) x(t) = u(t − 2) X(s) = 0−∞ x(t)e−st dt = 0−∞ u(t − 2)e−st dt = 2∞ e−st dt = e−2s s (b) x(t) = u(t +2) X(s) = 0−∞ u(t +2)e−st dt = 0−∞ e−st dt = 1 s (c) x(t) = e−2tu(t +1) 4X(s) = 0−∞ e−2tu(t +1)e−st dt = 0−∞ e−t(s+2) dt = 1 s + 2 (d) x(t) = e2tu(−t +2) X(s) = 0−∞ e2tu(−t +2)e−st dt = 0−2 et(2−s) dt = e2(2−s) − 1 2 − s (e) x(t) = sin(ωot) X(s) = 0−∞ 21j ejωot − e−jωot e−st dt = 12j 0−∞ et(jωo−s) dt − 0−∞ e−t(jωo+s) dt = 12j jω−o 1− s − jωo1+ s = ωo s2 + ω2 o (f) x(t) = u(t) − u(t − 2) X(s) = 0−2 e−st dt = 1 − e−2s s (g) x(t) = sin(πt), 0 < t < 1 0, otherwise X(s) = 0−1 21j ejπt − e−jπt e−st dt = π(1 + e−s) s2 + π2 56.29. Use the basic Laplace transforms and the Laplace transform properties given in Tables D.1 and D.2 to determine the unilateral Laplace transform of the following signals: (a) x(t) = dt d {te−tu(t)} a(t) = te−tu(t) ←−−−→ Lu A(s) = 1 (s +1)2 x(t) = d dta(t) Lu ←−−−→ X(s) = s (s +1)2 (b) x(t) = tu(t) ∗ cos(2πt)u(t) a(t) = tu(t) Lu ←−−−→ A(s) = 1 s2 b(t) = cos(2πt)u(t) Lu ←−−−→ s s2 + 4π2 x(t) = a(t) ∗ b(t) Lu ←−−−→ X(s) = A(s)B(s) X(s) = 1 s2(s2 + 4π2) (c) x(t) = t3u(t) a(t) = tu(t) Lu ←−−−→ A(s) = 1 s2 b(t) = −ta(t) Lu ←−−−→ B(s) = d dsA(s) = −s32 x(t) = −tb(t) Lu ←−−−→ X(s) = d dsB(s) = s64 (d) x(t) = u(t − 1) ∗ e−2tu(t − 1) a(t) = u(t) Lu ←−−−→ A(s) = 1 s b(t) = a(t − 1) Lu ←−−−→ B(s) = e−s s c(t) = e−2tu(t) ←−−−→ Lu C(s) = 1 s + 2 d(t) = e−2c(t − 1) ←−−−→ Lu D(s) = e−(s+2) s + 2 x(t) = b(t) ∗ d(t) Lu ←−−−→ X(s) = B(s)D(s) X(s) = e−2(s+1) s(s +2) 6(e) x(t) = 0t e−3τ cos(2τ)dτ a(t) = e−3t cos(2t)u(t) ←−−−→ Lu A(s) = s + 3 (s +3)2 + 4 −∞ t a(τ)dτ ←−−−→ Lu 1s −∞ 0− a(τ)dτ + A(ss) X(s) = s + 3 s((s +3)2 +4) (f) x(t) = t dt d (e−t cos(t)u(t)) a(t) = e−t cos(t)u(t) ←−−−→ Lu A(s) = s + 1 (s +1)2 + 1 b(t) = d dta(t) Lu ←−−−→ B(s) = s(s +1) (s +1)2 + 1 x(t) = tb(t) Lu ←−−−→ X(s) = − d dsB(s) X(s) = −s2 − 4s − 2 (s2 + 2s +2)2 6.30. Use the basic Laplace transforms and the Laplace transform properties given in Tables D.1 and D.2 to determine the time signals corresponding to the following unilateral Laplace transforms: (a) X(s) = s+2 1 s+3 1 X(s) = 1 s + 2 + −1 s + 3 x(t) = e−2t − e−3t u(t) (b) X(s) = e−2s ds d (s+1) 1 2 A(s) = 1 (s +1)2 Lu ←−−−→ a(t) = te−tu(t) B(s) = d dsA(s) Lu ←−−−→ b(t) = −ta(t) = −t2e−tu(t) X(s) = e−2sB(s) ←−−−→ Lu x(t) = b(t − 2) = −(t − 2)2e−(t−2)u(t − 2) (c) X(s) = (2s+1) 1 2+4 B(s a ) Lu ←−−−→ ab(at) 1 (s +1)2 + 4 Lu ←−−−→ 12 e−t sin(2t)u(t) x(t) = 1 4 e−0.5t sin(t)u(t) 7(d) X(s) = s ds d22 s21+9 + s+3 1 A(s) = 1 s2 + 9 Lu ←−−−→ a(t) = 1 3 sin(3t)u(t) B(s) = d dsA(s) Lu ←−−−→ b(t) = −ta(t) = − t 3 sin(3t)u(t) C(s) = d dsB(s) Lu ←−−−→ c(t) = −tb(t) = t2 3 sin(3t)u(t) D(s) = sC(s) Lu ←−−−→ d(t) = d dtc(t) − c(0−) = 23t sin(3t)u(t) + t2 cos(3t)u(t) E(s) = 1 s + 3 Lu ←−−−→ e(t) = e−3tu(t) x(t) = e(t) + d(t) = e−3t + 23t sin(3t) + t2 cos(3t) u(t) 6.31. Given the transform pair cos(2t)u(t) Lu ←−−−→ X(s), determine the time signals corresponding to the following Laplace transforms: (a) (s +1)X(s) sX(s) + X(s) Lu ←−−−→ d dtx(t) + x(t) = [−2 sin(2t) +cos(2t)] u(t) (b) X(3s) X(s a ) Lu ←−−−→ ax(at) x(t) = 1 3 cos(2 3 t)u(t) (c) X(s +2) X(s +2) Lu ←−−−→ e−2tx(t) x(t) = e−2t cos(2t)u(t) (d) s−2X(s) B(s) = 1 s X(s) Lu ←−−−→ −∞ t x(τ)dτ Lu ←−−−→ −∞ t cos(2τ)u(τ)dτ Lu ←−−−→ 0t cos(2τ)dτ 8B(s) Lu ←−−−→ 12 sin(2t) 1 s B(s) Lu ←−−−→ 0t 1 2 sin(2τ)dτ Lu ←−−−→ 1 − cos(2t) 4 u(t) (e) ds d e−3sX(s) A(s) = e−3sX(s) ←−−−→ Lu a(t) = x(t − 3) = cos(2(t − 3))u(t − 3) B(s) = d dsA(s) Lu ←−−−→ b(t) = −ta(t) = −t cos(2(t − 3))u(t − 3) 6.32. Given the transform pair x(t) Lu ←−−−→ 2s s2+2, where x(t) = 0 for t < 0, determine the Laplace transform of the following time signals: (a) x(3t) x(3t) Lu ←−−−→ 13 X(s 3 ) X(s) = 23s (3s)2 + 2 = 6s s2 +18 (b) x(t − 2) x(t − 2) Lu ←−−−→ e−2sX(s) = e−2s 2s s2 + 2 (c) x(t) ∗ dt d x(t) b(t) = d dtx(t) Lu ←−−−→ B(s) = sX(s) y(t) = x(t) ∗ b(t) Lu ←−−−→ Y (s) = B(s)X(s) = s[X(s)]2 Y (s) = s s22+ 2 s 2 (d) e−tx(t) e−tx(t) ←−−−→ Lu X(s +1) = 2(s +1) (s +1)2 + 2 9(e) 2tx(t) 2tx(t) Lu ←−−−→ −2 d ds X(s) = (4s2s2+2) − 82 (f) 0t x(3τ)dτ 0t x(3τ)dτ ←−−−→ Lu Y (s) = X3(s3s) Y (s) = 2 s2 +18 6.33. Use the s-domain shift property and the transform pair e−atu(t) ←−−−→ Lu s+1a to derive the unilateral Laplace transform of x(t) = e−at cos(ω1t)u(t). e−atu(t) ←−−−→ Lu 1 s + a x(t) = e−at cos(ω1t)u(t) = 12 e−at ejω1t + e−jω1t u(t) Using the s-domain shift property: X(s) = 1 2 (s − jω11) + a + (s + jω11) + a = 12 2(s + a) (s + a)2 + ω12 = (s + a) (s + a)2 + ω12 6.34. Prove the following properties of the unilateral Laplace transform: (a) Linearity z(t) = ax(t) + by(t) Z(s) = 0∞ z(t)e−st dt = 0∞ (ax(t) + by(t)) e−st dt = 0∞ ax(t)e−st dt + 0∞ by(t)e−st dt = a 0∞ x(t)e−st dt + b 0∞ y(t)e−st dt = aX(s) + bY (s) (b) Scaling z(t) = x(at) 10Z(s) = 0∞ x(at)e−st dt = 1 a 0∞ x(τ)e− as τ dτ = 1 a X(s a ) (c) Time shift z(t) = x(t − τ) Z(s) = 0∞ x(t − τ)e−st dt Let m = t − τ Z(s) = −∞τ x(m)e−s(m+τ) dm If x(t − τ)u(t) = x(t − τ)u(t − τ) Z(s) = 0∞ x(m)e−sme−sτ dm = e−sτX(s) (d) s-domain shift z(t) = esotx(t) Z(s) = 0∞ esotx(t)e−st dt = 0∞ x(t)e−(so−s)t dt = X(s − so) (e) Convolution z(t) = x(t) ∗ y(t) = 0∞ x(τ)y(t − τ) dτ; x(t), y(t) causal Z(s) = 0∞ 0∞ x(τ)y(t − τ) dτ e−st dt = 0∞ 0∞ x(τ)y(m) dτ e−sme−sτ dm = 0∞ x(τ)e−sτ dτ 0∞ y(m)e−sm dm = X(s)Y (s) 11(f) Differentiation in the s-domain z(t) = −tx(t) Z(s) = 0∞ −tx(t)e−st dt = 0∞ x(t)ds d (e−st) dt = 0∞ ds d x(t)e−st dt Assume: 0∞(.) dt and ds d (.) are interchangeable. Z(s) = d ds 0∞ x(t)e−st dt Z(s) = d ds X(s) 6.35. Determine the initial value x(0+) given the following Laplace transforms X(s): (a) X(s) = s2+51s−2 x(0+) = lim s→∞ sX(s) = s s2 + 5s − 2 = 0 (b) X(s) = s2+2 s+2 s−3 x(0+) = lim s→∞ sX(s) = s2 + 2s s2 + 2s − 3 = 1 (c) X(s) = e−2s s26+2 s2+s−s2 x(0+) = lim s→∞ sX(s) = e−2s 6s3 + s2 s2 + 2s − 2 = 0 6.36. Determine the final value x(∞) given the following Laplace transforms X(s): (a) X(s) = s22+5 s2+3 s+1 x(∞) = lim s→0 sX(s) = 2s3 + 3s s2 + 5s + 1 = 0 (b) X(s) = s3+2 s+2 s2+s x(∞) = lim s→0 sX(s) = s + 2 s2 + 2s + 1 = 2 12(c) X(s) = e−3s s2(ss+2) 2+12 x(∞) = lim s→0 sX(s) = e−3s 2s2 + 1 (s +2)2 = 14 6.37. Use the method of partial fractions to find the time signals corresponding to the following unilateral Laplace transforms: (a) X(s) = s2+3 s+3 s+2 X(s) = s + 3 s2 + 3s + 2 = A s + 1 + B s + 2 1 = A + B 3 = 2A + B X(s) = 2 s + 1 + −1 s + 2 x(t) = 2e−t − e−2t u(t) (b) X(s) = 2ss22+10 +5ss+6 +11 X(s) = 2s2 +10s +11 s2 + 5s + 6 = 2 − 1 (s +2)(s +3) 1 (s +2)(s +3) = A s + 2 + B s + 3 0 = A + B 1 = 3A + 2B X(s) = 2 − 1 s + 2 + 1 s + 3 x(t) = 2δ(t) + e−3t − e−2t u(t) (c) X(s) = s22+2 s−s1+1 X(s) = 2s − 1 s2 + 2s + 1 = A s + 1 + B (s +1)2 2 = A −1 = A + B x(t) = 2e−t − 3te−t u(t) (d) X(s) = s3+3 5ss+4 2+2s X(s) = 5s + 4 s3 + 3s2 + 2s = A s + B s + 2 + C s + 1 130 = A + B + C 5 = 3A + B + 2C 4 = 2A X(s) = 2 s + −3 s + 2 + 1 s + 1 x(t) = 2 − 3e−2t + e−t u(t) (e) X(s) = (s+2)( ss22−+2 3 s+1) X(s) = s2 − 3 (s +2)(s2 + 2s +1) = A s + 2 + B s + 1 + C (s +1)2 1 = A + B 0 = 2A + 3B + C −3 = A + 2B + 2C X(s) = 1 s + 2 + −2 (s +1)2 x(t) = e−2t − 2te−t u(t) (f) X(s) = s2+2 3s+2 s+10 X(s) = 3s + 2 s2 + 2s +10 = 3(s +1) − 1 (s +1)2 + 32 x(t) = 3e−t cos(3t) − 13e−t sin(3t) u(t) (g) X(s) = (s+2)( 4s2+8 s2+2 s+10 s+5) X(s) = 2 s + 2 + 2(s +1) (s +1)2 + 22 + −2 (s +1)2 + 22 x(t) = 2e−2t + 2e−t cos(2t) − e−t sin(2t) u(t) (h) X(s) = (s+2)( 3s2+10 s2+6 s+10 s+10) X(s) = 3s2 +10s +10 (s +2)(s2 + 6s +10) = A s + 2 + Bs + C s2 + 6s +10 3 = A + B 10 = 6A + 2B + C 10 = 10A + 2C X(s) = 1 s + 2 + 2(s +3) (s +3)2 + 1 − 6 (s +3)2 + 1 x(t) = e−2t + 2e−3t cos(t) − 6e−3t sin(t) u(t) 14(i) X(s) = 2s2+11 s2+5 s+16+ s+6e−2s X(s) = 2s2 +11s +16 + e−2s s2 + 5s + 6 = 2 + s + 4 s2 + 5s + 6 + e−2s s2 + 5s + 6 X(s) = 2 + −1 s + 3 + 2 s + 2 − e−2s 1 s + 3 + e−2s 1 s + 2 x(t) = 2δ(t) + 2e−2t − e−t u(t) + e−2(t−2) − e−3(t−2) u(t − 2) 6.38. Determine the forced and natural responses for the LTI systems described by the following differential equations with the specified input and initial conditions: (a) dt d y(t) + 10y(t) = 10x(t), y(0−) = 1, x(t) = u(t) X(s) = 1 s Y (s)(s + 10) = 10X(s) + y(0−) Y f(s) = 10X(s) s +10 = 10 s(s +10) = 1 s + −1 s +10 yf(t) = 1 − e−10t u(t) Y n(s) = y(0−) s +10 yn(t) = e−10tu(t) (b) dt d22 y(t) + 5 dt d y(t) + 6y(t) = −4x(t) − 3 dt d x(t), y(0−) = −1, dt d y(t) t=0− = 5, x(t) = e−tu(t) Y (s)(s2 + 5s +6) − 5 + s + 5 = (−4 − 3s) 1 s + 1 Y (s) = −1 (s +1)(s +2)(s +3) + s (s +2)(s +3) = Y f(s) + Y n(s) Y f(s) = −0.5 s + 1 + −2 s + 2 + 2.5 s + 3 yf(t) = −0.5e−t − 2e−2t + 2.5e−3t u(t) Y n(s) = −2 s + 2 + 3 s + 3 yn(t) = −2e−2t + 3e−3t u(t) (c) dt d22 y(t) + y(t) = 8x(t), y(0−) = 0, dt d y(t) t=0− = 2, x(t) = e−tu(t) Y (s)(s2 + 1) = 8X(s) + 2 15Y f(s) = 4 s + 1 + 4 s2 + 1 − 4s s2 + 1 yf(t) = 4 e−t +sin(t) − cos(t) u(t) Y n(s) = 2 s2 + 1 yn(t) = 2 sin(t)u(t) (d) dt d22 y(t) + 2 dt d y(t) + 5y(t) = dt d x(t), y(0−) = 2, dt d y(t) t=0− = 0, x(t) = u(t) Y (s)(s2 + 2s +5) = sX(s) + sy(0−) + 2y(0−) Y f(s) = 1 (s +1)2 + 22 yf(t) = 1 2 e−t sin(2t)u(t) Y n(s) = 2(s +2) (s +2)2 + 1 yn(t) = 2e−t cos(t)u(t) 6.39. Use Laplace transform circuit models to determine the current y(t) in the circuit of Fig. P6.39 assuming normalized values R = 1Ω and L = 1 2 H for the specified inputs. The current through the inductor at time t = 0− is 2 A. [Show More]

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