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Write the differential equation for each energy storage element. di2 = v1 dt di4 = v2 dt dv o = i5 dt ⎡i ⎤ ⎢ 2⎥ Therefore, the state vector is x = ⎢ i4 ⎥ ⎢ ⎥ ⎣vo ⎦ Now obtain v1, v2, and i5 in terms of the state variables. First find i1 in terms of the state variables. −v i + i1 + i3 + i5 + vo = 0 But i3 = i1 − i2 and i5 = i3 − i4 . Thus, −v i + i1 + (i1 − i2 ) + (i3 − i4 ) + vo = 0 Making the substitution for i3 yields −v i + i1 + (i1 − i2 ) + ((i1 − i2 ) − i4 ) + vo = 0 Solving for i1 50 Chapter 3: Modeling in the Time Domain i1 = 1 1 1 2 i2 + i4 − vo + vi 3 3 3 3 Thus, 2 1 1 2 v1 = vi − i1 = − i2 − i4 + vo + vi 3 3 3 3 Also, 1 1 1 1 i3 = i1 − i2 = − i2 + i4 − vo + vi 3 3 3 3 and 1 2 1 1 i 5 = i3 − i4 = − i2 − i4 − vo + vi 3 3 3 3 Finally, 1 2 2 1 v 2 = i5 + vo = − i2 − i 4 + vo + vi 3 3 3 3 Using v1, v2, and i5, the state equation is ⎡ 2 1 − ⎢− 3 ⎢ 3 • 1 2 ⎢ x = ⎢− − 3 3 ⎢ 1 2 ⎢− − ⎣ 3 3 ⎡2 ⎤ 1 ⎤ ⎥ ⎢ ⎥ 3 ⎥ ⎢ 3⎥ 2 ⎥ 1 x + ⎢⎢ ⎥⎥ vi ⎥ 3 ⎥ ⎢ 13 ⎥ 1⎥ ⎢ ⎥ − ⎣ 3⎦ 3⎦ y = [0 0 1]x 2.

The equivalent cascade transfer function is as shown below.

An input of Kr1 yields c = 5Kr1+7 ≠ Kc1. Therefore, not homogeneous. The system is not linear. 50. a. Let x = δx+0. Therefore, Solutions to Problems 43 .. δx + 3δx + 2δx =sin (0+ δx) d sinx δx = 0+ cosx ⎮ δx = δx dx ⎮ But, sin (0+ δx) = sin 0 + x =0 .. x =0 .. . Therefore, δx + 3δx + 2δx = δx. C ollecting term s, δx + 3δx + δx = 0 . b. Let x = δx+π. Therefore, .. δx+3δx+2δx =sin (π+δx) But, sin (π+δx) = sin π + .. d sinx ⎮ δx = 0+cosx ⎮ δx = −δx dx x=π x=π . . Therefore, δx+3δx+2δx = -δx. Collecting terms, δx+3δx+3δx = 0 .

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