Signal Processing and Linear Systems Solution Ch.13

Chapter 13 13.1-1 (a) hat w +10 vec y + 2y = f x_{1} = y and hence: x_{2} = hat y = dot z 1 Rightarrow dot z 2 = dot y hat x_{1} = x_{2} x_{2} = - 2x_{1} - 10x_{2} + f [ matrix dot x 1 \\ dot x 2 matrix ]= [[0, 1], [- 2, - 10]] * [[x_{1}], [x_{2}]] + [[0], [1]] * f Choose In matrix form we get (b) hat y +2e^ 4 hat y + log(y) = f Let us choose x_{1} = y and hence: hat x_{1} = x_{2} x_{2} = overline y = overline x 1 dot x 2 = - 2e ^ x_{1} * x_{2} -log x 1 +f It is easy to see that this set is nonlinear. (e) tilde y + phi 1 (y) tilde y + phi_{2}(y) * y = f x_{1} = y dot x_{1} = x_{2} dot x_{2} = - phi_{1}(tau_{1}) * x_{2} - phi_{2}(x_{1}) * x_{1} + f and x_{2} = overline y Let Then Also in this case we are dealing with a nonlinear set, since phi_{2}(x_{1}) and phi_{1}(x_{1}) are not constants. 13.2-1 Writing the loop equations we get: = f 3 Figure 513.2-1 177 f = x_{1} + 2i + 3i_{2} and where i 2 = dot f - x_{1} - dot x 2 2 -x 2 i= f - x_{1} - overline x 2 2 Also we have: 1 2 hat x 1 = f - x_{1} - hat x 2 2 -x 1 (1) Therefore dot x 1 = f - x_{1} - dot x 2 -2x 1 =-3x 1 - dot x 2 +f dot x 2 =31 2 =3[ f-x 1 - dot x 2 2 -x 2 ]= 3/2 * f - 3/2 * x_{1} - 3 2 dot x 2 -3x 2 Hence or 5/2 * x_{2} = - 3/2 * x_{1} - 3x_{2} + 3/2 * f dot x_{2} = - 3/5 * x_{1} - 6/5 * x_{2} + 3/5 * f We can also write: (2) Substituting equation (2) in equation (1) we obtain: dot x_{3} = - 3x_{1} + f - [- 3/5 * x_{1} - 6/5 * x_{2} + 3/5 * f] = - 12/5 * x_{1} + 6/5 * x_{2} + 2/5 * f Hence the state equations are: [ matrix dot x 1 \\ dot x 2 matrix ]= [[- 12/3, 6/8], [- 3/5, - 6/8]] * [[x_{1}], [x_{2}]] + [[3/8], [4/5]] * f(t) 3.2-2 In the Ist loop, the current i, can be computed as: 1/3 4 Figure $13.2-2 f= 1 3 i 1 +x 1 Rightarrow i_{1} = 3(f - x_{1}) We also have (using node equstion) 1 2 dot x_{1} = - 2x_{1} - x_{2} - 3x_{1} + 3f = - 5x_{1} - x_{2} + 3f Hence dot x_{1} = - 10x_{1} - 2x_{2} + 6f (1) (2) Writing the equations in the rightmost loop we get: x_{1} =x 2 + dot x 2 and dot x_{2} = x_{1} - x_{2} Hence from (1) and (2) the state equations are found as: [ matrix dot x 1 \\ dot x 2 matrix ]= [[- 10, - 2], [1, - 1]] * [[x_{1}], [x_{2}]] + [[6], [0]] * f y= dot x_{2} = x_{1} - x_{2} or v = [[1, - 1]] * [[x_{1}], [x_{2}]] The output equation is: 178 And finally x(t)=[ matrix x 1 (t)\\ z 2 (t) matrix ]= mathcal L ^ -1 |x(s)]=[ matrix langle6e^ -t -4e^ -2t )u(t)\\ (- 3e ^ (- t) + 4e ^ (-2t)) * u(t) matrix ] 13.3-2 x(s) = Phi(s) * x(0) + Phi(s) * BF(s) = Phi(s)|x(0)+BF(s)] \langlesI - A\rangle = [[s + 5, 6], [- 1, s]] and al-A Phi(s)=[ matrix s 1s+3)(s+2) & - 6/((s + 1)(s + 2)) \\ 1 (s+3)(s+2) & s+3 lfloor s+3)(s+2) matrix ] x(v) = Phi(v) * [x(0) + BF(s)] =[ matrix s [s+39(s+2) & -0 [s+3)(s+2) \\ 1 lceil s+3\ (s+2) & s+b lceil s+3\ (s+2) matrix ][ matrix 5+ 100 s^ 2 -10^ 4 \\ 4 matrix ] =[ matrix -36.02 s+2 + 39.00 s+1 - 19^ -2 +10^ 2 \\ 17.01/(s + 2) - 12.01/(s + 3) = 0/(s ^ 2 + 10 ^ 2) matrix ] -34.02+39.03 0.01 cos 100t 17.01e13.01 hence x(t) = \mathcal{L} ^ - 1 * (x(s)) +5+6 9-5 And hence: = (2) (2)1 13.3-3 x(s) = Phi(s) * [x(0) + BF(s)] +1 (-) = [])-(-)-(++) [+2] 0 1 (I-A)=-1 1 = [[1/(s + 2), 0], [1/([s + 1)(s + 2)), 1/(s + 1)]] x(0) + BF(s) = [[1/s], [- 1]] Also f(t)=u(t)= Rightarrow F(s) = 1/8 Hence: And thus: 0 BF(s) = [[1/s], [0]] x(s)=[ matrix 1 s+2 &0\\ 1 s+1)(s+2) & 1 s+1 matrix ] [[1/s], [- 1]] = [[1/(s(s + 2))], [1/(s(s + 1)(s + 2)) - 1/(s + 1)]] = [[1/(2s) - 1/(5(s + 2))], [1/(2s) - 2/(s + 1) - 1/(2(s + 2))]] And Hence: x(t)= mathcal L ^ -1 langle x(s))=[ matrix x 1 (t)\\ z 2 (t) matrix ]=[ matrix ( 1 2 - 1 2 e^ -2t )u(t)\\ (1/2 - 2e ^ (- t) + 1/2 * e ^ (-2t)) * u(t) matrix ] 187