Linear System Theory and Design Chapter 2

Chapter 2 2.1 (a) Linear (6) and (c) Nentnwar In 17). if we define overline y = y - y_{0} or shift the operciting point to 10. Yol, then the pair is, y) is meas. 2.2 Bevanse gles is not zero, for too, che ideal lowpass filter is not causal and cannot be built in the real world. 2.3 It is limeas, causal bat time varying. 2.4 If causal, then hat F (- ∞, ∞) =Hu (- infty, infty) =Hu (- infty, infty] =H P alpha u (- ∞ * ∞) Thus we have P_{x}*y = P_{w}*H_{4} = P_{w}*H*P_{w} Веспиле (P_{K}*HH)(d) = 0 for epsilon > alpha but (1 + P_{w}*u)(e) can be nonzero for ττα The Pa Hu & HP4 However, we do have (P_{alpha}*Hu)(t) = (H*P_{alpha}*u)(t) for tsa 2.5 Superposition property must hold for the input and the intial state, If 210) 70. the inttial states in (1) and (3) do not meet x(0) = x(0) - x(0) of chi(P) = x(theta) - x(theta) Thus (1) and (3) are forles In (2), we have x(c) = s(x(c) - x(c)) = x(c) Thus (2) is time. If x(0) = 0 then (x) sim(3) w true. 2.6 If omega_{t} = alpha then alpha^ 2 phi^ 4 (t) alpha u(e^ -i) = kappa * (u ^ 2 * (t))/(u(t - i)) + alpha y(toupenup^ 2 ) d i ^ 2 ( pi) omega j ( pi^ -i) + (omega_{i} ^ 2 * (pi))/(omega_{j}(pi - i)) ne rho u i ( pi)+ phi j ( pi))^ 2 omega j ( pi^ -i )+ phi k ( pi^ -i) Thus additurly dos not hold. 2,7 If uddeliwiły, zhon H( l,x 0 )= H(u + D, x_{0} + 0) = H(u, x_{0}) + H(0, 0) and H(0, 0) = 0 Les in be a positive in tiger then H(w,nx 0 )= H(w, x_{0}) + H(w, z_{0}) +...+H(w,z 0 ) = kappa*H (w,x * ) 0=A( pi*a + (- lambda * 4) , lambda x 0 +(-n+7 0 )) = H(nw, n*x_{e}) + H(- n * w_{e} - n*A_{e}*j) H (- nu_{1} - n*z_{4}) = - i * l(n*omega, n*z_{4}) = - n * H(omega, z_{4}) Thus additurly implies H( aleph 4 , chi chi l )= kappa*H(u, chi_{c}) for any positive or negatuts integer n. Let w = n/m where in and on are steges Ler 4mV and x_{0} = m*K_{0} . Than H(u, x_{c}) = mH(V, X_{0}) which inplies. H(V, X_{r}) = 1/m * H(u, x_{r}) + H(1/m * u, 1/m * t_{r}) Consider H(alpha*u_{1}*alpha*x_{1}) + H(eta/m * u, n/m * x_{0}) = nH(2/m * u, 1/m * x_{1}) = 1/m * H(u, x_{1}) = xH(u, x_{1}) Thus additurly implies H(a / 4, a*x_{d}) = xH / 4, x_{c} ) for any rational number a. 28 der x = t - t and gamma = t - tau Then t= (x - y) / 2 tau = (x - y) / z Consider (2g(t, tau))/(2x) = (2g((x + y)/3, (x - y)/2))/(2x) =lim Delta x 0 (f((x + Delta*x - y)/2, (x + Delta*x - y)/2) - f((x + y)/2, (x - y)/2))/(Delta*x) forall lim Delta x 0 9 2 + Delta x 2 , 2-y 2 + Delta x 2 ) - g * ((x + y)/3, (xy)/2) Delta x =lim Delta x 0 (g((x + y)/2, (x - y)/2) - g((x + y)/2, (x - y)/2))/(Delta*x) =0 Thus g(t, c) depends only on t-2. This can also be reem by setting. then g( pm,z)=g( pm - z ,0) alpha' = - tau 2.9 (i) t < 0 y(x) = 0 (ⅱ) 0 <= t <= 1 9ct - pi*y L 2 (4) 7 (t - epsilon) y(t) = (t ^ 2) / 2 1 <= t <= 2 (1) y(pi) = 1/2 - (z - pi)/2 ^ 2 + 1/2 - 2 * (k - 1)/2 ^ 2 =( - ((z - t) ^ 2)/2 - (*-1) ^ 2 j(4/3) * 2/d = 0.667 7 (1, 5) = 5/8 = c, epsilon = 5 (2 - r)! J_{4} ^ (t) (2)=0 2.10 Taking the Caplace transform and accumary jess untial conditions yield (x ^ 2 + zx - y) hat y (x)=(s-1) hat x (y) hat hat g (s)= hat g (s) hat mu (s) = (s - 1)/((s + 1)(s - i)) = 1/(5 + 1) Impulse repenss hat g (t)+ hat L ^ -1 [ hat f (t)]= e ^ (- j * t) 2.11 Let gees be the impulse reероне. Thea overline g(z) = integrate g(z) * 4(t - tau) dz from 0 to z = integrate g(z) dz from 0 to z Therefore we have y(t) = d dr tilde y (t) 2.12 [ matrix mathcal D jj (x)& mathcal D jk (x)\\ mathcal B 2j (x)& mathcal D 2k (x) matrix ][ matrix hat tilde g j (x)\\ hat tilde g j ^ (x) matrix ]=[ matrix mathcal N j (x)&N jk (x)\\ mathcal g jk omega N jk (x) matrix ][ matrix hat g jk (x)\\ hat g jk (x) matrix ] hat G (z)=[ matrix mathcal D ij (x)& partial ik (z)\\ mathcal D 2i ( omega)& partial 2i (z) matrix ]^ -1 [ matrix mathcal N n (z)&N ik (z)\\ N 2i (z)&N ik (z) matrix ] 2.13 4/4 * (t) = w(t - 1) r(z) = 1 for #00 alpha * 1 1 / u * (t) * r(t) + y(t) = 1u(t - i) y(e + 1) = 1 + y(e) Because 41030 for 0 = 0 < 1 iden y(t) = 1 for 1 <= E < 2 y(t) = 2 for 2 <= t < 3 Q = e : r(d)=0.5( r(t) + y(e ^ 1) ہ y(x + 1) =0.5(1+y(*)) Because g(sigma) = 0 t_{r} * 0 <= t <= 1 . eden y(t) = 0.5 for 1st&2 g(t) = 0, 5(1 + 0, 5) = 0, 75 * (1 + 0, 5) ^ (0, 2 <= 0 < 1) y(t) = 0.5(x + 0.75) = x * 0.75 * 0.75