e(t), E(s)E"(s),E(z) Σ r(t), R(s) Y(z) G (s) = G,(s)G(s) c(t), C(s) D(z) To =G,(s)(1-0STO/s b(t), B(s) H(s) For the system above: D(z)=(z/2 - 1)/z, Gr(s)=2/(s+1/2), H(s)=1/4, GHO(s)= (1-esTO/s, To=1/4 31. As in the class notes, for the system above define Gc(s) = G'(s) (1-esTo). Then, G'(s)= %3D a) s2(s+1/2) 2 b): s(s+1/2) d) none above (2s+1) 32. As in the class notes, for the system above, G'(z)= z(z-e-1/2) a) 4z(1-e-1/2) 4z(1-e-1/8) b) (z-1)(z-e-1/8) 4(z+1)(1-e-1/4) c) z(z-e-1/4) d) none above 33. The open-loop dc gain GOL(s)=B(s)/E(s) for the system above at 0 Hz is a) -2 b) -1 c) -1/2 d) none above 34. The closed-loop dc gain GCL(s)=C(s)/R(s) of the system is a) -4 b) -2 c) 2 d) none above 35. The closed-loop poles of GCL(z) are a) {0.28, 0.62} b) {-0.18, 0.44} c) {0.59, 0.85} d) none above

Use R dq/dt + q/C =E

Let x(t) is an input to a continuous time linear time system S and y(t) is the output. The input-output relation is given by, x(t)=e^ j3t longrightarrow y(t) = j * e ^ (j * 5t) x(t) = e ^ (- j * 3t) -> y(t) = - j * e ^ (- j * 5t) What is the output y(t) when the input x_{1}(t) = cos(5((3t)/5 - 2))

Find i_0 (t) for t > 0