For each of the linear time-invariant continuous-time systems with impulse response h(t) given as follows, determine if the system is BIBO stable. - (a) h(t) = [2t3 - 2t² + 3t − 2][u(t) − u(t − 10)] 1 (b) h(t) = for t≥ 1, h(t) = 0 for all t<1 (c) h(t) t = sin 2t for t≥ 0 (d) h(t) = e sin 2t for t≥ 0

4. A continuous-time signal x(t) is shown in Figure below. Sketch and label each of the following signals. |x(t) (a) x(t)u(1 - t); 2 (b) x(t)[u(t) – u(t – 1)]: (0) x(t)6 (t - }). 10 I 2

8. Given below is the signal m(t). Decompose this signal in terms of the unit step function u(t). -6 -5 -4 -2 -1 m(t) 2 1 0 -1 0 1 2 3 4 t Homework 2 5

Question #4 : Input output relationship of a linear time invariant continuous time system having zero initial conditions is given with the following differential equation d³y(t) d²y(t) - 5 dt³ dy(t) +9. dt - 5y(t) = x(t). dt2 Obtain output of the system for each given input. x1(t) = 3[u(t) – u(t – 2)]. a) b) c) x2(t) = 2e-3lu(t). x3(t) = x,(t) + x2(t – 2).

A continuous-time LTI system has the input x(t) and the impulse response h(t) as shown in figurebelow:a) Compute and plot y(t) for all time.b) Solve for the system output at t=3

a)A system with impulse response h(t) as below , Is the system BIBO stable (explain) u(t – 1) h(t) = t2

Convolve the following continuous-time signals, assuming that T₁=1 sec. and T, -1.5 sec Plot the result -1 T 2T, h(t) K

Consider the signal: x(t) = exp(-rt) cos(t) u(t) where u(t) is a step function. The signal x(t) will have Finite-energy signal when r>0 Finite-energy when r<0 Finite-energy for all values of r Infinite-energy regardless of the r value

The result of: u(t) 5u(t) * 8(t- 3) + 28(t – 1) * u(t + 3) is Select one: 5u(t- 3) + 2u(t) 5(t 3)u(t 3) + 2u(t) bu(t 3) + 2u(t | 2) 5(t 3)u(t- 3) + 2u(t + 2) Consinie

a)A system with impulse response h(t) as below , Is the system BIBO stable (1.5) u(t – 1) h(t) = t2