Solved Problems1. signals and systems Express the signals shown in Fig. 1 in terms of unit step functions. Fig. 1 Answer x(t ) u (t 1) 2u (t ) u (t 1) u (t 2) u (t 3) 2. Consider the system shown in Fig. 2. Determine whether it is (a) memoryless, ( b )causal, ( c ) linear, ( d ) time-invariant, or ( e ) stable. Fig. 2 (a) From Fig. 2 we have y (t ) T x (t ) x (t ) cos c (t ) ; Since the value of the output y ( t ) depends on only the present values of the input x( t ), the system is memoryless. (b) E.g . t 5, y (5) x(5) cos c (5) ; Since the output y ( t ) does not depend on the future values of the input x(t), the system is causal. (c) Let x(t ) 1 x(t ) 2 x(t ). Then y (t ) 1x1(t ) 2 x 2(t )cos c t 1x1 (t ) cos ct 2 x2 (t ) cos c t 1 y1 (t ) 2 y2 (t ) Thus, the system is linear. (d) Let But y1 (t ) x1 (t ) x (t t0 ) . Then y1 (t ) T x(t t0 ) x(t t0 ) cos c (t ) be the output produced by the shifted input y (t t0 ) x(t t0 ) cos c (t t0 ) y1 (t ) . Hence, the system is not time-invariant. Sopapun Suwansawang Solved Problems (e) Since cos c t 1. 3. we have signals and systems y (t ) x(t ) cos c t x(t ) Thus. where x(t ) u (t ) u (t (a) by an analytical technique. 3) and h(t ) u (t ) u (t 2) Sopapun Suwansawang . Evaluate y (t ) x(t ) h(t ). if the input x(t) is bounded. then the output y(t) is also bounded and the system is BIB0 stable. and (b) by a graphical method. x( )h(t ) for different values of t are sketched in figure below. For the other intervals. Functions h( ). x( ) and h(t ) overlap. We see that x( ) and h(t ) do not overlap for t 0 and t 5.Solved Problems signals and systems (b) by a graphical method. computing the area under the rectangular pluses for these intervals. Thus. x( ) and h(t ). we obtain Sopapun Suwansawang . and hence y (t ) 0 for t 0 and t 5 . Solved Problems signals and systems Sopapun Suwansawang . we get d 2 y (t ) dt 2 Or d 2 y (t ) dt 2 a1 a1 dy(t ) a2 y (t ) x(t ) dt dy(t ) a2 y (t ) x(t ) dt 5. Determine and sketch the output y[n] of this system to the input x[n]. (1). Write a differential equation that relates the output y(t) and the input x( t ). The impulse response h[n] of a discrete-time LTI system.(1) w(t ) is the input to the second integrator. we have w(t )) dy(t ) dt -------------. h[n] [n] [n 1] [n 2] [n 3] [n 4] [n 5] . (b) without using the convolution technique. (2) into Eq. x[n] [n 2] [n 4] x[n] h[n] x[n] [n] [n 1] [n 2] [n 3] [n 4] [n 5] x[n] x[n 1] x[n 2] x[n 3] x[n 4] x[n 5] Sopapun Suwansawang .(2) Substituting Eq. The continuous-time system consists of two integrators and two scalar multipliers. (a).Solved Problems signals and systems 4. e(t ) Since dw(t ) a1w(t ) a2 y(t ) x(t ) dt -------------. Solved Problems signals and systems y[n] [n 2] [n 4] [n 3] [n 5] [n 4] [n 6] [n 5] [n 7] [n 6] [n 8] [n 7] [n 9] [n 2] [n 3] 2 [n 6] 2 [n 7] [n 8] [n 9] y[n] 0. Write a difference equation that relates the output y[n] and the input x[n].0.1 6.0.2. Consider the discrete-time system.1.1.0. Sopapun Suwansawang .1.2. (3) w[n 1] y[n] x[n] 2 1 1 ---------------.(4) 1 1 ---------------.(2) y[n] 2w[n] w[n 1] Solving Eqs.(1) and (2) for w[n] and w[n 1] in term of x[n] and y[n] 1 ---------------.(1) w[n] x[n] w[n 1] 2 ---------------. we obtain 2 y[n] y[n 1] 4 x[n] 2 x[n 1] Sopapun Suwansawang . equating Eq.(5) w[n 1] y[n 1] x[n 1] 4 2 Thus. Write the input-output equation for the system. w[n] w[n 1] 1 ---------------.(4) w[n] y[n] x[n] 4 2 Changing n to (n 1) in Eq. we have 1 1 1 y[n] x[n] y[n 1] x[n 1] 2 4 2 Multiplying both sides of the above equation by 4 2 y[n] 4 x[n] y[n 1] 2 x[n 1] and rearranging terms.(5).Solved Problems signals and systems 7.(4) and Eq.