Solved_problems.pdf

Solved Problems 1.

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, we have

signals and systems

y (t )  x(t ) cos c t  x(t )

Thus, if the input x(t) is bounded, then the output y(t) is also bounded and the system is BIB0 stable. 3. Evaluate y (t )  x(t )  h(t ), where x(t )  u (t )  u (t (a) by an analytical technique, and (b) by a graphical method.

 3) and h(t )  u (t )  u (t  2)

Sopapun Suwansawang

Solved Problems

signals and systems

(b) by a graphical method. Functions h( ), x( ) and

h(t   ), x( )h(t   ) for different values of t are sketched in figure below. We see that x( ) and h(t   ) do not overlap for t  0 and t  5, and hence y (t )  0 for t  0 and t  5 . For the other intervals, x( ) and h(t   ) overlap. Thus, computing the area under the rectangular pluses for these intervals, we obtain

Sopapun Suwansawang

Solved Problems

signals and systems

Sopapun Suwansawang

Solved Problems

signals and systems

4. The continuous-time system consists of two integrators and two scalar multipliers. Write a differential

equation that relates the output y(t) and the input x( t ).

e(t )  Since

dw(t )  a1w(t )  a2 y(t )  x(t ) dt

-------------- (1)

w(t ) is the input to the second integrator, we have

w(t )) 

dy(t ) dt

-------------- (2)

Substituting Eq. (2) into Eq. (1), 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. The impulse response h[n] of a discrete-time LTI system. (a). Determine and sketch the output y[n] of this system to the input x[n]. (b) without using the convolution technique.

h[n]   [n]   [n  1]   [n  2]   [n  3]   [n  4]   [n  5] , 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

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]  0,0,1,1,0,0,2,2,1,1

6. Consider the discrete-time system. Write a difference equation that relates the output y[n] and the input x[n].

Sopapun Suwansawang

Solved Problems

signals and systems

7. Write the input-output equation for the system.

w[n]

w[n  1]

1 ---------------- (1) w[n]  x[n]  w[n  1] 2 ---------------- (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 ---------------- (3) w[n  1]  y[n]  x[n] 2 1 1 ---------------- (4) w[n]  y[n]  x[n] 4 2 Changing n to (n  1) in Eq.(4) 1 1 ---------------- (5) w[n  1]  y[n  1]  x[n  1] 4 2 Thus, equating Eq.(4) and Eq.(5), 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, we obtain

2 y[n]  y[n  1]  4 x[n]  2 x[n  1]

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