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)

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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

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signals and systems

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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

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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].

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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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