Homework Solution #2 solution

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Department
Electrical & Computer Engineering
Course
ENEE 322
Professor
Steven Marcus
Semester
Spring

Description
Homework #2 1.17 y(t) = x(sin(t)) (a) Since sin(t) 2 [▯1;1], y(t) only depends on x(▯);▯ 2 [▯1;1]. For example, y(▯▯) = x(sin(▯▯)) = x(0) Thus, y(t) is not causal. (b) Consider two arbitrary inputs x (t1 and x (t)2 The response to x (t) is y (t) = x (sin(t)): 1 1 1 The response to x (2) is y 2t) = x 2sin(t)). Let x 3t) be a linear combinition of x 1t) and x 2t), that is x3(t) = ax1(t) + bx2(t) where a and b are two arbitrary complex constants. If x3(t) is the input to the given system, then the corresponding output y3(t) is y3(t) = x3(sin(t)) = ax1(sin(t)) + bx2(sin(t)) = ay1(t) + by2(t) Therefore, the system is linear. 1.19 (b, d) 2 (b) y[n] = x [n ▯ 2] (i) Consider two arbitrary input x [1] and x [2]. x1[n] ! y1[n] = x1[n ▯ 2] x2[n] ! y2[n] = x2[n ▯ 2] Let x 3n] be a linear combinition of x 1n] and x 2n], that is x 3n] = ax 1n] + bx 2n] where a and b are arbitrary complex constants. If x3[n] is the input, the corresponding output y 3n] is 2 y3[n] = x3[n ▯ 2] 2 = (ax1[n ▯ 2] + bx2[n ▯ 2]) 2 2 2 2 = a x 1n ▯ 2] + b x 2n ▯ 2] + 2abx 1n ▯ 2]x 2n ▯ 2] 6= ay1[n] + by2[n] 1 Therefore, the system is not linear. (ii) Consider an arbitrary input x [1]. Let 2 y1[n] = x1[n ▯ 2] be the corresponding output. Consider a second input x [n] o2tained by time shifting: x2[n] = x 1n ▯ N] The output corresponding to the input x [n2 is 2 2 y2[n] = x2[n ▯ 2] = x 1n ▯ 2 ▯ N] Also note that 2 y1[n ▯ N] = x 1n ▯ 2 ▯ N] Thus, y1[n ▯ N] = y 2n] The system is time-invariant. (d) y(t) = Odfx(t)g (i) Linearity Consider two arbitrary inputs x (1) and x (2). x 1t) ▯ x1(▯t) x1(t) ▯! y 1t) = Odfx (1)g = 2 x (t) ▯! y (t) = Odfx (t)g = x 2t) ▯ x2(▯t) 2 2 2 2 Let x 3t) be a linear combinition of x 1t) and x 2t). That is, x3(t) = ax1(t) + bx2(t) where a and b are arbitrary scalars. If x (t) is the input to the given system,then the corresponding output y (t) 3 3 is y3(t) = Odfx 3t)g = Odfax (1) + bx (2)g 1 = 2 ((ax1(t) + bx2(t)) ▯ (ax1(▯t) + bx2(▯t))) = aOdfx (1)g + bOdfx (t2g = ay1(t) + by2(t) 2 Therefore, the system is linear. (ii) Time-Invariance Consider an arbitrary input x (1). Let x (t) ▯ x (▯t) y1(t) = Odfx 1t)g = 1 1 2 be the corresponding output. Consider a second input x (2) obtained by shifting x (1) in time: x2(t) = x1(t ▯ ▯) The output corresponding to the input x (t2 is y2(t) = Odfx (2)g x (t) ▯ x (▯t) = 2 2 2 = x 1t ▯ ▯) ▯ x 1▯t ▯ ▯) 2 Also note that x1(t ▯ ▯) ▯ x1(▯t + ▯) y 1t ▯ ▯) = 6= 2 (t) 2 Therefore, the system is not time-invariant. 1.27 (a,d,e,f) (a) y(t) = x(t ▯ 2) + x(2 ▯ t) (1) Since the value y(t) depends on the value x(t ▯ 2) and x(2 ▯ t), the system is not memoryless (2) Consider the input: x (t) = x (t ▯ ▯) 2 1 The corresponding output to x (t) is 2 y2(t) = x 2t ▯ 2) + x2(2 ▯ t) = x 1t ▯ 2 ▯ ▯) + x 12 ▯ t ▯ ▯) Note that y1(t ▯ ▯) = x1(t ▯ 2 ▯ ▯) + x1(2 ▯ t + ▯) 6= 2 (t) Thus, the system is not time-invariant. (3) Consider two arbitrary input x (t1 and x (t2, 3 x 1t) ▯! y1(t) = x1(t ▯ 2) ▯ 1 (2 ▯ t) x 2t) ▯! y2(t) = x2(t ▯ 2) ▯ 2 (2 ▯ t) Let x (t) be the linear combinition of x (t) and x (t) 3 1 2 x3(t) = ax1(t) + bx2(t) The response to the input x (t) is 3 y3(t) = ax1(t ▯ 2) + bx2(t ▯ 2) + ax1(t ▯ 2) + bx2(2 ▯ t) = a(x1(t ▯ 2) + x1(2 ▯ t)) + b(2 (t ▯ 2) + 2 (2 ▯ t)) = ay1(t) + by2(t) Thus, the system is linear. (4) Let t = 0, y(0) = x(▯2) + x(2) Thus, the system is not causal. ▯ 0; t < 0 (d) y(t) = x(t) + x(t ▯ 2) t ▯ 0 (1) Since the value of y(t) depends on x(t▯2), the system is not memoryless. (2) Consider an arbitrary input x 1t). Let ▯ 0; t < 0 y1(t) = x 1t) + x 1t ▯ 2); t ▯ 0 be the corresponding output. Consider a second input x 2t) obtained by shifting x 1t) in
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