BME343%20HW1_2009%20solution

BME343%20HW1_2009%20solution - BME343 HW1 1.1‐1 (b) E=...

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Unformatted text preview: BME343 HW1 1.1‐1 (b) E= (−1) dt + (1) dt = 2 + 1 = 3 2 2 0 2 2 3 2 3 ∫ ∫ 0 ∫ (d) E= (2) dt + (−2) dt = 8 + 4 = 12 . 2 2 2 ∫ 1.1‐5 (a) If x(t)=C*cos(ω0t+θ) 1 1 2 2 Px= lim ∫/ 2 C cos (ω0t + θ )dt = Tlim 2T T →∞ T →∞ −T X(t)=5+10cos(100t+π/3) Px= lim T /2 T /2 T /2 −T / 2 ∫ C 2 (1 + cos(2ω0t + 2θ )dt =C2/2. 1 102 52 dt + =75. T →∞ T ∫ 2 −T / 2 rms= 75 = 5 3 (e) x(t)= 10sin(5t)cos(10t) = 10 [sin(5t+10t)+sin(5t‐10t)] = 5[sin(15t)+sin(‐5t)] 2 From part (a), Px = 1.2‐3 52 52 + =25; rms=5 22 x1 (t ) = x(t + 1) + x(1 − t ) x2 (t ) = x( t +1 1− t ) + x( ) 2 2 t t −t t+2 2−t t −t x3 (t ) = x2 ( ) + x( ) + x( ) = x( ) + x( ) + x( ) + x( ) 2 2 2 4 4 2 2 4 t + 2 4 2−t 1 t x4 (t ) = x( ) + x( ) − x2 ( ) 3 2 3 2 3 2 4 t + 2 4 2−t 1 t + 2 1 2−t = x( ) + x( ) − x( ) − x( ) 3 2 3 2 3 4 3 4 x5 (t ) = x2 (2t ) + x(t + 1.5) + x(1.5 − t ) = x( 2t + 1 1 − 2t ) + x( ) + x(t + 1.5) + x(1.5 − t ) 2 2 = x(t + 0.5) + x(0.5 − t ) + x(t + 1.5) + x(1.5 − t ) 1.2‐5 (a) >> t=‐5:.01:5; >> u=inline('t>=0','t'); >> y=t.*(u(‐t‐1)‐u(‐t+1)); >> plot(t,y);xlabel('t'); ylabel('y(t)'); 1 Y (t) 0 -1 -5 -4 -3 -2 -1 0 t 1 2 3 4 5 y (t ) 2 v −1 1− v Set v=‐3t+1, t = = −3 3 y (t ) 1 1 − v 1 −(v − 1) x (v ) = = y( ) = y( ) 2 2 3 2 3 (b) y(t)=2x(‐3t+1) x(‐3t+1)= From y(t) to x(t): i) time scaling by x(t ) = 1 −(t − 1) y( ) 2 3 1 t : y ( ) 3 3 −t ii) time reversal: y ( ) 3 −(t − 1) iii) time shifting by 1: y ( ) 3 1 1 −(t − 1) iv) scale the signal by : y ( ) 22 3 1 1− t 1− t 1− t 1− t t−4 t+2 x(t)= { ⋅ [u (− − 1) − u (− + 1)]} = ⋅ [u ( ) − u( )] 23 3 3 6 3 3 >> t=‐5:.01:5; u=inline('t>=0','t'); >> x=(1‐t)/6.*(u((t‐4)./3)‐u((t+2)./3)); >> plot(t,x); xlabel('t');ylabel('x (t)'); 0.5 x (t) 0 -0.5 -5 -4 -3 -2 -1 0 t 1 2 3 4 5 1.3‐2 1 1 2 2 (a) true P = lim ∫/ 2 x (t )dt = T −T∫/ 2 x (t )dt T →∞ T −T (b) false: x(t)=u(t) is a power signal, Px=0.5; but not periodic. (c) true: Suppose x(t)=u(t)‐u(t‐1). y(t)=x(at)=u(at)‐u(at‐1) Ex= (1) dt =1; Ey= 0 T /2 T /2 ∫ 1 1/ a 2 ∫ (1) dt = a 2 0 1 (d) false: Power of x(at) is still P. Suppose x(t)=sin(t). y(t)=x(at)=sin(at). They have the same power. 1.4‐3 sin(t ) sin(0) δ (t ) = 2 δ (t ) = 0 2 0 +2 t +2 sin[π / 2(t − 2)] sin[π / 2(1 − 2)] 1 (d) δ (1 − t ) = δ (1 − t ) = − δ (1 − t ) 2 2 1 +4 5 t +4 (a) (e) 1 1 2+3j δ (ω + 3) = δ (ω + 3) = δ (ω + 3) jω + 2 j ⋅ (−3) + 2 13 1.4‐4 ∞ (a) −∞ ∞ ∫ δ (τ )x(t − τ )dτ = x(t ) ∫ δ (2t − 3) sin(π t )dt = sin(π ⋅ 2 ) = −1 ∫e x −1 (d) 3 −∞ ∞ (h) −∞ cos[ ( x − 5)]δ ( x − 3)dx = e3−1 cos[ (3 − 5)] = −e2 2 2 π π 1.4‐6 (b) x(t ) = u (t ) − δ (t − 1) − δ (t − 2) − δ (t − 3) ; −∞ ∫ x(t )dt = tu(t ) − u(t − 1) − u(t − 2) − u(t − 3) t >> t=0:.01:6; >> u=inline('t>=0','t'); >> y=t.*u(t)‐u(t‐1)‐u(t‐2)‐u(t‐3); >> plot(t,y); xlabel('t');ylabel('integration of x (t)'); 3 integration of x (t) 2 1 0 0 1 2 3 t 4 5 6 1.4‐10 (a) e −3t cos(3t ) = (b) e ; s=2 1.7‐1 (b) 2t 1 −3t +3 jt + e −3t −3 jt ] ; s= −3 ± 3 j [e 2 dy + 3ty (t ) = t 2 x(t ) dt dy1 + 3ty1 (t ) = t 2 x1 (t ) ; …………………(1) dt dy2 + 3ty2 (t ) = t 2 x2 (t ) ………………. (2) dt d (k1 y1 + k2 y2 ) + 3t[k1 y1 (t ) + k2 y2 (t )] = t 2 [k1 x1 (t ) + k2 x2 (t )] dt Therefore, k1 x1 (t ) + k2 x2 (t ) The system is linear. Suppose X1(t) y1(t); x2(t) y2(t) Multiply (1) and (2) with k1 and k2 respectively, add them together: k1 y1 (t ) + k2 y2 (t ) (d) dy + y 2 (t ) = x(t ) dt Suppose X1(t) y1(t); Multiply (1) by k1. dy1 + y12 (t ) = x1 (t ) ………….(1) dt dk1 y1 + k1 y12 (t ) = k1 x1 (t ) dt k1 y12 (t ) ≠ [k1 y1 (t )]2 , therefore, k1x(t) ≠> k1y(t). The system is nonlinear. (e) ( dy 2 ) + 2 y (t ) = x(t ) dt Suppose X1(t) y1(t); ( Multiply (1) by k1. dy1 2 ) + 2 y1 (t ) = x1 (t ) ……..(1) dt k1 ( dy1 2 ) + 2k1 y1 (t ) = k1 x1 (t ) ; dt dy d (k y ) k1 ( 1 )2 ≠ ( 1 1 ) 2 , Therefore, Therefore, k1x(t) ≠> k1y(t). dt dt The system is nonlinear. (f) dy dx + sin(t ) y (t ) = + 2 x(t ) dt dt The system is linear. Reason is the same as (b) 1.7‐2 (b) y(t)=x(‐t) y(t) is the time reversal of x(t). If x(t) is shifted in right, y(t) will be shifted in left. Thus the system is time‐varying. (d) y(t)=tx(t‐2) If x(t) is shifted by T, y1(t)=tx(t‐T‐2) ≠ y(t‐T). Thus the system is time‐varying. (e) y (t ) = −5 ∫ x(τ )dτ X(t) X(t‐T) 5 Suppose x(t) and x(t‐T) are the red curves. ‐10 ‐5 5 t ‐10+T ‐5 5 t When x(t) is shifted by T, y(t) is the integration between (‐5,5), which is not necessarily related to T. The system is time‐varying. 1.7‐7 (b) y(t)=x(‐t) When t<0, output is dependent on the future value of x(t). The system is non‐causal. (c) y(t)=x(at), a>1 When a=2. y(t)=x(2t), non‐causal system. 1.7‐10 ∞ (a) y (t ) = 0.5 −∞ ∫ x(τ )[δ (t − τ ) − δ (t + τ )]dτ = y(t ) = 0.5[ x(t ) − x(−t )] . The system get the odd potion of the input. (b) The system is a BIBO system. Because when x(t) is bounded, x(‐t) is bounded, y(t) is bounded. (c) The system is linear. If y1 (t ) = 0.5[ x1 (t ) − x1 (−t )] ; y2 (t ) = 0.5[ x2 (t ) − x2 (−t )] Then k1 y1 (t ) + k2 y2 (t ) = 0.5{[ k1 x1 (t ) + k2 x2 (t )] − [ k1 x1 (−t ) + k2 x2 (−t )]} The system is linear. (d) The system is not memoryless because the output depends on the input at t and –t. (e) The system is not causal because the output might depend on the future input. (When t<0) (f) The system is time‐varying. If x(t) is shifted by T, x(‐t) will shift in the opposite direction. ...
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