BME343%20HW1_2009%20solution

# BME343%20HW1_2009%20solution - BME343 HW1 1.1β1(b E=(β1...

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Unformatted text preview: BME343 HW1 1.1β1 (b) E= (β1) dt + (1) dt = 2 + 1 = 3 2 2 0 2 2 3 2 3 β« β« 0 β« (d) E= (2) dt + (β2) dt = 8 + 4 = 12 . 2 2 2 β« 1.1β5 (a) If x(t)=C*cos(Ο0t+ΞΈ) 1 1 2 2 Px= lim β«/ 2 C cos (Ο0t + ΞΈ )dt = Tlim 2T T ββ T ββ βT X(t)=5+10cos(100t+Ο/3) Px= lim T /2 T /2 T /2 βT / 2 β« C 2 (1 + cos(2Ο0t + 2ΞΈ )dt =C2/2. 1 102 52 dt + =75. T ββ T β« 2 βT / 2 rms= 75 = 5 3 (e) x(t)= 10sin(5t)cos(10t) = 10 [sin(5t+10t)+sin(5tβ10t)] = 5[sin(15t)+sin(β5t)] 2 From part (a), Px = 1.2β3 52 52 + =25; rms=5 22 x1 (t ) = x(t + 1) + x(1 β t ) x2 (t ) = x( t +1 1β t ) + x( ) 2 2 t t βt t+2 2βt t βt x3 (t ) = x2 ( ) + x( ) + x( ) = x( ) + x( ) + x( ) + x( ) 2 2 2 4 4 2 2 4 t + 2 4 2βt 1 t x4 (t ) = x( ) + x( ) β x2 ( ) 3 2 3 2 3 2 4 t + 2 4 2βt 1 t + 2 1 2βt = x( ) + x( ) β x( ) β x( ) 3 2 3 2 3 4 3 4 x5 (t ) = x2 (2t ) + x(t + 1.5) + x(1.5 β t ) = x( 2t + 1 1 β 2t ) + x( ) + x(t + 1.5) + x(1.5 β t ) 2 2 = x(t + 0.5) + x(0.5 β t ) + x(t + 1.5) + x(1.5 β t ) 1.2β5 (a) >> t=β5:.01:5; >> u=inline('t>=0','t'); >> y=t.*(u(βtβ1)βu(β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 β1 1β v Set v=β3t+1, t = = β3 3 y (t ) 1 1 β v 1 β(v β 1) x (v ) = = y( ) = y( ) 2 2 3 2 3 (b) y(t)=2x(β3t+1) x(β3t+1)= From y(t) to x(t): i) time scaling by x(t ) = 1 β(t β 1) y( ) 2 3 1 t : y ( ) 3 3 βt ii) time reversal: y ( ) 3 β(t β 1) iii) time shifting by 1: y ( ) 3 1 1 β(t β 1) iv) scale the signal by : y ( ) 22 3 1 1β t 1β t 1β t 1β t tβ4 t+2 x(t)= { β [u (β β 1) β u (β + 1)]} = β [u ( ) β u( )] 23 3 3 6 3 3 >> t=β5:.01:5; u=inline('t>=0','t'); >> x=(1βt)/6.*(u((tβ4)./3)β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 1.3β2 1 1 2 2 (a) true P = lim β«/ 2 x (t )dt = T βTβ«/ 2 x (t )dt T ββ T βT (b) false: x(t)=u(t) is a power signal, Px=0.5; but not periodic. (c) true: Suppose x(t)=u(t)βu(tβ1). y(t)=x(at)=u(at)βu(atβ1) Ex= (1) dt =1; Ey= 0 T /2 T /2 β« 1 1/ a 2 β« (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. 1.4β3 sin(t ) sin(0) Ξ΄ (t ) = 2 Ξ΄ (t ) = 0 2 0 +2 t +2 sin[Ο / 2(t β 2)] sin[Ο / 2(1 β 2)] 1 (d) Ξ΄ (1 β t ) = Ξ΄ (1 β t ) = β Ξ΄ (1 β t ) 2 2 1 +4 5 t +4 (a) (e) 1 1 2+3j Ξ΄ (Ο + 3) = Ξ΄ (Ο + 3) = Ξ΄ (Ο + 3) jΟ + 2 j β (β3) + 2 13 1.4β4 β (a) ββ β β« Ξ΄ (Ο )x(t β Ο )dΟ = x(t ) β« Ξ΄ (2t β 3) sin(Ο t )dt = sin(Ο β 2 ) = β1 β«e x β1 (d) 3 ββ β (h) ββ cos[ ( x β 5)]Ξ΄ ( x β 3)dx = e3β1 cos[ (3 β 5)] = βe2 2 2 Ο Ο 1.4β6 (b) x(t ) = u (t ) β Ξ΄ (t β 1) β Ξ΄ (t β 2) β Ξ΄ (t β 3) ; ββ β« x(t )dt = tu(t ) β u(t β 1) β u(t β 2) β u(t β 3) t >> t=0:.01:6; >> u=inline('t>=0','t'); >> y=t.*u(t)βu(tβ1)βu(tβ2)βu(tβ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 1.4β10 (a) e β3t cos(3t ) = (b) e ; s=2 1.7β1 (b) 2t 1 β3t +3 jt + e β3t β3 jt ] ; s= β3 Β± 3 j [e 2 dy + 3ty (t ) = t 2 x(t ) dt dy1 + 3ty1 (t ) = t 2 x1 (t ) ; β¦β¦β¦β¦β¦β¦β¦(1) dt dy2 + 3ty2 (t ) = t 2 x2 (t ) β¦β¦β¦β¦β¦β¦. (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 ) β¦β¦β¦β¦.(1) dt dk1 y1 + k1 y12 (t ) = k1 x1 (t ) dt k1 y12 (t ) β  [k1 y1 (t )]2 , therefore, k1x(t) β > 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 ) β¦β¦..(1) dt k1 ( dy1 2 ) + 2k1 y1 (t ) = k1 x1 (t ) ; dt dy d (k y ) k1 ( 1 )2 β  ( 1 1 ) 2 , Therefore, Therefore, k1x(t) β > 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) 1.7β2 (b) y(t)=x(β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 timeβvarying. (d) y(t)=tx(tβ2) If x(t) is shifted by T, y1(t)=tx(tβTβ2) β  y(tβT). Thus the system is timeβvarying. (e) y (t ) = β5 β« x(Ο )dΟ X(t) X(tβT) 5 Suppose x(t) and x(tβT) are the red curves. β10 β5 5 t β10+T β5 5 t When x(t) is shifted by T, y(t) is the integration between (β5,5), which is not necessarily related to T. The system is timeβvarying. 1.7β7 (b) y(t)=x(βt) When t<0, output is dependent on the future value of x(t). The system is nonβcausal. (c) y(t)=x(at), a>1 When a=2. y(t)=x(2t), nonβcausal system. 1.7β10 β (a) y (t ) = 0.5 ββ β« x(Ο )[Ξ΄ (t β Ο ) β Ξ΄ (t + Ο )]dΟ = y(t ) = 0.5[ x(t ) β x(βt )] . The system get the odd potion of the input. (b) The system is a BIBO system. Because when x(t) is bounded, x(βt) is bounded, y(t) is bounded. (c) The system is linear. If y1 (t ) = 0.5[ x1 (t ) β x1 (βt )] ; y2 (t ) = 0.5[ x2 (t ) β x2 (βt )] Then k1 y1 (t ) + k2 y2 (t ) = 0.5{[ k1 x1 (t ) + k2 x2 (t )] β [ k1 x1 (βt ) + k2 x2 (βt )]} The system is linear. (d) The system is not memoryless because the output depends on the input at t and βt. (e) The system is not causal because the output might depend on the future input. (When t<0) (f) The system is timeβvarying. If x(t) is shifted by T, x(βt) will shift in the opposite direction. ...
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