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20105ee102_1_hw2_soln

# 20105ee102_1_hw2_soln - 1 EE102 Fall Quarter 2010 Systems...

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1 EE102 Systems and Signals Fall Quarter 2010 Jin Hyung Lee Homework #2 Solutions Due: Wednesday, Oct 13, 2010 at 5 PM. 1. State whether the following systems are linear or nonlinear; time invariant or time variant; and why. (a) y ( t ) = x ( t ) sin( ωt + φ ) Solultion: Let x ( t ) = ax 1 ( t ) + bx 2 ( t ) , and check y ( t ) = x ( t ) sin( ωt + φ ) = ( ax 1 ( t ) + bx 2 ( t )) sin( ωt + φ ) = ax 1 ( t ) sin( ωt + φ ) + bx 2 ( t ) sin( ωt + φ ) = ay 1 ( t ) + by 2 ( t ) , so the system is linear . If we delay the input x ( t ) by τ x ( t - τ ) sin( ωt + φ ) However, the delayed output would be x ( t - τ ) sin( ω ( t - τ ) + φ ) which is not the same, so the system is time variant . (b) y ( t ) = x ( t ) x ( t - 1) We first check homogeneity, if we input x 1 ( t ) = ax ( t ) we get y 1 ( t ) = x 1 ( t ) x 1 ( t - 1) = ( ax ( t ))( a ( x ( t - 1)) = a 2 x ( t ) x ( t - 1) = a 2 y ( t ) so this system is non-linear . Next we check delay the input, x ( t - τ ) x ( t - τ - 1) = y ( t - τ ) so the system is time invariant . (c) y ( t ) = 1 + x ( t ) First we check homogeneity, 1 + a ( x ( t )) 6 = a (1 + x ( t ))

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2 so the system is non-linear . Next, if we delay the input 1 + x ( t - τ ) = y ( t - τ ) so the system is time invariant . (d) y ( t ) = cos( ωt + x ( t )) Solultion: We first check homogeneity cos( ωt + ax ( t )) 6 = a cos( ωt + x ( t )) so scaling the input by a does not result in a scaled output. This system is non-linear . Next, if we delay the input by τ cos( ωt + x ( t - τ )) 6 = cos( ω ( t - τ ) + x ( t - τ )) = y ( t - τ ) so this system is time variant . (e) y ( t ) = Z t -∞ x ( τ ) This time we’ll check superposition and homogeneity together, Z t -∞ ( ax 1 ( τ ) + bx 2 ( t )) = a Z t -∞ x 1 ( τ ) + b Z t -∞ x 2 ( t ) = ay 1 ( t ) + by 2 ( t ) so the system is linear . Next, if we delay the input by T (since τ is already used) Z t -∞ x ( τ - T ) = Z t - T -∞ x ( τ 0 ) 0 = y ( t - T ) where we have made the change of variables τ 0 = τ - T . So this system is time invariant . (f) y ( t ) = Z t/ 2 -∞ x ( τ ) Solultion: Again we’ll check superposition and homogeneity together, Z t/ 2 -∞ ( ax 1 ( τ ) + bx 2 ( t )) = a Z t/ 2 -∞ x 1 ( τ ) + b Z t/ 2 -∞ x 2 ( t ) = ay 1 ( t ) + by 2 ( t ) so the system is linear . Next, if we delay the input by T (since τ is already used) Z t/ 2 -∞ x ( τ - T ) = Z t/ 2 - T -∞ x ( τ 0 ) 0 = Z ( t - 2 T ) / 2 -∞ x ( τ 0 ) 0 = y ( t - 2 T ) where we have made the change of variables τ 0 = τ - T . Delaying the input by T delays the output by 2 T , so this system is time variant .
3 2. A periodic signal x ( t ) , with a period T , is applied to a linear, time-invariant system H . Show that the output y ( t ) y ( t ) = H ( x ( t )) is also periodic, with period T . Solution: If we delay the input x ( t ) by T the output is y ( t - T ) = H ( x ( t - T )) = H ( x ( t )) = y ( t ) where the first equality is due to the fact that the system is time invariant, the second is due to x ( t ) being periodic with period T , and the final equality is due to the definition of y ( t ) . Hence y ( t - T ) = y ( t ) and y ( t ) is periodic with period T . 3. Sample and hold system. A sample and hold (S/H) system, with sample time h , is described by y ( t ) = x ( h b t/h c ) , where b a c denotes the largest integer that is less than or equal to a .

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