# exam1solns - BME 311 1. [10] Linearity and time-invariance....

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BME 311 EXAM #1 SOLUTIONS February 22, 2006 1. [10] Linearity and time-invariance. (a) y ( t ) = sin( x ( t )) i. x 1 ( t ) y 1 ( t ) = sin( x 1 ( t )) , x 2 ( t ) y 2 ( t ) = sin( x 2 ( t )) a 1 x 1 ( t ) + a 2 x 2 ( t ) y ( t ) = sin( a 1 x 1 ( t ) + a 2 x 2 ( t )) 6 = a 1 sin( x 1 ( t )) + a 2 sin( x 2 ( t )) = a 1 y 1 ( t ) + a 2 y 2 ( t ) the system is nonlinear . ii. x 1 ( t ) y 1 ( t ) = sin( x 1 ( t )) x 2 ( t ) = x 1 ( t - t 0 ) y 2 ( t ) = sin( x 2 ( t )) = sin( x 1 ( t - t 0 )) y 1 ( t - t 0 ) = sin( x 1 ( t - t 0 )) = y 2 ( t ) the system is time-invariant . (b) y [ n ] = n k = -∞ x [ k - 1] i. x 1 [ n ] y 1 [ n ] = n k = -∞ x 1 [ k - 1] , x 2 [ n ] y 2 [ n ] = n k = -∞ x 2 [ k - 1] a 1 x 1 [ n ] + a 2 x 2 [ n ] y [ n ] = n k = -∞ ( a 1 x 1 [ k - 1] + a 2 x 2 [ k - 1]) = a 1 n k = -∞ x 1 [ k - 1]+ a 2 n k = -∞ x 2 [ k - 1] = a 1 y 1 [ n ]+ a 2 y 2 [ n ] the system is linear . ii. x 1 [ n ] y 1 [ n ] = n k = -∞ x 1 [ k - 1] x 2 [ n ] = x 1 [ n - n 0 ] y 2 [ n ] = n k = -∞ x 2 [ k - 1] = n k = -∞ x 1 [ k - 1 - n 0 ] y 1 [ n - n 0 ] = n - n 0 k = -∞ x 1 [ k - 1] = n k = -∞ x 1 [ k - 1 - n 0 ] = y 2 [ n ] the system is time-invariant . (c) y ( t ) = d 2 d t 2 x ( t ) i. x 1 ( t ) y 1 ( t ) = d 2 d t 2 x 1 ( t ) , x 2 ( t ) y 2 ( t ) = d 2 d t 2 x 2 ( t ) a 1 x 1 ( t ) + a 2 x 2 ( t ) y ( t ) = d 2 d t 2 ( a 1 x 1 ( t ) + a 2 x 2 ( t )) = a 1 d 2 d t 2 x 1 ( t ) + a 2 d 2 d t 2 x 2 ( t ) = a 1 y 1 ( t ) + a 2 y 2 ( t ) the system is linear . ii. x 1 ( t ) y 1 ( t ) = d 2 d t 2 x 1 ( t ) x 2 ( t ) = x 1 ( t - t 0 ) y 2 ( t ) = d 2 d t 2 x 2 ( t ) = d 2 d t 2 x 1 ( t - t 0 ) y 1 ( t - t 0 ) = d 2 d t 2 x 1 ( t - t 0 ) = y 2 ( t ) the system is time-invariant . (d)

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## This note was uploaded on 04/30/2008 for the course BIOMEDE 311 taught by Professor Steel during the Winter '06 term at University of Michigan.

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exam1solns - BME 311 1. [10] Linearity and time-invariance....

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