Assign1_sol - 1 (sin( t )). Let x 2 ( t ) = x 1 ( t-t )....

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EE3210 Assignment 1 (Solution) 1. (a) Periodic, fundamental period = π . (b) Periodic, fundamental period = 8. 2. (a) This is a periodic signal with period ω π / 2 = T . [] 2 ) 2 2 cos( 1 2 ) ( cos 1 2 2 / 2 / 2 2 / 2 / 2 2 A dt t T A dt t A T P T T T T x = + + = + = θω (b) This is a periodic signal with period / 2 = T . 2 2 / 2 / 2 2 / 2 / 2 1 1 C dt C T dt Ce T P T T T T t j y = = = 3. Let the system response to the inputs x 1 ( t ) and x 2 ( t ) be y 1 ( t ) and y 2 ( t ), respectively. Then ) ( ) ( 3 1 1 1 t x t y dt dy = + and ) ( ) ( 3 2 2 2 t x t y dt dy = + . Multiplying the first equation by a and the second by b , and adding them yields [] [] ) ( ) ( ) ( ) ( 3 ) ( ) ( 2 1 2 1 2 1 t bx t ax t by t ay t by t ay dt d + = + + + . Therefore, when the input is ) ( ) ( 2 1 t bx t ax + , the system response is . Consequently, the system is linear. ) ( ) ( 2 1 t by t ay + 4. The system is NOT time invariant. To see this, let the system response to the inputs x 1 ( t ) and x 2 ( t ) be y 1 ( t ) and y 2 ( t ), respectively. Then y 1 ( t ) = x
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Unformatted text preview: 1 (sin( t )). Let x 2 ( t ) = x 1 ( t-t ). Note that y 2 ( t ) = x 2 (sin( t )) = x 1 (sin( t )-t ), which is NOT equal to y 1 ( t-t ). 5. Let the input and the output of the inverse system be y [ n ] and z [ n ]. Input-Output Relationship: z [ n ] = y [ n ] – y [ n-1]. D y [ n ] z [ n ] +-1 6. (a) Choose a = b = 1. (b) Choose b = a , and x 2 ( t ) = x 1 ( t ), OR choose b = 0. 7. No. There are many counter-examples. Here is one of them: x 1 ( t ) = sin t and x 2 ( t ) = -sin t . Their sum is equal to zero, thus having zero power. But the power of x 1 ( t ) and x 2 ( t ) are both positive and non-zero....
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Assign1_sol - 1 (sin( t )). Let x 2 ( t ) = x 1 ( t-t )....

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