a1_solution

# a1_solution - Instructors Solutions to Assignment 1 Problem...

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Instructors Solutions to Assignment 1 Problem 1.2:

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Assignment 1 (Fall 2009) CSE3451: Signals and Systems 2 Problem 1.3: Problem 1.5:
Assignment 1 CSE3451: Signals and Systems 3 Problem 1.6: we note that both complex exponential terms are periodic with the same period K = 8. Signal x 3[ k ] is, therefore, periodic with an overall period of 8.

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Assignment 1 (Fall 2009) CSE3451: Signals and Systems 4 Problem 1.10: Problem 1.22: (i) ( ) ( ) ( ) ( ) 1 5 4 5 4 5 4 t t dt t dt t dt ! ! ! " " " #" #" #" # # = # = # = \$ \$ \$ . (ii) ( ) ( ) ( ) ( ) 6 6 6 1 5 4 5 4 5 4 t t dt t dt t dt ! ! ! "# "# "# " " = " = " = \$ \$ \$ .
Assignment 1 CSE3451: Signals and Systems 5 (iii) ( ) ( ) ( ) ( ) 6 6 6 1 5 4 5 4 5 0 t t dt t dt t dt ! ! ! " " " # # = # = # = \$ \$ \$ . (iv) ( ) ( ) ( ) ( ) ( ) ( ) 3 10 10 2 4 2 3 4 9 3 3 9 2 / 3 5 3 / 4 5/6 5 ( ) 5 t t dt t t dt t t dt ! ! ! " " " #" #" #" # # = # # = # # \$ \$ \$ which simplifies to ( ) ( ) 10 10 460 10 460 4 2 3 3 9 9 81 9 81 115/ 27 5 t dt t dt ! ! " " # # #" #" \$# % & ( = ) # # = # = ( * + , , !"#"\$ . (v) ( ) ( ) ( ) ( ) ( ) ( ) exp 1 sin 5 4 (1 ) exp 1 sin 5 4 ( 1) t t t dt t t t dt ! " ! " # # \$# \$# \$ + \$ = \$ + \$ % % which simplifies to ( ) ( ) ( ) ( ) exp 0 sin 6 4 ( 1) sin 6 4 ( 1) sin 3 2 1 t dt t dt ! " ! " ! # # \$# \$# = \$ = \$ = = \$ % % . (vi) ( ) ( ) ( ) ( ) ( ) 2 1 2 1 2 1 1 sin 3 4 ( 1) sin 3 4 1 sin 3 4 t t t t t e t dt t e t dt t e ! " ! " ! # # \$ + \$ + \$ + = \$ \$# \$# % & % & % & + \$ + = + + = + ( ( ( ) ) which simplifies to ( ) ( ) 3 3 3 1 2 sin 3 4 sin 3 4 e e e ! ! = " + = " = " . (vii) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 5 6 10 sin 3 4 5 6 10 sin 3 4 t u t u t t t dt u t u t t ! " ! # = \$# \$ \$ \$ \$ = \$ \$ \$ % & % & ( ( ) which simplifies to ( ) ( ) ( ) [ ] ( ) 5 6 5 10 sin 3 5 4 0 0 sin 15 4 0 u u ! ! = " " " = " = # \$ % & . (viii) By noting that only the impulses located at t = 20 ( m = 4), t = 1 5 ( m = 3), t = 10 ( m = 2), t = 5 ( m = 1), t = 0 ( m = 0 ), t = 5 ( m = 1 ), t = 1 0 ( m = 2 ), t = 15 ( m = 3 ), and t = 20 ( m = 4) lie within the integration range of ( 21 t 21), the integral reduces to dt m t t dt m t t I m m ! " ! " # # = # \$ #\$ = % % & ( ( ) * # + = % % & ( ( ) * # + = 21 21 4 4 21 21 ) 5 ( ) 5 ( . Changing the order of summation and integration, we obtain ( ) . 0 4 3 2 1 0 1 2 3 4 5 5 ) 5 ( 4 4 4 4 21 21 = + + + + + ! ! ! ! = = ! " = # # \$ ! = ! = ! m m m dt m t t I

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Assignment 1 (Fall 2009) CSE3451: Signals and Systems 6 Problem 1.26 Problem 2.9 (i) ( ) ( 2) y t x t = ! (a) Linearity: Since 1 2 1 1 1 2 2 2 1 2 1 2 ( ) ( 2) ( ) ( ) ( 2) ( ) ( ) ( ) ( 2) ( 2) ( ) ( ) x x x t x t y t x t x t y t t t x t x t y t y t ! " ! " ! " + # # \$ = # \$ = \$ + \$ = + therefore, the system is a linear system. (b) Time Invariance: For inputs x 1 ( t ) and x 2 ( t ) = x 1 ( t T ), the outputs are given by ( ) 1 1 1 2 1 2 1 2 ( ) ( 2) ( ) ( ) ( ) ( 2) ( 2) x t x t y t x t x t T x t x t T y t ! " = = " ! " = " " = and 1 1 2 ( ) ( 2) ( ) y t T x t T y t ! = ! ! = , the system is time invariant. (c) Stability: Assume that the input is bounded | x ( t )| M . Then, the output