# Solution Set 1

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Signals and Systems Solutions for Problem Set 1.
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Problem Set #1Submitted by:  Umar-Robert Qattan UID:  704506988 Date:  June 30, 20141. Show whether the systems  S  1  and  S  2  are linear or nonliner; time-invarient ortime-varying: S  1  :  y  =  T  1 [ x 1 ( t )] , y ( t ) =    t 0 τx ( τ  ) dτ   +  x ( t ) , t ≥ 0Let  x 1 ( t ) and  x 2 ( t ) be two unique input signals: x 1 ( t ) =    t 0 τ  [ x 1 ( τ  ) dτ   +  x 1 ( t )] (1) x 2 ( t ) =    t 0 τ  [ x 2 ( τ  ) dτ   +  x 2 ( t )] (2)To check for linearity of the system, apply superposition to  x 1 ( t ) and  x 2 ( t ): y 1 ( t ) +  y 2 ( t ) =  T  1 [ x 1 ( t ) +  x 2 ( t )]=    t 0 τx 1 ( τ  ) dτ   +  x 1 ( t ) +    t 0 τx 2 ( τ  ) dτ   +  x 2 ( t )=    t 0 τ  [ x 1 ( τ  ) +  x 2 ( τ  )] dτ   +  x 1 ( t ) +  x 2 ( t )(3)Compare the above superpositioned expression with the below transformation of the sum of two unique input signals: T  1 [ x 1 ( t ) +  x 2 ( t )] =    t 0 τ  [ x 1 ( τ  ) +  x 2 ( τ  )] dτ   +  x  p ( t )=    t 0 τx 1 ( τ  ) dτ   +    t 0 τx 2 ( τ  ) dτ   +  x  p ( t )(4)In order to satisfy linearity of   S  1 , the right hand expressions from equations (3) and (4) mustbe equal. And, so,  x 1 ( t ) +  x 2 ( t ) must equal  x  p ( t ).1  To show whether S  1  is not only Linear, but also Time-Invariant, let there be an advancementin time (time-delay) in  S  1  of   λ  seconds:: x 1 ( t − λ )  ⇒  y 1 ( t − λ ) =  T  1 [ x 1 ( t − λ )] (5) x 2 ( t − λ )  ⇒  y 2 ( t − λ ) =  T  1 [ x 2 ( t − λ )] (6) y 1 ( t − λ ) =    t 0 ( τ   − λ ) x 1 ( τ   − λ ) dτ   +  x 1 ( t ) (7) y 2 ( t − λ ) =    t 0 ( τ   − λ ) x 2 ( τ   − λ ) dτ   +  x 2 ( t ) (8) y 1 ( t − λ ) +  y 2 ( t − λ ) =    t 0 ( τ   − λ ) x 1 ( τ   − λ ) dτ   +    t 0 ( τ   − λ ) x 2 ( τ   − λ ) dτ  + x 1 ( t − λ ) +  x 2 ( t − λ )(9)To further show whether  S  1  is a Time-Invariant system, apply the transformation  y ( t ) = T  1 [ x ( t )] to the sum of the input signals,  x 1 ( t ) and  x 2 ( t ): T  1 [ x 1 ( t − λ ) +  x 2 ( t − λ )] =    t 0 ( τ   − λ )[ x 1 ( τ   − λ ) +  x 2 ( τ   − λ )] dτ   +  x 1 ( t ) +  x 2 ( t )=    t − λ − λ ρ [ x 1 ( ρ ) +  x 2 ( ρ )] dτ   +  x 1 ( t ) +  x 2 ( t )=  y 1 ( t − τ  ) +  y 2 ( t − τ  )(10)Because the sum of the transformed input signals  x 1 ( t ) and  x 2 ( t ) in S  1  is equal to the trans-formation of the sum of the input signals as shown by the R.H.S expressions in equations(9) and (10),  S  1  is a Time-Invariant system.Finally,  S  1  is an LTI because it is both Linear and Time-Invariant.2
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