Electrical Engineering 120 - Spring 2000 - Varaiya - Mid 1

# Electrical Engineering 120 - Spring 2000 - Varaiya - Mid 1...

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EECS 120. Solutions to Midterm No. 1, February 17, 2000. 1. 20 points Find the expression for the frequency response from x to y in terms of H 1 ,H 2 3 for the system depicted in: (a) Part (a) of Figure 1 . Ans Fix ω Reals and take as input the signal x : t 7→ X ( ω ) e jωt .S i n c e H 1 is LTI, the other signals are of the form w : t 7→ W ( ω ) e and y : 7→ Y ( ω ) e . Moreover, W ( ω )= X ( ω )+ Y ( ω ) ,Y ( ω H 1 ( ω ) W ( ω ) , from which we obtain the desired frequency response, H 4 ( ω Y ( ω ) X ( ω ) = H 1 ( ω ) 1 H 1 ( ω ) (b) Part (b) of Figure 1 . Ans Rewrite part (b) of the ±gure as (c), and introduce the signal names u,v . Take as input the signal x : t 7→ X ( ω ) e in c e H 1 2 3 are all LTI, u,v,y are all exponential, u : t 7→ U ( ω ) e ,v : t 7→ V ( ω ) e ,y : t 7→ Y ( ω ) e . From the previous part we know V ( ω H 4 ( ω ) U ( ω ). And then, following the same argument as in the previous part, Y ( ω ) X ( ω ) = H 4 ( ω ) H 3 ( ω ) 1 H 2 ( ω ) H 3 ( ω ) H 4 ( ω ) . And substituting for H 4 from the previous part, Y ( ω ) X ( ω ) = H 4 ( ω ) H 3 ( ω ) 1 H 2 ( ω ) H 3 ( ω ) H 4 ( ω ) = H 1 ( ω ) H 3 ( ω ) 1 H 1 ( ω )

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Electrical Engineering 120 - Spring 2000 - Varaiya - Mid 1...

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