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ECE108HW%235Sol_06

# ECE108HW%235Sol_06 - ECE 108 Hw#5 Solution 1 Problem 3.22(a...

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ECE 108 Hw#5 — Solution 1. Problem 3.22 (a) Fig (b) Here 0 2 6 and 6 3 T π π ω = = = Then () ( ) ( ) ( ) 6 6 6 6 2 ( 1) 1 2 f t t t t t δ δ δ δ = + + − + and ( ) ( ) 2 2 2 2 0 1 1 1 1 6 6 6 6 1 1 1 1 6 6 6 6 1 1 cos 2 cos 3 3 o o o o o o o o jk jk jk jk k jk jk jk jk o X e e e e e e e e k k ω ω ω ω ω ω ω ω ω ω = + = + = Hence ( ) ( ) ( ) [ ] 0 2 2 2 2 2 2 2 1 cos 2 cos 3 3 2 cos cos for 0 3 3 k k k o o o o S S X k k k k jk k k k k ω ω ω ω ω π π π = =− =− = 0 1 2 X =

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2 3.22 (a) Figure (c) in the textbook. Here 0 2 3 and 3 T π ω = = Then () () () () ( ) 2 3 3 2 3 3 2 d x t dg t f t t t dt dt δ δ = = =− + + Therefore 0 0 0 0 0 2 3 2 2 2 2 2 2 0 0 0 0 1 1 1 9 1 for 0 4 1 2 jk k jk k k jk jk k k k F e F e G jk jk G F e e X k jk k k k X ω ω π ω ω ω ω ω ω π =− + − + = = − + = = =− = =
3 2. Given the RC circuit shown (a) Determine the Transfer Function of the System, assuming the input is and the output ( ) i v t is ( ) 0 v t First write the differential equation () ( ) () () 0 0 1 t dv t v t i d i t C C dt τ τ −∞ = = () () ( ) 1 t i v t Ri t i d C τ τ −∞ = + Substitute the expression for i ( t ) into the second equation () () ( ) () () 0 0 0 0 1 t i dv t dv dv t v t RC C d RC v t dt C d dt τ τ τ −∞ = + = + Put this differential equation in standard form () () () 0 0 1 i dv t v t v t dt RC + = Then ( ) 1 1 1 1 1 H s RC s s RC = = + +

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4 (b) For , determine the Amplitude and Phase Characteristic of this system 0 2 ω π = ( ) 0 0 1 1 1 1 2 H jk jk j k ω ω π = = + + ( ) ( ) ( ) 0 2 2 1 0 1 1 4 tan 2 H H jk k k k ω π θ ω π = + =−
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