ECE2260F08_HW4p1soln

ECE2260F08_HW4p1soln - 2260 F 08 HOMEWORK#4 prob 1 solution...

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2260 F 08 HOMEWORK #4 prob 1 solution 1. v o ( t ) + + v i ( t ) R 1 = 1.2 k ω 1000 2000 β = 1 M 450 r /s ω 0 = 1 M 1600 r /s = 625 r /s 1.0 0.75 H ( j ω ) 0 Given the resistor connected as shown and using not more than one each R, L, and C in the dashed-line box, design a circuit to go in the dashed-line box that will produce the band-pass |H( j ω )| vs. ω shown above. That is: max H ( j ω ) = 3 4 and occurs at ω 0 = 1 M 1600 r/s = 625 r/s bandwidth β = 1 M 450 r/s H ( j ω ) = 0 at ω = 0 and lim ω→∞ H ( j ω ) = 0
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S OL ' N : This is a band-pass filter with a center frequency at ω 0 and a bandwidth of β . It differs from a standard band-pass only in having a maximum magnitude response of 0.75 instead of 1.0. We can achieve this reduction by placing a 3.6 k resistor from top to bottom just to the right of R 1 : v o ( t ) + + v i ( t ) R 1 = 1.2 k R 2 = 3.6 k Taking the Thevenin equivalent of the input source and the two R 's, we have the following equivalent circuit: v o ( t ) + + R Thev = 1.2 k ||3.6 k = 900 3.6k 3.6k + 1.2k v i ( t ) = 3 4 v i ( t ) Now we use a band-pass configuration with
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