ECE2260F09_HW3p2soln

ECE2260F09_HW3p2soln - 2260 F 09 HOMEWORK #3 prob 2...

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2260 F 09 HOMEWORK #3 prob 2 solution E X : + v o v i + R 1 = 20 k Ω R 2 = 30 k Ω R 3 = 3 k Ω L = 500 μ H a) Determine the transfer function V o /V i . Hint: use a Thevenin equivalent on the left side. b) Plot |V o /V i | versus ω . c) Find the cutoff frequency, ω c . S OL ' N : a) The technique employed in this solution focuses on writing the transfer function, H ( j ω ), as a product of a real-valued scaling factor and a simple RLC circuit whose transfer function, H 1 ( j ω ), involves only one L , one C , and one R . H ( j ω ) = k H 1 ( j ω ) where k real number The virtue of this approach is that cutoff frequencies are unaffected when a transfer function is scaled by a real number. Since the magnitude of the transfer function at the cutoff frequency is defined relative to the maximum magnitude of the transfer function, and the magnitudes are scaled by the same real number at every frequency, the cutoff frequency is unaffected by a real scaling factor multiplying a transfer function. Thus, the cutoff frequency of H ( j ω ) is the same as the cutoff frequency of transfer function H 1 ( j ω ). This idea will be exploited in part (c), but it also helps to clarify the derivation of the transfer function. Our first step is to take a Thevenin equivalent of the input voltage, V i , and R 1 and R 2 .
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+ V V i + R 1 R 2 + V + R Th = R 1 R 2 V Th = V i R 2 R 1 + R 2 We find v Th as the voltage across R 2 when nothing is connected across R 2 . (In other words, we remove L and R 3 when determining the Thevenin equivalent, and v Th is the open-circuit voltage across R 2 .) We have a simple voltage-divider.
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ECE2260F09_HW3p2soln - 2260 F 09 HOMEWORK #3 prob 2...

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