pop quiz 3 solutions

# V in v out sc 1 v out bracketrightbigg g m 1 g m 2 s

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V in - V out ) sC 1 - V out bracketrightbigg = g m 1 g m 2 s 2 C 1 C 2 V in - g m 1 g m 2 s 2 C 1 C 2 V out - g m 2 sC 2 V out Collecting the terms of V out and V in we get V out parenleftbigg 1 + g m 1 g m 2 s 2 C 1 C 2 + g m 2 sC 2 parenrightbigg = g m 1 g m 2 s 2 C 1 C 2 V in Multiplying both sides by s 2 and simplifying we get H ( s ) = V out ( s ) V in ( s ) = g m 1 g m 2 C 1 C 2 s 2 + g m 2 C 2 s + g m 1 g m 2 C 1 C 2 = ω 2 0 s 2 + ω 0 Q s + ω 2 0 (b) Matching coefficients from the last step shown in part (a). We note that ω 2 0 = g m 1 g m 2 C 1 C 2 ω 0 = radicalbigg g m 1 g m 2 C 1 C 2 ω 0 Q = g m 2 C 2 Q = ω 0 C 2 g m 2 = radicalbigg g m 1 g m 2 C 1 C 2 C 2 g m 2 = radicalBigg g m 1 C 2 g m 2 C 1 (c) This is a lowpass filter since it matches the standard form of a LPF as given in the last step of (a). We can also note the lowpass characteristics by examining the properties of H ( s ) at s = 0 and as s → ∞ . Dr. Vahe Caliskan 1 of 2 Handed out: April 12, 2013

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ECE 412 Introduction to Filter Synthesis Pop Quiz #3 Solutions University of Illinois at Chicago Spring 2013 Extra Workspace Dr. Vahe Caliskan 2 of 2 Handed out: April 12, 2013
• Spring '11
• Vahe
• The Current, Low-pass filter, Vout

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