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A frequency filter (different from the previous problem) is shown below. K Vout -0000 a) Derive an expression for Vout/V, as a function of the angular frequency w of the voltage source. b) Show that when w→0, Vout/V. 1. This means that the low frequencies pass. c) Show that when woo, Vout/V→ 0. This means that the high frequencies do not pass.

A frequency filter (different from the previous problem) is shown below. K Vout -0000 a) Derive an expression for Vout/V, as a function of the angular frequency w of the voltage source. b) Show that when w→0, Vout/V. 1. This means that the low frequencies pass. c) Show that when woo, Vout/V→ 0. This means that the high frequencies do not pass.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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Problem 10 A frequency filter (different from the previous problem) is shown below. V a) Derive an expression for Vout/V, as a function of the angular frequency w of the voltage source. b) Show that when w→0, Vout/Vs →1. This means that the low frequencies pass. c) Show that when woo, Vout/Vs → 0. This means that the high frequencies do not pass.
Transcribed Image Text:Problem 10 A frequency filter (different from the previous problem) is shown below. V a) Derive an expression for Vout/V, as a function of the angular frequency w of the voltage source. b) Show that when w→0, Vout/Vs →1. This means that the low frequencies pass. c) Show that when woo, Vout/Vs → 0. This means that the high frequencies do not pass.
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