RC Low-Pass Filter Frequency Analysis

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Vocabulary flashcards covering the frequency response, transfer function, cut-off values, and high-frequency behavior of an RC low-pass filter based on the lecture notes.

Last updated 7:16 PM on 10/5/26
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8 Terms

1
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Transfer Function H(j\text{\textomega}) of an RC Low-Pass Filter

H(j\text{\textomega}) = \frac{1}{1 + j\text{\textomega} RC}

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Module |H(j\text{\textomega})|

|H(j\text{\textomega})| = \frac{1}{\sqrt{1 + (\text{\textomega} RC)^2}}

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Argument \text{ extphi}(\text{ extomega})

\text{\textphi}(\text{\textomega}) = -\arctan(\text{\textomega} RC)

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Time Constant RCRC

For R=2 kΩR = 2\,\text{k}\Omega and C=10 nFC = 10\,\text{nF}, RC=2000×10−8=2×10−5 sRC = 2000 \times 10^{-8} = 2 \times 10^{-5}\,\text{s}

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Cut-off Pulsation \text{ extomega}c

\text{\textomega}_c = \frac{1}{RC} = \frac{1}{2 \times 10^{-5}} = 50\,000\,\text{rad/s}

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Cut-off Frequency fc

f_c = \frac{\text{\textomega}_c}{2\pi} \approx \frac{50\,000}{2\pi} \approx 7\,960\,\text{Hz} \approx 8\,\text{kHz}

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Gain in dB at Cut-off Frequency

At \text{\textomega} = \text{\textomega}_c, ∣H∣=12|H| = \frac{1}{\sqrt{2}}, resulting in Gain(dB)=20log⁡(12)≈−3 dB\text{Gain(dB)} = 20 \log\left(\frac{1}{\sqrt{2}}\right) \approx -3\,\text{dB}

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High-Frequency Behavior (\text{ extomega} >> \text{ extomega}c)

For \text{\textomega} \gg \text{\textomega}_c, |H| \approx \frac{1}{\text{\textomega} RC}. The gain decreases with a slope of −20 dB/decade-20\,\text{dB/decade}, the phase tends toward −90∘-90^\circ, passing low frequencies and attenuating high frequencies.