Bode diagram slope contribution: In a log magnitude plot, a single pole factor (1 + s/ω) in the transfer function contributes a slope of how many decibels per octave at high frequency?
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A-20 dB per octave
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B-10 dB per octave
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C-6 dB per octave
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D-2 dB per octave
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E-40 dB per octave
Answer
Correct Answer: -6 dB per octave
Explanation
Introduction / Context:Bode diagrams are frequency response plots showing system behavior on a logarithmic scale. The slope of the magnitude plot indicates how quickly the response falls with frequency. A single pole introduces a characteristic slope change.
Given Data / Assumptions:
- We are analyzing a single pole contribution to a Bode magnitude plot.
- Frequency scaling is in octaves (doubling of frequency).
Concept / Approach:
A single pole contributes –20 dB per decade beyond its corner frequency. Since one decade = 10× frequency and one octave = 2× frequency, the per-octave slope must be converted accordingly.
Step-by-Step Solution:
Slope = –20 dB per decade.1 decade ≈ 3.32 octaves (since log2(10) ≈ 3.32).Therefore slope per octave = –20 / 3.32 ≈ –6 dB per octave.Verification / Alternative check:
Textbook Bode plot rules confirm: each pole beyond the break frequency introduces a –20 dB/decade slope or equivalently –6 dB/octave.
Why Other Options Are Wrong:
- –20 dB/octave: too steep; this is per decade, not octave.
- –10 dB/octave: incorrect conversion.
- –2 dB/octave: too shallow.
- –40 dB/octave: corresponds to two poles, not one.
Common Pitfalls:
- Confusing decade with octave.
- Mixing the effect of a single pole with that of multiple poles.
Final Answer:
–6 dB per octave.