A student plots a system's Bode magnitude response using a linear frequency axis. The gain is 60 dB at 1 rad/s and falls to 0 dB at 1,000 rad/s, with a single pole at ωc = 10 rad/s. Why is this plot nearly useless for analysis?
ADecibel values cannot be displayed on a linear frequency axis
BThe interesting behavior near the corner frequency occupies an extremely narrow region of the plot; most of the horizontal space shows a flat region at one end and a compressed roll-off at the other
CGain values above 40 dB cannot be accurately represented without a log-magnitude axis
DThe phase plot cannot be overlaid on a magnitude plot with a linear frequency axis
A log frequency axis compresses the ratio between frequencies, not their absolute difference. On a linear axis spanning 0–1000 rad/s, the corner frequency at 10 rad/s occupies just 1% of the plot width — everything interesting (the transition from flat to rolling-off) is squeezed into a tiny sliver on the left. A logarithmic axis allocates equal visual space to each decade (1–10, 10–100, 100–1000), spreading the interesting features evenly. This is the fundamental motivation for Bode's design: most systems have dynamics spanning many decades of frequency, and only a log axis reveals them clearly.
Question 2 Multiple Choice
A second-order system has two poles at the same corner frequency ωc (a double pole). What slope does the Bode magnitude asymptote approach at frequencies much greater than ωc?
A−20 dB/decade, same as a single pole
B−40 dB/decade, because each pole contributes −20 dB/decade and they add on the log scale
C−6 dB/decade, because the poles interact and partially cancel
DThe slope depends on the damping ratio, not just the number of poles
Each first-order pole contributes −20 dB/decade slope above its corner frequency. For a double pole at the same corner frequency, both poles start contributing simultaneously, giving a total slope of −40 dB/decade. This additivity on the log scale is one of the key advantages of Bode plots: gains multiply in the linear domain, but they add in the log (dB) domain, so you can simply sum the contributions of individual poles and zeros. More generally, an nth-order pole or zero cluster contributes ±20n dB/decade in the asymptotic region.
Question 3 True / False
On a logarithmic frequency axis, a first-order pole contributes a slope of −20 dB/decade above its corner frequency — this asymptote is a straight line.
TTrue
FFalse
Answer: True
Above the corner frequency ωc, a first-order pole term (1 + jω/ωc) ≈ jω/ωc, so its magnitude is |ω/ωc| = ω/ωc. In dB: 20 log₁₀(ω/ωc). On a log-ω axis, this is 20 × (log₁₀ω − log₁₀ωc), which is a linear function of log₁₀ω — a straight line with slope −20 dB per decade of frequency. This piecewise-linear structure is the mathematical reason Bode's asymptotic approximation works: the log scale turns the nonlinear magnitude function into a slope that can be sketched with a straightedge.
Question 4 True / False
The asymptotic straight-line approximation of a Bode magnitude plot is exact at the corner frequency — the primary errors occur far from the corner.
TTrue
FFalse
Answer: False
The maximum error of the asymptotic approximation occurs right at the corner frequency, not far from it. At ωc, the true gain is |1 + j1| = √2, which is 20 log₁₀(√2) ≈ 3 dB above the asymptote (which predicts 0 dB at that point). The approximation is actually most accurate far from the corner frequency, where one or the other asymptote dominates. The 3 dB error at the corner is a known, systematic deviation that engineers account for when using asymptotic Bode plots — the 'corner frequency' is also called the '−3 dB frequency' for this reason.
Question 5 Short Answer
What property of logarithmic scales makes the Bode plot's asymptotic straight-line approximation possible? Explain why the same simplification would not work on a linear frequency axis.
Think about your answer, then reveal below.
Model answer: Logarithmic scales turn multiplicative relationships into additive ones. A factor of (1 + jω/ωc) in a transfer function becomes 20 log₁₀|1 + jω/ωc| in dB. When ω >> ωc, this is approximately 20 log₁₀(ω/ωc) = 20(log₁₀ω − log₁₀ωc). Plotted against log₁₀ω (the log frequency axis), this is a linear function — a straight line with slope +20. On a linear frequency axis, the same quantity would be 20 log₁₀(ω/ωc) plotted against ω itself, which is a logarithmic curve, not a line. The straightness of the asymptote is entirely a consequence of plotting a log-domain quantity against a log-domain axis.
The deeper point is that Bode plots exploit the algebraic structure of transfer functions: poles and zeros multiply in the frequency domain, but they add in the log domain. On log-log axes, each pole or zero contributes an independent straight-line segment, and the total response is simply the sum of the segments. This linearity collapses on any other scale choice.