Questions: Frequency Response and Bode Plots

Question 1

A circuit's transfer function has a single real pole at ω_p = 1000 rad/s. Well above ω_p, the asymptotic Bode magnitude plot falls at approximately:

• A. -10 dB/decade
• B. -20 dB/decade
• C. -40 dB/decade
• D. -6 dB/octave only, which is not the same as -20 dB/decade

Answer: (B)

Each real pole contributes a -20 dB/decade slope above its break frequency. This is equivalent to -6 dB/octave (since one octave is a factor of 2 in frequency, and log₁₀(2) ≈ 0.301, so 20 × 0.301 ≈ 6 dB per factor-of-2). A common confusion: -6 dB/octave and -20 dB/decade are the same rate expressed in different units.

Question 2

At the -3 dB frequency of a low-pass filter, the output signal power is half the input signal power.

• T. True
• F. False

Answer: True

The -3 dB point means |H(jω)| = 1/√2 ≈ 0.707 in voltage ratio. Since power is proportional to voltage squared, power ratio = (1/√2)² = 1/2. So the -3 dB frequency is correctly called the half-power frequency. A misconception is that -3 dB represents a sharp cutoff — in reality, signals above this frequency are attenuated but not eliminated.

Question 3

Why is the frequency axis on a Bode plot logarithmic rather than linear?

Model answer: A logarithmic frequency axis allows a single plot to span many decades of frequency (e.g., 1 Hz to 1 MHz), and it causes the asymptotic magnitude segments to appear as straight lines with slopes in dB/decade. This makes pole and zero locations easy to identify visually and enables rapid hand sketching from the factored transfer function.

On a linear scale, low frequencies would be compressed into a tiny region and high frequencies would dominate the plot. The log scale gives equal visual space to each decade, matching how human perception and circuit behavior both span wide ranges. The straight-line asymptotes arise because each pole factor contributes a term proportional to log(ω), which plots as a straight line.