Bode plots and excitation signals Bode magnitude and phase plots are generated from a system’s steady-state response to which class of input?

Introduction / Context:
Bode plots summarize a linear time-invariant (LTI) system’s frequency response by showing gain and phase versus frequency. To obtain these characteristics, we conceptually excite the system with inputs at single frequencies and measure the steady-state output amplitude and phase shift. Identifying the correct excitation class ensures the right mental model for frequency-domain analysis.

Given Data / Assumptions:

• System is LTI and stable for the frequencies of interest.
• Frequency response G(jω) is defined via sinusoidal steady-state.
• Magnitude and phase are plotted over a logarithmic frequency axis.

Concept / Approach:
By definition, the frequency response is the ratio of the steady-state output to input when the input is a sinusoid of frequency ω: u(t) = A sin(ω t). For an LTI system, the steady-state output is also sinusoidal at the same frequency but with different amplitude and a phase lag/lead. Sampling this response across ω yields the data for the Bode magnitude and phase plots. Although other test signals (impulse, step, ramp) can be used to derive frequency response indirectly via transforms, the foundational definition uses sinusoidal input.

Step-by-Step Solution:

Verification / Alternative check:
Experimental frequency response analysis conducts chirp or swept-sine tests; computationally, FFT of sinusoidal steady-state also yields G(jω).

Why Other Options Are Wrong:

• Impulse/Step/Ramp: Useful for time-domain characterisation; frequency response is defined via sinusoidal excitation.

Common Pitfalls:
Confusing “obtained from” with “related to”: while impulse/step responses can be transformed to frequency domain, the canonical construction of Bode data is sinusoidal steady-state.

Final Answer:
Sinusoidal