Why analyze frequency and phase?
A complex system can be probed one sine wave at a time.
Calculate amplitude reduction and phase lag for one stable first-order model.
Lead and lag networks act differently at different frequencies. We can measure that effect by testing one sine wave at a time.
Review the previous lesson: Lead and lag compensationFirst, picture it
Push a swing slowly and it follows; push near its natural rhythm and it moves more; push too quickly and it may barely respond. The amplitude ratio and timing lag depend on the input frequency.
Build the idea step by step
Apply a repeating sine input to a stable LTI system and wait for its startup motion to fade. The remaining output repeats at the same frequency. Its amplitude may be larger or smaller, and its peaks may arrive earlier or later. The amplitude ratio is the magnitude; the timing shift expressed as part of a cycle is phase. A phase of minus 90 degrees means one-quarter of a cycle late, not a motor shaft turned by 90 degrees. Repeat the test at different frequencies to see which motions pass through.
Read the equations
At a sinusoidal steady state, magnitude scales amplitude and phase shifts timing relative to the input.
What the symbols mean
- plant or loop frequency-response function
- angular frequency, in radians per second
- output phase relative to input, in degrees or radians
- imaginary unit
Use this sinusoidal relation after transients have faded, for a stable linear model. Saturation can distort a sine into another shape. Angular frequency is in radians per second; dividing by 2π gives cycles per second, or hertz.
For a first-order low-pass plant, high frequencies are attenuated and increasingly delayed. This is a shape sketch, not numerical data from the lab.
One sine through a first-order plant
Use a stable normalized plant with a 1 second time constant. Apply a sine wave of amplitude 2 at 1 radian per second.
At this frequency the gain magnitude is about 0.707 and phase is minus 45 degrees.
After startup transients fade, output amplitude is about 1.414 and its peaks occur later than the input peaks.
More detail and model assumptions
Ask how one frequency passes through
Feed a stable linear plant a small sinusoid u(t)=A sin(ωt). After transients decay, the output has the same frequency but a different amplitude and phase: y(t)=A|G(jω)| sin(ωt+φ). Repeat across ω to see which motions pass, which fade and which lag. This is the steady-sinusoid meaning of frequency response.
Angular frequency ω uses rad/s; ordinary frequency f uses cycles/s, with ω=2πf. A ‘10’ on a rad/s axis is not 10 Hz. Also, phase is a timing relation between two signals at that frequency; −90° does not mean a shaft is permanently turned −90°.
Keep the linearization in view
The motor workbench evaluates a nominal discrete loop with a selected sample/hold period. Its Bode and Nyquist curves do not include voltage clipping, noise, the changed actual inertia or anti-windup switching. Those effects remain visible in the time simulation, so disagreement between the two views is a clue about assumptions, not automatically a calculation error.
Frequency analysis is most informative when the input is small enough that the local model remains valid. If a large target step saturates the actuator, use both views: frequency plots to reason about the design model and time traces to report what that limited run actually did.
Connect the example to the lab
A motor speed loop can track a slow target change yet fail to follow rapid oscillations. That can be desirable if rapid variations are measurement noise.
Compare nominal and limited behaviorMake a prediction before moving a control.
- PREDICTWhy might a smooth Bode curve disagree with a clipped time run?
- CHANGE ONE THINGOpen the motor's frequency workbench, then increase the target until applied voltage reaches ±12 V.
- OBSERVECompare requested and applied voltage while checking whether the nominal Bode curve moved.
- EXPLAINName the assumption that fails during clipping.
Phase is frequency dependent. A single 'delay angle' cannot describe an entire system.
Check your understanding+
Does −90° phase mean a shaft physically rotated backward by 90°?
No. It means the sinusoidal output lags the input by one quarter of that signal's period.Why the next lesson follows
A Bode plot repeats this magnitude-and-phase calculation across many frequencies.