Bode plots: gain and phase
Two aligned plots show how a loop responds across slow and fast inputs.
First, picture it
A Bode magnitude plot answers 'how much does this frequency get through?' The phase plot answers 'how late does it arrive?' The logarithmic axis compresses a huge frequency range into a readable map.
What the model says
Magnitude is 20 log₁₀|L(jω)| dB and phase is ∠L(jω), usually for an explicitly named open loop L. Zero dB means |L|=1, not zero signal. At gain crossover ωgc, the open-loop magnitude equals 1. At phase crossover ωpc, its phase reaches −180° modulo 360°. These landmarks feed gain and phase margins, subject to the loop's assumptions.
M_dB(ω)=20 log₁₀|L(jω)|; 0 dB ⇔ |L|=1A +20 dB amplitude ratio means 10×, not 20×.
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.
Make it concrete
If a loop is +6 dB at some frequency, its sinusoidal output amplitude is roughly twice its input amplitude at that frequency.
Do not confuse plant G, controller C, open loop L=CG and closed loop T=L/(1+L). Their Bode plots answer different questions.
Check your understanding+
What does 0 dB mean on a loop-gain Bode plot?
The magnitude of the open-loop transfer function is one at that frequency.