How Do Bode Plots Show Frequency Response?

TL;DR

A Bode plot shows an LTI system’s steady-state frequency response using separate gain and phase plots against frequency on a logarithmic axis. Gain is expressed in decibels using 20 log base 10 of the amplitude ratio, while phase is traditionally shown in degrees. Both quantities can be calculated by substituting s = jω into the system’s transfer function.

Transcript

welcome back to control system lectures as always if you have any questions or comments on what I'm about to say please leave it in the comment section below and I'll address it in this lecture I want to present an introduction to frequency response methods and in particular I'm going to describe the bod plot how it's generated and used to visualiz... Read More

Key Insights

• An LTI system is constructed from constant multiplication, differentiation, integration, and addition, with division and subtraction represented through reciprocal multiplication and negative addition. These operations preserve the shape and frequency of a sinusoidal input while allowing its amplitude and phase to change.
• A sinusoidal input is especially useful for frequency-response analysis because its output from an LTI system remains sinusoidal at the same frequency. The system’s effect can therefore be characterized completely at each frequency by the output-to-input amplitude gain and the associated phase shift.
• A Bode plot is a graphical representation of frequency response using two separate plots. One shows gain and the other shows phase, and both place frequency on a logarithmic horizontal axis so behavior can be visualized across the entire frequency spectrum.
• Decibel gain is calculated from an amplitude ratio using 20 log base 10 of that ratio. The factor of 20 follows from power being proportional to amplitude squared and from applying logarithm properties to the power-ratio expression multiplied by 10.
• The example system combines a path with constant gain two and a path that integrates the input. For a sinusoid at 0.5 radians per second, the resulting output amplitude is approximately 2.83, corresponding to about 9 dB, with a phase shift of -0.785 radians or -45 degrees.
• Steady-state frequency response is obtained from a transfer function by setting the Laplace variable s equal to jω. This substitution removes the real component associated with transient behavior and allows gain and phase to be calculated directly as functions of angular frequency.
• The example transfer function is (2s + 1)/s, which becomes 2 - j/ω after substituting s = jω. Its real component remains two, while its imaginary component changes with frequency, creating frequency-dependent magnitude and phase values.
• Complex-plane geometry provides gain and phase without trigonometric signal-combination identities. The magnitude is the distance from the origin to the transfer function’s complex value, calculated with the Pythagorean theorem, while phase is its angle from the positive real axis, calculated using arctangent.

Questions & Answers

Q: What is a Bode plot used for in control systems?

A Bode plot is used to visualize an LTI system’s steady-state frequency response across the frequency spectrum. It consists of two graphs plotted against frequency on a logarithmic horizontal axis. The first graph shows gain in decibels, and the second shows phase in degrees. Together, they reveal how the system changes the amplitude and timing of sinusoidal inputs at different frequencies.

Q: Why do sinusoidal inputs retain their frequency in an LTI system?

Sinusoidal inputs retain their frequency because the mathematical operations used to construct an LTI system preserve the shape and frequency of a sine wave. These operations include multiplication by a constant, differentiation, integration, and addition. Consequently, the output remains a sinusoid at the input frequency, although its amplitude can change and its phase can shift in time.

Q: How is amplitude gain converted to decibels for a Bode plot?

Amplitude gain is converted to decibels by taking the base-10 logarithm of the output-to-input amplitude ratio and multiplying the result by 20. The factor of 20 appears because sinusoidal power is proportional to amplitude squared. Applying logarithm properties moves the exponent two outside the logarithm, changing 10 times the logarithm of amplitude squared into 20 times the logarithm of amplitude.

Q: How can frequency response be calculated from a transfer function?

Frequency response can be calculated by replacing the Laplace variable s with jω in the system’s transfer function. Frequency-response analysis concerns steady-state behavior after transient effects have died out, so the real component of s is set to zero. The resulting complex expression provides a magnitude that represents amplitude gain and an angle that represents phase shift for every selected frequency ω.

Q: What is the transfer function of the example system?

The example system sends its input through two parallel paths, one multiplying it by two and the other integrating it, before adding the results. Its transfer function is (2s + 1)/s, which can also be written as 2 + 1/s. For steady-state frequency analysis, substituting s = jω changes this expression to 2 - j/ω.

Q: How are gain and phase found from a complex transfer function value?

Gain and phase are found by plotting the complex transfer function value on real and imaginary axes. Gain is the length of the line from the origin to that point, calculated as the square root of the real component squared plus the imaginary component squared. Phase is the angle between that line and the positive real axis, which can be calculated using a sign-aware arctangent operation.

Q: What gain and phase does the example produce at 0.5 radians per second?

At an input frequency of 0.5 radians per second, the example system produces an output amplitude of approximately 2.83 for an input amplitude of one. This corresponds to approximately 9 dB of gain. The phase shift is -0.785 radians, which is -45 degrees. The output remains sinusoidal and retains the input frequency of 0.5 radians per second.

Q: How does the example system behave as frequency increases?

For the example frequency response 2 - j/ω, the real component stays fixed at two while the magnitude of the imaginary component decreases as frequency increases. The gain therefore becomes shorter on the complex-plane diagram and approaches an amplitude gain of two, or 6 dB, as frequency approaches infinity. The phase simultaneously approaches zero degrees as the imaginary component approaches zero.

Summary & Key Takeaways

• A sinusoidal input to a linear time-invariant system produces a sinusoidal output at the same frequency. The output can differ only in amplitude and phase. This property makes sinusoidal signals especially useful for characterizing how an LTI system responds across a spectrum of input frequencies.

• A Bode diagram separates frequency response into gain and phase plots, with frequency displayed on a logarithmic horizontal axis. Amplitude gain is converted to decibels using 20 log base 10 of the amplitude ratio, while phase shift is conventionally expressed in degrees.

• Frequency response can be obtained efficiently from a transfer function by replacing s with jω for steady-state analysis. For the example transfer function (2s + 1)/s, this substitution gives 2 - j/ω. Its complex magnitude determines gain, and its angle relative to the positive real axis determines phase.