Drawing Bode Plot From Transfer Function | Second-Order - Double Zero & Complex Poles | Example #3
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- Опубликовано: 21 окт 2022
- In this video, we will discuss how to draw the Bode plot from a given transfer function. This is Example 3 in this series where we discuss double real zero and complex poles, and required corrections in magnitude peaking and phase. We will workout step by step how to convert the transfer function into a standard form and how to generate and use the magnitude and phase slopes to draw the Bode plot.
We will use the asymptotic approximations to draw the magnitude and phase plot for the Bode plot. Finally, we will compare these drawings using MATLAB simulations and discuss the deviations.
For more videos on drawing Bode plot from transfer function, see the playlist: Drawing the Bode plot from Transfer Function: • Drawing the Bode plot ...
We follow a logical procedure and provide the solutions step by step such that it clear for you what and why we carry out a specific action.
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The error calculation using the 20log(2*zeta) does not apply when you have zeros. It only apply to prototype 2nd order systems. The error in this case is 2.7dB not 1.93dB
The assumption made for analytic calculations, is that the system is all-pole second-order system. The formulas for the peak time, damping ratio, etc. are also valid for all-pole second-order system only. The approximate values are calculated using these formulas for other systems, also including zeros. Yes, the zeros can have a significant effect depending on their location with respect to the dominant poles of the system.
Отличный урок! Спасибо большое!
Спасибо!
Thank you!
You are welcome!
Good video. Just a remark, for the resonance peak i use the expression 1/(2*zeta*sqrt(1 - zeta^2)) which give the exact peak you show in your matlab plot. Thanks!
Thanks for your message. Indeed, you can calculate the resonance peak using the given formula. Thanks for the info.
@@CanBijles No you can't! The above formula only applies to prototype system w/o zeros...
@@ebarbie5016 The assumption made for analytic calculations, is that the system is all-pole second-order system. The formulas for the peak time, damping ratio, etc. are also valid for all-pole second-order system only. The approximate values are calculated using these formulas for other systems, also including zeros. Yes, the zeros can have a significant effect depending on their location with respect to the dominant poles of the system.
Why do we have peaking when zeta is low or why peaking beacuse of conjugate poles. Can you explain peaking graph when zeta is low??
Zeta is called the damping ratio. When zeta is between 0 and 1 you will have overshoot in the time domain for step input. When it is 1 or larger there is no overshoot.
In the frequency domain, the zeta will determine the peaking given in the Bode plot. If the zeta is smaller than 1/sqrt(2), you have gain peaking in the frequency domain. The proof for this requires some detailed derivation and most Control Systems books will explain this in detail. I found this after a quick search: engineeronadisk.com/book_modeling/bodea4.html
why are the double real zeroes positive? Shouldn't it be negative 100?
The poles or zeros which are negative on the complex plane will be designated by their frequency, and this is always a positive value. For example, a pole at s = -10 means that the frequency of this pole is 10 rad/sec. Similarly for the zero.
@@CanBijles thank you!
@@calvinguo3181 You are welcome! See the complete playlist here: ruclips.net/p/PLuUNUe8EVqlmb8rHE6lIsfshHWfHNdn7J