Drawing Bode Plot From Transfer Function | Second-Order - Double Zero & Complex Poles | Example #3

[@ebarbie5016]

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

[@CanBijles]

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.

[@SMV1972]

Отличный урок! Спасибо большое!

[@CanBijles]

Спасибо!

[@aaronsarinana1654]

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!

[@CanBijles]

Thanks for your message. Indeed, you can calculate the resonance peak using the given formula. Thanks for the info.

[@ebarbie5016]

@@CanBijles No you can't! The above formula only applies to prototype system w/o zeros...

[@CanBijles]

​​@@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.

[@Rin-xy4tt]

Thank you!

[@CanBijles]

You are welcome!

[@Sandeepk-jk1fu]

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??

[@CanBijles]

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

[@calvinguo3181]

why are the double real zeroes positive? Shouldn't it be negative 100?

[@CanBijles]

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.

[@calvinguo3181]

@@CanBijles thank you!

[@CanBijles]

@@calvinguo3181 You are welcome! See the complete playlist here: kzbin.info/aero/PLuUNUe8EVqlmb8rHE6lIsfshHWfHNdn7J