Bode diagrams 2 - frequency response gain and phase for transfer functions

Demonstrates how to solve for the frequency response parameters of a system from atransfer function model and hence shows that the gain and phase have simple analytic dependence upon the system parameters.
Lectures aimed at engineering undergraduates. Presentation focuses on understanding key prinicples, processes and problem solving rather than mathematical rigour.

[@christopherimpey6208]

so gratefull for this i finally understand how to translate the transfer function into frequency response and phase response

[@mehdighourchian7276]

very very helpful. Thank you very much.

[@ZoranNinkovic]

Great lecture, thank you for your time and effort

[@MoSamArafat]

Sir, May I suggest creating playlists for videos discussing the same topic?
Great Videos, thank you very much!

[@johnrossiter2323]

Mohamed Samir These exist on the website with the table of contents. See the introduction video for the link. Anthony

[@rajnishraj8893]

Really good video.....

[@mieszkokudzia5656]

Great job!

[@abhideepsingh4484]

Sir, time frame 7:16, phase difference pertaining to numerator should be atan(2w/1) - .....

[@abhideepsingh4484]

+Abhideep Singh sorry sir, you had later corrected in the video. I did not listen to the full video.

[@tomahan044]

i like this

[@TechTins_Projects]

At 3:03 I always assumed you had to multiply top and bottom by the complex conjugate to get the denominator as real But here you don't do that but instead take some kind of short cut with just taking the magnitude of the denominator. Can any one explain why this is valid? This has confused me a lot.

[@johnrossiter2323]

There also some resources on this site on complex numbers. Perhaps they will help with this.Anthony

[@boxingcroatia]

very nice

[@patrickneckolaishen605]

At 3:07 when you square 3jw, why isn't it -9w ?

[@johnrossiter2323]

Patrick Neckolaishen If you do (3jw)^2 then you will get 9*j^2*w^2 which is -9w^2.

[@SooDiStUrB3d]

Dean Jin T_T And then he proceeds and substitutes the numbers into the wrong formula.
But at least I understood.

[@rezadwiutomo7000]

I think the reason why he puts 9w^2 instead of -9w^2 because it has been in modulus equation so that you don't need to consider imaginary j anymore. therefore, it'll be sqrt((3w)^2+(2-w^2)) = sqrt(9w^2+(2-w^2)^2)

[@karenroman290]

@@johnrossiter2323 I think there might be a typo, because when you move 3jw under the square root it should be -9w^2

[@jeffshrees6791]

YUCK