Bode diagrams 10 - sketching with asymptotic information

Builds on the previous video by showing how some asymptotic information in the Bode plot can be obtained with minimal or no computation. This asymptotic information can be used as the basis for surprisingly accurate Bode diagram sketching for systems with multiple simple poles and zeros and requires minimal extra computations.
Lectures aimed at engineering undergraduates. Presentation focuses on understanding key prinicples, processes and problem solving rather than mathematical rigour.

[@andreialexandru7027]

Brilliant! Simply Brilliant! Thank You, Sir!

[@tomahan044]

for each video today of this amazing channeli've been saing i like this.
realy appreciate this, you put this online for free and thats amazing

[@adnanalhomssi3706]

Billion Thanks from heart , wish you were my Instructor

[@shouchengliu210]

Thank you sooooooo much!

[@cansuayan2845]

sağol john.

[@guoitdreuy]

Hi I was wondering for example 2, for G(jw) = jw/6, I underhand that the jw part gives rise to the slope. What about the denominator of 6? Does this not cause a shift or offset of -15.6dB? Is it ignored? Thanks

[@johnrossiter2323]

guoitdreuy log(w/g)=log(w)-log(6) so yes there is a shift but this does not affect the slope. Following the asymptotes from left to right automatically ensures the shift is taken account off correctly in the asymptote plot.
Anthony

[@guoitdreuy]

+John Rossiter Ah right OK, thanks a lot! Euan

[@minabagheri3735]

on 1:26 at the bottom of the page for calculating the modulus shouldn't it be 1/w instead of w?

[@ThomasTurkington]

Thanks so much Anthony. Just a question if you don´t mind... At 3:44, should it not be -24 dB/dec, since it´s [20log 3/5] - [20log w] = -4.4 - 20 = approx -24 dB/dec ? And similarly at 4:16 the slope should be approx -30 dB/dec, taking into account the [20log 3] - [40log w]?
Thanks!!

[@johnrossiter2323]

You need to distinguish between slope and value. The slope is only affected by the terms with an w in them.Anthony

[@karenroman290]

15:23 Why didn't you add the 1/s into the table? Wouldn't that affect the slope value

[@johnrossiter2323]

There is no corner frequency so impact on slope and gain is the same throughout.Anthony

[@nice-one]

19:27 on wat basis do we skect -20db/dec slope where does that line gonna meet on y axis, how to draw slope of 20 db/dec?

[@johnrossiter2323]

+Sumit Shah I suggest just using a ruler and graph paper. It should then be easy to sketch the correct slope.Anthony

[@nice-one]

19:25 -20db/dec line meets above 10db in yaxis , what exctaly is tht point and how can we find ?

[@johnrossiter2323]

+Sumit Shah if you are doing a sketch an estimate is good enough. These are asymptotes and not the exact Bode plot anyway.Anthony

[@Jnglfvr]

These tutorials are hampered by the failure to convert the transfer functions to time constant form. Doing so would avoid the tedious process of having to calculate a separate gain for each factor. The low frequency gain is factored out and 20*log10(K) is plotted and added to the other gain plots which are 0 dB until the corner frequency and then rise (zero) or fall (pole) by 20 dB/decade for each factor.

[@keenwasp]

There is a mistake in 3:45.
I think you meant 20log (3/(5w)) instead of 20log (3/(2w))

[@johnrossiter2323]

+keenwasp Fair point thanksAnthony