d-q theory : AC to DC transformations

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[@RamKumar-hm2vi]

He is not a Professor bhai, he is a GOD of DQ Transformation. Such a Easy explanation in 13 minutes video

[@imadsalih8740]

Appreciate...how interesting and easy to understand when the teacher is really a teacher...

[@hewadsaad3014]

And it only took 13-14 minuts

[@chuckjls]

Such an excellent and clear explanation. Thank-you, Professor.

[@prateeknaryal4031]

very nice explanation by Umanand Sir. Big fan from a very long time.

[@shahbazkhan-jj7ve]

Best explanation of these topic so far..

[@Aaqib_Qureshi]

Best explanation. I am a big fan of Umanand sir. Hope to work with him one day

[@Attexi]

It's a really good explanation. Thanks!

[@CarlosHenrique-gx9hx]

Thank you so much, greetings from Brazil

[@emen1371]

Great explanation. Thanks

[@shashwatsingh2682]

Excellent explanation!

[@esbenmadsen5220]

There is an error in this video. On the drawing seen at 1:40 b is defined as sin(x+(2/3)*pi), however, on the phase plot b and c are switched. You can see this because when a is 0 (sin(0)=0), b should be a positive value (sin(0+(2/3)*pi)>0), and likewise c should be negative. The answer is that b and c are switched from the vector plot to the phase plot. The result is that with the transformation that is presented, the ab vector should be going clockwise.

[@rajensama]

You are right but most of the video , not only in this one, has done so. Phase B and C are switched.

[@rahulvirmani2235]

Very Nice and simple explanation sir....

[@bezibedie4259]

Well explain sir. Please can you the zero component of the DQZ transform ?

[@vijayapriyas1026]

Good morning sir,Thank you so much for your valuable explanation, may I know how Rab (alpha beta) is transferred into Rdq. I don't know the concept behind that sir.

[@حميدالحاتمي-ع2غ]

Why there is no factor of 2/3 in the Clark transformation?

[@praveens4278]

Why acb is used instead of abc..in Clark transformation..

[@grpagobo]

thanks prof. appreciate the clear explanation.

[@arthurenricocecidossantos1212]

At 5:42 you magically changed the phase sequence from abc (as you can see at 5:24) to acb. Please explain why we need to do that in order to the transformation work corectly

[@abhisekrout976]

Thank you very well explained!!

[@lukecurran1198]

thank you

[@durgaprasadchary5733]

If we transfer ABC to dqo, can we control active nd reactive components

[@xichenjiang7799]

The best thing I learned: how to draw three-phase.

[@xichenjiang7799]

@Hilario Shehan scammers

[@miftahulislam1291]

Amazing explanation Sir

[@AxElKo440]

Good lesson, but I don't understand where do I get ρ angle in the AlphaBeta to DQ transformation. Could you advise me please sir?

[@obaidurrahman1303]

alphabetha is a fixed plance where as dq is rotating, and hence the angle is added to give a point of reference of the rotation

[@بشارمحمد-د3ز]

Thank you...with regards

[@adelsalam9735]

THANKS

[@yashrajsharma1]

Thankyou

[@ginebro1930]

Thanks!

[@drnsksai]

really good

[@HOOHA333]

Hi All,I did not get one thing here, Why does alpha and beta frame needs to have imaginary and real components ?

[@smadnan6283]

Thnku sir

[@dineshth6711]

Thank you, sir. Very good explanation

[@monikalohana5812]

can anyone plz guide me that why negative e^-je is taken when converting from alpha beta to dq axis ?

[@dineshth6711]

@@monikalohana5812 Watch this video ma (from 15:00 minutes) kzbin.info/www/bejne/i3SYiZ2lfbOEn9U

[@pedrotorres8317]

Thank you for the clear explanation!

[@farhan.mukhtiar]

Sir i can't understand the 0.5 point.

[@Manster-minds]

Life saver

[@rubenboros]

Thank you!

[@sagexiao8281]

thank u, sir!

[@ammarali4197]

Thank you so much

[@rajeshbnarasimha934]

super sir...

[@명승호-v5w]

thank you !

[@rajeshjami3317]

thank you sir

[@monikalohana5812]

can anyone plz guide me that why negative e^-je is taken when converting from alpha beta to dq axis ?

[@atmanandmaya2090]

because dq-axis reference frame lag space vector...….see dq-axis theory vedio of design of photovoltaic systems

[@curly0117]

@@atmanandmaya2090 shouldn't it be cos(-p) +j*sin(-p) at 10:10?

[@bradlymcnell9466]

how does that negative .5 became positive while multiplying with cos 60. It should be negative 0.25 right? so negative .25 and negative .25 sums up and give -0.5 which overall gives 0.5 and not 1.5 Am i wrong?

[@lutosronnie]

If you look at the original phasors of b, or c, cos 60 of these phasors would lie along the x-axis but facing in the opposite direction to phase a. Therefore, the resultant is a - b - c. But b and c are negative 0.25. So the resultant becomes 1 - (-0.25) - (-0.25) = 1.5

[@maheshpadmanabh6564]

We use vector addition and not algebraic sum.

[@amitkumarpandey8956]

please use Hindi language

[@VaibhavPrakashSrivastava]

Thanks

[@brammekeful]

Thanks!

[@kangkandas2468]

thank you sir

[@leoli8164]

Thank you!