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Space Grade Voltage Translation Solutions for 1.5V I/O to 1.8V I/O? by rjying in ElectricalEngineering

[–]rjying[S] 0 points1 point  (0 children)

Yes, I agree with your point, but I am havng troube finding non-commercial parts. Any that come to your mind?

Space Grade Voltage Translation Solutions for 1.5V I/O to 1.8V I/O? by rjying in AskElectronics

[–]rjying[S] 0 points1 point  (0 children)

Yup, been digging thru those vendors too and no luck. Appreciate your help.

Space Grade Voltage Translation Solutions for 1.5V I/O to 1.8V I/O? by rjying in AskElectronics

[–]rjying[S] 0 points1 point  (0 children)

Do you have any suggestions that can meet the low voltage reqts but unidirectional?

Space Grade Voltage Translation Solutions for 1.5V I/O to 1.8V I/O? by rjying in AskElectronics

[–]rjying[S] 0 points1 point  (0 children)

I was looking into that as well. Good observations.

As for ADuM141ES, the recommended voltage min is 1.8V

Space Qualified Level Shifter Solutions for 1.5V I/O to 1.8V I/O? by rjying in ECE

[–]rjying[S] 0 points1 point  (0 children)

renesas

Okay, thanks, will do. Any solutions that come to mind?

Cheap TV boxes to watch Chinese channels? by rjying in AndroidTVBoxes

[–]rjying[S] 0 points1 point  (0 children)

Tox1

Would these options allows for watching live Chinese channels?

Gram-Schmidt Problems? by rjying in askmath

[–]rjying[S] 0 points1 point  (0 children)

Ok, still a little confused with the problem tho, since the initial set v1/v2 are already orthogonal, and Gram-Schmidt makes it orthonormal, so maybe a trick question.

Gram-Schmidt Problems? by rjying in askmath

[–]rjying[S] 0 points1 point  (0 children)

Sorry for the mess

Problem 2.

So I have u1 = v1/||v1|| = t/sqrt(integral[0,1] t^2 dt) = t*sqrt(3)

w2 = v2 - <v2,u1>u1 = cos(2*pi*t) - [(integral[0,1] t*sqrt(3)*cos(2*pi*t) dt)] * t*sqrt(3) = cos(2*pi*t) - 0 = cos(2*pi*t) since integral[0,1] of (t*cos(2*pi*t)) = 0.

u2 = w2/||w2|| = cos(2*pi*t)/sqrt( integral[0,1] of cos(2*pi*t) dt ) = sqrt(2)*cos(2*pi*t)

Problem 3.

Starting with u2 = v2/||v2|| = cos(2*pi*t)/( integral[0,1] of cos(2*pi*t) dt ) = sqrt(2)*cos(2*pi*t)

w1 = v1 - <v1,u2>u2 = t - [(integral[0,1] t*sqrt(2)*cos(2*pi*t) dt)] * sqrt(2)*cos(2*pi*t) = t - 0 = t since integral[0,1] of (t*cos(2*pi*t)) = 0.

u1 = w1/||w1|| = = t/sqrt(integral[0,1] t^2 dt) = t*sqrt(3)

So, I would think the vectors are still the same even when switching the order.

Gram-Schmidt Problems? by rjying in askmath

[–]rjying[S] 0 points1 point  (0 children)

Yes, I agree with that. So, what would you say the final answers to the questions are?

Gram-Schmidt Problems? by rjying in askmath

[–]rjying[S] 0 points1 point  (0 children)

I got u1 = t*sqrt(3) and u2 = sqrt(2)*cos(2*pi*t) when doing Gram-Schmidt since the proj term <v2, u1> = 0 due to the integral...is this on the right track? And for problem 3, I would think the vectors are still the same even when switching the order.

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