Anyway I am quite sure of my solution

Bye

I will ask the EE guys

@heather problems, frankly

@EmilioPisanty it depends on my mood - if I'm trying to waste time, it's fabulous! otherwise, yeah, it can be annoying.

Raw text was an ok way to build tables and diagrams in the 90s

It's not that you can waste time doing it, is that there's no way at all to do a bunch of basic stuff that's not ridiculously time intensive

A realistic npn transitor example (electriciantraining.tpub.com/14179/css/…). Usually no more than 2 to 5 percent of the total current is base current (IB) while the remaining 95 to 98 percent is collector current (IC)

$I_C+I_B=I_E$

Base currents governs the magnitude of collector current as I said.

That is true.

Which leads to amplification.

But base currennt is tiny.

@YashasSamaga When did I say it isn't ?

-_-

Btw @DanielSank did you see Ruslan's second autoionization question?

@2017 go back and read the previous msgs from the start

and that isn't problem

@YashasSamaga At the first even you made a silly error when you said that 19.6 is the answer.

the question is unsolvable

It isn't

I realized the mistake later

the question wording is wrong

@YashasSamaga Why?

Phew! From half an hour I am asking you why the question is wrong!!!

too hard to type in one hand

@2017 it is not solvable

show me ur sol

@YashasSamaga Again -_-

Tell me why!!

Ok bye

Have dinner

See you tomorrow

I told u the mistake long ago

huh

@YashasSamaga You didnt

@EmilioPisanty Yeah I saw that.

.

Can't answer it though.

i don't think.

Read what I replied next

I told you what exactly was wrong

NO^

Ic+Ib=Ie

Why should Ic=98+emitter current?

Makes no sense.

I said Ic/Ib=98/2

When did I say Ib is 98 percent of Ic?

Phew.

I still didn't find anything wrong.

And also confirmed my solution with my textbook.

Have a good day/night :) See you. We can quarrel more tomorrow!

a challenge, @DanielSank - how would you explain a quantum computer using only the 1000 most common words? If interested, you can use this.

(if not, of course, that's fine as well =)

:35696987

> It is a thing that decides answer to hard problem using not obvious way of how the world works.

lol

So non-specific.

lol

mine was

> A computer that uses what we know about small pieces of matter to do things a normal computer can't do. Some things this different computer can do are taking very, very big numbers and finding numbers can only be grouped into groups of one and themselves and that the very very large number can be grouped into, or finding a name or place within a very large group of names or places.

Wow

good

eh, not really

it's rather vague

Better than mine.

yours is awesome =P

it really does summarize quantum computing rather effectively.

sort of

It's really vague though.

@YashasSamaga Give me the complete question

It is not clear whether this is small-signal or large-signal

What is the bias of the transistor

beta is only defined for small-signal

and in DC calculations, we only use the circuitary around the transistor to calculate Ib and Ic (DC)

Is it the DC current or the signal being amplified?

I can guess (from the typical values for a BJT) that this is the small-signal current.

@Mostafa The situation is something like this ^

Once again I find that I have been unable to resist trying to 'help' one of our ... avid amateur physicists (?) ... stop screwing around and get serious.

If 100 emitter electrons are injected into base

I should know better, but I just can't seem to help myself.

Then 98 of it crosses over to collecter

And 2 of it passes as base current

@2017 This is the total current in the photo.

@Mostafa What?

The circuit is only a DC circuit for biasing the transistor

It doesn't have any signal.

@Mostafa Or it has a boring constant signal with exactly the same (lack of) information content.

I assumed it to be a DC source.

Yeah, the question isn't clear enough

But that is the best guesstimate I could come up with.

@2017 Is that photo from the question itself (or your guess)?

@Mostafa Na, the question doesn't have any photo

Tbh, as they mentioned "injection of electrons into the base" it seems like a common emitter configuration. The base emitter voltage is anyway the control factor which injects electrons from the emitter into the base.

@2017 In that case, it's almost clear that it is the AC signal (not total current); especially since it's a power transistor.

And then, the base AC signal would be just that remaining %2 current.

(Which is 0.4 mA)

and I assumed the transistor is working in its active forward region.

@Mostafa "it's almost clear that it is the AC signal ". Why not a DC signal like a cell applied for a short period of time? Either ways we get the same result.

What makes you think that it must be an AC signal ?

First, why do you think we get the same result? even assuming its AC signal, you have to assume the transistor is in its active forward region.

@2017 because 20mA is too low for collector current

Especially for a power transistor

@Mostafa you may be interested in a project that @BernardoMeurer and I are starting.

@Mostafa Can't there be a low potential/voltage DC supply which produces low collector current? I agree with your point though.

Bernardo is frustrated by the lack of quantitative analysis in debates about the relative quality of various pieces of audio equipment.

For example, people on the internet (and manufacturer advertisements) will boast a certain signal to noise ratio for a DAC, but it is unclear how that affects audio quality or what levels of SNR are really needed.

So, we are starting a blog.

Bernardo is an audiophile. He knows the hardware and he knows the sorts of claims people make. I happen to be reasonably good at noise calculations and other related calculational elements. I also know how to actually measure things in the lab.

We aim to provide a series of posts in which we inform readers with real analysis and real data, e.g. comparing various DAC's to one another with actual waveforms and spectral anslysis.

As you are in EE, I thought you might be interested.

github.com/digitalphase/digitalphase.github.io/tree/master

@DanielSank I'm not sure that link goes where you think it goes

ahaha woops

github.com/digitalphase/digitalphase.github.io/tree/master

Can you fix it @ACuriousMind?

Thy will be done ;)

thanks

on earth as it is in heaven

Give us this day... wait isn't this a physics forum?

@Mostafa Yes, the information is available, but it's very hard to digest even for me.

@DanielSank our daily bread

@2017 The transistor has several different operation modes

@heather Yes yes.

@2017 If you apply the wrong voltage, it may not even turn on. Or it can damage the junctions, .... only under very special conditions it enters the useful mode (active forward region). This is why we need to put a lot of bias circuitry around the transistor (or any other IC which doesn't have them inside itself)

(based on Oppenheim's DSP book)

Just fell victim to the question-closed-while-writing-and-answer thing. :b(

Thankfully the script notified me, so I have saved the answer to a emacs buffer pending improvement to the question.

@dmckee Oh, was that my fault?

@dmckee So you prefer emacs over Vim? :/

@ACuriousMind Yeah. But you are right that the particular question wasn't any good.

I could always write the good version of the question.

@Mostafa I made that decision back when there was just vi, so I don't really know ... I never used vim.

And I know enough vi to do the really important things with a system: starting with 'get emacs installed'.

Ahhh, a new edition of Editor Wars! ::gets popcorn::

War! What's it good for?

I think gnu-emacs was in version 14 or 15 back then, and you had to decide between gnu and X emacen.

vim is cool.

@ACuriousMind don't forget some candy - Reese's Pieces are always nice.

And a soda?

@heather No idea what those are

@ACuriousMind ET's favorite candy, of course.

^

::feels old::

Phone home.

Also amused at how little attention that movie has gotten since it passed out of the public consciousness a few months after release.

Talk about a one-hit wonder.

@dmckee Man, I saw the German dub of that movie over a decade ago, not gonna remember the type of candy that was in it

I still haven't watched it.

And I have to say that "peanut butter candy" does not sound that enticing to me

You got your peanut butter on my chocolate.

Back to the editor thing ... I made that decision when Eight Megs And Continuously Swapping was a funny backronym of the name, because it had some truth to it.

@ACuriousMind Wait for the AMA with dmckee. I'm asking a modified version of this in every future AMA. Although it did have Emacs or Vim in its first edition. Not gonna talk more on this until then.

@ACuriousMind but it's SO GOOD

@ACuriousMind WHAT

this is a tragedy

and must be rectified

@heather I...don't think they exist here

@ACuriousMind ::jaw drops::

I am so, so sorry

But my bosses had gotten a truckload of ex-SSC computers and we had scavanged the memory and disks from two of every three to buff-out the others, so I had twenty megabytes in my desktop machine.

I was swimming in memory and could afford both X and emacs!

@heather what's reese's pieces?

@ACuriousMind ^

My opinion on recent events:

@DanielSank that's one hell of a broad category

Vim, Reece's Pieces are good, yes we shall overcome the theorists.

@DanielSank oh, that kind of event and that degree of recent

@EmilioPisanty peanut butter and chocolate in a candy shell.

@DanielSank sounds tasty enough, but why all the diehard fandomness?

en.wikipedia.org/wiki/Reese's_Pieces?wprov=sfsi1

@Mostafa ???

@Mostafa We have patience...

We will outlast you

@EmilioPisanty it's not diehard fandomness, it's shock that no one's aware of it's awesomeness.

3/10 for Reece's Pieces

Ya know what's good? Actual, savory peanut butter on its own!

@Danu yeah, they're good, not amazing.

Not this weird messed up artificial (extremely sweet) stuff that I encountered in the US

Messed up?

It tastes like some R&D guys had too much time to play around with it

look, regular PB is great, but Reese's Pieces are great too...

@Danu Ew.

@ACuriousMind don't listen to him

Damn it's Reese's, lol

Reece's sounds so much more logical

@ACuriousMind Have you ever had it? I think some supermarkets here also sell the US kind, or a variation thereof. Very sugary.

Btw @ACuriousMind what's the situation with the holidays next week

Do I need to do groceries tomorrow for the next 3 days?

...what holidays?

@Mostafa What inspired that post?

@Danu If you mean Aschermittwoch etc., these are not law-mandated holidays and shops should be open

Yea... Munich

@ACuriousMind you can also call it Ash Wednesday, you know

@DanielSank The trend toward experimental/applied discussions in the past few days...and the growing number of them (us)

@EmilioPisanty I actually didn't know what it was called in English!

@Mostafa I see.

And I was too lazy to google :P

@ACuriousMind Ash Wednesday tends to translate pretty transparently

Hmm, maybe stores are open after all

Good :-)

Carnival Tuesday somewhat less so

hence Mardi Gras and Shrove Tuesday

What about Green Thursday?

@ACuriousMind what's Green Thursday?

The day after Ash Wednesday

Gründonnerstag in German

@ACuriousMind first time I see a culture give that a specific name

other than "the second day of Lent"

no, wait

I'm an idiot, it's not the day after Ash Wednesday, it's the day before Good Friday

@ACuriousMind ah, well, then that's something else entirely

> Maundy Thursday (also known as Holy Thursday, Covenant Thursday, Great and Holy Thursday, Sheer Thursday, and Thursday of Mysteries)

so a variety of names even in English, it appears

Green Thursday is the day before good Friday?

What will they think of next :-/

Yo @ACuriousMind this true?

> Some countries, such as Germany, have laws prohibiting certain acts, such as dancing and horse racing, that are seen as profaning the solemn nature of the day.

(Good Friday, that)

@EmilioPisanty Yes, it's called the *Tanzverbot*/prohibiton of dancing and there are regular (small) protests against it on the days where it applies

@ACuriousMind that's ridiculous

just imagine what this chatroom's US contingent would make of that if they hear of it

@EmilioPisanty I think we've had that discussion here already in the past (and also about most shops having to close on holidays)

@ACuriousMind yeah, we talked about shops, but nothing this ridiculous came out at the time

It's a conservative country.

@skillpatrol what?

what?

@EmilioPisanty I agree it's ridiculous, but it currently doesn't seem to be on the agenda of any party to lobby to change it

Yo how is $\langle \phi(0) \phi(t) \rangle$ defined for some random quantum system?

Doesn't help that these sorts of laws are on the state level, so you'd have to pass through all 16 state legislatures to abolish it in the whole country.

Like, supposing I'm in state $\Psi$, how do I write that out more explicitly?

$\langle \Psi | \phi \phi | \Psi(t) \rangle$ ?

@ACuriousMind So it's at state level but every Bundesland has it?

@DanielSank What is $\phi(t)$?

@EmilioPisanty Yep, although the specific days on which it applies vary slightly

@ACuriousMind Can you just guess what I'm talking about?

If I knew what $\phi(t)$ were I wouldn't have to ask.

@DanielSank With you I never know whether you're talking signal processing, QM, or something else entirely ;)

I'm trying answer roughly the question of how two time separated measurements on a quantum system are correlated.

@ACuriousMind Quantum.

Sorry, didn't realize I was leaving out critical information.

Ok, for some reason when I try to Heisenbergify it, I don't like the result. Trying again...

No, it's wrong

aha!

Ah

how do you make a variable upright in LaTeX?

You have $\langle \psi(0)\vert \phi \exp(\mathrm{i}Ht)\phi \lvert \psi(t)\rangle$

^ That's the right expression for the correlator, yes?

Wait no, I don't like what you just wrote.

Yes, that's $\langle \phi(0)\phi(t)\rangle$ as Schrödinger-y as you'll get it

Nice sign change.

@heather \mathrm?

@DanielSank I always get those damn signs wrong :P

Ok let's go deeper. How can we see that the expression written there tells me what happens if I have a state, do a measurement, wait $t$, and do another measurement?

We know that "wait" is mathematically $\exp(-i H t / \hbar)$.

We don't know what "measure" is though.

We need an operator representation of "measure".

yup, that worked @ACuriousMind, thanks!

I know that exists but I don't know what it looks like.

Also @EmilioPisanty.

Projection, no?

@0celo7 sort of, but what exactly does the operator for a measurement look like?

@DanielSank It does?

@ACuriousMind If it doesn't then all is lost.

@heather you never make a variable upright

as in: variables should always be in italics

@DanielSank Well, it's a correlation function, so it tells you something about the relation between $\phi(0)$ and $\phi(t)$, but I don't see that it does exactly what you said there

@heather you never make a variable upright

as in: variables should always be in italics

@ACuriousMind I don't see it either, but I suspect it does.

unless it's a multi-letter symbol

Let's prove or disprove it!

@EmilioPisanty variable was poor phrasing - it was a subscript

@heather I'm pretty confident you know what you're doing there - it was a cheap dig =P

@DanielSank Well, purely formally, it's the covariance between the random variables $\phi(0)$ and $\phi(t)$. But that's the covariance between "take this state and measure $\phi$" and "take this state, wait $t$, then measure $\phi$", not "take it, measure $\phi$, wait $t$, measure $\phi$" as you said.

@EmilioPisanty okay =)

@ACuriousMind Surely those are related.

Ah, but they are indeed different.

@DanielSank Not necessarily so - if you state is far from an eigenstate of $\phi$, you can't really tell what you get after you measure it the first time, so the latter is a far more complicated thing

This is really interesting. Suppose I have a Fock state of a harmonic oscillator. That thing has no time dependence of $\langle x \rangle$.

So what is the meaning of the covariance of {measure x} and {wait t and then measure x} and why does that covariance oscillate?

actually, let me verify that it oscillates.

@ACuriousMind For a harmonic oscillator, $U(t) a U(t)^\dagger = a \exp(-i \omega t)$, yes?

$U$ is propogator.

Hmm, I seem to remember this.

@DanielSank Yep

I get $$\langle \Phi|\phi U^\dagger(t) \phi U(t) | \Phi \rangle = n \cos(\omega t) + \exp(i \omega t)/2$$

(assuming $\phi = (a+a^\dagger)/\sqrt{2}$)

This is all oddly familiar...

I think it's a standard exercise to compute that thing for the QHO

It's not very informative on its own, though

@ACuriousMind Yeah this all comes up when you study spectral density of a bath of harmonic oscillators, e.g. in the Caldeira Leggett model.

I have notes on that somewhere.

The main significance should be that it's the Green's function for the equation of motion (general QFT principle: Two-point functions are Green's functions for the classical e.o.m.)

@ACuriousMind Indeed.

I think the issue that a quantum measurement changes the state means that you can't directly imbue it with a meaning of the sort you were initially looking for

Hello! I am working on some physics homework, and I have a question that asks "To stretch a spring 6.00 cm from its unstretched length, 15.0 J of work must be done, what is the force constant of this spring?" My logic was that if Work=Force (Distance parallel), and Force (of spring) = k(Distance), then W=k(D)^2. But alas, I am getting the question wrong. Why would this equation not work?