text:
# LieElement @ LieElement -- formal multiplication of Lie elements

## Synopsis

• Operator: @
• Usage:
u=x (symbol @) y
• Inputs:
• x, an instance of the type LieElement, $x$ is of type $L$, where $L$ is of type LieAlgebra
• y, an instance of the type LieElement, $y$ is of type $L$
• Outputs:
• u, an instance of the type LieElement, u is of type $L$, the formal Lie product of $x$ and $y$

## Description

The "at sign" $@$ is used as infix notation for a "formal" Lie multiplication (and also formal multiplication by scalars) where no simplifications are performed. (The formal addition is written as ++ and / is used as formal subtraction.) In this sense, it is different from the use of SPACE as multiplication operator, which always gives an object of normal form as output. The formal operations are useful when relations are introduced in a big free Lie algebra, since then it might be too hard to compute the normal form of the relations, which is not needed in order to define a quotient Lie algebra. For an example, see Minimal models, Ext-algebras and Koszul duals.

 i1 : L = lieAlgebra{a,b} o1 = L o1 : LieAlgebra i2 : (b@b)@a/3@b@a@b++2@a@b@b o2 = (b b a) - (b b a) - 3 (b a b) + 2 (a b b) o2 : L i3 : (b b) a - 3 b a b + 2 a b b o3 = 3 (b b a) o3 : L


text:
# Synopsis: Less wiggle room for the gravitational constant

Ultracold atoms trapped in optical lattices allow high-precision measurement of the gravitational constant.

Ultracold atoms have proven their value as test beds for creating new quantum states of matter and simulating how electrons behave in solids. Trapped and cooled atoms have also taken a starring role in high-precision measurements of fundamental constants and in atomic clocks. Writing in Physical Review Letters, Nicola Poli and colleagues at the University of Florence, Italy, report their use of ultracold strontium atoms to push forward our knowledge of Earth’s gravitational constant.

Building on prior proposals to use atoms captured in optical traps, Poli et al. transfer roughly one million strontium-88 atoms at $0.6\phantom{\rule{0.333em}{0ex}}\mu \text{K}$ into a vertically oriented optical lattice. Since the atoms move in a periodic potential, they undergo Bloch oscillations, similar to electrons in a solid-state lattice under the influence of an applied electric field. Bloch oscillations arise when the induced momentum change of the particle caused by a field interacts with the band structure of the lattice. In Poli et al.’s experiments, however, the applied field is gravity. By imaging the cloud of strontium atoms undergoing Bloch oscillations, the authors were able to measure the acceleration due to Earth’s gravitational field, $g$, to within $140\phantom{\rule{0.333em}{0ex}}\text{ppb}$ and compare the result with measurements by the best conventional gravimeter.

The authors say that their hope is that by optimizing the technique, measurements might become possible that test general relativity, theories of quantum gravity, and deviations from Newtonian gravity. – David Voss

### Announcements

More Announcements »

## Previous Synopsis

Biological Physics

Spintronics

## Related Articles

Atomic and Molecular Physics

### Viewpoint: Superfluids Hit the Street

A flow pattern dubbed the von Kármán vortex street, which is renowned for its aesthetic beauty and extreme power, has been created in a superfluid. Read More »

Atomic and Molecular Physics

### Viewpoint: Lamb Shift Spotted in Cold Gases

Cold atomic gases exhibit a phononic analog of the Lamb shift, in which energy levels shift in the presence of the quantum vacuum. Read More »

Atomic and Molecular Physics

### Synopsis: Quantum Droplets Swell to a Macrodrop

Experiments with ultracold magnetic atoms reveal liquid-like quantum droplets that are 20 times larger than previously observed droplets.    Read More »

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# Warning: A Word Of Advice

Discussion in 'Chemistry' started by Reiku, Jun 27, 2008.

1. ### ReikuBannedBanned

Messages:
11,238
Warning Parents!!! You May Kill Your Child!! So take heed;

You should not make a child or a teen smoke a superfluous amount of cigarettes if you catch them smoking one. It seems to be something many parents have done in hope it would deter them for life.

Cigarettes contain a poison called Plutonium-210 or given chemically as an elementary symbol in chemistry $Po^{210}$, a HIGHLY radioactive material. Even in small doses, it gives off more radioactivity
Where $102.91 kJ.mol^{-1}$ is the heat of evaporation when concerning the radioactive strength of the radiation magnitude emitted from a system, and it is found to be quite high when considering Plutonium-210.

If we consider the magnitude when comparing its mass content, polonium-210 is around $250,000$ times more toxic than hydrogen cyanide (the actual $LD_{so}$ for $Po^{210}$ is about $1micg$ for an $80$ kg person compared to about $250mg$ for hydrogen cyanide). That’s how powerful it is.

And even though it is a small amount, by making your child smoke the amount I have heard parents make them go through as a lesson not to, you are in fact putting the child’s life at risk, never mind for the sake of health in the future.

If you make your child smoke one after the other, in let’s a say a quantity of seven packets, you are in fact subjecting their bodies to levels of Plutonium which is a radioactive substance, and since their body capacity well out-weighs that of the volume of an average adult, you are subjecting them very possibly to magnitudes that can cause a similar, agonizing death with that what happened just last year... For those who are not sure what the side effects are, take your memories back to the Russian spy who was assassinated using the very same chemicals in reasonably small amounts just last year. And, as you might, he died as a result.

SO PLEASE!!!! I beg you, parents of the future, and parents of the present, do not use this as a way to stop your child from ever smoking again, because you may accidentally kill your child from over-exposure to the radioactive properties found in a single cigarette.

References

Cooperative Research Forum :: Iran :: . Dirty bomb and Iran war
6 Apr 2007 ... Ex-KGB spy Litvinenko died in London on 11/23. ..... shedding its supply of plutonium and enriched uranium - capable of producing dozens of ...
www.cooperativeresearch.org/phorum5/read.php?7,723 - 80k - Cached - Similar pages

George W. Bush - Terrorist in the White House - Main Page
12 Apr 2004 ... Egyptian 'Mossad spy' dies in UK PDF Aljazeera JUNE 28, 2007 ...... and up went the purity of the smaller Plutonium 239 core. ...
www.nogw.com/ - 349k - Cached - Similar pages

On Cigarettes CONTENTS of radioactive plutonium:

P
Plutonium is a fission fuel for Nuclear Energy and weapons ( Atomic Bomb; .... gas grills, certain home and portable stoves, and certain cigarette lighters. ...
www.hydrocut.com/Terms/P.html - 72k - Cached - Similar pages

Science in Christian Perspective
Alpha-emitters such as polonium and fissile plutonium 239 can be ... the main cancer-causing agent in cigarettes, or large quantities of CU2, NO, and SO2, ...
www.asa3.org/aSA/PSCF/1980/JASA6-80Irish.html - 27k - Cached - Similar pages

(Abstract of a possible experimentation that can prove or disprove the above)

~ Excerpt: The ONLY way to falsify the above (or at least, from what I understand) is by summing the quantity of radiation from one cigarette alone, and then make a total sum, of what was given hypothetically several packets.

This total sum, should give some identification to how it would effect and quite possibly kill a child, through the body-mass content over the sum of the P-210. This, if measured correctly, we can then scientifically determined whether this torment on the child could cause death itself, in a slow, agonizing, as painful is to much of a remedial word to describe the effects of such radioactivity working against the subjects internal defense mechanisms, which in result, makes turns them into a vegetabalized state, and with very little movement, since motor-functions are disabled through the processes.

3. ### OrleanderOH JOY!!!!Valued Senior Member

Messages:
25,817
what kind of moron does that? I thought it was just a Hollywood movie/TV thing.

5. ### draqonBannedBanned

Messages:
35,006
so...Reiku...everyone who smokes cigarettes is also radioactive?

why is this not in pseudoscience?

7. ### ReikuBannedBanned

Messages:
11,238

Exactly. In my country, many parents have subjected their children to this unfrotuate set of actions.

And no... it;s not Hollywood material.

8. ### ReikuBannedBanned

Messages:
11,238
Did you not understand my point about mass-body ratio?

9. ### draqonBannedBanned

Messages:
35,006
well...that 250,000 thing...were did you get it?

10. ### ReikuBannedBanned

Messages:
11,238
A wiki source, so that i could make sure that my recollection of the magnitude was correct.

Messages:
25,817

12. ### ReikuBannedBanned

Messages:
11,238
Actually, i didn't not only read it, it was also broadcast on BBC television as a report from a medical doctor. So, yes... i will indeed believe it.

13. ### OrleanderOH JOY!!!!Valued Senior Member

Messages:
25,817
Reiku, are you confusing plutonium with polonium?

Messages:
3,724
Yeah, I think Reiku means Polonium-210, not Plutonium-210. I don't have this information memorized, but I'm pretty sure that 210 doesn't fall within the range of Plutonium's isotopes.

Messages:
10,296
Absolutely correct! The "great master of physics" obviously pays no more attention to detail in chemistry than he does in physics!!! UGH!!!!:bugeye:

(He also doesn't know the difference in the English words "bare" and "bear" - as in "bear with me" - even though English is his first language.)

16. ### ReikuBannedBanned

Messages:
11,238
I apologize. I mixed the two up.

But, the OP still holds true.

Messages:
10,296
Yes, you are correct. Plutonium isotopes range from Pu-228 to Pu-247.

Obviously, the OP has no idea what he's talking about. (As usual.)

18. ### ReikuBannedBanned

Messages:
11,238
I got two very similar words confused. It doesn't remove the principle and stated facts given. Grow up...

Messages:
10,296
Your original premise is still false - even after considering the silly chemical mistake. Nicotine is, and always has been the mot poisonous chemical found in cigarettes and tar is the most carcinogenic.

Show proof that your OP is correct and that my statements (above) are incorrect.

20. ### ReikuBannedBanned

Messages:
11,238
No. There is the same radioactive material found even in ciggarettes. My conclusions are backed by actual science.

21. ### ReikuBannedBanned

Messages:
11,238
If you are denying this, then ask any med-doc on-line.

Messages:
10,296
Invalid response.

Show proof of your claim. You volunteered the statements, it's up to you to prove your claim.

23. ### OilIsMasteryBannedBanned

Messages:
3,288
Putin murdered Litvinenko with Polonium (Po) not Plutonium (Pu)...


text:
# Advent of Code 2021 day 23

The solution to day 23 was the longest program so far. At 320 lines of code, this solution is 50% longer than the next longest (day 15, another search problem) and twice as long as day 16, the longest non-search solution. The program was long because it had to contain a lot of domain knowledge about the problem setting.

## Understanding the domain

Fundamentally, the problem was similar to many previous ones. The arrangement of amphipods in the burrow represents a state, and there are legal moves that transform one state to another. The problem is to search the space of legal move sequences to turn the initial state into a goal state.

I'd normally handle this by allowing each amphipod to move one space and just allowing the search tree to become deep. But in this case, the additional rules on movement meant I had to take a different approach.

The problem has a constraint that, once an amphipod stops in the hallway, it can't move again until it enters its destination room. I could have handled that by expanding the notion of a problem state to include tracking which amphipod, if any, was currently moving. Instead, I decided to pre-compute "moves" a being the whole trail from start to end, and keeping those moves as atomic.

This meant there were three types of moves:

1. From a room to the hallway
2. From the hallway to a room
3. From a room directly to a room (but not back to the same room)

Including the room-room moves eliminates some duplicate search states if there are several hall spaces between the source and destination rooms. It doesn't matter which one is used to break the journey from room to room, so I eliminate that choice by including them all in one move.

The other piece of domain knowledge I included was ensuring that when an amphipod enters a room, it always moves to the furthest available space. All the spaces must be filled eventually, so it makes little sense to have unfilled spaces at the far end.

## Data structures

There ended up being a lot to keep track of, so I created a bunch of data structures to keep track of everything. I also used lenses a lot to access all the different parts of these structures.

Amphipods and positions have explicit types, and the position of the amphipods is wrapped in a MoveState. I also rename the lenses for the V2 constructor, as I'm using (row, column) coordinates.

data Amphipod = A | B | C | D deriving (Show, Read, Eq, Ord, Enum)

type Coord = V2 Int -- r, c
_r :: Lens' (V2 Int) Int
_r = _x
_c :: Lens' (V2 Int) Int
_c = _y

type MoveState = M.Map Coord Amphipod

A Step is a single valid move of an amphipod, including the distance travelled and the spaces moved through; the latter is used to check there's nothing in the way of a move. Steps relates the starting position (the key) with the set of final positions.

An AppliedMove combines the positions of amphipods after a move and the Step that got there.

data Step = Step
{ _destination :: Coord
, _distance :: Int
, _transits :: S.Set Coord
, _entryRequirement :: Maybe Amphipod
} deriving (Show, Eq, Ord)
makeLenses ''Step

type Steps = M.Map Coord (S.Set Step)

data AppliedMove = AppliedMove
{ _afterMove :: MoveState
, _appliedStep :: Step
}
deriving (Show, Eq, Ord)
makeLenses ''AppliedMove

Finally, a Burrow holds all the possible Steps and information about where the hall and rooms are. That gets fed into the Reader monad, for use inside the search routines.

data Burrow = Burrow
{ _possibleSteps :: Steps
, _roomColumns :: M.Map Amphipod Int
, _hallRow :: Int
} deriving (Show, Eq)
makeLenses ''Burrow

## Building the burrow

mkBurrow reads the text from the input file and returns the Burrow (including all the moves) and the initial arrangement of amphipods. Mainly it's just walking over the text file, pulling out bits, and recording their positions.

mkBurrow :: String -> (Burrow, MoveState)
mkBurrow text = (burrow, initState) -- (burrow, initState)
where rows = lines text
hall = mkHall (rows!!1)
rooms = mkRooms $drop 2 rows roomCols = S.map (^. _c)$ M.keysSet rooms
hall' = S.filter ((S.notMember roomCols) . (^. _c)) hall
routes = mkRoutes hall' rooms
roomColMap = M.fromList $zip [A .. D]$ S.toAscList roomCols
burrow = Burrow { _possibleSteps = routes, _roomColumns = roomColMap, _hallRow = 1}
initState = mkInitialState rows

mkHall :: String -> S.Set Coord
mkHall text = S.fromList hallCoords
where hallCols = filter ((/= '#') . snd) $zip [0..] text hallCoords = map ((V2 1) . fst) hallCols mkRooms :: [String] -> M.Map Coord Amphipod mkRooms text = M.unions rooms where rooms = map mkRoom$ zip [2..] text

mkRoom :: (Int, String) -> M.Map Coord Amphipod
mkRoom (r, text) = M.fromList roomCoords
where roomCols = filter ((elem ("ABCD." :: String)) . snd) $zip [0..] text roomCoords = zip (map ((V2 r) . fst) roomCols) [A .. D]  The interesting part comes in mkRoutes. It finds all the moves of each separate type (as described above) the combines them with M.unionsWith. mkRoutes :: S.Set Coord -> M.Map Coord Amphipod -> Steps mkRoutes halls rooms = M.unionsWith (S.union) [hallRoutes, roomHallRoutes, roomRoomRoutes] where hallRoutes = S.foldr' (mkHallRoute rooms) M.empty halls roomHallRoutes = S.foldr' (mkRoomHallRoute halls) M.empty (M.keysSet rooms) roomRoomRoutes = S.foldr' (mkRoomRoomRoute hallRow rooms) M.empty (M.keysSet rooms) hallRow = S.findMin$ S.map (^. _r) halls

(As a little note to self, I spent a fair bit of time running in circles updating the various Maps without using S.union to combine the multiple sets for each key!)

As an example, making the routes from the hall spaces to the room spaces, mkHallRoute, is a nested fold. First I create the routes for each hall space then, for each one, the routes from that hall space to each room space.

mkHallRoute :: M.Map Coord Amphipod -> Coord -> Steps -> Steps
mkHallRoute rooms here routes = M.foldrWithKey' (mkHallRoute1 here) routes rooms

mkHallRoute1 :: Coord -> Coord -> Amphipod -> Steps -> Steps
mkHallRoute1 here@(V2 hr hc) there@(V2 tr tc) entry routes = M.insert here (S.insert step existingRoutes) routes
where step = Step { _destination = there
, _distance = (S.size transits)
, _transits = transits
, _entryRequirement = Just entry
}
cMin = min hc tc
cMax = max hc tc
transits = S.delete here $S.fromList$ [V2 hr c | c <- [cMin..cMax]] ++ [V2 r tc | r <- [hr..tr]]
existingRoutes = M.findWithDefault S.empty here routes

The other types of route creation follow a very similar pattern.

## Searching

Most of the search is standard A* search. The interesting bits are the problem-specific parts.

successors finds all the legal applied moves from a particular arrangment of amphipods. It does that by, for each amphipod, finding all the legal moves of that amphipod.

successors :: MoveState -> BurrowContext (S.Set AppliedMove)
successors moveState =
do steps <- asks (^. possibleSteps)
let succs = M.foldrWithKey' (legalSteps steps moveState) S.empty moveState
return succs

legalSteps :: Steps -> MoveState -> Coord -> Amphipod -> S.Set AppliedMove -> S.Set AppliedMove
legalSteps steps state here amphipod acc = S.union appliedSteps acc
where allSteps = steps ! here
occupiedSpaces = M.keysSet state
freeSpaces st = S.null $S.intersection occupiedSpaces (st ^. transits) freeSteps = S.filter freeSpaces allSteps validTargetSteps = S.filter (\st -> fromMaybe amphipod (st ^. entryRequirement) == amphipod) freeSteps openRoomSteps = S.filter (openRoom state) validTargetSteps highestRowSteps = S.filter (highestRow (S.map (^. destination) openRoomSteps)) openRoomSteps appliedSteps = S.map (\s -> AppliedMove { _afterMove = (applyStep state here s) , _appliedStep = s } ) highestRowSteps  legalSteps finds the legal moves by finding all the moves (allSteps = steps ! here) then filtering out the illegal ones over several stages: • freeSteps are steps where every spaced moved through is empty • validTargetSteps are those to a place this amphipod could go (the hall or its own room) • openRoomSteps are where the target room is "open", i.e. does not have any non-destined amphipods • highestRowSteps are to the furthest unoccupied space in a column ### Heuristic It's A* search, so it's guided by a heuristic. The one I use relaxes the constraints of "can't pass through other amphipods" and "only one amphipod per space in the destination room". estimateCost :: MoveState -> BurrowContext Int estimateCost state = do rCols <- asks (^. roomColumns) hRow <- asks (^. hallRow) let amphipodCosts = M.mapWithKey (estimateACost rCols hRow) state return$ sum $M.elems amphipodCosts estimateACost :: M.Map Amphipod Int -> Int -> Coord -> Amphipod -> Int estimateACost rCols hRow (V2 r c) amphipod = (singleStepCost amphipod) * dist where targetCol = rCols ! amphipod dist = if c == targetCol then 0 else (r - hRow) + (abs (c - targetCol)) + 1  It may not be much, but it's enough to halve the runtime of the program. ## Profiling My first attempts at this took several minutes to run part 2. This seemed like too long a time. I profiled the code and found that 96% of the runtime was in legalSteps (which wasn't a huge surprise). But the profiler, by default, works at the level of functions, so didn't give me any more detail than that. I manually inserted some additional call centres at all the filter stages of legalSteps, with annotations like {-# SCC freeSteps #-}, like this: legalSteps :: Steps -> MoveState -> Coord -> Amphipod -> S.Set AppliedMove -> S.Set AppliedMove legalSteps steps state here amphipod acc = S.union appliedSteps acc where allSteps = steps ! here freeSpaces st = S.null$ S.intersection (M.keysSet state) (st ^. transits)
freeSteps = {-# SCC freeSteps #-} S.filter freeSpaces allSteps
validTargetSteps = {-# SCC validTargetSteps #-} S.filter (\st -> fromMaybe amphipod (st ^. entryRequirement) == amphipod) freeSteps
openRoomSteps = {-# SCC openRoomSteps #-} S.filter (openRoom state) validTargetSteps
highestRowSteps = {-# SCC highestRowSteps #-} S.filter (highestRow (S.map (^. destination) openRoomSteps)) openRoomSteps
appliedSteps = S.map (\s -> AppliedMove
{ _afterMove = (applyStep state here s)
, _appliedStep =  s
}
) highestRowSteps

That showed that most of the time was spent finding the freeSteps. On a hunch that GHC wasn't bright enough to cache the result of M.keysSet state, I pulled that out into a separate named variable:

occupiedSpaces = M.keysSet state
freeSpaces st = S.null \$ S.intersection occupiedSpaces (st ^. transits)
freeSteps = S.filter freeSpaces allSteps

Just that halved the runtime.

## Code

You can get the code from my locally-hosed Git repo, or from Gitlab.


text:
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We would need the design only done in photoshop, and to be given to us in PSD. There is no need to code/css/ the sites - we will have that done, however if you are able to do this all then it would benefit you in the long run as we require more projects but to start, just one site that allows us to see how you operate.

Please send examples to stand a better chance, or a link to a portfolio.

DO NOT BID IF YOU HAVE NOT GOT SOMETHING LIKE THIS TO SHOW AS EXAMPLES! YOUR BID WILL BE IGNORED

Thanks!

Kemahiran: Reka Bentuk Grafik, Photoshop, Reka Bentuk Laman Web

Tentang Majikan:
( 1 ulasan ) Cape Town, South Africa

ID Projek: #353697

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text:
Engineering Mechanics: Statics & Dynamics (14th Edition)

$\omega_{AB}=2 rad/s$
We can determine the required angular velocity as follows: We know that $\vec{v_C}=\vec{v_B}+\vec{\omega_{BC}}\times \vec{r_{C/B}}$..eq(1) As $\vec{v_C}=-4\hat j$ and $\vec{r_{C/B}}=3cos30\hat i+3sin 30\hat j=2.597\hat i+1.5\hat j$ similarly, $\vec{v_B}=-(\omega_{AB}r_{AB})\hat j=-2\omega_{AB}\hat j$ We plug in the known values in eq(1) to obtain: $-4\hat j=-2\omega_{AB}\hat j+\omega_{BC} \hat k\times (2.598\hat i+1.5\hat j)$ $\implies -4\hat j=-1.5\omega_{BC}\hat i+(-2\omega_{AB}+2.598\omega_{BC})\hat j$ Comparing the $i$ component, we obtain: $0=-1.5\omega_{BC}$ $\implies \omega_{BC}=0 rad/s$ and comparing the $j$ component, we obtain: $-4=-2\omega_{AB}+2.598\omega_{BC}$ $\implies -4=-2\omega_{AB}+0$ This simplifies to: $\omega_{AB}=2 rad/s$


text:
# [Problem Set 4 Networks] Problem 3: Hopfield model

Link of the iPython notebook for the code

# PROBLEM 3: Hopfield Model

We will study the properties of a Hopfield network with $N = 64$ neurons, whose dynamics obeys the following differential equation:

$\dot{\textbf{x}} = - \textbf{x} + f(W \textbf{x}) + σ η(t)$

where

• $\textbf{x}$ is the firing rates vector

• $W$ is the $N × N$ weight matrix

• $η(t)$ is a Gaussian noise of magnitude $σ = 0.1$

• the activation function $f ≝ \sign$ for the sake of simplicity

## 1. Storing a pattern

A pattern $\textbf{p} ∈ 𝔐_{N,1}(\lbrace -1, 1 \rbrace)$ is stored in the Hopfield network by setting:

$W ≝ \frac 1 N \textbf{p} \textbf{p}^\T$

Here are some examples of patterns, visualized as $8×8$ images:

To begin with, let us focus on storing the battery pattern $\textbf{p}_{\text{bat}}$:

We set

$W ≝ \frac 1 N \, \textbf{p}_{\text{bat}} \textbf{p}^\T_{\text{bat}}$

Then, we simulate the network with resort to the Euler method (with time-increment $dt = Δt ≝ 0.1$ (arbitrary units)):

$\textbf{x}(t+Δt) = (1 - Δt)\, \textbf{x}(t) + Δt \, \sign(W \textbf{x}(t)) + \sqrt{Δt} \, σ η(t)$

which yields, for random starting patterns:

It can be noticed that the network relaxes to two fixed points, in our examples:

• the battery pattern: $\textbf{p}_{\text{bat}}$
• and its opposite pattern: $-\textbf{p}_{\text{bat}}$

From a theoretical standpoint, this can be explained by the fact that both $\textbf{p}{\text{bat}}$ and $-\textbf{p}{\text{bat}}$ are fixed points of the function

$\textbf{x} ⟼ \sign(W\textbf{x})$

Indeed:

\begin{align*} \sign\left(W \textbf{p}_{\text{bat}}\right) & = \sign\left(\frac 1 N \left(\textbf{p}_{\text{bat}} {\textbf{p}_{\text{bat}}}^\T\right) \textbf{p}_{\text{bat}}\right) \\ & = \sign\left(\frac 1 N \textbf{p}_{\text{bat}} \left({\textbf{p}_{\text{bat}}}^\T \textbf{p}_{\text{bat}}\right)\right) &&\text{(because the matrix product is associative)}\\ &= \sign\Bigg(\frac 1 N \textbf{p}_{\text{bat}} \overbrace{\Bigg(\sum\limits_{ i=1 }^N \underbrace{ \left[\textbf{p}_{\text{bat}}\right]_i^2}_{= \; 1}\Bigg)}^{= \; N}\Bigg) \\ &= \sign\left(\textbf{p}_{\text{bat}}\right) \\ &= \textbf{p}_{\text{bat}} &&\text{(because } \textbf{p}_{\text{bat}} ∈ 𝔐_{N,1}(\lbrace -1, 1 \rbrace)\\ & &&\;\text{ so } \sign\left(\left[\textbf{p}_{\text{bat}}\right]_i\right) = \left[\textbf{p}_{\text{bat}}\right]_i \quad ∀1≤i≤N \text{ )} \end{align*}

And:

\begin{align*} \sign\left(W (-\textbf{p}_{\text{bat}})\right) & = \sign\left(-W \textbf{p}_{\text{bat}}\right)\\ &= -\sign\left(W \textbf{p}_{\text{bat}}\right) &&\text{(because } \sign \text{ is an odd function)}\\ &= -\textbf{p}_{\text{bat}} \end{align*}

Therefore, leaving the noise aside, by setting:

$φ ≝ \textbf{x} ⟼ -\textbf{x} + \sign(W\textbf{x})$

so that the corresponding dynamics without noise is given by:

$\dot{\textbf{x}} = φ(\textbf{x})$

it follows that:

$φ(\textbf{p}_{\text{bat}}) = φ(-\textbf{p}_{\text{bat}}) = \textbf{0}$

As a result: $\textbf{p}{\text{bat}}$ and $-\textbf{p}{\text{bat}}$ are fixed points of the corresponding deterministic (i.e. without noise) dynamics, as a result of which the stochastic dynamics relaxes to them.

## 2. Storing two patterns

Now, let us add a new pattern $\textbf{p}_{\text{com}}$:

so that

$W = \frac 1 N \left(\textbf{p}_{\text{bat}}\textbf{p}_{\text{bat}}^\T + \textbf{p}_{\text{com}}\textbf{p}_{\text{com}}^\T\right)$

And we simulate the network on random starting patterns:

Again, we can show that:

• the battery pattern $\textbf{p}{\text{bat}}$ and its opposite pattern: $-\textbf{p}{\text{bat}}$
• the computer pattern $\textbf{p}{\text{com}}$ and its opposite pattern: $-\textbf{p}{\text{com}}$

are stored, owing to the fact that they are fixed points of the corresponding deterministic dynamics.

Indeed (for the sake of convenience, we set $\textbf{p} ≝ \textbf{p}{\text{bat}}$ and $\textbf{q} ≝ \textbf{p}{\text{com}}$):

\begin{align*} \sign\left(W \textbf{p}\right) & = \sign\left(\frac 1 N \left(\textbf{p} \textbf{p}^\T + \textbf{q} \textbf{q}^\T\right) \textbf{p}\right) \\ & = \sign\Big(\underbrace{\frac 1 N \left(\textbf{p} \textbf{p}^\T\right) \textbf{p}}_{\rlap{= \; \textbf{p} \; \text{(cf. previously)}}} + \frac 1 N \left(\textbf{q} \textbf{q}^\T\right) \textbf{p}\Big)\\ & = \sign\Big(\textbf{p} + \frac 1 N \, \textbf{q} \underbrace{\left(\textbf{q}^\T \textbf{p}\right)}_{\rlap{= \; ⟨\textbf{q},\, \textbf{p}⟩}}\Big)\\ \end{align*}

But:

$\vert ⟨\textbf{q},\, \textbf{p}⟩ \vert \, = \, \Bigg\vert \sum\limits_{ i = 1}^N \underbrace{[\textbf{q}]_i [\textbf{p}]_i}_{∈ \lbrace -1, 1 \rbrace} \Bigg\vert \, ≤ \, N$

And for $ε ∈ \lbrace -1, 1 \rbrace$:

\begin{align*} & ⟨\textbf{q},\, \textbf{p}⟩ = ε N \\ ⟺ \quad & \sum\limits_{ i = 1}^N [\textbf{q}]_i [\textbf{p}]_i = εN \\ ⟺ \quad & \sum\limits_{ i = 1}^N \underbrace{([\textbf{q}]_i [\textbf{p}]_i - ε)}_{≤ 0 \text{ if } ε = 1 \text{ else } ≥ 0} = 0 \\ ⟺ \quad & ∀1 ≤ i ≤ N, \quad [\textbf{q}]_i [\textbf{p}]_i - ε = 0\\ ⟺ \quad & ∀1 ≤ i ≤ N, \quad [\textbf{q}]_i [\textbf{p}]_i = ε\\ ⟺ \quad & ∀1 ≤ i ≤ N, \quad [\textbf{q}]_i = ε/[\textbf{p}]_i = ε[\textbf{p}]_i &&\text{ (because } [\textbf{p}]_i ∈ \lbrace -1, 1 \text{ )} \rbrace\\ \end{align*}

NB: alternatively, we could have used the Cauchy-Schwarz inequality:

$\vert ⟨\textbf{q},\, \textbf{p}⟩ \vert ≤ \underbrace{\Vert \textbf{q} \Vert_2 \, \Vert \textbf{p} \Vert_2}_{= \sqrt{N}\sqrt{N} = N} \quad \text{ and } \quad \Big(\vert ⟨\textbf{q},\, \textbf{p}⟩ \vert = N ⟺ \textbf{q} = ± \textbf{p} \Big)$

Here, as:

$\textbf{q} ≠ \textbf{p} \quad \text{ and } \quad \textbf{q} ≠ -\textbf{p}$

it comes that:

$\vert ⟨\textbf{q},\, \textbf{p}⟩ \vert < N$

Hence:

\begin{align*} \sign\left(W \textbf{p}\right) & = \sign\Big(\textbf{p} + \underbrace{\frac{\overbrace{⟨\textbf{q},\, \textbf{p}⟩}^{\rlap{∈ [-(N-1)\,, \,N-1] \; }}}{N}}_{∈ \, ]-1, 1[} \; \textbf{q}\Big)\\ &\overset{⊛}{=} \sign\left(\textbf{p}\right) &&\text{(because each } \frac {⟨\textbf{q},\, \textbf{p}⟩} N \, [\textbf{q}]_i ∈ \left[-1 + \frac 1 N, \, 1-\frac 1 N \right]\\ & && \quad \text{: it can't change the sign of } [\textbf{p}]_i ∈ \lbrace -1, 1 \rbrace\text{ )} \\ &= \textbf{p} \end{align*}

And likewise, by symmetry:

$\sign\left(W \textbf{q}\right) = \textbf{q}$

Once we know that $\textbf{p}$ and $\textbf{q}$ are fixed points, the same holds true for their opposites, as $\sign$ is an odd function (the proof is the same as previously).

So $±\textbf{p}$ and $±\textbf{q}$ are all fixed points of the corresponding deterministic dynamics, as a result of which the stochastic dynamics relaxes to them.

NB: in the computations, the key point is the equality $⊛$, which stems from $\textbf{q}$ being different from $±\textbf{p}$ (by Cauchy-Schwarz).

On top of that, it can be noted that the network relaxes to the fixed-point pattern which is the closest to the starting pattern! For example, consider the second-to-last and the third-to-last examples in figure 2.c.2:

• for both of these examples, in the starting pattern, we can distinguish what looks like the base of a (future) screen
• but the the second-to-last example converges to a bright-screen computer, whereas the third-to-last converges to a dark-screen one. This can be accounted for by the fact that there are more dark pixels in the third-to-last starting pattern than in the second-to-last one!

Therefore, we begin to see why Hopfield networks can be thought of as modeling an associative memory-like process: the starting patterns are similar to “visual clues”, to which the network associates the best (i.e. the closest) matching stored pattern.

## 3. The general rule

The general rule defining the weight matrix is:

$W ≝ \frac 1 N \sum_{i=1}^M \textbf{p}_i \textbf{p}_i^\T$

where the $\textbf{p}_i$ are the patterns to be stored.

### 3 patterns: linear combinations stored

Let us add yet another pattern: $\textbf{p}_{\text{cof}}$:

so that, in compliance with the general rule:

$W = \frac 1 N \left(\textbf{p}_{\text{bat}}\textbf{p}_{\text{bat}}^\T + \textbf{p}_{\text{com}}\textbf{p}_{\text{com}}^\T + \textbf{p}_{\text{cof}}\textbf{p}_{\text{cof}}^\T\right)$ Then, we simulate the network with random initial conditions:

It appears that the networks relaxes to the following linear combinations of $\lbrace \textbf{p}_{\text{bat}}, \textbf{p}_{\text{com}}, \textbf{p}_{\text{cof}} \rbrace$ too:

• $\textbf{p}_{\text{bat}} - \textbf{p}_{\text{com}} - \textbf{p}_{\text{cof}}$
• $\textbf{p}_{\text{bat}} - \textbf{p}_{\text{com}} + \textbf{p}_{\text{cof}}$

So: on top of the patterns from which the weight matrix is constructed and their opposites, linear combinations thereof may also be stored.

### Storing capacity

Finally, we will investigate how many patterns we can store in this $N=64$ neurons network by:

• adding new patterns, one after another
• and then simulating the network with each pattern as initial condition to see if the pattern is stored or not

As a consequence:

• up until 6 (included) patterns: all the patterns are successfully stored
• from 7 pattern on, some are not properly stored, and it gets worse and worse as the number of patterns increases:

• with 7 patterns: the computer screen is “prettified by bright pixels” (looking like light reflection!), but not stored unaltered
• with 8 patterns: on top of the light reflection on the computer screen, the copy icon is not stored at all: the network converges to an out-of-shape computer screen. In the same vein, the phone pattern ends up being deformed.
• with 9 patterns: the battery pattern is not stored anymore, the computer base is shifted, the cup of coffee, the thumbs up, the phone and the arrow are distorted, and the copy icon keeps converging toward the computer (they must look too similar for the full network’s pattern memory).

On the whole: the $64$-neurons Hopfield network can perfectly store $6$ patterns (and almost $7$ patterns, if it were not for the computer pattern ending up with a few bright pixels on the screen).

Tags:

Updated:


text:
This demo plays a constant amplitude pulse to output 1 of the OPX and feeds it back to input 1 of the OPX. We then record this signal.

## Config#

The config file declares 1 quantum element called qe1 which has both an input and an output. The quantum element input is connected to OPX output number 1. The quantum element output is connected to OPX input number 1.

Crucially, the element also defines two parameters: time_of_flight and smearing these must be declared to perform a measurement. These parameters are described in detail on the QUA documentation site.

The quantum element has one operation called readoutOp. The readoutOp operation plays a constant amplitude pulse called readoutPulse which has a duration 1 $\mu S$.

readoutPulse also declares a digital_marker. This is a digital signal which tells the OPX when to perform the measurement.

##### info

Note that failing to declare a digital_marker will not fail program compilation, but will prevent data from being acquired.

## Program#

The QUA program calls the measure command with the readoutOp. The third argument to measure is the name of the stream where the ADC measured samples are saved.

We run the program on the simulator for 500 clock cycles and take the simulated samples from the simulation job.

The acquired ADC stream is taken from result_handles. Note the change of name of the raw ADC steam from the name we specified in the program (raw_adc) to the name we use to get the stream (raw_adc_input1). This is an idiosyncrasy of the raw ADC interface which doesn't not appear in other QUA data saving mechanisms.

## Post processing#

We generate two subplots showing the simulated samples on top and the measured stream on the bottom. The x scale is in sample number. While the sample rate is 1/nanosecond, we do not attempt to align top and bottom to specific time stamps to keep this example as simple as possible.


text:
# zbMATH — the first resource for mathematics

##### Examples
 Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.

##### Operators
 a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
##### Fields
 any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article)
Rational curves and surfaces with rational offsets. (English) Zbl 0872.65011

Summary: Given a rational algebraic surface in the rational parametric representation $s\to \left(u,v\right)$ with unit normal vectors

$n\to \left(u,v\right)=\left(s{\to }_{u}×s{\to }_{v}\right)/\parallel s{\to }_{u}×s{\to }_{v}\parallel ,$

the offset surface at distance $d$ is

$s{\to }_{d}\left(u,v\right)=s\to \left(u,v\right)+dn\to \left(u,v\right)·$

This is in general not a rational representation, since $\parallel s{\to }_{u}×s{\to }_{v}\parallel$ is in general not rational. We present an explicit representation of all rational surfaces with a continuous set of rational offsets $s{\to }_{d}\left(u,v\right)$. The analogous question is solved for curves, which is an extension of Farouki’s Pythagorean hodograph curves to the rationals. Additionally, we describe all rational curves $c\to \left(t\right)$ whose arc length parameter $s\left(t\right)$ is a rational function of t. Offsets arise in the mathematical description of milling processes and in the representation of thick plates, such that the presented curves and surfaces possess a very attractive property for practical use.

##### MSC:
 65D17 Computer aided design (modeling of curves and surfaces) 68U07 Computer aided design 53A07 Higher-dimensional and -codimensional surfaces in Euclidean $n$-space


text:
# How to edit in-text citation in plainnat style?

I manage my bibliography with JabRef and everything is OK but I have a problem with my plainnat.bst file. Possibly, I want my in-text citation to be like: [Author et al., year] or [Autor, Autor et al., year] when authors are more than 2.

I think I have to change the FUNCTION {format.lab.names} code but I don't know how to do it. Can you please help me ?

This is my code (plainnat default):

FUNCTION {format.lab.names}
{ 's :=
s #1 "{vv~}{ll}" format.name$s num.names$ duplicate$#2 > al { pop$ " et~al." * }
{ #2 <
'skip${ s #2 "{vv }{ll}" format.name$ "others" =
{ " et~al." * }
{ " e " * s #2 "{vv~}{ll}" format.name$* } if$
}
if$} if$
}

Yes, Mario, you are perfecly right...I just forgot to say that I'm already using natbib!

However, today I've been editing my plainnat.bst file and I managed to write author's name in smallcaps, year of articles in bold and right now my in-text citation looks like this (I would have posted a picture but I don't have 10 reputations yet):

[Imperi, F. et al., 2009]

and in the bibliography:

Imperi, F., Ciccosanti, F., Basulto Perdomo, A., Tiburzi, F., Mancone, C., Alonzi, T., Ascenzi, P., Piacentini, M., Visca, P., e Fimia, G.M. Analysis of the periplasmic proteome of Pseudomonas aeruginosa, a metabolically versatile opportunistic pathogen. Proteomics, 9:1901–1915, 2009. (Citato alle pagine 1, 5, e 6.)

I'm nearly satisfied with it, but I just want to know if there's a way to add also the second name of this (and every) article in the in-text citations, like:

[Imperi, F., Ciccosanti, F. et al., 2009]

-
Welcome to TeX.sx! – texenthusiast Apr 11 '13 at 20:38
I'm a bit puzzled. I used plainnat.bst for a document recently and I've just checked: author lists are abbreviated in the main part of the text by default when they have too many authors with natbib+plainnat.bst. Can you give a minimal example of your problem (.tex example file + .bib file)? – DCTLib Apr 12 '13 at 1:15

You can solve this by using the natbib package:

For example:

\documentclass{article}

\usepackage[english]{babel}
\usepackage{url}
\usepackage{natbib}

\begin{document}
According to \citep{mario2013}

\bibliographystyle{plainnat}
\bibliography{refs}

\end{document}

where the refs.bib file is:

@article{mario2013,
author={mario},
title={Marito and Friends},
journal={My Journal},
year = {2013},
}

Will produce an output like this:

For more information please read the natbib documentation.

-


text:
Probabilistic hazard assessment

Methodology

A Probabilistic Seismic Hazard Assessment (PSHA) quantifies the contributions of all sources for all possible magnitudes within the affected area. In the same way as a Deterministic Seismic Hazard Assessment (DSHA), the PSHA stages identify all earthquake sources capable of producing significant ground-motion at the site and enable the selection of a minimum number of independent ground-motion models that capture the expected range of all possible ground-motions in the target region. The fundamental difference is the inclusion of random uncertainty in the calculation of the spatial and temporal distribution of earthquakes, as well as the (aleatory) uncertainty associated with the ground-motion models.

In a probabilistic framework, ground-motion intensities (PGA, Sa) are expressed in terms of annual exceedance probability (AEP). The annual exceedance probability (AEP) is the rate of exceedance of a specific level of seismic intensity due to the potential occurrence of earthquakes from that source over a one-year period. The methodology employed for estimating the AEP is based on the Cornell-McGuire methodology85 which combines earthquake occurrence models, seismic source zone models, magnitude-recurrence relationships, and ground-motion intensity models through the total probability theorem.

Assuming the occurrence of earthquakes follows a Poisson process (i.e. that average time between events is known, but the exact timing of events is random), the probability that a certain strong motion intensity $$I$$ exceeds a certain value $$i^\ast$$ during a time interval $$T_e$$ is given by $$P_{T_e}\left[I>i^\ast\right]=1-\mathrm{exp}(-\lambda_I (i^\ast)\,T_e )$$ where $$\lambda_I (i^\ast)$$ is the annual rate of exceedance and $$T_e$$ is the exposure time. For a given source, the annual rate of exceedance of an intensity target level $$i^\ast$$ $$\lambda_I(i^\ast )$$ can be expressed by the Total Probability Theorem (TPT):

$\lambda_I^{(s)}(i^\ast)\simeq \nu_o^{(s)}\sum_{N_R}\sum_{N_M}P[I>i^\ast|m_j,r_k]^{(s)} f_M^{(s)} (m_j)\,\Delta m\,f_{R|M}^{(s)} (r_k |m_j) \Delta r$

where $$M_{max}^{(s)}$$ is the MCE of the source (s) and $$\nu_o=P[M>M_o^{(s)}]$$ is the annual frequency of occurrence of earthquakes with magnitude greater than a minimum value $$M_o$$. Using the TPT the annual exceedance rate of earthquakes with magnitude greater than $$m^\ast$$ can be expressed in terms of an asymptotically truncated Gutenberg-Richter magnitude-frequency distribution. The discrete mass probability for magnitudes $$P[M=m_j]^{(s)}=f_M^{(s)}(m_j)\,dM$$ can be obtained from the first derivative of the magnitudes cumulative-density function $$F_M^{(s)} (m_j)$$ valid for $$M_{max}^{(s)}<m^\ast<M_o^{(s)}$$: $\lambda^{(s)}[M>m^\ast]=\lambda_M^{(s)}(m^\ast)=\nu_o^{(s)}(1-F_M^{(s)}(m^\ast))\simeq \nu_o^{(s)}\dfrac{\mathrm{e}^{-\beta\left(m^\ast-M_o^{(s)}\right)}}{1-\mathrm{e}^{-β \left(M_{max}^{(s)}-M_o^{(s)} \right) }}$ The discrete mass probability function for distances $$f_{R⁄M}^{(s)} (r^\ast |m^\ast ) dr^\ast=P[R=r^\ast |m^\ast ]^{(s)}$$ can be obtained numerically from the histogram of an array of epicentres inside the boundaries of each source.

The Return Period $$T_R$$ is defined as the inverse of the total AEP $T_R=1/\lambda_I (i^\ast) =-T_e/\mathrm{ln}(1-P_{T_e} [I>i^* ])$

Uncertainty Analysis

The design earthquake obtained by the PSHA methodology incorporates the spatial and temporal variability directly, by means of the total probability theorem. Because some GMPE include in their formulation other source parameters, such as the rake angle of the fault plane, the dip of the fault plane, the faulting type (strike-slip, normal, reverse), the PSHA model implemented in OpenQuake86 also incorporates the epistemic uncertainty in the unknown source parameters through a logical tree of intensity models, recurrences and maximum rupture area sizes. The different branches of the model determine different spectral ordinates $$S_a(T)$$ in rock. Using a simple algorithm, the median and variance of a PDF probability density function of the seismic intensity of the PSHA model are estimated, where the median of the intensities represents the most probable value taken by the recurrence parameters, the hypocentre, and the maximum size of the fault.

Results

Exceedance Probabilities (AEP)

The annual exceedance probability (AEP) curves for PGA are shown in Figure 22. All values were obtained from ground-motion models assuming rock site conditions (NEHRP site class AB). The inverse of the annual exceedance probability is the return period (TR) and the hazard curves in terms of TR are shown in Figure 23.

Peak Ground Accelerations (PGA)

Table 11 and Table 12 present the mean and 84th percentile of PGA values expected at GGM for different return periods and site conditions. The maximum design earthquake in rock assumes an AEP of 1:10,000 years and results in $$PGA\approx$$ 0.194 $$\mathrm{g}$$ (mean) and $$PGA\approx$$ 0.268 $$\mathrm{g}$$ (84th percentile). For site class “C,” it reported $$PGA\approx$$ 0.23 $$\mathrm{g}$$ (mean) and $$PGA\approx$$ 0.359 $$\mathrm{g}$$ (84h percentile).

 Peak-Ground Accelerations (PGA) [g] - (mean) NEHRP TR=500 TR=1000 TR=2500 TR=5000 TR=10000 AB 0.021 0.038 0.076 0.124 0.194 B 0.022 0.039 0.083 0.139 0.222 C 0.028 0.048 0.095 0.151 0.230 D 0.035 0.060 0.116 0.183 0.276
 Peak-Ground Accelerations (PGA) [g] - (+84%) NEHRP TR=500 TR=1000 TR=2500 TR=5000 TR=10000 AB 0.028 0.051 0.104 0.171 0.268 B 0.033 0.061 0.128 0.215 0.344 C 0.043 0.075 0.148 0.235 0.359 D 0.055 0.093 0.181 0.286 0.433

Uniform Hazard Spectrum (UHS)

The Uniform Hazard Spectrum (UHS) is the set of spectral ordinates (for all structural periods of interest) that have the same probability of exceedance. Figure 24 presents the mean UHS spectral ordinates for different exceedance probabilities (return periods) in rock. Spectral ordinates at AEP=1:10,000 yr represents the maximum design earthquakes $$S_a(T_n)$$ for rock at the project site, obtained from a probabilistic methodology. Figure 25 shows mean values and other percentiles for the 10,000-years uniform hazard spectra. Mean uniform hazard spectra for other site conditions than rock in terms of NEHRP classes are shown in Figure 26. The plots for other percentiles are available in the project’s dashboard PSHA.

1. Robin K. McGuire, Seismic Hazard and Risk Analysis (Earthquake Engineering Research Institute, 2004).↩

2. Pagani et al., “OpenQuake Engine.”↩


text:
# Power System Stability

 Question 1
In the single machine infinite bus system shown below, the generator is delivering the real power of 0.8pu at 0.8 power factor lagging to the infinite bus. The power angle of the generator in degrees (round off to one decimal place) is _________
 A 12.8 B 28.4 C 20.5 D 32.6
GATE EE 2019      Power System Stability
Question 1 Explanation:
\begin{aligned} X &=0.25+0.2+0.4||0.4 \\ &=0.45+0.2=0.65pu \\ P&=V_{pu} \times I_{pu} \cos \phi \\ 0.8 &=1 \times I_{pu} \times 0.8 \\ I_{pu} &= 1pu\\ E &= V+jI_aX_s\\ &=1+1\angle -36.86^{\circ} \times j0.65 \\ &= 1.484 \angle 20.51^{\circ}pu\\ \delta &= 20.51^{\circ} \end{aligned}
 Question 2
Consider a lossy transmission line with $V_{1} \; and \; V_{2}$ as the sending and receiving end voltages, respectively. Z and X are the series impedance and reactance of the line, respectively. The steady-state stability limit for the transmission line will be
 A greter than $|\frac{V_{1}V_{2}}{X}|$ B less than $|\frac{V_{1}V_{2}}{X}|$ C equal to $|\frac{V_{1}V_{2}}{X}|$ D equal to $|\frac{V_{1}V_{2}}{Z}|$
GATE EE 2018      Power System Stability
Question 2 Explanation:
With only x:
$P_{max}=\left | \frac{V_1V_2}{x} \right |$

With Lossy Tr, Line
$P=\left | \frac{V_1V_2}{z} \right | \cos (\beta -\delta )-\left | \frac{AV_2^2}{z} \right |\cos (\beta -\delta )$

Therefore, with Lossy Line $P_{max} \lt \left | \frac{V_1V_2}{x} \right |$
 Question 3
A 3-phase, 2-pole, 50 Hz, synchronous generator has a rating of 250 MVA, 0.8 pf lagging. The kinetic energy of the machine at synchronous speed is 1000 MJ. The machine is running steadily at synchronous speed and delivering 60 MW power at a power angle of 10 electrical degrees. If the load is suddenly removed, assuming the acceleration is constant for 10 cycles, the value of the power angle after 5 cycles is ________ electrical degrees.
 A 1.27 B 2.7 C 12.7 D 127
GATE EE 2017-SET-2      Power System Stability
Question 3 Explanation:
\begin{aligned} S&=250 MVA \\ \cos \phi &=0.8 \\ K.E. &=1000MJ \\ P_e&= 60MW\\ \delta _0&=10^{\circ} \\ t&=10\; cycles\\ &=\frac{10}{50}=0.2\; sec \\ t&=5\; cycle =0.1sec \\ \text{Load}&\text{ is removed,} \\ P_e&=0 \\ P_a&=P_m-P_e \\ P_m &=60MW \\ M&=\frac{HS}{180f}=\frac{K.E.}{180f} \\ &= \frac{1000}{180 \times 50}=0.111\\ \frac{d^2 \delta }{dt^2}&=\frac{P_a}{M}=545.45 \text{ ele.degree}/s^2 \\ \text{Integrating}&\text{ the above eq.} \\ \int \left ( \frac{d^2 \delta }{dt^2} \right )dt&=\int \frac{P_a}{M}dt=545.45 \times t\\ \text{Integrating}&\text{ tit once again with dt} \\ &=545.45 \times t(dt)\\ \delta &=\frac{545.45 \times t^2}{2}=2.7^{\circ}\\ \text{So, new}&\text{ value of power angle}\\ &=10^{\circ}+2.7^{\circ}=12.7^{\circ} \end{aligned}
 Question 4
The figure shows the single line diagram of a power system with a double circuit transmission line. The expression for electrical power is $1.5 sin\delta$, where $\delta$ is the rotor angle. The system is operating at the stable equilibrium point with mechanical power equal to 1 pu. If one of the transmission line circuits is removed, the maximum value of $\delta$ as the rotor swings, is 1.221 radian. If the expression for electrical power with one transmission line circuit removed is $P_{max} sin\delta$, the valueof $P_{max}$, in pu is _________.
 A 0.729 B 1.22 C 2.6 D 4.8
GATE EE 2017-SET-1      Power System Stability
Question 4 Explanation:

Using equal area criteria:
$A_1=A_2$
$\int_{\delta _0}^{\delta _c}(P_{m0}-P_{max1} \sin \delta )d\delta =\int_{\delta _c}^{\delta _2}(P_{max1} \sin \delta -P_{m0})d\delta$
By solving above integration,
$P_{max1}=\frac{P_{m0}(\delta _2-\delta _0)}{\cos \delta _0- \cos \delta _2}$
Given data:
\begin{aligned} P_{m0} &=1pu \\ P_e&= 1.5 \sin (\delta _0)=P_{m0}=1pu\\ \delta _0&41.8^{\circ}=0.7295\; rad \\ \delta _2 &=1.221\; rad =96.95^{\circ} \end{aligned}
Substitute above values in above equation,
$P_{max1}=\frac{1(1.221-0.7295)}{\cos 41.8- \cos 69.95}=1.220 p.u.$
 Question 5
The synchronous generator shown in the figure is supplying active power to an infinite bus via two short, lossless transmission lines, and is initially in steady state. The mechanical power input to the generator and the voltage magnitude E are constant. If one line is tripped at time $t_1$ by opening the circuit breakers at the two ends (although there is no fault), then it is seen that the generator undergoes a stable transient. Which one of the following waveforms of the rotor angle $\delta$ shows the transient correctly?

 A A B B C C D D
GATE EE 2015-SET-2      Power System Stability
Question 5 Explanation:
Initial value of $\delta =\delta _0$ at $t =t_1$ with both lines are in service of one of the line circuit breakes opened the $\delta$ is increased for same time and stabilizes to new value of $\delta$.
 Question 6
A 50 Hz generating unit has H-constant of 2 MJ/MVA. The machine is initially operating in steady state at synchronous speed, and producing 1 pu of real power. The initial value of the rotor angle $\delta$ is 5$^{\circ}$, when a bolted three phase to ground short circuit fault occurs at the terminal of the generator. Assuming the input mechanical power to remain at 1 pu, the value of $\delta$ in degrees, 0.02 second after the fault is _________.
 A 0.5 B 2.8 C 4.2 D 5.9
GATE EE 2015-SET-1      Power System Stability
Question 6 Explanation:
Let $3-\phi$ fault occurs at t=0
$\Rightarrow \; P_e=0\text{ for }t\geq 0^+$
Now, swing equation for $t\geq 0^+$
\begin{aligned} \frac{2H}{\omega _s}\frac{d^2\delta }{dt^2} &=P_m \\ \frac{d^2\delta }{dt^2}&= \frac{P_m \omega _s }{2H}\\ \frac{d\delta }{dt}&= \frac{P_m \omega _s }{2H}t+C\\ C=\left.\begin{matrix} \frac{d\delta }{dt} \end{matrix}\right| _{t=0^+}&=\omega (t=0^+)-\omega _s\\ C=\omega _s-\omega _s&=0\;(\omega (t-0^+)=\omega _s) \\ \frac{d\delta }{dt}&=\frac{P_m \omega _s }{2H}t \\ \Rightarrow \; \delta &= \frac{P_m \omega _s }{4H}t^2+C'\\ C'&= \delta |_{t=0^+}=\delta _0\\ \delta &= \frac{P_m \omega _s }{4H}t^2+\delta _0\\ \text{Now in problem,}\\ \delta (t=0.02)&\\ &=\frac{1 \times 2 \times 180 \times 50 \times 0.02^2 +5^{\circ}}{4 \times 2}\\ &=5.9^{\circ} \end{aligned}
 Question 7
The figure shows the single line diagram of a single machine infinite bus system.

The inertia constant of the synchronous generator H = 5 MW-s/MVA. Frequency is 50 Hz. Mechanical power is 1 pu. The system is operating at the stable equilibrium point with rotor angle $\delta$ equal to 30$^{\circ}$. A three phase short circuit fault occurs at a certain location on one of the circuits of the double circuit transmission line. During fault, electrical power in pu is $P_{max} \sin \delta$ . If the values of $\delta \; and \; d\delta/dt$ at the instant of fault clearing are 45$^{\circ}$ and 3.762 radian/s respectively, then $P_{max}$ (in pu) is ______.
 A 0.11 B 0.23 C 0.48 D 0.64
GATE EE 2014-SET-3      Power System Stability
Question 7 Explanation:

\begin{aligned} \frac{2H}{\omega _s}\frac{d^2\delta }{dt^2}&=(P_m-P_e) \\ \text{Let, } \omega _r&=\frac{d\delta }{dt} \\ \frac{2H}{\omega _s}\frac{d\omega _r }{dt} &=(P_m-P_e) \\ \text{Multiplying with } &\omega _r \text{ in both side}\\ \frac{H}{\omega _s}\left ( 2\omega _r \frac{d\omega _r }{dt}\right )&= (P_m-P_e)\frac{d\delta }{dt}\\ \Rightarrow \; \frac{H}{\omega _s} 2d(\omega _r)^2&=(P_m-P_e)d\delta \\ \Rightarrow \; \frac{H}{\omega _s}\int_{\omega _{r_1}}^{\omega _{r_2}}d(\omega _r^2) &= \int_{\delta _1}^{\delta _2}(P_m-P_e)d\delta \\ \text{where, } & \\ \omega _{r_1}&=\left.\begin{matrix} \frac{d\delta }{dt} \end{matrix}\right|_{\delta =\delta _1} =\omega (\delta =\delta _1)-\omega _s\\ \omega _{r_2}&=\left.\begin{matrix} \frac{d\delta }{dt} \end{matrix}\right|_{\delta =\delta _2} =\omega (\delta =\delta _2)-\omega _s\\ \frac{H}{\omega _s}(\omega _{r_2}^2 -\omega _{r_1}^2)&=\int_{\delta _1}^{\delta _2}(P_m-P_e)d\delta \\ \text{In problem, }&\\ \frac{d\delta }{dt}=0, &\frac{d\delta }{dt}=3.762\\ \frac{5}{314}(3.762^2-0^2)&=\int_{30^{\circ}}^{45^{\circ}}(1-P_{max} \sin \delta )d\delta \\ \Rightarrow \;\; P_{max}&=0.23p.u. \end{aligned}
 Question 8
There are two generators in a power system. No-load frequencies of the generators are 51.5 Hz and 51 Hz, respectively, and both are having droop constant of 1Hz/MW. Total load in the system is 2.5 MW. Assuming that the generators are operating under their respective droop characteristics, the frequency of the power system in Hz in the steady state is ______.
 A 25 B 50 C 75 D 100
GATE EE 2014-SET-2      Power System Stability
Question 8 Explanation:
Let no-load frequency of generator-1 be 51.5 Hz and no-load frequency of generator-2 be 51 Hz.
Given total load = 2.5 Mwand drop of both machines =1 Hz/MW

Let machine-1 shares a load of $P_1$ MW them machine-2 will share a load of $(2.5-P_1)$ MW
Le the steady state frequency of the system be f.
\begin{aligned} \text{Now, }\frac{51.5-f}{P_1}&=1=droop \\ P_1&= (51.5-f)\;\;...(i)\\ \text{Also, } \left ( \frac{51-f}{2.5-P_1} \right )&=1=droop \\ 51-f &=2.5-P_1 \\ P_1 &=f-48.5 \;\;...(ii) \\ \text{equating eq. }&\text{(i) and (ii), we get,} \\ 51.5-f &=f-48.5 \\ 2f&=100 \\ f&=50Hz \end{aligned}
 Question 9
A synchronous generator is connected to an infinite bus with excitation voltage $E_f= 1.3$ pu. The generator has a synchronous reactance of 1.1 pu and is delivering real power (P) of 0.6 pu to the bus. Assume the infinite bus voltage to be 1.0 pu. Neglect stator resistance. The reactive power (Q) in pu supplied by the generator to the bus under this condition is _____.
 A 0.8 B 0.11 C 0.45 D 1.82
GATE EE 2014-SET-2      Power System Stability
Question 9 Explanation:
Given, $E_f=1.3pu, X_s=1.1pu, V=1 pu, P=0.6pu$
Neglecting stator resistance i.e. $\theta _s=90^{\circ}$
Output power delivered,
\begin{aligned} P &=\frac{VE_f}{X_s}\sin \delta \\ \sin \delta &=\frac{PX_s}{VE_f} \\ &= \frac{0.6 \times 1.1}{1 \times 1.3}\approx 0.5\\ \therefore \;\; \delta &=30^{\circ}=\text{Power angle} \\ \therefore \;\; \cos \delta &= \cos 30^{\circ}=\frac{\sqrt{3}}{2} \end{aligned}
Reactive power is given by,
\begin{aligned} Q&=\left [ \frac{VE_f}{X_s} \cos \delta -\frac{V^2}{X_s} \right ]\\ &=\left [ \frac{1 \times 1.3}{1.1} \times \frac{\sqrt{3}}{2}-\frac{1^2}{1.1} \right ]\\ Q&=(1.023-0.909)=0.114pu\\ &\approx 0.11pu \end{aligned}
 Question 10
A cylinder rotor generator delivers 0.5 pu power in the steady-state to an infinite bus through a transmission line of reactance 0.5 pu. The generator no-load voltage is 1.5 pu and the infinite bus voltage is 1 pu. The inertia constant of the generator is 5MW-s/MVA and the generator reactance is 1 pu. The critical clearing angle, in degrees, for a three-phase dead short circuit fault at the generator terminal is
 A 53.5 B 60.2 C 70.8 D 79.6
GATE EE 2012      Power System Stability
Question 10 Explanation:

\begin{aligned} P_e&= \frac{|E||V|}{(X_d+X_T)}\sin \delta _0\\ 0.5 &=\frac{1.5 \times 1}{(1+0.5)}\sin \delta _0 \\ \delta _0 &=\sin ^{-1}(0.5)=30^{\circ}=\frac{\pi}{6} \end{aligned}
Critical clearing angle
\begin{aligned} \delta _{cr}&=\cos ^{-1}[(\pi-2\delta _0) \sin \delta _0 -\cos \delta _0]\\ &=\cos ^{-1}\left [ \left ( \pi-\frac{\pi}{3} \right ) \times \frac{1}{2} -\frac{\sqrt{3}}{2}\right ]\\ &= \cos ^{-1} (0.18)=79.56^{\circ} \end{aligned}
There are 10 questions to complete.


text:
• Normalisation and assessment of neutron diffraction data from liquids

• # Fulltext

https://www.ias.ac.in/article/fulltext/pram/005/06/0328-0338

• # Keywords

Neutron diffraction; liquids; structure factor

• # Abstract

A couple of normalisation criteria have been proposed to assess liquid structure data over the past decade. In this paper, a critical examination of these and other plausible criteria is made. Neutron diffraction data from liquid krypton measured by Clayton and Heaton is used to study these criteria. It is shown that if the structure factorS (Q)=1+γ (Q) satisfies the Krogh-Moe relation, namely$$\rho _{0 = \frac{1}{{2\pi ^2 }}} {\text{ }}\mathop \smallint \limits_{\text{0}}^\infty {\text{ }}Q^{\text{2}} {\text{ }}\gamma {\text{ (}}Q{\text{) }}dQ$$ whereQ is the wavevector transfer andρ0 the mean atomic number density,S (Q) is properly normalised. Subjecting such data to the other criteria would reveal, however, the quality of data in so far as systematic errors and/or termination errors are concerned. If the data is measured with reasonable accuracy to sufficiently large values of the wavevector transfer all these criteria would assess the quality of the data to almost the same extent. This is established in this paper by improving and extending the liquid krypton data by procedures similar to that of Kaplow, Strong and Averbach and subjecting this revised data to the various criteria. It is concluded, thereby, that (i) one can study the neutron diffraction data from liquids using any of the criteria mentioned in the paper to improve the quality of the data and (ii) for normalisation purposes the simple relation {ie328-2} would suffice.

• # Author Affiliations

1. Nuclear Physics Division, Bhabha Atomic Research Centre, Bombay - 400085

• # Pramana – Journal of Physics

Volume 96, 2022
All articles
Continuous Article Publishing mode

• # Editorial Note on Continuous Article Publication

Posted on July 25, 2019