Whats The Wattage In Bkk?

[Guest endure]

Since we engulf in peacounting: As we talk Alternating Currents, the voltage is at any given time anywhere between zero and 360, depending on which point of the sinus curve you measure. The integral over a full period, however, is 220 to 240 volts.

<{POST_SNAPBACK}>

Shouldn't that be between zero and 340? (240 * 1.414)?

[chownah]

Since we engulf in peacounting: As we talk Alternating Currents, the voltage is at any given time anywhere between zero and 360, depending on which point of the sinus curve you measure. The integral over a full period, however, is 220 to 240 volts.

<{POST_SNAPBACK}>

Shouldn't that be between zero and 340? (240 * 1.414)?

<{POST_SNAPBACK}>

Not sure how serious this discussion is but shouldn't it be 220*1.414? And also what is this integral you are doing all about anyway?

[Tywais]

Since we engulf in peacounting: As we talk Alternating Currents, the voltage is at any given time anywhere between zero and 360, depending on which point of the sinus curve you measure. The integral over a full period, however, is 220 to 240 volts.

Shouldn't that be between zero and 340? (240 * 1.414)?

Not sure how serious this discussion is but shouldn't it be 220*1.414? And also what is this integral you are doing all about anyway?

Was wondering that too when I responded earlier that the integral of a sine is zero whether it is a full period or infinite number of periods and nowhere can come up with 220 to 240. If he is referring to zero to 360 degrees it is still zero over that period and the peak is 311 Volts (+ and -) along that axis.

[quadricorrelator]

I am retired.  I used to be an analog microchip designer.

<{POST_SNAPBACK}>

Lucky you retired before all that digital stuff took over!!

<{POST_SNAPBACK}>

It seems like analog should be dead, but somehow it is still alive. Many thought analog was on its way out 26 years ago (when I first started). But, the need for analog designers always seems to come up: power supply design for notebook computers is a hot field, cell phone radio frequency amplifiers is a hot field, disk drive design has quite a bit of analog circuitry in it, front end amplifier circuitry for fiber optics requires analog, phase lock loop design for communications, and on and on.

Data received from the world is analog. Such systems always need analog signal processing.

Maxim Corporation, Linear Technology Corporation, and the analog division of National Semiconductor are all analog companies which have been very profitable over a long period of time. But, they have recruiting difficulties.

There is a shortage of analog designers relative to the demand. Universities don't want to teach it because the world needs much greater numbers of digital designers and computer programmers. Students don't want to study it because it is not glamorous (at least not like digital design), it has a lot of mathematics in it, it takes about three or four years of industrial experience before you are independent enough to do a design on your own (partially because it is a mature field it takes time to learn everything already known), and it doesn't fit most people's temperment because it requires a very narrow focus (you spend your career learning more and more about the same basic devices; transistor, capacitor, resistor, and inductor).

But, when I was making my career decision, I didn't even consider digital design. It just wasn't for me. I like the math of analog circuitry. I liked the narrow focus. I liked the idea of going deeper and deeper into the same field.

I think there is quite a bit of job security for analog designers. I don't know how the pay scales compare for equivalent levels of competence.

<{POST_SNAPBACK}>

quadricorrelator Good Post - thank you, I for one found it interesting, proving your never to old to learn something new. One of the most interesting points for me was the "narrow focus" there are many people who try to breakaway from a natural or taught narrow focus not knowing (I include myself) there are fields where it can be advantageous.

mijan24

<{POST_SNAPBACK}>

Thank you for the kind words.

Yes, I tried to fit my personality to the kind of work available. I guess we all must try to find a way we can fit in the work place. Actually, I had great difficulty finding a way to fit in the work place for a long time.

Too Old to Learn: I hope I am never too old to learn. I hope I always enjoy learning. These pleasures will probably out last the enjoyment (or capability to participate in) of physical activities.

Right now I am pursuing my interest in Mathematical Logic, Set Theory, and Manifolds.

I recently became curious in the collapse of the hedge fund "Long Term Capital Management" in 1998. The fund was formed by a bunch of geniuses (two were Nobel Prize winners) who thought they could use purely mathematical techniques (developed by the Nobel Prize winners) to get great returns in the stock market. They were very successful in the first few years, but then their system collapsed very dramatically. They were bailed out by 15 financial institutions, each contributing 300 million dollars in order to prevent a a much more profound problem which might have been very wide spread. You can read about this if you do a google search. It is very dramatic.

I am studying the mathematics behind their methods, even though they failed. It is fascinating to me. I am amazed that advanced mathematics could be applied to finance to make money. It is interesting that the equations are analogous to some of the equations in Quantum Mechanics. They use Brownian motion (a random process describing how a particle suspended in a liquid is bounced around randomly by molecules) to model the random fluctuations in stock prices.

They thought they had found a way to hedge out risk completely use the Black-Scholes formula. Myron Scholes was one of the Nobel Prize winners at the hedge fund.

If I understand correctly, their methods failed because they made false mathematical assumptions. They assumed that future market behavior would be the same as historical market behavior. They tried to fit historical data to a "log normal" curve, but that turns out to be a mistake.

But, I am still learning about it.

Too old to have a romantic relatinnship: Do you have thoughts on this? I suppose it depends on the individual. It is a concern of mine right now.

[daleyboy]

Five bloody pages of wattage, slow week for everyone is it?

[dr_Pat_Pong]

What Watt ?

[daleyboy]

What Watt ? 

Is that said in a posh English accent doc? Ok yah, what what, old chum

[quadricorrelator]

Since we engulf in peacounting: As we talk Alternating Currents, the voltage is at any given time anywhere between zero and 360, depending on which point of the sinus curve you measure. The integral over a full period, however, is 220 to 240 volts.

<{POST_SNAPBACK}>

Shouldn't that be between zero and 340? (240 * 1.414)?

<{POST_SNAPBACK}>

Not sure how serious this discussion is but shouldn't it be 220*1.414? And also what is this integral you are doing all about anyway?

<{POST_SNAPBACK}>

Was wondering that too when I responded earlier that the integral of a sine is zero whether it is a full period or infinite number of periods and nowhere can come up with 220 to 240. If he is referring to zero to 360 degrees it is still zero over that period and the peak is 311 Volts (+ and -) along that axis.

<{POST_SNAPBACK}>

1. Integral over one period of sine: The integral over one period of a sine function is zero. The integral of a curve is the area of the curve above the x axis (horizontal axis) - the area of the curve below the x axis (horizontal axis).

The sine function has the same area above and below the axis so they cancel out.

Integral (sin(kx) dx) from x = a/k to x = a/k + 2*pi/k

= 1/k * Integral (sin(kx) dkx) from x = a/k to x = a/k + 2*pi/k

= 1/k * Integral (sin(u) du) from u = a to u = a + 2*pi (by the change of variables theorem)

= 1/k * (-cos(u)) from u = a to u = a + 2*pi

= 1/k * (-cos(a) - (-cos(a + 2*pi))

= 1/k *( cos(a+2*pi) - cos(a))

= 1/k * (cos (a) *cos(2*pi) -sin(a)*sin(2*pi)) - cos(a))

= 1/k * (cos(a) *1 - sin(a)* 0 - cos(a) )

= 1/k * (cos (a) - cos(a) )

= 0

2. Integration over an infinite number of periods:

The brief answer is that this integral does not converge so it doesn't exist. To see this, you have to go back to the definition of the "improper integral". An improper integral is one whose integration limits go to infinity.

Defintion:

Integral (f(x) dx) from x = 0 to x = infinity

is defined as

limit as u goes to infinity of Integral (f(x)dx) from x = 0 to x = u

But, now we must define the limit of a function as u goes to infinity.

Definition: L = Limit (g(u)) as u goes to infinity if and only if for all E, there exist an H, such that if u>H, then the absolute value of (g(u) - L) is less than E.

Now we can apply these definitions to try to evaluate our integral.

Integral (sin(x) dx) from x = 0 to x = infinity

= limit as u goes to infinity (Integral (sin(x) dx) ) from x = 0 to x = u

= limit as u goes to infinity (-cos(x)) from x = 0 to x = u

= limit as u goes to infinity (cos(u) - cos(0))

= limit as u goes to infinity (cos (u) - 1)

Does such a limit exist? I will assume that such a limit did exist, and the arrive a contradiction. This will show that the original assumption "that the limit does exist" is incorrect.

Suppose the limit was L. Then, for any E, we could find an H such that if u>H, then -E <(cos(u) - 1) - L<E

so that (1+ L)-E < cos(u) < (L+1) + E

This inequality must hold for any value of E, so we can pick an E which is arbitrarily small. Let's pick E = .01, then there exists an H such that u>H implies that

.99 +L < cos(u) < 1.01 + L

so the maximum of cos(u) - minimum of cos (u) can only be .02 for values of u greater than H.

Let us examine some values of u which are greater than H. Let us examine the values of cos(u) for H<u<H + 2*pi. Over this region, cos(u) reaches a maximum of 1, and a minimum of -1. But, we already deduced that the maximum (cos(u)) - minimum (cos(u)) for u>h was .02. This contradiction implies that our original assumption (that the limit exists) is wrong.

3. How to evaluate power delivered to a resistor from a sinusoidal voltage source.

Suppose you have a voltage source across a resistor. Let's say the voltage source is described by Vsin(kt). Let's say the resistor has value R.

From previous messages in this thread, we know that the instaneous power delivered to any device is Power = voltage * current.

The current through the resistor is current = voltage / resistance.

So, the instantaneous power (power at any point in time) = voltage * current

= (voltage * voltage)/resistance

= (V*V*sin(kt)*sin(kt))?R

So the power is changing all the time in this resistor. Sometimes it is 0, but sometimes it is (V*V)/R. We might want to know the average power delivered to the resistor. To do this we must find the area under the power curve for one cycle, and then divide by the cycle time.

The time of one cycle (period) is (2*pi)/k.

Average power= k/(2*pi) * Integral (V*V*sin(kt)*sin(kt))/R dt) from t=0 to t=2*pi/k

= (k/(2*pi))*V*V/R Integral (sin(kt)sin(kt)dt) from t=0 to t=2*pi/k

= (k/(2*pi))*V*V/R*(1/k) Integral (sin(kt)sin(kt)dkt) from t=0 to t=2*pi/k

= (k/(2*pi))*V*V/R*(1/k) Integral (sin(u)sin(u)du) from u=0 to u=2*pi

= (k/(2*pi))*V*V/R*(1/k) Integral ( ((1-cos(2*u))/2)du) from u=0 to u=2*pi

= (k/(2*pi))*V*V/R*(1/k)(1/2) Integral ( ((1-cos(2*u))/2)d2*u) from u=0 to u=2*pi

= (k/(2*pi))*V*V/R*(1/k)(1/2) Integral ( ((1-cos(y))/2)dy) from y=0 to y=pi

(change of variable y=2*u)

= (k/(2*pi))*V*V/R*(1/k)(1/2) ((y + sin(y)/2) from y=0 to y=pi

= (k/(2*pi))*V*V/R*(1/k)(1/2) ((y + sin(y)/2) from y=0 to y=pi

= (k/(2*pi))*V*V/R*(1/k)(1/2) (pi)/2

= V*V/(2*R)

4. Line voltage maximum: The line voltage is not specified in terms of the maximum voltage. When they say 220 volts, they are refering to the RMS voltage. This means that they take the square of the voltage, then they take the mean of that over one cycle, and then they take the square root of that.

So, let's see if we can figure out the peak voltage using this idea.

Suppose our line voltage is Vsin(kt) (k is going to be 2*pi*60)

We want to know what V is. But, the power company tells us only the RMS value.

square root (mean(square(Vsin(kt) = 220.

We need to solve this equation for V.

square root (mean (V*V*sin(kt)*sin(kt)dt))) = 220 from t = 0 to t=2*pi/k

square root (V*V*(k/(2*pi)) Integral (sin(kt)sin(kt)dt))) = 220 from t = 0 to t=2*pi/k

but from the previous calculation we know that this integral is k/4, so

square root (V*V*(k/(2*pi)* (4/k)) = 220

square root V*V/2 = 220

V/1.4 = 220

V = 308

so the voltage swings between -308 volts and +308 volts.

[daleyboy]

Is this REALLY necessary? Does anyone REALLY care?

[toastwars]

I don't!

[Glauka]

Is this REALLY necessary? Does anyone REALLY care?

<{POST_SNAPBACK}>

[chuchok]

What Watt ? 

<{POST_SNAPBACK}>

Wat

[SiamOne]

Since we engulf in peacounting: As we talk Alternating Currents, the voltage is at any given time anywhere between zero and 360, depending on which point of the sinus curve you measure. The integral over a full period, however, is 220 to 240 volts.

<{POST_SNAPBACK}>

Shouldn't that be between zero and 340? (240 * 1.414)?

<{POST_SNAPBACK}>

Not sure how serious this discussion is but shouldn't it be 220*1.414? And also what is this integral you are doing all about anyway?

<{POST_SNAPBACK}>

Was wondering that too when I responded earlier that the integral of a sine is zero whether it is a full period or infinite number of periods and nowhere can come up with 220 to 240. If he is referring to zero to 360 degrees it is still zero over that period and the peak is 311 Volts (+ and -) along that axis.

<{POST_SNAPBACK}>

1. Integral over one period of sine: The integral over one period of a sine function is zero. The integral of a curve is the area of the curve above the x axis (horizontal axis) - the area of the curve below the x axis (horizontal axis).

The sine function has the same area above and below the axis so they cancel out.

Integral (sin(kx) dx) from x = a/k to x = a/k + 2*pi/k

= 1/k * Integral (sin(kx) dkx) from x = a/k to x = a/k + 2*pi/k

= 1/k * Integral (sin(u) du) from u = a to u = a + 2*pi (by the change of variables theorem)

= 1/k * (-cos(u)) from u = a to u = a + 2*pi

= 1/k * (-cos(a) - (-cos(a + 2*pi))

= 1/k *( cos(a+2*pi) - cos(a))

= 1/k * (cos (a) *cos(2*pi) -sin(a)*sin(2*pi)) - cos(a))

= 1/k * (cos(a) *1 - sin(a)* 0 - cos(a) )

= 1/k * (cos (a) - cos(a) )

= 0

2. Integration over an infinite number of periods:

The brief answer is that this integral does not converge so it doesn't exist. To see this, you have to go back to the definition of the "improper integral". An improper integral is one whose integration limits go to infinity.

Defintion:

Integral (f(x) dx) from x = 0 to x = infinity

is defined as

limit as u goes to infinity of Integral (f(x)dx) from x = 0 to x = u

But, now we must define the limit of a function as u goes to infinity.

Definition: L = Limit (g(u)) as u goes to infinity if and only if for all E, there exist an H, such that if u>H, then the absolute value of (g(u) - L) is less than E.

Now we can apply these definitions to try to evaluate our integral.

Integral (sin(x) dx) from x = 0 to x = infinity

= limit as u goes to infinity (Integral (sin(x) dx) ) from x = 0 to x = u

= limit as u goes to infinity (-cos(x)) from x = 0 to x = u

= limit as u goes to infinity (cos(u) - cos(0))

= limit as u goes to infinity (cos (u) - 1)

Does such a limit exist? I will assume that such a limit did exist, and the arrive a contradiction. This will show that the original assumption "that the limit does exist" is incorrect.

Suppose the limit was L. Then, for any E, we could find an H such that if u>H, then -E <(cos(u) - 1) - L<E

so that (1+ L)-E < cos(u) < (L+1) + E

This inequality must hold for any value of E, so we can pick an E which is arbitrarily small. Let's pick E = .01, then there exists an H such that u>H implies that

.99 +L < cos(u) < 1.01 + L

so the maximum of cos(u) - minimum of cos (u) can only be .02 for values of u greater than H.

Let us examine some values of u which are greater than H. Let us examine the values of cos(u) for H<u<H + 2*pi. Over this region, cos(u) reaches a maximum of 1, and a minimum of -1. But, we already deduced that the maximum (cos(u)) - minimum (cos(u)) for u>h was .02. This contradiction implies that our original assumption (that the limit exists) is wrong.

3. How to evaluate power delivered to a resistor from a sinusoidal voltage source.

Suppose you have a voltage source across a resistor. Let's say the voltage source is described by Vsin(kt). Let's say the resistor has value R.

From previous messages in this thread, we know that the instaneous power delivered to any device is Power = voltage * current.

The current through the resistor is current = voltage / resistance.

So, the instantaneous power (power at any point in time) = voltage * current

= (voltage * voltage)/resistance

= (V*V*sin(kt)*sin(kt))?R

So the power is changing all the time in this resistor. Sometimes it is 0, but sometimes it is (V*V)/R. We might want to know the average power delivered to the resistor. To do this we must find the area under the power curve for one cycle, and then divide by the cycle time.

The time of one cycle (period) is (2*pi)/k.

Average power= k/(2*pi) * Integral (V*V*sin(kt)*sin(kt))/R dt) from t=0 to t=2*pi/k

= (k/(2*pi))*V*V/R Integral (sin(kt)sin(kt)dt) from t=0 to t=2*pi/k

= (k/(2*pi))*V*V/R*(1/k) Integral (sin(kt)sin(kt)dkt) from t=0 to t=2*pi/k

= (k/(2*pi))*V*V/R*(1/k) Integral (sin(u)sin(u)du) from u=0 to u=2*pi

= (k/(2*pi))*V*V/R*(1/k) Integral ( ((1-cos(2*u))/2)du) from u=0 to u=2*pi

= (k/(2*pi))*V*V/R*(1/k)(1/2) Integral ( ((1-cos(2*u))/2)d2*u) from u=0 to u=2*pi

= (k/(2*pi))*V*V/R*(1/k)(1/2) Integral ( ((1-cos(y))/2)dy) from y=0 to y=pi

(change of variable y=2*u)

= (k/(2*pi))*V*V/R*(1/k)(1/2) ((y + sin(y)/2) from y=0 to y=pi

= (k/(2*pi))*V*V/R*(1/k)(1/2) ((y + sin(y)/2) from y=0 to y=pi

= (k/(2*pi))*V*V/R*(1/k)(1/2) (pi)/2

= V*V/(2*R)

4. Line voltage maximum: The line voltage is not specified in terms of the maximum voltage. When they say 220 volts, they are refering to the RMS voltage. This means that they take the square of the voltage, then they take the mean of that over one cycle, and then they take the square root of that.

So, let's see if we can figure out the peak voltage using this idea.

Suppose our line voltage is Vsin(kt) (k is going to be 2*pi*60)

We want to know what V is. But, the power company tells us only the RMS value.

square root (mean(square(Vsin(kt) = 220.

We need to solve this equation for V.

square root (mean (V*V*sin(kt)*sin(kt)dt))) = 220 from t = 0 to t=2*pi/k

square root (V*V*(k/(2*pi)) Integral (sin(kt)sin(kt)dt))) = 220 from t = 0 to t=2*pi/k

but from the previous calculation we know that this integral is k/4, so

square root (V*V*(k/(2*pi)* (4/k)) = 220

square root V*V/2 = 220

V/1.4 = 220

V = 308

so the voltage swings between -308 volts and +308 volts.

<{POST_SNAPBACK}>

Thank you, I find it interesting, and do care.

I was looking at the previous post really, about the maths genii (?) who tried to beat the market. Of course, like everybody else, they were successful when the market was bouyant. The admission that their maths was incorrect is no surprise, the market movements are so random as to be incoherent.

Maybe their incoherency formulae (better) would work.

[mijan24]

Although the level of quadricorrelators maths far out-weighs my own I find it enlightening and the story on the investors comes back to "assume nothing"

Yes I care quadricorrelator your posts are exercising many a mind - I can imagine some old cogs cranking slowly as the cob-webs clear and their owners reach for a - for those to old to enjoy anyhing but try a

[dr_Pat_Pong]

Although the level of quadricorrelators maths far out-weighs my own I find it enlightening and the story on the investors comes back to "assume nothing"

Yes I care :welcome :quadricorrelator your posts are exercising many a mind - I can imagine some old cogs cranking slowly as the cob-webs clear and their owners reach for a - for those to old to enjoy anyhing but    try a

<{POST_SNAPBACK}>

[Crossy]

Don't you just love the way a simple typo in the OP has sent this thread into orbit?

Stirred up the old grey-matter, remembering when I studied all this stuff in Uni. XX years ago.

[Tywais]

Don't you just love the way a simple typo in the OP has sent this thread into orbit?

Well this is ThaiVisa, I think there is nothing forum members enjoy more then to jump on an error.

[chang35baht]

Do any of you guys know where I can buy a meter to measure the amount of knowledge I have consumed from reading this thread.

K = Pr x T(measured in minutes) / DU(don`t understand) x STML(short term memory loss).

Thanx guys.

[Jai Dee]

Don't you just love the way a simple typo in the OP has sent this thread into orbit?

Stirred up the old grey-matter, remembering when I studied all this stuff in Uni. XX years ago.

More like XXXX beers ago for me...

[quadricorrelator]

Although the level of quadricorrelators maths far out-weighs my own I find it enlightening and the story on the investors comes back to "assume nothing"

Yes I care quadricorrelator your posts are exercising many a mind - I can imagine some old cogs cranking slowly as the cob-webs clear and their owners reach for a - for those to old to enjoy anyhing but    try a

<{POST_SNAPBACK}>

Thanks for the welcome. It's good to connect with like minds.

The story of the Long Term Capital Management captured my imagination. I read a riveting book about it called "When Genius Failed" (I bought it at Kinokunya). The core members of the company were shy, socially underdeveloped, and self-confessed nerds, who really didn't even care that much about money (they liked making it, but they didn't care about spending it, and they were mostly interesting in proving their theories were right). They were getting paid more than anyone else in the company too.

It was interesting that the "nerds" were former students of the Nobel Prize winners, and they were in their early thirties at the time. But, in fact, they had far more power and made more money that their professors. The Nobel Prize winners developed the theory, but were muscled out by their students who were more practical. In fact, the conviction of the core traders gave them power to bully executives at brokerage houses like Saloman, Merryl, Bear-Stearns, etc. Eventually they bumped heads with Soros and Buffet (when the company started to fail). Even with all their conviction and supreme financial confidence, they still had trouble finding a date for Saturday night.

According to this book, the hedge fund failed because their predictions were based on historical data, but within three years, the market did something which it had never done before.

Mathematically speaking, they were fitting the probability density curves with a lognormal curve, which is just the "Bell Curve" for IQ etc. This curve drops off very very rapaidly when you go out greater than 5 sigma (5 standard deviations) from the mean. But, the actually market data does fit the curve when you get that far out. But, the theory made use of this part of the Bell Curve. As a result, it failed.

The book points out that Scholes (one of the Nobel Prize winners professors) advisor at University of Chicago, Fama, showed that the market violates the Bell Curve probability density function about every three or four years. According to this book, the guys should have known better.

The same core guys have formed a new hedge fund called JWM Partners (John Merriwether) just down the street from their old company.

[quadricorrelator]

Since we engulf in peacounting: As we talk Alternating Currents, the voltage is at any given time anywhere between zero and 360, depending on which point of the sinus curve you measure. The integral over a full period, however, is 220 to 240 volts.

<{POST_SNAPBACK}>

Shouldn't that be between zero and 340? (240 * 1.414)?

<{POST_SNAPBACK}>

Not sure how serious this discussion is but shouldn't it be 220*1.414? And also what is this integral you are doing all about anyway?

<{POST_SNAPBACK}>

Was wondering that too when I responded earlier that the integral of a sine is zero whether it is a full period or infinite number of periods and nowhere can come up with 220 to 240. If he is referring to zero to 360 degrees it is still zero over that period and the peak is 311 Volts (+ and -) along that axis.

<{POST_SNAPBACK}>

1. Integral over one period of sine: The integral over one period of a sine function is zero. The integral of a curve is the area of the curve above the x axis (horizontal axis) - the area of the curve below the x axis (horizontal axis).

The sine function has the same area above and below the axis so they cancel out.

Integral (sin(kx) dx) from x = a/k to x = a/k + 2*pi/k

= 1/k * Integral (sin(kx) dkx) from x = a/k to x = a/k + 2*pi/k

= 1/k * Integral (sin(u) du) from u = a to u = a + 2*pi (by the change of variables theorem)

= 1/k * (-cos(u)) from u = a to u = a + 2*pi

= 1/k * (-cos(a) - (-cos(a + 2*pi))

= 1/k *( cos(a+2*pi) - cos(a))

= 1/k * (cos (a) *cos(2*pi) -sin(a)*sin(2*pi)) - cos(a))

= 1/k * (cos(a) *1 - sin(a)* 0 - cos(a) )

= 1/k * (cos (a) - cos(a) )

= 0

2. Integration over an infinite number of periods:

The brief answer is that this integral does not converge so it doesn't exist. To see this, you have to go back to the definition of the "improper integral". An improper integral is one whose integration limits go to infinity.

Defintion:

Integral (f(x) dx) from x = 0 to x = infinity

is defined as

limit as u goes to infinity of Integral (f(x)dx) from x = 0 to x = u

But, now we must define the limit of a function as u goes to infinity.

Definition: L = Limit (g(u)) as u goes to infinity if and only if for all E, there exist an H, such that if u>H, then the absolute value of (g(u) - L) is less than E.

Now we can apply these definitions to try to evaluate our integral.

Integral (sin(x) dx) from x = 0 to x = infinity

= limit as u goes to infinity (Integral (sin(x) dx) ) from x = 0 to x = u

= limit as u goes to infinity (-cos(x)) from x = 0 to x = u

= limit as u goes to infinity (cos(u) - cos(0))

= limit as u goes to infinity (cos (u) - 1)

Does such a limit exist? I will assume that such a limit did exist, and the arrive a contradiction. This will show that the original assumption "that the limit does exist" is incorrect.

Suppose the limit was L. Then, for any E, we could find an H such that if u>H, then -E <(cos(u) - 1) - L<E

so that (1+ L)-E < cos(u) < (L+1) + E

This inequality must hold for any value of E, so we can pick an E which is arbitrarily small. Let's pick E = .01, then there exists an H such that u>H implies that

.99 +L < cos(u) < 1.01 + L

so the maximum of cos(u) - minimum of cos (u) can only be .02 for values of u greater than H.

Let us examine some values of u which are greater than H. Let us examine the values of cos(u) for H<u<H + 2*pi. Over this region, cos(u) reaches a maximum of 1, and a minimum of -1. But, we already deduced that the maximum (cos(u)) - minimum (cos(u)) for u>h was .02. This contradiction implies that our original assumption (that the limit exists) is wrong.

3. How to evaluate power delivered to a resistor from a sinusoidal voltage source.

Suppose you have a voltage source across a resistor. Let's say the voltage source is described by Vsin(kt). Let's say the resistor has value R.

From previous messages in this thread, we know that the instaneous power delivered to any device is Power = voltage * current.

The current through the resistor is current = voltage / resistance.

So, the instantaneous power (power at any point in time) = voltage * current

= (voltage * voltage)/resistance

= (V*V*sin(kt)*sin(kt))?R

So the power is changing all the time in this resistor. Sometimes it is 0, but sometimes it is (V*V)/R. We might want to know the average power delivered to the resistor. To do this we must find the area under the power curve for one cycle, and then divide by the cycle time.

The time of one cycle (period) is (2*pi)/k.

Average power= k/(2*pi) * Integral (V*V*sin(kt)*sin(kt))/R dt) from t=0 to t=2*pi/k

= (k/(2*pi))*V*V/R Integral (sin(kt)sin(kt)dt) from t=0 to t=2*pi/k

= (k/(2*pi))*V*V/R*(1/k) Integral (sin(kt)sin(kt)dkt) from t=0 to t=2*pi/k

= (k/(2*pi))*V*V/R*(1/k) Integral (sin(u)sin(u)du) from u=0 to u=2*pi

= (k/(2*pi))*V*V/R*(1/k) Integral ( ((1-cos(2*u))/2)du) from u=0 to u=2*pi

= (k/(2*pi))*V*V/R*(1/k)(1/2) Integral ( ((1-cos(2*u))/2)d2*u) from u=0 to u=2*pi

= (k/(2*pi))*V*V/R*(1/k)(1/2) Integral ( ((1-cos(y))/2)dy) from y=0 to y=pi

(change of variable y=2*u)

= (k/(2*pi))*V*V/R*(1/k)(1/2) ((y + sin(y)/2) from y=0 to y=pi

= (k/(2*pi))*V*V/R*(1/k)(1/2) ((y + sin(y)/2) from y=0 to y=pi

= (k/(2*pi))*V*V/R*(1/k)(1/2) (pi)/2

= V*V/(2*R)

4. Line voltage maximum: The line voltage is not specified in terms of the maximum voltage. When they say 220 volts, they are refering to the RMS voltage. This means that they take the square of the voltage, then they take the mean of that over one cycle, and then they take the square root of that.

So, let's see if we can figure out the peak voltage using this idea.

Suppose our line voltage is Vsin(kt) (k is going to be 2*pi*60)

We want to know what V is. But, the power company tells us only the RMS value.

square root (mean(square(Vsin(kt) = 220.

We need to solve this equation for V.

square root (mean (V*V*sin(kt)*sin(kt)dt))) = 220 from t = 0 to t=2*pi/k

square root (V*V*(k/(2*pi)) Integral (sin(kt)sin(kt)dt))) = 220 from t = 0 to t=2*pi/k

but from the previous calculation we know that this integral is k/4, so

square root (V*V*(k/(2*pi)* (4/k)) = 220

square root V*V/2 = 220

V/1.4 = 220

V = 308

so the voltage swings between -308 volts and +308 volts.

<{POST_SNAPBACK}>

Thank you, I find it interesting, and do care.

I was looking at the previous post really, about the maths genii (?) who tried to beat the market. Of course, like everybody else, they were successful when the market was bouyant. The admission that their maths was incorrect is no surprise, the market movements are so random as to be incoherent.

Maybe their incoherency formulae (better) would work.

<{POST_SNAPBACK}>

Thanks. Since you seem interested, I wanted to mention a link in which economists, and even the Noble Prize winners of the failed company are interviewed about Long Term Capital Management's failure. I found it fascinating. It doesn't go into the mathematical details, but I found it stimulating and dramatic.

http://www.bbc.co.uk/science/horizon/1999/midas_script.shtml

I am studying the mathematics they used right now (the mathematics is useful, but not for beating the stock market). It falls under the category of "Options Pricing Theory". There are many lectures and books about it on the internet. Just do a google search on Black-Scholes.]

Here is a site with many free (and legal) finance books you can download:

http://www.econphd.net/notes.htm

It raises the question in my mind, "If these geniuses can't find a way to win at the stock market, then what change to the rest of us have?" Maybe there is someone out there who has found a system, but why should I think I would be that lucky person?

[stumonster]

It raises the question in my mind, "If these geniuses can't find a way to win at the stock market, then what change to the rest of us have?" 

maybe they forgot to factor into the equations the amount of insider trading/ stock manipulation / creative accounting that goes on.

I avoided this thread because I though it would just contain many posts bagging the original posters misuse of terminology - I am now glad that curiosity as to why there were so many replies led me to read it - thanks for your posts quadricorrelator , and there will always be a requirement for analog techs as the world is analog and no matter how many slices your ADC takes out of the waveform when you require accuracy it has to be analog.

[Boon Mee]

... when you require accuracy it has to be analog.

I've noticed that with the newer cars now the dashboard displays are a lot more analog. Always liked the analog meters vs. digital for displaying oil/gas/temps etc but then I've been call 'retro' for this...

[stumonster]

... when you require accuracy it has to be analog.

I've noticed that with the newer cars now the dashboard displays are a lot more analog. Always liked the analog meters vs. digital for displaying oil/gas/temps etc but then I've been call 'retro' for this...

analog displays ( even digitally produced ) are quicker to be comprehended by humans than a digital readout.

[RDN]

... when you require accuracy it has to be analog.

<{POST_SNAPBACK}>

Are we going to start discussing vinyl analogue LPs versus digital CDs now?

[stumonster]

Are we going to start discussing vinyl analogue LPs versus digital CDs now?  :D

...ah surface area - the base of life as we know it....

[quadricorrelator]

Are we going to start discussing vinyl analogue LPs versus digital CDs now?  :D

...ah surface area - the base of life as we know it....

<{POST_SNAPBACK}>

Random noise in analog circuits same same random fluctations in stock market:

It is difficult to escape from the analog world. Even digital circuits are really analog when you get down to the basic devices which make up the circuits.

Digital circuits are made up of transistors, and those are analog devices. The transistors are built up into digital structures, but the designers who do that work are analog designers.

What I find most interesting is that digital circuits are not 100% digital. You think of a digital signal as going between two voltages. But, actually, the voltages are not pure. The voltages have tiny random fluctations on them. This is unavoidable, and the noise is intrinsic to the basic devices. The noise current in a transistor goes as the square root of its operating current. Even a resistor has voltage noise which goes as the square root of its resistance (or can be modeled as current noise inversely proportional to the square root of its resistance).

But, these tiny random fluctuations have a probability distribution, so that big fluctuations are unlikely. However, they do happen occasionally, and they cause errors. A "1" can be mistaken for a "0" etc.

Interestingly enough, the probability characteristics for electrical noise are the same as the random fluctuations in stock prices. They are both modeled as lognormal probability densities.

The electrical fluctuations are so small that you might only get an error 1 out of every 10 billion bits. But, at a 1 Gigabit data rate, this would give an error every 10 seconds! That would be horrible if it were an error in your bank account.

As data rates go up, the problem will get worse and worse. Also, the problem gets worse because the noise fluctations are also proportional to the square root of the bandwidth of the circuit (related to data rate).

The problem is solved partially by allowing errors and then using coding techniques to detect and correct the errors. In the stock market, this kind of redundancy won't work because people don't give back your money, so that you can get another chance to invest.

One last thought: There is one more similarity between stock market statistics and electrical statistics. The volatility of an asset increases as the square root of the amount of time it is held. Luckily, the mean return increases exponentially (compound interest means exponential growth), so if you wait long enough, your mean portfolio value grows faster than its volatility.

We often put resistors in parallel to reduce the noise fluctuations. This is equivalent to diversifying your porfolio in the stock market. The average return stays the same, but the volatility goes down.

[chownah]

"Even a resistor has voltage noise which goes as the square root of its resistance "

What gives rise to this noise?

[Crossy]

"Even a resistor has voltage noise which goes as the square root of its resistance "

What gives rise to this noise?

Random movement of electrons.

On high value resistors it can be significant, that's why you see labels saying 'low noise' on what is essentially a passive component.

[Tywais]

"Even a resistor has voltage noise which goes as the square root of its resistance "

What gives rise to this noise?

Random movement of electrons.

On high value resistors it can be significant, that's why you see labels saying 'low noise' on what is essentially a passive component.

And it gets worse as the temperature of the resistor increases. More heat, more electron movement which also causes the resistor to change resistance.