']['error
Banned
+630|6947|The Netherlands
hey guys,


i modified my AGP-slot to an PCI-E

he electromagnetic spectrum is the range of all possible electromagnetic radiation. Also, the "electromagnetic spectrum" (usually just spectrum) of an object is the range of electromagnetic radiation that it emits, reflects, or transmits.

The electromagnetic spectrum, shown in the table, extends from just below the frequencies used for modern radio (at the long-wavelength end) to gamma radiation (at the short-wavelength end), covering wavelengths from thousands of kilometres down to fractions of the size of an atom. It is commonly said that EM waves beyond these limits are uncommon, although this is not actually true. The 22-year sunspot cycle, for instance, produces radiation with a period of 22 years, or a frequency of 1.4*10-9 Hz. At the other extreme, photons of arbitrarily high frequency may be produced by colliding electrons with positrons at appropriate energy. 1024 Hz photons can be produced today with man-made accelerators. In our universe the short wavelength limit is likely to be the Planck length, and the long wavelength limit is the size of the universe itself (see physical cosmology), though in principle the spectrum is infinite.

Electromagnetic energy at a particular wavelength λ (in vacuum) has an associated frequency f and photon energy E. Thus, the electromagnetic spectrum may be expressed equally well in terms of any of these three quantities. They are related according to the equations:

    \lambda = \frac{c}f \,\!

and

    E=hf \,\!

or

    E=hc/\lambda \,\!

where:

    * c is the speed of light, 299792458 m/s (c \approx 3 \cdot 10^8 \ \mbox{ m}/\mbox{s} = 300,000 \ \mbox{km}/\mbox{s}).
    * h is Planck's constant, (h \approx 6.626069 \cdot 10^{-34} \ \mbox{J} \cdot \mbox{s} \approx 4.13567 \ \mathrm{\mu} \mbox{eV}/\mbox{GHz}).


\hat K_j (1) f(1)= \phi_j(1) \int { \frac{\phi_j^*(2)f(2)}{r_{12}}dv_2}

where \hat K_j (1) is the one-electron Exchange operator,

f(1),f(2) are the one-electron wavefunction being acted upon by the Exchange operator as functions of the positions of electron 1 and electron 2,

φj(1),φj(2) are the one-electron wavefunction of the j-th electron as functions of the positions of electron 1 and electron 2,

r12 is the distance between electrons 1 and 2.



    out = νEc + ρEs
    Ec = AEi
    Ei = βEc + ξEs

The model is maybe called "Black's model".

Here Es is a source quantity and Eout is an output quantity. In electronics, each of Es and Eout can be either a voltage or a current.

Since we are dealing with physical quantities, β * A is dimensionless. ρ has the same dimension as ξ * A * ν.

Now, a feedback amplifier consists of a gain part and a feedback network. The (usually passive) feedback network makes the total gain precise.

The asymptotic behaviour of the model comes when A goes towards infinity. Then ξ, β and ν determine the overall gain, when ρ is ignored (which is usually the case).

"Structured Electronic Design" (see references) makes a simplification, which perhaps is not correct. They say that the transfer function of the system is

    A_t = \frac{A}{1-A * \beta} + \rho

but that is only true if ξ * ν = 1, in which case At becomes dimensionless. The error they make is ξ and ν cannot be disregarded unless they are dimensionless. They also claim that β is the transfer function of the feedback network. This is usually not the case.

Imagine a voltage amplifier where A is taken as the transconductance of a transistor, and where the gain is set by a voltage divider. Then the dimension of A is conductance (current/voltage), the dimension for β is impedance (voltage/current), whereas the transfer function of the feedback network is dimensionless, and cannot be β.

Resistors

To find the total resistance of all the components, add together the individual resistances of each component:

    A diagram of several resistors, connected end to end, with the same amount of current going through each

    R_\mathrm{total} = R_1 + R_2 + \cdots + R_n

    for components in series, having resistances \ R_1, \ R_2, etc.


To find the current, \ I use Ohm's law I = \frac{V}{R_{total}}

To find the voltage across any particular component with resistance \ R_i, use Ohm's law again. V_i = I \cdot R_i

    Where \ I is the current, as calculated above.

Note that the components divide the voltage according to their resistances, so, in the case of two resistors:

    \frac{V_1}{V_2} = \frac{R_1}{R_2}

[edit]

Inductors

Inductors follow the same law, in that the total inductance of non-coupled inductors in series is equal to the sum of their individual inductances:

    A diagram of several inductors, connected end to end, with the same amount of current going through each

    L_\mathrm{total} = L_1 + L_2 + \cdots + L_n

However, in some situations it is difficult to prevent adjacent inductors from influencing each other, as the magnetic field of one device couples with the windings of its neighbours. This influence is defined by the mutual inductance M. For example, if you have two inductors in series, there are two possible equivalent inductances:

    \ L_\mathrm{total} = (L_1 + M) + (L_2 + M) or
    \ L_\mathrm{total} = (L_1 - M) + (L_2 - M)

Which formula is the correct one, depends how the magnetic fields of both inductors influence each other.

When there are more than two inductors, it gets more complicated, since you have to take into account the mutual inductance of each of them and how each coils influences the other.

So for three coils, there are three mutual inductances (M12,M13 and M23) and eight possible equations.
[edit]

Capacitors

Capacitors follow a different law. The total capacitance of capacitors in series is equal to the reciprocal of the sum of the reciprocals of their individual capacitances:

    A diagram of several capacitors, connected end to end, with the same amount of current going through each

    {1\over{C_\mathrm{total}}} = {1\over{C_1}} + {1\over{C_2}} + \cdots + {1\over{C_n}}

The working voltage of a series combination of identical capacitors is equal to the sum of voltage ratings of individual capacitors provided that equalizing resistors are used to ensure equal voltage division.
[edit]

Parallel circuits

Voltages across components in parallel with each other are the same in magnitude and they also have identical polarities. Hence, the same voltage variable is used for all circuits elements in such a circuit.

To find the total current, I, use Ohm's Law on each loop, then sum. (See Kirchhoff's circuit laws for an explanation of why this works). Factoring out the voltage (which, again, is the same across parallel components) gives:

    I_\mathrm{total} = V \cdot \left(\frac{1} {R_1} + \frac{1} {R_2} + \cdots + \frac{1} {R_n}\right)

[edit]

Notation

The parallel property can be represented in equations by two vertical lines "||" (as in geometry) to simplify equations. For two resistors,

    R_\mathrm{total} = R_1 \| R_2 = {R_1 R_2 \over R_1 + R_2}

[edit]

Resistors

To find the total resistance of all the components, add together the individual reciprocal of each resistance of each component, and take the reciprocal of the sum:

    A diagram of several resistors, side by side, both leads of each connected to the same wires

    {1 \over R_\mathrm{total}} = {1 \over R_{1}} + {1 \over R_{2}} + \cdots + {1 \over R_{n}}

    for components in parallel, having resistances R1, R2, etc.

The above rule can be calculated by using Ohm's law for the whole circuit

    Rtotal = V / Itotal

and substituting for Itotal

To find the current in any particular component with resistance Ri, use Ohm's law again.

    Ii = V / Ri

Note, that the components divide the current according to their reciprocal resistances, so, in the case of two resistors:

    I1 / I2 = R2 / R1

[edit]

Inductors

Inductors follow the same law, in that the total inductance of non-coupled inductors in parallel is equal to the reciprocal of the sum of the reciprocals of their individual inductances:

    A diagram of several inductors, side by side, both leads of each connected to the same wires

    {1\over{L_\mathrm{total}}} = {1\over{L_1}} + {1\over{L_2}} + \cdots + {1\over{L_n}}

Once again, if the inductors are situated in each others' magnetic fields, one has to take into account mutual inductance. If the mutual inductance between two coils in parallel is M then the equivalent inductor is:

    {1 \over L_\mathrm{total}} = {1 \over (L_1 + M)} + {1 \over (L_2 + M)} or
    {1 \over L_\mathrm{total}} = {1 \over (L_1 - M)} + {1 \over (L_2 - M)}

And once again, which formula is the correct one, depends how the magnetic fields of both inductors influence each other.

The principle is the same for more than two inductors, but you now have to take into account the mutual inductance of each inductor on each other inductor and how they influence each other. So for three coils, there are three mutual inductances (M12,M13 and M23) and eight possible equations.
[edit]

Capacitors

Capacitors follow a different law. The total capacitance of capacitors in parallel is equal to the sum of their individual capacitances:

    A diagram of several capacitors, side by side, both leads of each connected to the same wires

    C_\mathrm{total} = C_1 + C_2 + \cdots + C_n

The working voltage of a parallel combination of capacitors is always limited by the smallest working voltage of an individual capacitor.

Last edited by ']['error (2006-09-12 10:15:13)

elmo1337
Banned
+186|6862|The real world
Ye I saw it as him home, it ran so good on bf2! He had 12x aa while I only get 6x with my x1900xt!
max
Vela Incident
+1,652|6870|NYC / Hamburg

i also saw that once in a closed briefing from ati. its the future of the gfx card industry
once upon a midnight dreary, while i pron surfed, weak and weary, over many a strange and spurious site of ' hot  xxx galore'. While i clicked my fav'rite bookmark, suddenly there came a warning, and my heart was filled with mourning, mourning for my dear amour, " 'Tis not possible!", i muttered, " give me back my free hardcore!"..... quoth the server, 404.
(T)eflon(S)hadow
R.I.P. Neda
+456|7132|Grapevine, TX
You've got to be kidding
jamie_v1.1
Member
+35|6805|Leeds, UK
Eh???, my head hurts.....


Laymans terms anyone???
m3thod
All kiiiiiiiiinds of gainz
+2,197|6974|UK

']['error wrote:

Way too long quote!

m3thod: i just got my botty spanked for quoting too much
Translated by m3thod:

Buy a PCI-e motherboard.

Last edited by m3thod (2006-09-14 12:08:49)

Blackbelts are just whitebelts who have never quit.
m3thod
All kiiiiiiiiinds of gainz
+2,197|6974|UK
I have analsyed this further as i was bored and

Wrong Wrong Wrong

The multiplex conguatulator needs to amplify the resonance to at least 5X that of the gigaplex flux capacitor.

Thus the dimension is truly a paradigm and can be defined in a total state of equilibrium.
Blackbelts are just whitebelts who have never quit.
max
Vela Incident
+1,652|6870|NYC / Hamburg

m3thod wrote:

I have analsyed this further as i was bored and

Wrong Wrong Wrong

The multiplex conguatulator needs to amplify the resonance to at least 5X that of the gigaplex flux capacitor.

Thus the dimension is truly a paradigm and can be defined in a total state of equilibrium.
but you have forgotten that a normal flux capacitor is used to control the radiation of the slot, therefore no amplification of the mulitiplex is needed

Last edited by max (2006-09-12 10:23:45)

once upon a midnight dreary, while i pron surfed, weak and weary, over many a strange and spurious site of ' hot  xxx galore'. While i clicked my fav'rite bookmark, suddenly there came a warning, and my heart was filled with mourning, mourning for my dear amour, " 'Tis not possible!", i muttered, " give me back my free hardcore!"..... quoth the server, 404.
ShellShock.PwN
Member
+31|7090|Barrie Ontario

m3thod wrote:

... equilibrium.
Good  movie

Last edited by ShellShock.PwN (2006-09-12 10:24:16)

jamie_v1.1
Member
+35|6805|Leeds, UK
lol
siciliano732
Member
+202|6952|New York
wtf did you just say? lol...i have pci-e but...i still would like to know what your talking about
-=AudigyXi=-
Member
+18|6959
Just hook a mr fusion on top of ur tower.
']['error
Banned
+630|6947|The Netherlands
its true, my agp mobo now runs an 7900GTX
siciliano732
Member
+202|6952|New York
how do you do it though....cuz the way it was explained above is beyond me
Widjerd
I like sausage
+18|6844|Bristol UK
probably an adapter
OrangeHound
Busy doing highfalutin adminy stuff ...
+1,335|6952|Washington DC

Wrong section ... jokes belong in Not BF2S - Not BF2. 
Harmonics
A stabbing fatality
+22|6806
I think I'd rather buy an PCIe board.
scottomus0
Teh forum ghey!
+172|6940|Wigan. Manchester. England.
What The Fuck
Penetrator
Certified Twat
+296|6811|Bournemouth, South England
Good shpeil on the geek-speak, can I get an adaptor, or should I buy a PCI-e Mobo and shove the old one up your ass?
BigmacK
Back from the Dead.
+628|7054|Chicago.
Impressive work.
-=raska=-
Canada's French Frog
+123|6929|Quebec city, Canada
ok now tell me how to make a pci-e card fit into an agp slot ?
Penetrator
Certified Twat
+296|6811|Bournemouth, South England

-=raska=- wrote:

ok now tell me how to make a pci-e card fit into an agp slot ?
Push. Very hard... He never said that PCI-e WORKED in the AGP slot, just said it was there...
Superslim
BF2s Frat Brother
+211|6995|Calgary

pt_22 wrote:

for those that have been had by this... go play bf2 and press alt+f4 and you will get a special prize in game...
both old and weak
stryyker
bad touch
+1,682|7023|California

So.... When does the Delorean go back into time?
jimmanycricket
EBC Member
+56|6958|Cambridge, England
a pci express MB costs like £50 working at five pounds an hour you could afford one in ten hours that has a warrenty and didnt run the risk of failier, plus i bet it took more then ten hours to amke, so i think this is worthy of  a "you just wasted Many useful hours of your life for a compleatly irrelevent and useless task." congratulations.

have a cookie...

Board footer

Privacy Policy - © 2025 Jeff Minard