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From Wikipedia

Amplitude

Amplitude is the magnitude of change in the oscillating variable with each oscillation within an oscillating system. For example, sound waves in air are oscillations in atmospheric pressure and their amplitudes are proportional to the change in pressure during one oscillation. If a variable undergoes regular oscillations, and a graph of the system is drawn with the oscillating variable as the vertical axis and time as the horizontal axis, the amplitude is visually represented by the vertical distance between the extrema of the curve.

In older texts the phase is sometimes very confusingly called the amplitude.

Concepts

Peak-to-peak amplitude

Peak-to-peak amplitude is the change between peak (highest amplitude value) and trough (lowest amplitude value, which can be negative). With appropriate circuitry, peak-to-peak amplitudes can be measured by meters or by viewing the waveform on an oscilloscope. Peak-to-peak is a straightforward measurement on an oscilloscope, the peaks of the waveform being easily identified and measured against the graticule. This remains a common way of specifying amplitude, but sometimes other measures of amplitude are more appropriate.

Peak amplitude

In audio system measurements, telecommunications and other areas where the measurand is a signal that swings above and below a zero value but is not sinusoidal, peak amplitude is often used. This is the absolute value of the signal.

Semi-amplitude

Semi-amplitude means half the peak-to-peak amplitude. It is the most widely used measure of orbital amplitude in astronomy and the measurement of small semi-amplitudes of nearby stars is important in the search for exoplanets. For a sine wave, peak amplitude and semi-amplitude are the same.

Some scientists use "amplitude" or "peak amplitude" to mean semi-amplitude, that is, half the peak-to-peak amplitude.

Root mean square amplitude

Root mean square(RMS) amplitude is used especially inelectrical engineering: the RMS is defined as the square root of the mean over time of the square of the vertical distance of the graph from the rest state.

For complex waveforms, especially non-repeating signals like noise, the RMS amplitude is usually used because it is both unambiguous and has physical significance. For example, the average power transmitted by an acoustic or electromagnetic wave or by an electrical signal is proportional to the square of the RMS amplitude (and not, in general, to the square of the peak amplitude).

For alternating currentelectrical power, the universal practice is to specify RMS values of a sinusoidal waveform. The peak-to-peak voltage of a sine wave is nearly 3 times the RMS value, but is rarely used. Some common meter types used in electrical engineering are calibrated for RMS amplitude, but actually operate on a DC input. Both digital voltmeters and moving coil meters are in this category. Such meters require the AC input to be first rectified. They are not true RMS meters, but rather, are reading proportional to either rectified average or peak amplitude. The RMS calibration is only correct for a sine wave input since the ratio between peak, average and RMS values is dependent on waveform. Until recently, true RMS meters were mostly used only in radio frequency measurements. These instruments based their measurement on detecting the heating effect in a load resistor with a thermistor. The advent of microprocessor controlled meters capable of calculating RMS by sampling the waveform has made true RMS measurement commonplace.

Ambiguity

In general, the use of peak amplitude is simple and unambiguous only for symmetric periodic waves, like a sine wave, a square wave, or a triangular wave. For an asymmetric wave (periodic pulses in one direction, for example), the peak amplitude becomes ambiguous. This is because the value is different depending on whether the maximum positive signal is measured relative to the mean, the maximum negative signal is measured relative to the mean, or the maximum positive signal is measured relative to the maximum negative signal (the peak-to-peak amplitude) and then divided by two. In electrical engineering, the usual solution to this ambiguity is to measure the amplitude from a defined reference potential (such as ground or 0 V). Strictly speaking, this is no longer amplitude since there is the possibility that a constant (DC component) is included in the measurement.

Pulse amplitude

In telecommunication, pulse amplitude is the magnitude of a pulse parameter, such as the voltage level, current level, field intensity, or power level.

Pulse amplitude is measured with respect to a specified reference and therefore should be modified by qualifiers, such as "average", "instantaneous", "peak", or "AC input and a DC output of roughly twice the peak input voltage. They are a variety of voltage multiplier circuit and a voltage doubler is often, but not always, a single stage of a general form of such a circuit. The term is usually applied to circuits consisting of rectifying diodes and capacitors only; other means of doubling voltages are not included.

Villard circuit

The Villard circuit consists simply of a capacitor and a diode. While it has the great benefit of simplicity, its output has very poor ripple characteristics. Essentially, the circuit is a diode clamp circuit. The capacitor is charged on the negative half cycles to the peak AC voltage (Vpk). The output is the superposition of the input AC waveform and the steady DC of the capacitor. The effect of the circuit is to shift the DC value of the waveform. The negative peaks of the AC waveform are "clamped" to 0 V (actually −VF, the small forward bias voltage of the diode), so the positive peaks of the output waveform are 2Vpk. The peak-to-peak ripple is an enormous 2Vpk and cannot be smoothed unless the circuit is effectively turned into one of the more sophisticated forms.

Greinacher circuit

The Greinacher voltage doubler is a significant improvement over the Villard circuit for a small cost in increased components. The ripple is much reduced, being nominally zero under open-circuit load conditions, but, when current is being drawn, depends on the resistance of the load and the value of the capacitors used. The circuit works by following a Villard cell stage with what is in essence a peak detector or envelope detector stage. The peak detector cell has the effect of removing most of the ripple while preserving the peak voltage in the output.

This circuit was first invented by Heinrich Greinacher in 1913 (published 1914) in order to provide the 200–300 V he needed for his newly invented ionometer, the 110 V AC supplied by the Zurich power stations of the time being insufficient. He later (1920) extended this idea into a cascade of multipliers. This cascade of Greinacher cells is often inaccurately referred to as a Villard cascade. It is also called a Cockcroft–Walton multiplier after the particle accelerator machine built by John Cockcroft and Ernest Walton, who independently rediscovered the circuit in 1932.

The concept in this topology can be extended to a voltage quadrupler circuit by using two Greinacher cells of opposite polarities driven from the same AC source. The output is taken across the two individual outputs. Note that, like a bridge circuit, it is not possible to simultaneously ground both the input and output of this circuit

Bridge circuit

The Delon circuit uses a bridge topology for voltage doubling. This form of circuit was, at one time, commonly found in cathode ray tube television sets where it was used to provide an e.h.t. voltage supply. Generating voltages in excess of 5 kV with a transformer has safety issues in terms of domestic equipment and in any case is not economic. However, black and white television sets required an e.h.t. of 10 kV and colour sets even more. Voltage doublers were used to either double the voltage on an e.h.t winding on the mains transformer or were applied to the waveform on the line flyback coils.

The circuit consists of two half-wave peak detectors, functioning in exactly the same way as the peak detector cell in the Greinacher circuit above. Each of the two peak detector cells operates on opposite half-cycles of the incoming waveform. Since their outputs are in series, the output is twice the peak input voltage.

A full-wave version of this circuit has the advantage of lower peak diode currents, improved ripple and better load regulation but requires a centre-tap to the transformer as well as more components.


Power supply

A power supply is a device that supplies electricalenergy to one or more electric loads. The term is most commonly applied to devices that convert one form of electrical energy to another, though it may also refer to devices that convert another form of energy (e.g., mechanical, chemical, solar) to electrical energy. A regulated power supply is one that controls the output voltage or current to a specific value; the controlled value is held nearly constant despite variations in either load current or the voltage supplied by the power supply's energy source.

Every power supply must obtain the energy it supplies to its load, as well as any energy it consumes while performing that task, from an energy source. Depending on its design, a power supply may obtain energy from:

A power supply may be implemented as a discrete, stand-alone device or as an integral device that is hardwired to its load. In the latter case, for example, low voltage DC power supplies are commonly integrated with their loads in devices such as computers and household electronics.

Constraints that commonly affect power supplies include:

  • The amount of voltage and current they can supply.
  • How long they can supply energy without needing some kind of refueling or recharging (applies to power supplies that employ portable energy sources).
  • How stable their output voltage or current is under varying load conditions.
  • Whether they provide continuous or pulsed energy.

Power supply types

Power supplies for electronic devices can be broadly divided into linear and switching power supplies. The linear supply is usually a relatively simple design, but it becomes increasingly bulky and heavy for high-current equipment due to the need for large mains-frequency transformers and heat-sinked electronic regulation circuitry. Linear voltage regulators produce regulated output voltage by means of an active voltage divider that consumes energy, thus making efficiency low. A switched-mode supply of the same rating as a linear supply will be smaller, is usually more efficient, but will be more complex.

Battery

A battery is an alternative to a line-operated power supply; it is independent of the availability of mains electricity, suitable for portable equipment and use in locations without mains power. A battery consists of several electrochemical cells connected in series to provide the voltage desired. Batteries may be primary (able to supply current when constructed, discarded when drained) or secondary (rechargeable; can be charged, used, and recharged many times)

The primary cell first used was the carbon-zinc dry cell. It had a voltage of 1.5 volts; later battery types have been manufactured, when possible, to give the same voltage per cell. Carbon-zinc and related cells are still used, but the alkaline battery delivers more energy per unit weight and is widely used. The most commonly used battery voltages are 1.5 (1 cell) and 9V (6 cells).

Various technologies of rechargeable battery are used. Types most commonly used are NiMH, and lithium ion and variants.

DC power supply

An AC powered unregulated power supply usually uses a transformer to convert the voltage from the wall outlet (mains) to a different, nowadays usually lower, voltage. If it is used to produce DC, a rectifier is used to convert alternating voltage to a pulsating direct voltage, followed by a filter, comprising one or more capacitors, resistors, and sometimes inductors, to filter out (smooth) most of the pulsation. A small remaining unwanted alternating voltage component at mains or twice mains power frequency (depending upon whether half- or full-wave rectification is used)—ripple—is unavoidably superimposed on the direct output voltage.

For purposes such as charging batteries the ripple is not a problem, and the simplest unregulated mains-powered DC power supply circuit consists of a transformer driving a single diode in series with a resistor.

Before the introduction of solid-state electronics equipment used valves (vacuum tubes) which required high voltages; power supplies used step-up transformers, rectifiers, and filters to generate one or more direct voltages of some hundreds of volts, and a low alternating voltage for filaments. Only the most advanced equipment used expensive and bulky regulated power supplies.

AC power supply

An AC power supply typically takes the voltage from a wall outlet (mains supply, often 230v in Europe) and lowers it to the desired voltage (eg 9vac). As well as lowering the voltage some filtering may take place. An example use for an AC power supply is powering certain guitar effects pedals (e.g. the Digitech Whammy pedal) although usually these pedals require DC.

Linear regulated power supply

The voltage produced by an unregulated power supply will vary depending on the load and on variations in the AC supply voltage. For critical electronics applications a

Vibrating string

A vibration in a string is a wave. Usually a vibrating string produces a sound whose frequency in most cases is constant. Therefore, since frequency characterizes the pitch, the sound produced is a constant note. Vibrating strings are the basis of any string instrument like guitar, cello, or piano.

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Wave

The speed of propagation of a wave in a string (v) is proportional to the square root of the tension of the string (T) and inversely proportional to the square root of the linear density (\mu) of the string:

v = \sqrt{T \over \mu}.

Derivation

Let \Delta x be the length of a piece of string, m its mass, and \mu its linear mass. If the horizontal component of tension in the string is a constant, T, then the tension acting on each side of the string segment is given by

T_{1x}=T_1 \cos(\alpha) \approx T.
T_{2x}=T_2 \cos(\beta)\approx T.

If both angles are small, then the tensions on either side are equal and the net horizontal force is zero. From Newton's second law for the vertical component, the mass of this piece times its acceleration, a, will be equal to the net force on the piece:

\Sigma F_y=T_{2y}-T_{1y}=T_2 \sin(\beta)-T_1 \sin(\alpha)=\Delta m a\approx\mu\Delta x \frac{\partial^2 y}{\partial t^2}.

Dividing this expression by T and substituting the first and second equations obtains

\frac{\mu\Delta x}{T}\frac{\partial^2 y}{\partial t^2}=\frac{T_2 \sin(\beta)}{T_2 \cos(\beta)}-\frac{T_1 \sin(\alpha)}{T_1 \cos(\alpha)}=\tan(\beta)-\tan(\alpha)

The tangents of the angles at the ends of the string piece are equal to the slopes at the ends. Using this fact and rearranging provides

\frac{1}{\Delta x}\left(\left.\frac{\partial y}{\partial x}\right|^{x+\Delta x}-\left.\frac{\partial y}{\partial x}\right|^x\right)=\frac{\mu}{T}\frac{\partial^2 y}{\partial t^2}

In the limit that \Delta x approaches zero, the left hand side is the definition of the second derivative of y:

\frac{\partial^2 y}{\partial x^2}=\frac{\mu}{T}\frac{\partial^2 y}{\partial t^2}.

This is the wave equation for y(x,t), and the coefficient of the second time derivative term is equal to v^{-2}; thus

v=\sqrt{T\over\mu},

where v is the speed of propagation of the wave in the string. (See the article on the wave equation for more about this). However, this derivation is only valid for vibrations of small amplitude; for those of large amplitude, \Delta x is not a good approximation for the length of the string piece, the horizontal component of tension is not necessarily constant, and the horizontal tensions are not well approximated by T.

Frequency of the wave

Once the speed of propagation is known, the frequency of the sound produced by the string can be calculated. The speed of propagation of a wave is equal to the wavelength \lambda divided by the period \tau, or multiplied by the frequency f :

v = \frac{\lambda}{\tau} = \lambda f.

If the length of the string is L, the fundamental harmonic is the one produced by the vibration whose nodes are the two ends of the string, so L is half of the wavelength of the fundamental harmonic. Hence:

f = \frac{v}{2L} = { 1 \over 2L } \sqrt{T \over \mu}

where T is the tension, \mu is the linear density, and L is the length of the vibrating part of the string. Therefore:

  • the shorter the string, the higher the frequency of the fundamental
  • the higher the tension, the higher the frequency of the fundamental
  • the lighter the string, the higher the frequency of the fundamental

Moreover, if we take the nth harmonic as having a wavelength given by \lambda_n = 2L/n, then we easily get an expression for the frequency of the nth harmonic:

f_n = \frac{nv}{2L}

And for a string under a tension T with density \mu, then

f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}}

Observing string vibrations

One can see the waveforms on a vibrating string if the frequency is low enough and the vibrating string is held in front of a CRT screen such as one of a television or a computer (not of an oscilloscope). This effect is called the stroboscopic effect, and the rate at which the string seems to vibrate is the difference between the frequency of the string and the refresh rate of the screen. The same can happen with a fluorescent lamp, at a rate which is the difference between the frequency of the string and the frequency of the alternating current. (If the refresh rate of the screen equals the frequency of the string or an integer multiple thereof, the string will appear still but deformed.) In daylight, this effect does not occur and the string will appear to be still, but thicker and lighter, due to persistence of vision.

A similar but more controllable effect can be obtained using a stroboscope. This device allows the frequency of the xenon flash lamp to be exactly matched to the frequency of vibration of the string; in a darkened room, this clearly shows the waveform. Otherwise, one can use bending or, perhaps more easily, by adjusting the machine heads, to obtain the same frequency, or a multiple of, the AC frequency to achieve the same effect. For example, in the case of a guitar, the bass string pressed to the third fret gives a G at 97.999 Hz; with a slight adjustment, a frequency of 100 Hz can be obtained, exactly one octave above the alternating current


From Yahoo Answers

Question:For example, equation for a standing wave: y(x,t) = A[sin(kx)]sin(wt) I've also seen it written as A[sin(kx)]cos(wt) and I've seen y(x,t) = Acos(kx-wt) written as: y(x,t) = Asin(kx-wt) Can someone please explain. Thanks a lot. Also, how would I know when to use which one in a wave problem?

Answers:In both examples, substituting cos(wt) for sin(wt), phase-shifts the wave by 90 degrees. Either equation can be used, and in either case you'll need to add a phase-shift term to satisfy the "initial conditions" or "boundary conditions" in the specific problem being solved. Here's a generalized equation for a sinewave (applicable to both traveling waves and sinewaves): A sine wave (or harmonic wave or sinusoid), with an amplitude u, is described by the equation: u (x, t) = Acos (kx - t + ) , where A is the semi-amplitude of the wave, half the peak-to-peak amplitude, often called simply the amplitude the maximum distance from the highest point of the disturbance in the medium (the crest) to the equilibrium point during one wave cycle. The equation variables are defined as follows: x is the space coordinate, t is the time coordinate, k is the wavenumber (spatial frequency), is the temporal frequency, and is a phase offset.

Question:I need to hit (6,7) and (10, -3) with a sine wave that must complete at least two periods where 0 < x < 6 before it hits the first point Can someone please explain how I can do this? I also need to include amplitude, period, a phase shift, and vertical displacement

Answers:Let's try to hit (9, 3) and (13, -1), four or more periods where 0 < x < 9 before reaching (9, 3). Follow this and you can do the one you presented. 1) Notice the difference in the y values is equal to 4. If we make (9, 3) be at the top of a wave and (13, -1) be at the bottom of a wave, then 4 is equal to twice the amplitude. Amplitude = 2. 2) Halfway between 3 and -1 (y values) is 1. Therefore, the vertical displacement is one unit up. 3) If the sine wave was to simply drop from (9, 3) to (13, -1), half of one period would be 13 - 9 = 4 and one period would be 8 (with four periods being 32). But, we need four or more periods from x = 0 to 9. If the sine wave was to drop form (9, 3), bounce back up, bounce back down and hit (13, -1) it was move three times as fast and one period would be 8/3. Four periods would then be 32/3. But, 32/3 > 9. Got to move even faster. (9, 3), bounce back up, bounce back down, up again, down again, (13, -1) has a period of 8/5. Four periods would be 32/5 and 32/5 < 9. Good enough. So, the period will be 8/5. Let's now develop a sine wave with three of the four factors you requested and then modify it with a phase shift. The general equation is y = A sin [(x - s)p] + V .... A = amplitude ... p = period divided by 2pi ... s = phase shift .... V = vertical displacement In this example, A = 2, V = 1 and p = (8/5) / 2pi = 4 / 5pi . As one fourth of one period is 2/5, then the sine wave begins going up off of y = 1 where x = 9 - 2/5 = (45 - 2) / 5 = 43 / 5. Get closer to x = 0 with a sine wave beginning by stepping back by periods (length 8/5). Step back five periods to reach x = (43 - 5(8) ) / 5 = (43 - 40) / 5 = 3 / 5. So, the phase shift s = 3 / 5 or - 3 / 5. Test 3 / 5. If it checks, you're done. Otherwise, use - 3 / 5.

Question:Please if anyne know how to solve this help me!!!, If the third harmonic of standing wave in a vibrating string is 600Hz, what is the fundamental frequency?

Answers:think its 200mhz Think of the third harmonic as a standing wave with three humps. First one going up then one going down then one going back up. In other words a sine wave plus a half of sine wave. Notice that the first sine wave is a full wavelentght, this is added to a half of sine wave (half wave lenght). So if we are unsing a string and say the full length of the string is L then: L = wavelenght + wavelenght/2 Now for the second part The fundamental frequency occurs when there is only one hump for the standing wave. In other words only a half of sine wave. ie a half of waveleght. So if we are using the same string as above and say the full length is L then: L = (fundamental) / 2 Notice now that you have the same legnth in both equation...since they are the same length..you may set these two things equal to eachothere (fundamental) = + / 2 And rember that you are dealing with frequencies so just change the wavelenght over to frequency with the forumula V = * f Do alittle algebra and your final equaltion will then be f(fundamental) = f/ 3 600/3 Mhz ...there ya go

Question:Is the circuit used for rectification the RC-filter rectifier?

Answers:AC =alternating current, like sin wave . DC , is direct current, has frequency =0 but in AC frequency isnt zero. Rectification is the process of using diodes (it alllows current only one way) to modifiy a sign wave. Say if you want only positive half of a sign wave, then using diodes u can accomplish this.

From Youtube

GENERATOR - YOU DO THE ACTUAL IN/OUT POWER MATH :You do all the math. I first measured the AC output current with a true RMS multimeter through one wire of the 120 volt lit neon lamp.. That lamp was my load across my magnet flux pickup coil Secondly, I measured the AC voltage across the lit 120 volt neon load. Both of those values times each other should indicate power out of the magnet / coil itself. That AC power figure in itself has no relation to the Bedini/Davro coil or circuit The third reading was the DC voltage actually feeding the Bedini/Davro circuit/coil that spins the magnet. 6.1 DC volts I did not video the DC current measurement because I had to place the meter in series to measure the current. It would have make the video to long. That DC current was .81 milliamps. All of this was done with the magnet flux pickup coil laying horizontally.If supported vertically I ran the risk I would accidentilly roll it off the Bedini coil.while making this video. I have higher readings with the flux coil in the vertical position The magnet used is still a 1x1 inch diameter Diametric ring magnet. Next I am going to feed all AC power into a minature full wave bridge rectifier and then compare DC in power to DC out power. That might make readings easier. I was told the multimeter itself might be to close to the spinning magnet to make an accurate AC current reading? A half hour ago I lit up 40 LED lamps strung in series connected to the AC coil output. Each lamp takes .20 milliamps to power up. 3 volts each. The AC voltage ...