Saturday, March 11, 2017

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Friday, March 10, 2017

Adobe after effect cs6..tutorial..how to make frezze..video fream..

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So you want to create a simple but dynamic logo animation in after effects without using the third party plugin, then you came to the right place. In this tutorial, we are going to create a very simple and nice looking Logo Animation for your YouTube channel or video. Watch this tutorial and make some really cool animation.

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Parallel Resonance Circuit

In many ways a parallel resonance circuit is exactly the same as the series resonance circuit we looked at in the previous tutorial.

Both are 3-element networks that contain two reactive components making them a second-order circuit, both are influenced by variations in the supply frequency and both have a frequency point where their two reactive components cancel each other out influencing the characteristics of the circuit. Both circuits have a resonant frequency point.

The difference this time however, is that a parallel resonance circuit is influenced by the currents flowing through each parallel branch within the parallel LC tank circuit. A tank circuit is a parallel combination of L and C that is used in filter networks to either select or reject AC frequencies. Consider the parallel RLC circuit below.

Parallel RLC Circuit

Let us define what we already know about parallel RLC circuits.

A parallel circuit containing a resistance, R, an inductance, L and a capacitance, C will produce a parallel resonance (also called anti-resonance) circuit when the resultant current through the parallel combination is in phase with the supply voltage. At resonance there will be a large circulating current between the inductor and the capacitor due to the energy of the oscillations, then parallel circuits produce current resonance.

A parallel resonant circuit stores the circuit energy in the magnetic field of the inductor and the electric field of the capacitor. This energy is constantly being transferred back and forth between the inductor and the capacitor which results in zero current and energy being drawn from the supply. This is because the corresponding instantaneous values of IL and IC will always be equal and opposite and therefore the current drawn from the supply is the vector addition of these two currents and the current flowing in IR.

In the solution of AC parallel resonance circuits we know that the supply voltage is common for all branches, so this can be taken as our reference vector. Each parallel branch must be treated separately as with series circuits so that the total supply current taken by the parallel circuit is the vector addition of the individual branch currents.

Then there are two methods available to us in the analysis of parallel resonance circuits. We can calculate the current in each branch and then add together or calculate the admittance of each branch to find the total current.

We know from the previous series resonance tutorial that resonance takes place when VL = -VC and this situation occurs when the two reactances are equal, XL = XC. The admittance of a parallel circuit is given as:

Resonance occurs when XL = XC and the imaginary parts of Y become zero. Then:

Notice that at resonance the parallel circuit produces the same equation as for the series resonance circuit. Therefore, it makes no difference if the inductor or capacitor are connected in parallel or series. Also at resonance the parallel LC tank circuit acts like an open circuit with the circuit current being determined by the resistor, R only. So the total impedance of a parallel resonance circuit at resonance becomes just the value of the resistance in the circuit and Z = R as shown.

At resonance, the impedance of the parallel circuit is at its maximum value and equal to the resistance of the circuit. Also at resonance, as the impedance of the circuit is now that of resistance only, the total circuit current, I will be “in-phase” with the supply voltage, VS.

We can change the circuit’s frequency response by changing the value of this resistance. Changing the value of R affects the amount of current that flows through the circuit at resonance, if both L and C remain constant. Then the impedance of the circuit at resonance Z = RMAX is called the “dynamic impedance” of the circuit.

Impedance in a Parallel Resonance Circuit

Note that if the parallel circuits impedance is at its maximum at resonance then consequently, the circuits admittance must be at its minimum and one of the characteristics of a parallel resonance circuit is that admittance is very low limiting the circuits current. Unlike the series resonance circuit, the resistor in a parallel resonance circuit has a damping effect on the circuits bandwidth making the circuit less selective.

Also, since the circuit current is constant for any value of impedance, Z, the voltage across a parallel resonance circuit will have the same shape as the total impedance and for a parallel circuit the voltage waveform is generally taken from across the capacitor.

We now know that at the resonant frequency, ƒr the admittance of the circuit is at its minimum and is equal to the conductance, G given by 1/R because in a parallel resonance circuit the imaginary part of admittance, i.e. the susceptance, B is zero because BL = BC as shown.

Susceptance at Resonance

From above, the inductive susceptance, BL is inversely proportional to the frequency as represented by the hyperbolic curve. The capacitive susceptance, BC is directly proportional to the frequency and is therefore represented by a straight line. The final curve shows the plot of total susceptance of the parallel resonance circuit versus the frequency and is the difference between the two susceptance’s.

Then we can see that at the resonant frequency point were it crosses the horizontal axis the total circuit susceptance is zero. Below the resonant frequency point, the inductive susceptance dominates the circuit producing a “lagging” power factor, whereas above the resonant frequency point the capacitive susceptance dominates producing a “leading” power factor.

So at the resonant frequency, ƒr the current drawn from the supply must be “in-phase” with the applied voltage as effectively there is only the resistance present in the parallel circuit, so the power factor becomes one or unity, ( θ = 0o ).

Also as the impedance of a parallel circuit changes with frequency, this makes the circuit impedance “dynamic” with the current at resonance being in-phase with the voltage since the impedance of the circuit acts as a resistance. Then we have seen that the impedance of a parallel circuit at resonance is equivalent to the value of the resistance and this value must, therefore represent the maximum dynamic impedance (Zd) of the circuit as shown.

Current in a Parallel Resonance Circuit

As the total susceptance is zero at the resonant frequency, the admittance is at its minimum and is equal to the conductance, G. Therefore at resonance the current flowing through the circuit must also be at its minimum as the inductive and capacitive branch currents are equal ( IL = IC ) and are 180o out of phase.

We remember that the total current flowing in a parallel RLC circuit is equal to the vector sum of the individual branch currents and for a given frequency is calculated as:

At resonance, currents IL and IL are equal and cancelling giving a net reactive current equal to zero. Then at resonance the above equation becomes.

Since the current flowing through a parallel resonance circuit is the product of voltage divided by impedance, at resonance the impedance, Z is at its maximum value, ( =R ). Therefore, the circuit current at this frequency will be at its minimum value of V/R and the graph of current against frequency for a parallel resonance circuit is given as.

Parallel Circuit Current at Resonance

The frequency response curve of a parallel resonance circuit shows that the magnitude of the current is a function of frequency and plotting this onto a graph shows us that the response starts at its maximum value, reaches its minimum value at the resonance frequency when IMIN = IR and then increases again to maximum as ƒ becomes infinite.

The result of this is that the magnitude of the current flowing through the inductor, L and the capacitor, C tank circuit can become many times larger than the supply current, even at resonance but as they are equal and at opposition ( 180o out-of-phase ) they effectively cancel each other out.

As a parallel resonance circuit only functions on resonant frequency, this type of circuit is also known as an Rejecter Circuit because at resonance, the impedance of the circuit is at its maximum thereby suppressing or rejecting the current whose frequency is equal to its resonant frequency. The effect of resonance in a parallel circuit is also called “current resonance”.

The calculations and graphs used above for defining a parallel resonance circuit are similar to those we used for a series circuit. However, the characteristics and graphs drawn for a parallel circuit are exactly opposite to that of series circuits with the parallel circuits maximum and minimum impedance, current and magnification being reversed. Which is why a parallel resonance circuit is also called an Anti-resonance circuit.

Bandwidth & Selectivity of a Parallel Resonance Circuit

The bandwidth of a parallel resonance circuit is defined in exactly the same way as for the series resonance circuit. The upper and lower cut-off frequencies given as: ƒupperand ƒlower respectively denote the half-power frequencies where the power dissipated in the circuit is half of the full power dissipated at the resonant frequency 0.5( I2 R ) which gives us the same -3dB points at a current value that is equal to 70.7% of its maximum resonant value, ( 0.707 x I )2 R

As with the series circuit, if the resonant frequency remains constant, an increase in the quality factor, Q will cause a decrease in the bandwidth and likewise, a decrease in the quality factor will cause an increase in the bandwidth as defined by: BW = ƒr /Q or BW = ƒupper - ƒlower

Also changing the ratio between the inductor, L and the capacitor, C, or the value of the resistance, R the bandwidth and therefore the frequency response of the circuit will be changed for a fixed resonant frequency. This technique is used extensively in tuning circuits for radio and television transmitters and receivers.

The selectivity or Q-factor for a parallel resonance circuit is generally defined as the ratio of the circulating branch currents to the supply current and is given as:

Note that the Q-factor of a parallel resonance circuit is the inverse of the expression for the Q-factor of the series circuit. Also in series resonance circuits the Q-factor gives the voltage magnification of the circuit, whereas in a parallel circuit it gives the current magnification.

Bandwidth of a Parallel Resonance Circuit

Parallel Resonance Example No1

A parallel resonance network consisting of a resistor of 60Ω, a capacitor of 120uF and an inductor of 200mH is connected across a sinusoidal supply voltage which has a constant output of 100 volts at all frequencies. Calculate, the resonant frequency, the quality factor and the bandwidth of the circuit, the circuit current at resonance and current magnification.

Resonant Frequency, ƒr

Inductive Reactance at Resonance, XL

Quality factor, Q

Bandwidth, BW

The upper and lower -3dB frequency points, ƒH and ƒL

Circuit Current at Resonance, IT

At resonance the dynamic impedance of the circuit is equal to R

Current Magnification, Imag

Note that the current drawn from the supply at resonance (the resistive current) is only 1.67 amps, while the current flowing around the LC tank circuit is larger at 2.45 amps. We can check this value by calculating the current flowing through the inductor (or capacitor) at resonance.

Parallel Resonance Tutorial Summary

We have seen that Parallel Resonance circuits are similar to series resonance circuits. Resonance occurs in a parallel RLC circuit when the total circuit current is “in-phase” with the supply voltage as the two reactive components cancel each other out. At resonance the admittance of the circuit is at its minimum and is equal to the conductance of the circuit. Also at resonance the current drawn from the supply is also at its minimum and is determined by the value of the parallel resistance.

The equation used to calculate the resonant frequency point is the same for the previous series circuit. However, while the use of either pure or impure components in the series RLC circuit does not affect the calculation of the resonance frequency, but in a parallel RLC circuit it does.

In this tutorial about parallel resonance, we have assumed that the components are purely inductive and purely capacitive with negligible resistance. However in reality the coil will contain some resistance. Then the equation for calculating the parallel resonant frequency of a circuit is therefore modified to account for the additional resistance.

Resonant Frequency using Impure Components

RMS Voltage Tutorial

In our tutorial about the AC Waveform we looked briefly at the RMS Voltage value of a sinusoidal waveform and said that this RMS value gives the same heating effect as an equivalent DC power and in this tutorial we will expand on this theory a little more by looking at RMS voltages and currents in more detail.

The term “RMS” stands for “Root-Mean-Squared”. Most books define this as the “amount of AC power that produces the same heating effect as an equivalent DC power”, or something similar along these lines, but an RMS value is more than just that. The RMS value is the square root of the mean (average) value of the squared function of the instantaneous values. The symbols used for defining an RMS value are VRMS or IRMS.

The term RMS, ONLY refers to time-varying sinusoidal voltages, currents or complex waveforms were the magnitude of the waveform changes over time and is not used in DC circuit analysis or calculations were the magnitude is always constant. When used to compare the equivalent RMS voltage value of an alternating sinusoidal waveform that supplies the same electrical power to a given load as an equivalent DC circuit, the RMS value is called the “effective value” and is generally presented as: Veff or Ieff.

In other words, the effective value is an equivalent DC value which tells you how many volts or amps of DC that a time-varying sinusoidal waveform is equal to in terms of its ability to produce the same power. For example, the domestic mains supply in the United Kingdom is 240Vac. This value is assumed to indicate an effective value of “240 Volts rms”. This means then that the sinusoidal rms voltage from the wall sockets of a UK home is capable of producing the same average positive power as 240 volts of steady DC voltage as shown below.

RMS Voltage Equivalent

So how do we calculated the RMS Voltage of a sinusoidal waveform. The RMS voltage of a sinusoid or complex waveform can be determined by two basic methods.

Graphical Method – which can be used to find the RMS value of any non-sinusoidal time-varying waveform by drawing a number of mid-ordinates onto the waveform.
Analytical Method – is a mathematical procedure for finding the effective or RMS value of any periodic voltage or current using calculus.

RMS Voltage Graphical Method

Whilst the method of calculation is the same for both halves of an AC waveform, for this example we will consider only the positive half cycle. The effective or rms value of a waveform can be found with a reasonable amount of accuracy by taking equally spaced instantaneous values along the waveform.

The positive half of the waveform is divided up into any number of “n” equal portions or mid-ordinates and the more mid-ordinates that are drawn along the waveform, the more accurate will be the final result. The width of each mid-ordinate will therefore be no degrees and the height of each mid-ordinate will be equal to the instantaneous value of the waveform at that time along the x-axis of the waveform.

Graphical Method

Each mid-ordinate value of a waveform (the voltage waveform in this case) is multiplied by itself (squared) and added to the next. This method gives us the “square” or Squared part of the RMS voltage expression. Next this squared value is divided by the number of mid-ordinates used to give us the Mean part of the RMS voltage expression, and in our simple example above the number of mid-ordinates used was twelve (12). Finally, the square root of the previous result is found to give us the Root part of the RMS voltage.

Then we can define the term used to describe an rms voltage (VRMS) as being “the square root of the mean of the square of the mid-ordinates of the voltage waveform” and this is given as:

and for our simple example above, the RMS voltage will be calculated as:

So lets assume that an alternating voltage has a peak voltage (Vpk) of 20 volts and by taking 10 mid-ordinate values is found to vary over one half cycle as follows:

Voltage	6.2V	11.8V	16.2V	19.0V	20.0V	19.0V	16.2V	11.8V	6.2V	0V
Angle	18o	36o	54o	72o	90o	108o	126o	144o	162o	180o

The RMS voltage is therefore calculated as:

Then the RMS Voltage value using the graphical method is given as: 14.14 Volts.

RMS Voltage Analytical Method

The graphical method above is a very good way of finding the effective or RMS voltage, (or current) of an alternating waveform that is not symmetrical or sinusoidal in nature. In other words the waveform shape resembles that of a complex waveform. However, when dealing with pure sinusoidal waveforms we can make life a little bit easier for ourselves by using an analytical or mathematical way of finding the RMS value.

A periodic sinusoidal voltage is constant and can be defined as V(t) = Vm.cos(ωt) with a period of T. Then we can calculate the root-mean-square (rms) value of a sinusoidal voltage (V(t)) as:

Integrating through with limits taken from 0 to 360o or “T”, the period gives:

Dividing through further as ω = 2π/T, the complex equation above eventually reduces down too:

RMS Voltage Equation

Then the RMS voltage (VRMS) of a sinusoidal waveform is determined by multiplying the peak voltage value by 0.7071, which is the same as one divided by the square root of two ( 1/√2 ). The RMS voltage, which can also be referred to as the effective value, depends on the magnitude of the waveform and is not a function of either the waveforms frequency nor its phase angle.

From the graphical example above, the peak voltage (Vpk) of the waveform was given as 20 Volts. By using the analytical method just defined we can calculate the RMS voltage as being:

VRMS = Vpk x 0.7071 = 20 x 0.7071 = 14.14V

Note that this value of 14.14 volts is the same value as for the previous graphical method. Then we can use either the graphical method of mid-ordinates, or the analytical method of calculation to find the RMS voltage or current values of a sinusoidal waveform. Note that multiplying the peak or maximum value by the constant 0.7071, ONLY applies to sinusoidal waveforms. For non-sinusoidal waveforms the graphical method must be used.

RMS Voltage Summary

Then to summarise. When dealing with alternating voltages (or currents) we are faced with the problem of how do we represent a voltage or signal magnitude. One easy way is to use the peak values for the waveform. Another common method is to use the effective value which is also known by its more common expression of Root Mean Square or simply the RMS value.

The root mean squared, RMS value of a sinusoid is not the same as the average of all the instantaneous values. The ratio of the RMS value of voltage to the maximum value of voltage is the same as the ratio of the RMS value of current to the maximum value of current. Most multi-meters, either voltmeters or ammeters, measure RMS value assuming a pure sinusoidal waveform. For finding the RMS value of non-sinusoidal waveform a “True RMS Multimeter” is required.

The RMS value of a sinusoidal waveform gives the same heating effect as a DC current of the same value. That is if a direct current, I passes through a resistance of R ohms, the DC power consumed by the resistor as heat will therefore be I2R watts. Then if an alternating current, i = Im.sinθ flows through the same resistance, the AC power converted into heat will be: I2rms.R watts.

Then when dealing with alternating voltages and currents, they should be treated as RMS values unless otherwise stated. Therefore an alternating current of 10 amperes will have the same heating effect as a direct current of 10 amperes and a maximum value of 14.14 amperes.

Having now determined the RMS value of an alternating voltage (or current) waveform, in the next tutorial we will look at calculating the Average value, VAV of an alternating voltage and finally compare the two.

Average Voltage Tutorial

Having looked at the RMS Voltage value of an alternating waveform in a previous tutorial, we can now look at calculating another value using either the mid-ordinate rule or analytical rule to find a waveforms “average” or mean voltage.

The process used to find the Average Voltage of an alternating waveform is very similar to that for finding its RMS value, the difference this time is that the instantaneous values are not squared and we do not find the square root of the summed mean.

The average voltage (or current) of a periodic waveform whether it is a sine wave, square wave or triangular waveform is defined as: “the quotient of the area under the waveform with respect to time”. In other words, the averaging of all the instantaneous values along time axis with time being one full period, (T).

For a periodic waveform, the area above the horizontal axis is positive while the area below the horizontal axis is negative. The result is that the average or mean value of a symmetrical alternating quantity is zero because the area above the horizontal axis (the positive half cycle) is the same as the area below the axis (the negative half cycle) and cancel each other out in the sum of the two areas as a negative cancels a positive producing zero average voltage.

Then the average or mean value of a symmetrical alternating quantity, such as a sine wave, is the average value measured over only half a cycle since over a complete cycle the average value is zero regardless of the peak amplitude.

The electrical terms Average Voltage and Mean Voltage or or even average current, can be used in both an AC and DC circuit analysis or calculations. The symbols used for representing an average value are defined as: VAV or IAV.

Average Voltage Graphical Method

Again consider only the positive half cycle from the previous RMS voltage tutorial. The mean or average voltage of a waveform can be found again with a reasonable amount of accuracy by taking equally spaced instantaneous values.

The positive half of the waveform is divided up into any number of “n” equal portions or mid-ordinates. The width of each mid-ordinate will therefore be no degrees (or t seconds) and the height of each mid-ordinate will be equal to the instantaneous value of the waveform at that point along the x-axis of the waveform.

The Graphical Method

Each mid-ordinate value of the voltage waveform is added to the next and the summed total, V1 to V12 is divided by the number of mid-ordinates used to give us the “Average Voltage”. Then the average voltage (VAV) is the mean sum of mid-ordinates of the voltage waveform and is given as:

and for our simple example above, the average voltage is therefore calculated as:

So as before lets assume again that an alternating voltage of 20 volts peak varies over one half cycle as follows:

Voltage	6.2V	11.8V	16.2V	19.0V	20.0V	19.0V	16.2V	11.8V	6.2V	0V
Angle	18o	36o	54o	72o	90o	108o	126o	144o	162o	180o

The Average voltage value is therefore calculated as:

Then the Average Voltage value using the graphical method is given as: 12.64 Volts.

Average Voltage Analytical Method

As said previously, the average voltage of a periodic waveform whose two halves are exactly similar, either sinusoidal or non-sinusoidal, will be zero over one complete cycle. Then the average value is obtained by adding the instantaneous values of voltage over one half cycle only. But in the case of an non-symmetrical or complex wave, the average voltage (or current) must be taken over the whole periodic cycle mathematically.

The average value can be taken mathematically by taking the approximation of the area under the curve at various intervals to the distance or length of the base and this can be done using triangles or rectangles as shown.

Approximation of the Area

By approximating the areas of the rectangles under the curve, we can obtain a rough idea of the actual area of each one. By adding together all these areas the average value can be found. If an infinite number of smaller thinner rectangles were used, the more accurate would be the final result as it approaches 2/π.

The area under the curve can be found by various approximation methods such as the trapezoidal rule, the mid-ordinate rule or Simpson’s rule. Then the mathematical area under the positive half cycle of the periodic wave which is defined as V(t) = Vp.cos(ωt) with a period of T using integration is given as:

Where: 0 and π are the limits of integration since we are determining the average value of voltage over one half a cycle. Then the area below the curve is finally given as Area = 2VP. Since we now know the area under the positive (or negative) half cycle, we can easily determine the average value of the positive (or negative) region of a sinusoidal waveform by integrating the sinusoidal quantity over half a cycle and dividing by half the period.

For example, if the instantaneous voltage of a sinusoid is given as: v = Vp.sinθ and the period of a sinusoid is given as: 2π, then:

Which is therefore given as the standard equation for the Average Voltage of a sine wave as:

Average Voltage Equation

Then the average voltage (VAV) of a sinusoidal waveform is determined by multiplying the peak voltage value by the constant 0.637, which is two divided by pi (π). The average voltage, which can also be referred to as the mean value, depends on the magnitude of the waveform and is not a function of either the frequency or the phase angle.

Referring to our graphical example above, the peak voltage, (Vpk) was given as 20 Volts. Using the analytical method the average voltage is therefore calculated as:

VAV = Vpk x 0.637 = 20 x 0.637 = 12.74 volts

Which is the same value as for the graphical method.

To find the peak value from a given average voltage value, just rearrange the formula and divide by the constant. For example, what is the sinusoidal peak value, Vpk if the average value is 65 volts.

Vpk = VAV ÷ 0.637 = 65 ÷ 0.637 = 102 volts

Note that multiplying the peak or maximum value by the constant 0.637 ONLY applies to sinusoidal waveforms.

Average Voltage Summary

Then to summarise. When dealing with alternating voltages (or currents), the term Average value is generally taken over one complete cycle, whereas the term Mean value is used for one half of the periodic cycle.

The average value of a whole sinusoidal waveform over one complete cycle is zero as the two halves cancel each other out, so the average value is taken over half a cycle. The average value of a sine wave of voltage or current is 0.637 times the peak value, (Vp or Ip. This mathematical relationship between the average values applies to both AC current and AC voltage.

Sometimes it is required to be able to calculate the value of the direct voltage or current output from a rectifier or pulse type circuit such as a PWM motor circuit because the voltage or current, although not reversing, is changing continuously. Since there are no phase reversals the average value is used and the RMS (root-mean-square) value is unimportant for this type of application.

The main differences between an RMS Voltage and an Average Voltage, is that the mean value of a periodic wave is the average of all the instantaneous areas taken under the curve over a given period of the waveform, and in the case of a sinusoidal quantity, this period is taken as one-half of the cycle of the wave. For convenience the positive half cycle is generally used.

The effective value or root-mean-square (RMS) value of the waveform is the effective heating value of the wave compared to a steady DC value and is the square root of the mean of the squares of the instantaneous values taken over one complete cycle.

For a pure sinusoidal waveform ONLY, both the average voltage and the RMS voltage (or currents) can be easily calculated as:

Average value = 0.637 × maximum or peak value, Vpk

RMS value = 0.707 × maximum or peak value, Vpk

One final comment about using Average Voltage and RMS Voltage. Both values can be used to represent the “Form Factor” of a sinusoidal alternating waveform. Form factor is defined as being the shape of an AC waveform and is the RMS voltage divided by the average voltage (form factor = rms value/average value).

So for a sinusoidal or complex waveform the form factor is given as: ( π/(2√2) ) which is approximately equal to the constant, 1.11. Form factor is a ratio and therefore has no electrical units. If the form factor of a sinusoidal waveform is known, then the average voltage can be found using the RMS voltage value and vice-versa.

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Saturday, March 11, 2017

Biye Pagol | Bangla Comedy Natok | Arfan Nisho | Anika Kabir Shokh | Dr....

Friday, March 10, 2017

Adobe after effect cs6..tutorial..how to make frezze..video fream..

Clean Logo Animation in After Effects - After Effects Tutorial - No Thir...