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SINUSOIDAL STEADY STATE ANALYSISAnalyzing ac circuits usually requires three steps.Steps to analyze AC Circuits:

1. Transform the circuit to the phasor or frequency domain.2. Solve the problem using circuit techniques (nodal analysis, mesh analysis, superposition, etc.).3. Transform the resulting phasor to the time domain.REVIEW: Given the following circuit with values in frequency space, write the actual element values in the time domain:

NODAL ANALYSIS

The basis of nodal analysis is Kirchhoff’s current law. Since KCL is valid for phasors, as demonstrated in previously, we can analyzeac circuits by nodal analysis. The following examples illustrate this.Example:Find i x in the circuit shown using nodal analysis.

Convert the circuit to the frequency domain and :

Thus, the frequency-domain equivalent circuit is as shown

Now apply KCL at node 1

At node 2

But I x = V 1 / − j2.5. Substituting this gives

Simplifying we get:11V 1 + 15V 2 = 0We now have two equations in V1 and V2. We can solve this system of equations be substitution or using a matrix.

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The current I x is given by

Transforming this to the time domain,i x = 7.59 cos(4t + 108.4◦) A

MESH ANALYSIS

Kirchhoff’s voltage law (KVL) forms the basis of mesh analysis. The validity of KVL for ac circuits was shown previously and isillustrated in the following examples.Example:Determine current I o in the circuit below using mesh analysis.

Apply KVL to mesh 1

Applying KVL to mesh 1, we obtain(8 + j10 − j2)I1 − ( − j2)I2 − j10I3 = 0

Apply KVL to mesh 2

For mesh 2,(4 − j2 − j2)I2 − ( − j2)I1 − ( − j2)I3 + 20 , 90◦ = 0

Given that for mesh 3, I3 = 5, use this system of equations to solve for I1 and I2.

(8 + j8)I1 + j2I2 = j50 j2I1 + (4 − j4)I2 = − j20 − j10In matrix form as

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Since I0 = - I2 then we know I0.

SUPERPOSITION THEOREMSince ac circuits are linear, the superposition theorem applies to ac circuits the same way it applies to dc circuits. The theorembecomes important if the circuit has sources operating at different frequencies. In this case, since the impedances depend onfrequency, we must have a different frequency-domain circuit for each frequency. The total response must be obtained by addingthe individual responses in the time domain. It is incorrect to try to add the responses in the phasor or frequency domain.Why? Because the exponential factor e jωt is implicit in sinusoidal analysis, and that factor would change for every angular frequencyω. It would therefore not make sense to add responses at different frequencies in the phasor domain. Thus, when a circuit hassources operating at different frequencies, one must add the responses due to the individual frequencies in the time domain.Example:Use superposition to find I0 in the circuit below:

Let Io = Io’ + Io’’ where Io’ and Io’’ are due to the voltage and current sources, respectively.To find Io’, consider the circuit at right. If we let Z be the parallel combination of − j2and 8 + j10, then

And the current is

To find I’’, use circuit at right. For mesh 1:(8 + j8)I 1 − j10I 3 + j2I 2 = 0For mesh 2,(4 − j4)I 2 + j2I 1 + j2I 3 = 0For mesh 3,I 3 = 5Substituting(4 − j4)I 2 + j2I 1 + j10 = 0Expressing I 1 in terms of I 2 givesI 1 = (2 + j2)I 2 − 5Substituting, we get

(8 + j8)[(2 + j2)I 2 − 5] − j50 + j2I 2 = 0Solving for I2:

The total current is then the sum of these two currents:

SOURCE TRANSFORMATIONSource transformation in the frequency domain involves transforming a voltage source in series with an impedance to a currentsource in parallel with an impedance, or vice versa. As we go from one source type to another, we must keep the followingrelationship in mind:

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THEVENIN AND NORTON EQUIVALENT CIRCUITSThevenin’s and Norton’s theorems are applied to ac circuits in the same way as theyare to dc circuits. The only additional effort arises from the need to manipulatecomplex numbers. The frequency-domain version of a Thevenin equivalent circuit isdepicted in (a), where a linear circuit is replaced by a voltage source in series withan impedance. The Norton equivalent circuit is illustrated in (b), where a linearcircuit is replaced by a current source in parallel with an impedance. Keep in mindthat the two equivalent circuits are related asV Th = Z N I N , Z Th = Z N

 just as in source transformation. V Th is the open-circuit voltage while I N is the

short-circuit current.If the circuit has sources operating at different frequencies, the Thevenin or Nortonequivalent circuit must be determined at each frequency. This leads to entirelydifferent equivalent circuits, one for each frequency, not one equivalent circuit withequivalent sources and equivalent impedances.Example:Find the Thevenin equivalent of the circuit as seen from terminals a-b.

Redraw the circuit by combining series impedences:

 To find V Th , we apply KCL at node 1 to find I0. Then apply KVL to the right hand loop.

15 = I o + 0.5I o ⇒ I o = 10 A

Applying KVL to the loop, we obtain− I o (2 − j4) + 0.5I o (4 + j3) + V Th = 0orV Th = 10(2 − j4) − 5(4 + j3) = − j55

Thus, the Thevenin voltage isV Th = 55, − 90◦VTo find Zth, remove the independent course and connect a arbitrary fixed current source (In this case 3A since it makes the matheasy) to terminals a and b and redraw the circuit:

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Now apply KCL at the node and KVL to the outer loop. Find Zth as the ratio of the Voltage to the Current.

At the node, KCL gives3 = I o + 0.5I o ⇒ I o = 2 A

Applying KVL to the outer loop givesV s = I o (4 + j3 + 2 − j4) = 2(6 − j)The Thevenin impedance is

--------------------------Alternative current, the current intensity is fluctating, it wont remains constant, where as, in Direct Current, current remainsconstant, we know tat heat generated during one transmission of power is directly proportionally to current pass thro' it, here in

AC, heat generated in transmission=(current)square, so, current s not constant so hear produced by AC transmission is low, whereas in DC, heat generated in transmission =(current)square, which s constant and has high value so, passing DC for long distance swaste of power-------------------------------

A wave pulse is a disturbance that moves through a medium.A periodic wave is a periodic disturbance that moves through a medium. The medium itself goes nowhere. The individual atomsand molecules in the medium oscillate about their equilibrium position, but their average position does not change. As theyinteract with their neighbors, they transfer some of their energy to them. The neighboring atoms in turn transfer this energy totheir neighbors down the line. In this way the energy is transported throughout the medium, without the transport of any matter. 

The animation on the right portrays a medium as a series of particles connected by springs. As one individual particle is disturbed,and then returns to its initial position, it transmits the disturbance to the next interconnected particle. This disturbance continuesto be passed on to the next particle. The result is that energy is transported from one end of the medium to the other end of themedium without the actual transport of matter. Each particle returns to its original position.

Periodic waves are characterized by a frequency, a wavelength, and by their speed. The wave frequency f is the oscillationfrequency of the individual atoms or molecules. The period T = 1/f is the time it takes any particular atom or molecule to gothrough one complete cycle of its motion. The wavelength λ is the distance along the direction of propagation between two atoms

which oscillate in phase.In a periodic wave, a pulse travels a distance of one wavelength λ in a time equal to one period T. The speed v of the wave can beexpressed in terms of these quantities.v = λ/T = λf 

Amplitude modulation (AM) is a technique used in electronic communication, most commonly for transmittinginformation via a radio carrier wave. AM works by varying the strength of the transmitted signal in relation to theinformation being sent. For example, changes in signal strength may be used to specify the sounds to bereproduced by a loudspeaker, or the light intensity of television pixels. Contrast this with frequency modulation, inwhich the frequency is varied, and phase modulation, in which the phase is varied.In the mid-1870s, a form of amplitude modulation—initially called "undulatory currents"—was the first method to

successfully produce quality audio over telephone lines. Beginning with Reginald Fessenden's audio demonstrationsin 1906, it was also the original method used for audio radio transmissions, and remains in use today by manyforms of communication—"AM" is often used to refer to the mediumwave broadcast band (see AM radio).

Waves and Wavelike Motion

Waves are everywhere. Whether we recognize it or not, we encounter waves on a daily basis. Sound waves, visible light waves,radio waves, microwaves, water waves, sine waves, cosine waves, stadium waves, earthquake waves, waves on a string, and slinkywaves and are just a few of the examples of our daily encounters with waves. In addition to waves, there are a variety of phenomena in our physical world that resemble waves so closely that we can describe such phenomenon as being wavelike. Themotion of a pendulum, the motion of a mass suspended by a spring, the motion of a child on a swing, and the "Hello, GoodMorning!" wave of the hand can be thought of as wavelike phenomena. Waves (and wavelike phenomena) are everywhere!We study the physics of waves because it provides a rich glimpse into the physical world that we seek to understand and describeas students of physics. Before beginning a formal discussion of the nature of waves, it is often useful to ponder the variousencounters and exposures that we have of waves. Where do we see waves or examples of wavelike motion? What experiences dowe already have that will help us in understanding the physics of waves?

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For many people, the first thought concerning waves conjures up a picture of a wave moving across the surface of an ocean, lake,pond or other body of water. The waves are created by some form of a disturbance, such as a rock thrown into the water, a duckshaking its tail in the water or a boat moving through the water. The water wave has a crest and a trough and travels from onelocation to another. One crest is often followed by a second crest that is often followed by a third crest. Every crest is separated bya trough to create an alternating pattern of crests and troughs. A duck or gull at rest on the surface of the water is observed to bobup-and-down at rather regular time intervals as the wave passes by. The waves may appear to be plane waves that travel togetheras a front  in a straight-line direction, perhaps towards a sandy shore. Or the waves may be circular waves that originate from thepoint where the disturbances occur; such circular waves travel across the surface of the water in all directions. These mentalpictures of water waves are useful for understanding the nature of a wave and will be revisited later when we begin our formal

discussion of the topic.

 Analogue modulation: time domain (waveforms), frequency domain (spectra), amplitude modulation (am), frequencymodulation (fm), phase modulation (pm)

2 Digital modulation: waveforms and spectra, Frequency Shift Keying (FSK), Binary Phase Shift Keying (BPSK)[including Gaussian Minimum Shift Keying (GMSK)], Quadrature

Phase Shift Keying (QPSK) [including π/4QPSK]3 Error coding: General principles of block, convolutional, parity, interleaving4 Compression: Regular Pulse Excitation – Linear Predictive Coding – Long Term Prediction (RPE-LPC-LTP)

BASIC AC GENERATIONFrom the previous discussion you learned that a current-carrying conductor produces a magnetic field around itself. In module 1,under producing a voltage (emf) using magnetism, you learned how a changing magnetic field produces an emf in a conductor.That is, if a conductor is placed in a magnetic field, and either the field or the conductor moves, an emf is induced in the conductor.

This effect is called electromagnetic induction.CYCLEFigures 1-7 and 1-8show a suspended loop of wire (conductor) being rotated (moved) in a clockwise direction through the magnetic field between thepoles of a permanent magnet. For ease of explanation, the loop has been divided into a dark half and light half. Notice in (A) of thefigure that the dark half is moving along (parallel to) the lines of force. Consequently, it is cutting NO lines of force. The same istrue of the light half, which is moving in the opposite direction. Since the conductors are cutting no lines of force, no emf isinduced. As the loop rotates toward the position shown in (B), it cuts more and more lines of force per second (inducing an ever-increasing voltage) because it is cutting more directly across the field (lines of force). At (B), the conductor is shown completingone-quarter of a complete revolution, or 90°, of a complete circle. Because the conductor is now cutting directly across the field,the voltage induced in the conductor is maximum. When the value of induced voltage at various points during the rotation from (A)to (B) is plotted on a graph (and the points connected), a curve appears as shown below.Figure 1-7. - Simple alternating-current generator.Figure 1-8. - Basic alternating-current generator.

As the loop continues to be rotated toward the position shown below in (C), it cuts fewer and fewer lines of force. The inducedvoltage decreases from its peak value. Eventually, the loop is once again moving in a plane parallel to the magnetic field, and noemf is induced in the conductor.

The loop has now been rotated through half a circle (one alternation or 180°). If the preceding quarter-cycle is plotted, it appearsas shown below.When the same procedure is applied to the second half of rotation (180° through 360°), the curve appears as shown below. Noticethe only difference is in the polarity of the induced voltage. Where previously the polarity was positive, it is now negative.The sine curve shows the value of induced voltage at each instant of time during rotation of the loop. Notice that this curve contains360°, or two alternations. TWO ALTERNATIONS represent ONE complete CYCLE of rotation.Assuming a closed path is provided across the ends of the conductor loop, you can determine the direction of current in the loop byusing the LEFT-HAND RULE FOR GENERATORS. Refer to figure 1-9. The left-hand rule is applied as follows: First, place your lefthand on the illustration with the fingers as shown. Your THUMB will now point in the direction of rotation (relative movement of thewire to the magnetic field); your FOREFINGER will point in the direction of magnetic flux (north to south); and your MIDDLE FINGER(pointing out of the paper) will point in the direction of electron current flow.Figure 1-9. - Left-hand rule for generators.

By applying the left-hand rule to the dark half of the loop in (B) in figure 1-8, you will find that the current flows in the directionindicated by the heavy arrow. Similarly, by using the left-hand rule on the light half of the loop, you will find that current thereinflows in the opposite direction. The two induced voltages in the loop add together to form one total emf. It is this emf which causesthe current in the loop.When the loop rotates to the position shown in (D) of figure 1-8, the action reverses. The dark half is moving up instead of down,

and the light half is moving down instead of up. By applying the left-hand rule once again, you will see that the total induced emf and its resulting current have reversed direction. The voltage builds up to maximum in this new direction, as shown by the sinecurve in figure 1-8. The loop finally returns to its original position (E), at which point voltage is again zero. The sine curverepresents one complete cycle of voltage generated by the rotating loop. All the illustrations used in this chapter show the wire loopmoving in a clockwise direction. In actual practice, the loop can be moved clockwise or counterclockwise. Regardless of thedirection of movement, the left-hand rule applies.If the loop is rotated through 360° at a steady rate, and if the strength of the magnetic field is uniform, the voltage produced is asine wave of voltage, as indicated in figure 1-9. Continuous rotation of the loop will produce a series of sine-wave voltage cycles or,in other words, an ac voltage.As mentioned previously, the cycle consists of two complete alternations in a period of time. Recently the HERTZ (Hz) has beendesignated to indicate one cycle per second. If ONE CYCLE PER SECOND is ONE HERTZ, then 100 cycles per second are equal to100 hertz, and so on. Throughout the NEETS, the term cycle is used when no specific time element is involved, and the term hertz(Hz) is used when the time element is measured in seconds.

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Q.16 When a conductor is rotated in a magnetic field, at what points in the cycle is emf (a) at maximum amplitude and (b) atminimum amplitude?Q.17 One cycle is equal to how many degrees of rotation of a conductor in a magnetic field?Q.18 State the left-hand rule used to determine the direction of current in a generator.Q.19 How is an ac voltage produced by an ac generator?