Till now we focused on the propagation of a uniform plane wave in an unbounded medium either free space (or) dielectric.

we now consider a monochromatic uniform plane wave that travels through one medium and then enters another medium of infinite extent.

at this stage we assume that the interface between the two media is normal to the direction of propagation of the incoming wave.

The wave that is propagating in the first medium is called incident wave. Assume the direction of propagation of the incoming wave along positive z-direction.

The interface (or) the boundary is a plane z=0 in this case.

if direction of propagation was along +ve x-axis plane would be x=0 plane.
if direction of propagation was along +ve y-axis plane would be y=0 plane.

The wave reflected back into the same medium is called reflected wave and the wave that is propagating into the second medium is the transmitted wave.

incident and reflected waves are in opposite directions to each other.

Incident waves:-

.

.

. the d.o.p of H is .

Reflected waves:-

.

.

. the d.o.p of H is .

Transmitted waves:-

.

.

. the d.o.p of H is .

Now the Transmission and reflection coefficients are defined as follows

.

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similarly the transmission coefficient is

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Derivation of coefficients:-

By using the boundary conditions,

the tangential components of E are continuous

i.e, .

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the tangential components of H are discontinuous

i.e, . let us assume at the Boundary.

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by solving the above two equations the transmission and reflection coefficients using electric field strength are

and .

similarly the transmission and reflection coefficients using magnetic field strength are

whenever an EM Wave travelling in one medium impinges second medium the wave gets partially transmitted and partially reflected depending up on the type of the second medium.

Assume the first case in Normal incidence that is Normal incidence on a Perfect conductor.

i.e an EM wave propagating in free space strikes suddenly a conducting Boundary which means the other medium is a conductor.

The figure shows a plane Wave which is incident normally upon a boundary between free space and a perfect conductor.

assume the wave is propagating in positive z-axis and the boundary is z=0 plane.

The transmitted wave since the electric field intensity inside a perfect conductor is zero.

The incident and reflected waves are in the medium 1 that is free space.

The energy transmitted is zero so the energy absorbed by the conductor is zero and entire wave is reflected to the same medium

now incident wave is

in free space for medium 1

( a wave propagating in positive z-direction) and the reflected wave is ( a wave propagating in positive z-direction).

.

by using tangential components .

The resultant wave is .

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the above equation is in phasor notation , converting the above equation into time-harmonic (or) sinusoidal variations

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This is the wave equation which represents standing wave , which is the contribution of incident and reflected waves. as this wave is stationary it does not progress.

it has maximum amplitude at odd multiples of and minimum amplitude at multiples of .

Similarly The resultant Magnetic field is

The resultant wave is .

.

the above equation is in phasor notation , converting the above equation into time-harmonic (or) sinusoidal variations

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this wave is a stationary wave it has minimum amplitude at odd multiples of and maximum amplitude at multiples of .

In order to find out the various types of materials in magnetic fields and their behaviour we use the knowledge of the action of magnetic field on a current loop with a simple model of an atom

Magnetic materials are classified on the basis of presence of magnetic dipole moments in the materials.

a charged particle with angular momentum always contributes to the permanent contributions to the angular moment of an atom
1. orbital magnetic dipole moment.
2. electron spin moment.
3. Nuclear spin magnetic moment.

Orbital Magnetic dipole Moment:-

The simple atomic model is one which assumes that there is a central positive nucleus surrounded by electrons in various circular orbits.

an electron in an orbit is analogous to a small current loop and as such experiences a torque in an external magnetic field, the torque tending to align the magnetic field produced by the orbiting electron with the external magnetic field.

Thus the resulting magnetic field at any point in the material would be greater than it would be at that point when the other moments were not considered.

so there are Quantum numbers which describes the orbital state of notion of electron in an atom there are n,l and ml
n-principal Quantum number, which determines the energy of an electron.
l-Orbital Quantum number which determines the angular momentum of orbit.
ml-magnetic Quantum number which determines the component of magnetic moment along the direction of an electric field.

electron spin Magnetic Moment:-

The angular momentum of an electron is called spin of the electron. as electron is a charged particle the spin of the electron produces magnetic dipole moment because electron is spinning about it’s own axis and thus generates a magnetic dipole moment.

is the value of electron spin when we consider an atom those electrons which are in shells which are not completely filled with contribute to a magnetic moment for the atom.

Nuclear spin Magnetic Moment:-

a third contribution of the moment of an atom is caused by nuclear spin this provides a negligible effect on the overall magnetic properties of material

That is the mass of the nucleus is much larger than an electron thus the dipole moments due to nuclear spin are very small.

so the total magnetic dipole moment of an atom is nothing but the summation of all the above mentioned .

The solution to the above equation is of the form .

where is known as relaxation time and defined as the time it takes a charge placed in the interior of a material to drop to = 36.8 percent of it’s initial value.

is the initial charge density (i.e, at t=0) the equation shows that as a result of introducing charge at some interior point of the material there is a decay of volume charge density this decay is associated with the charge movement from the interior point at which it was introduced to the surface of the material.

– is the time constant known as the relaxation time (or) rearrangement time.

So far, we have considered the existence of the electric field in a Homogeneous medium. If the field exists in a region consisting of two different media, the conditions that the field must satisfy at the interface separating the media are called “Boundary conditions”.

These conditions are helpful in determining the field on one side of the boundary if the field on the other side is known.

The conditions will be dictated by the types of material the media are made of.

We shall consider the Boundary conditions at an interface separating

Di-electric and Di-electric .

Conductor and Di-electric.

Conductor and free space.

To determine the boundary conditions, we need to use two Maxwell’s equations.

– Gauss’s law in Electrostatics.

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The Electric field intensity could be considered as result of two components tangential and normal components.

.

Di-electric and Di-electric :-

In order to find out the , at the interface between two different magnetic materials boundary conditions are required.

Consider two Di-electric materials having permeabilities and as shown in the figure

To apply a path is required, which is a closed one in a plane normal to the boundary surface.

Here a closed path is abcda on the surface of the boundary

Then .

.

(since On the boundary)

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, So the tangential components of are discontinuous where as are continuous.

In order to apply , a surface is required which is nothing but Gaussian surface (or) a Pillbox enclosing some charge.

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( -Because on the boundary , so there exists no side of surface).

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since ,

From equations (1) and (2) the normal components of and are discontinuous at the boundary .

Consider a closed rectangular path in xy-plane which encloses a current element and the current flows in z-direction.

at center point P is .

By applying Gauss’s law to differential volume element leads to a concept of divergence as similar to that by applying ampere’s law to a current element in a closed path leads to curl.

It states that the magnetic field intensity dH produced, at the point P by the differential current element I dl

is proportional to the product I dl and the the angle between the element and the line joining the point P to the element.

and is inversely proportional to the square of the distance R between P and the current element.

then the direction of can be determined by right hand rule with the right hand thumb pointing in the direction of the current and the fingers encircling the wire in the direction of .

i.e, .

.

where k is the constant of proportionality , .

A/m.

A/m.

is perpendicular to the plane that contains and .

A/m.

then the total magnetic field strength measured at a point P is given by

A/m.

closed path is taken since the current can flow only in closed path and this is called as integral form of Biot-Savart’s law.

as similar to different charge distributions in electro-statics , there exists different current elements like line, surface and volume in the study of static magnetic fields.

A/m. —-for a line current element.

A/m. —-for a surface current element.

A/m. —-for a volume current element.

the dot and cross products between dl and I represents either H is out of (or) into the page(plane) .

consider an infinite sheet in the z=0 plane, which has uniform current density A/m .

Let us suppose the current is flowing in the positive y direction.

the sheet of current is assumed to be in rectangular co-ordinate system

Let us suppose the conductor is carrying a current I , by right hand thumb rule magnetic field is produced around the conductor is right angles to the direction of I.

In this case of infinite sheet , the current is in the y-direction there is no component of H along the direction of y and also the z components cancel each other because of opposite direction of the fields produced so only x components of H exists.

from Ampere’s Circuit law .

the component

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similarly .

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as

this will be changed to

In general for a finite sheet of current density A/m Magnetic field is generalised as .

Magnetic forces are required to study the force , a magnetic field exerts on charged particles, current elements and loops which is used in electrical devices in ammeters, volt meters, Galvano meters.

There are 3 ways in which force due to magnetic fields can be experienced.

The force can be due to a moving charged particle in a Magnetic field.

on a current element in an external B field.

between two current elements.

Force on a charged particle:-

we know that .

.

where is the electric force on a stationary (or) moving electric charge in an electric field and is related to . where and are in the same direction.

a magnetic field can exert force only on a moving charge , suppose a charge Q is moving with velocity u (or) v in a magnetic field (B) is

.

from the equations is independent of velocity of the charge and performs work on the charge which changes its kinetic energy but depends on the charge velocity and is normal to it so work done it does not cause increase in the kinetic energy of the charge.

is small compared to except at high velocities.

so a charge which is in movement has both electric and magnetic fields.

Then .

.

This is known as Lorentz’s force equation. It relates mechanical force to electrical force.

if the mass of the charges particle is m

Then .

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.

.

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The solution to this equation is important in determining the motion of charged particles in in such cases the energy transfer is only by means of electric field.

Force on a current element:-

Consider a current carrying conductor , in order to find out the force acting on the current carrying element by the magnetic field .

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is nothing but a elemental charge dQ moving with the velocity .

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The line integral is for the current is along the closed path.

i.e,

The magnetic field produced by the current element does not exert force on the element itself just as a point charge does not exert force on itself.

So the magnetic field that exerts force on must be from the another element in other words the magnetic field is external to the current element .

Similarly we have force equations for other current elements and as follows

and .

So the magnetic field is defined as the force per unit current element

i.e, .(or) similar to .

so the describes the force properties of a magnetic field.

Force between two current elements (Ampere’s force law):-

Consider two current loops and then by Biot- Savart’s law both current elements produces respective magnetic fields so we may find the force on element s due to the field produced by .

Field produced by current element is is .

So the force applied on the element is by the field .

..

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This is similar to coulomb’s law in electrostatics. Here it is law of force between two current elements and is analogous to coulomb’s law

.

Then the force acting on loop 2 due to the field produced by the current element t is nothing but

Note:- This is nothing but the Ampere’s force law that is the force between two current carrying conductors is given by it.

Consider a conductor of finite length placed along z-axis as shown in the figure

The conductor has a finite length AB , where A and B are located at distances Z_{1} and Z_{2} above the origin with it’s upper and lower ends respectively subtending angles and at P.

P is the point at which is to be determined.

Consider a differential element along the Z-axis at a distance Z from the origin.

where .

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as , and .

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Case 1 :-

when the conductor is semi-finite that is A is located at origin and B at .

i.e, and .

then .

Case 2:-

when conductor is infinite in length A is at and B at implies and .

Consider as infinitely long straight conductor placed along z-axis carrying a current I .

In order to determine at some point P. we allow a closed path which passes through the point P and encloses the current carrying conductor symmetrically such path is known as Amperian path.

To apply Ampere’s law the conditions t be satisfied are

The field is either tangential (or) Normal to the path at each point of the closed path.

The magnitude of must be same at all points of the path where is tangential.

Now, is given by .

The path we are assuming is in the direction of so .

.

Ampere’s law is used to find out at P

i.e, from Ampere’s circuit law .

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from Ampere’s law .

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Ampere’s law is applied to find the value of at any point P in it’s field.

Ampere’s Circuit law states that the line integral of the tangential component of around a closed path is the same as the net current (I_{enc}) enclosed by the path.

i.e, .

This is similar to Gauss’s law and can be applied to determine when the current distribution is symmetrical it’s a special case of Biot-savart’s law.

Proof:-

Consider a circular loop which encloses a current element . Let the current be in upward direction then the field is in anti- clock wise .

The current which is enclosed by the circular loop is of infinite length then at any point A is given by

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which is known as the integral form of Ampere’s circuit law.

The function of a receiver in a binary Communication system is to distinguish between two transmitted signals (or) () in the presence of noise.

The performance of Receiver is usually measured in terms of the probability of error P_{e} an the receiver is said to be optimum if it yields the minimum probability of error.

i.e, optimum receiver is the one with minimum probability of error P_{e} .

optimum receiver takes the form of Matched filter when the noise at the receiver input is white noise.

optimum receiver (or) optimum filter:-

The block diagram of optimum receiver is as shown in the figure below

the decision boundary is set to .

Probability of error of optimum filter:-

The probability of error can be obtained as similar to Integrate and dump receiver. Here we will consider noise as Gaussian Noise.

The output of optimum filter is .

The output of sampler is

suppose if Binary ‘1’ is transmitted then the input is , to find the probability of error this transmitted ‘1’ should be received as ‘0’.

this is possible when the condition is true.

1 will be received as 0 .

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similarly a Binary ‘0’ will be received as ‘1’ if and only if

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the conditions are summarized in the table

Noe the Probability Distribution Function of Gaussian noise with zero mean and standard deviation is given by

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Probability of error= probability ‘1’ will be received as ‘0’ =probability ‘0’ will be received as ‘1’.

When a CT band limited signal is sampled at , then the successive cycles of the spectrum of the sampled signal overlap with each other as shown below

Some aliasing is produced in the signal this is due to under sampling.

aliasing is the phenomenon in which a high frequency component in the frequency spectrum of the signal takes as a low frequency component in the spectrum of the sampled signal.

Because of aliasing it is not possible to reconstruct x(t) from g(t) by low pass filtering.

The spectral components in the overlapping regions and hence the signal is distorted.

Since any information signal contains a large no.of frequencies so the decision of sampling frequency is always become a problem.

A signal is first passed through LPF before sampling.

i.e, it is band limited by this LPF which is known as pre-alias filter.

To avoid aliasing

Pre-alias filter must be used to limit the band width of the signal to Hz.

Sampling frequency must be .

Pre-alias filter means before sampling is passed through a LPF to make a perfect band limited signal.

Reconstruction filter (Low Pass Filter) Procedure to reconstruct actual signal from sampled signal:-

Low Pass Filter is used to recover original signal from it’s samples. This is also known as interpolation filter.

An LPF is that type of filter which passes only low frequencies up to cut-off frequency and rejects all other frequencies above cut-off frequency.

For an ideal LPF, there is a sharp change in the response at cut-off frequency as shown in the figure.

i.e, Amplitude response becomes suddenly zero at cut-off frequency which is not possible practically that means an ideal LPF is not physically realizable.

i.e, in place of an ideal LPF a practical filter is used.

In case of a practical filter, the amplitude response decreases slowly to zero (this is one of the reason why we choose )

This means that there exists a transition band in case of practical Low Pass Filter in the reconstruction of original signal from its samples.

Signal Reconstruction (Interpolation function):-

The process of reconstructing a Continuous Time signal x(t) from it’s samples is known as interpolation.

Interpolation gives either approximate (or) exact reconstruction (or) recovery of CT signal.

One of the simplest interpolation procedures is known as zero-order hold.

Another procedure is linear interpolation. In linear interpolation the adjacent samples (or) sample points are connected by straight lines.

We may also use higher order interpolation formula for reconstructing the CT signal from its sample values.

If we use the above process (Higher order interpolation) the sample points are connected by higher order polynomials (or) other mathematical functions.

For a Band limited signal, if the sampling instants are sufficiently large then the signal may be reconstructed exactly by using a LPF.

In this case an exact interpolation can be carried out between sample points.

Mathematical analysis:-

A Band limited signal x(t) can be reconstructed completely from its samples, which has higher frequency component f_{m} Hz.

If we pass the sampled signal through a LPF having cut-off frequency of f_{m} Hz.

From sampling theorem

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g(t) has a multiplication factor . To reconstruct x(t) (or) X(f) , the sampled signal must be passed through an ideal LPF of Band Width of Hz and gain .

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If sampling is done at Nyquist rate , then Nyquist interval is .

therefore .

h(t) = 0. at all Nyquist instants , when g(t) is applied at the input to this filter the output will be x(t) .

Each sample in g(t) results a sinc pulse having amplitude equal to the strength of sample. If we add all these sinc pulses that gives the original signal x(t) .

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This is known as interpolation formula

It is assumed that the signal x(t) is strictly band limited but in general an information signal may contain a wide range of frequencies and can not be strictly band limited this means that the maximum frequency in the signal can not be predictable.

then it is not possible to select suitable sampling frequency f_{s} .

The range of values of the complex variable s for which Laplace Transform converges is called the Region of Convergence (ROC).

i.e, The region of Convergence (or) existence of signal’s Laplace transform X(S) is the set of values of s for which the integral defining the direct L T/F X(S) converges.

The ROC is required for evaluating the inverse L T/F of x(t) from X(S).

i.e, the operation of finding the inverse T/F requires an integration in the complex plane.

i.e, .

The path of integration is along S-plane that is along with varying from and moreover , the path of integration must lie in the ROC for X(S).

for example the signal , this is possible if so the path of integration is shown in the figure

Thus to obtain from , the integration is performed through this path for the function . such integration in the complex plane requires a back ground in the theory of functions of complex variables.

so we can avoid this integration by compiling a Table of L T/F’s . so for inverse L T/F’s we use this table instead of performing complex integration.

specific constraints on the ROC are closely associated with time-domain properties of x(t).

Properties of ROC/ constraints (or) Limitations:-

1.The ROC of X(S) consists of strips parallel to the axis in the S-plane.

i.e, The ROC of X(S) consists of the values of s for which Fourier T/F of converges this is possible if is fully integrable thus the condition depends only on . Hence ROC is the strips (bands) which is only in terms of values of .

2.

3. For Rational Laplace T/F’s , the ROC does not contain any poles. This is because X(S) is finite at poles and the integral can not be converge at this point.

4. If x(t) is of finite duration and absolutely integrable, then the ROC is the entire S-plane.

5. If x(t) is right-sided and if the line is in the ROC, then all values of s for which will also be in the ROC.

i.e, if the signal is right-sided then for ROC : .

6. If x(t) is left-sided and if the line is in the ROC, then all values of s for which will also be in the ROC.

7. If x(t) is two-sided and if the line is in the ROC, then the ROC consists of a strip in the s-plane that includes the line .

for the both sided signal , the ROC lies in the region . This ROC is the strip parallel to axis in the s-plane.

8. If the L T/F X(S) of x(t) is rational, then it’s ROC is bounded by poles (or) extends to infinity in addition no poles of X(s) are contained in the ROC.

If the function has two poles , then ROC will be area between these two poles for two sided signal, if for single sided signal the area extends from one pole to infinity.

But is does not include any pole.

9. If the L T/F X(S) of x(t) is rational, then if x(t) is right-sided. The ROC is the region in the s-plane to the right of the right most pole and if x(t) is left-sided, the ROC is the region in the s-plane to the left of the left most pole.

we have already defined the signal as any ordinary function of time. To understand more about signal we consider it as a problem. A problem is better understood (or) better remembered if it can be associated with some familiar phenomenon.

we always search for analogies while studying a new problem.

i.e, In the study of abstract problems analogies are very helpful. Particularly if the problem can be shown to be analogous to some concrete phenomenon.

It is then easier to gain some insight into the new problem from the knowledge of the analogous phenomenon.

There is a perfect analogy that exists between vectors and signals which leads to a better understanding of signal analysis. we shall now briefly review the properties of vectors.

Vectors:-

A vector is specified by magnitude and direction .

Let us consider two vectors and . It is possible to find out the component of one vector along the other vector.

In order to find out the component of vector along . Let us assume it as , which is only the magnitude.

how do we represent physically the component of one vector along ? This is possible by finding the projection of one vector on to the other.

i.e, by drawing a perpendicular from to

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There exists two other possibilities.

but these are not suitable. the error vectors are more in these cases.

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If is the angle between two vectors and , the component of along is

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The component of along is .

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If two vectors are orthogonal .

i.e, .

Signals:-

The concept of vector comparison & orthogonality can be extended to signals.

i.e, a signal is nothing but a single-valued function of independent variable. Assume two signals and , now to approximate in terms of over .

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Now, we choose in order to achieve the best approximation.

i.e, which keeps the error as minimum as possible.

One possible way for minimizing error is to choose minimize the average value of .

i.e, as .

But the process of averaging gives a false indication.

i.e, for example while approximating a function with a null function is

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indicates that during to without any error

i.e, .

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Average value of error is .

This seems to be error is zero but actually there exists some error.

To avoid this false indication, we choose to minimize the average of the square of the error

i.e, Mean Square Error .

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To find value which keeps error minimum .

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Which is similar to where denotes the inner product between two Real signals

A Co-axial cable is a Transmission line, in which two conductors are placed co-axially and are separated by some dielectric material with dielectric constant (or) permittivity ( ).

a conductor is in the form of a cylinder with some radius, let the radius of inner conductor is ‘a’ meters and that of outer conductor be ‘b’ meters.

Now connect this co-axial conductor to a supply of ‘V’ volts , after applying ‘V’ assume positive charges are distributed on and negative charges on .

Now, a field is induced between and because of flux lines, to find out at any point P between these two conductors

location of P is out of the conductor an inside the conductor .

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assume a cylindrical co-ordinate system and axis of cable coincides with z-axis this is similar to a line charge distribution placed along the z-axis.