February 2023 - ece4uplp

aliasing effect in Sampling

Effect of under sampling (aliasing effect):-

When a CT band limited signal is sampled at $f_{s} < 2f_{m}$ , 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 $f_{m}$ Hz.
Sampling frequency must be

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

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Capacitance of a Co-axial cable

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 ( $\epsilon$ ).

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 $M_{2}$ and negative charges on $M_{1}$ .

Now, a field is induced $\overrightarrow{E}$ between $M_{2}$ and $M_{1}$ because of flux lines, to find out $\overrightarrow{E}$ at any point P between these two conductors

location of P is out of the conductor $M_{2}$ an inside the conductor $M_{1}$ .

$\therefore \overrightarrow{E}_{at P} = \overrightarrow{E}_{\ due \ to \ inner \ conductor \ M_{1}}$ .

assume a cylindrical co-ordinate system $\rho ,\ \phi , \ z$ and axis of cable coincides with z-axis this is similar to a line charge distribution $\rho_{L}$ placed along the z-axis.

$\rho _{L}=\frac{Q}{L}$ .

$\therefore \overrightarrow{E}_{at P} = \frac{\rho_ {L}}{2\pi \epsilon _{o}\rho }\overrightarrow{a}_{\rho }$ .

$\therefore V = -\int_{1}^{2}\overrightarrow{E}.\overrightarrow{dl}$ .

$V = -\int_{1}^{2} \frac{\rho_ {L}}{2\pi \epsilon _{o}\rho }\overrightarrow{a}_{\rho }.(d\rho \overrightarrow{a}_{\rho }+d\phi \overrightarrow{a}_{\phi }+dz \overrightarrow{a}_{z })$ .

$V = -\int_{b}^{a} \frac{\rho_ {L}}{2\pi \epsilon _{o}\rho }\overrightarrow{a}_{\rho }.(d\rho \overrightarrow{a}_{\rho })$ .

$V = - \frac{\rho_ {L}}{2\pi \epsilon _{o} }(\ln a-\ln b)$ .

$V = \frac{\rho_ {L}}{2\pi \epsilon _{o} }(\ln b-\ln a)$ .

$V = \frac{\rho_ {L}}{2\pi \epsilon _{o} }\ln (\frac{b}{a})$ .

$V = \frac{Q}{2\pi \epsilon _{o}L }\ln (\frac{b}{a}) \ \because \ \rho _{L} = \frac{Q}{L}$ .

$\therefore C_{co-axial} =\frac{Q}{V} = \frac{2\pi \epsilon_{o}L}{\ln (\frac{b}{a})}$ .

L- length of the conductors.

b-radius of the outer conductor.

a- radius of the inner conductor.

$\epsilon$ – permittivity of the medium.

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Capacitance of a spherical conductor

choose two spherical conductor systems as shown in the figure with the inner conductor – $M_{2}$ -radius-a and outer conductor – $M_{1}$ -radius-b .

Now induced field is directed from $M_{2}$ to $M_{1}$ . then the potential difference between the two conductors is

$V= -\int_{1}^{2}\overrightarrow{E} .\overrightarrow{dl}$ .

by assuming a point P between the two conductors such that P is out of inner spherical conductor ( $M_{2}$ ) an inside the outer conductor ( $M_{1}$ ).

$\therefore \overrightarrow{E}_{at P} = \frac{Q}{4\pi \epsilon _{o}r^{2} }\overrightarrow{a}_{r}$ .

$\therefore V = -\int_{1}^{2}\overrightarrow{E}.\overrightarrow{dl}$ .

$V = -\int_{1}^{2} \frac{Q}{4\pi \epsilon _{o}r^{2} }\overrightarrow{a}_{r}.(dr \overrightarrow{a}_{r }+d\phi \overrightarrow{a}_{\phi }+d\theta \overrightarrow{a}_{\theta })$ .

$V = -\int_{r=b}^{a} \frac{Q}{4\pi \epsilon _{o}r^{2} }\overrightarrow{a}_{r}.(dr \overrightarrow{a}_{r }+d\phi \overrightarrow{a}_{\phi }+d\theta \overrightarrow{a}_{\theta })$ .

$V = - \frac{Q}{4\pi \epsilon _{o} }(\frac{1}{a}-\frac{1}{b})$ .

$V = \frac{Q}{4\pi \epsilon _{o} }(\frac{1}{b}-\frac{1}{a})$ .

$\therefore C_{spherical} =\frac{Q}{V} = \frac{4\pi \epsilon _{o}}{(\frac{1}{b}-\frac{1}{a}) }$ .

b-radius of the outer conductor.

a- radius of the inner conductor.

$\epsilon$ – permittivity of the medium.

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Capacitance (C)- Farads

When two conductors are embedded in a homogeneous dielectric medium as shown in the figure

$M_{2}$ – conductor 2 carries a positive charge (+Q) and $M_{1}$ -conductor 1 carries a negative charge (-Q). Assume the two charges are equal.

$\therefore$ The resultant charge is zero.

we know that charge is carried on the surface as a surface charge density $\rho _{s}$ and $\overrightarrow{E}$ is normal to the conducting surface.

since $M_{2}$ carries positive charge the flux is directed from $M_{2}$ to $M_{1}$ . flux starts from positive charge and ends at negative charge.

i.e, flux lines leaving one conductor must terminate at the surface of the other conductor.

$\therefore$ a field is induced between the two conductors $M_{2}$ and $M_{1}$ and the medium between these two conductors is a dielectric material.

Now, work must be done to carry a positive charge from $M_{1}$ to $M_{2}$ , that is opposite to the induced field.

$V_{21}= -\int_{2}^{1}\overrightarrow{E_{induced}}.\overrightarrow{dl}$

$V_{21}=V_{1}-V_{2}$ .

$V= -\int_{2}^{1}\overrightarrow{E_{induced}}.\overrightarrow{dl}$ .

Now, a system as shown in the above figure will have a special physical property called capacitance.

Definition of Capacitance of a conductor:-

The capacitance of a conductor is defined as the physical property of the conductor which is the ability to store Electrical Energy

i.e, $C=\frac{Q}{V}$ Farads

(or)

Capacitance is the ration of charge of the conductor (either m $M_{1}$ (or) $M_{2}$ ) to the voltage applied between the two conductors.

Procedure to find Capacitance:-

Assume Q and determine V in terms of Q , or else assume V and determine Q in terms of V.
choose a suitable Co-ordinate system.
assume the two conducting plates are carrying charges +Q and -Q.
Determine using Gauss’s law (or) by using Coloumb’s law and find $V= - \int \overrightarrow{E}.\overrightarrow{dl}$ .
Finally calculate $C=\frac{Q}{V}$ .

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Capacitance of Parallel Plate Capacitor

Choose two parallel conducting plates with charge densities separated by a distance ‘d’ meters as shown in the figure.

Assume charges are distributed uniformly on the plates.

Now apply a voltage source ‘V’ to these plates, then all positive charges are accumulated on conductor $M_{2}$ similarly negative charges are accumulated on conductor $M_{1}$ .

$\therefore$ the charges give rise to a field $\overrightarrow{E}$ in between them called as induced electric field.

To find out capacitance, choose a co-ordinate system x, y and z as shown in the figure

$\therefore V =-\int \overrightarrow{E}.\overrightarrow{dl}$ .

Noe the potential difference is $V=V_{2}-V_{1}$ .

$V_{12}=V_{2}-V_{1}$ .

$\therefore V =-\int_{1}^{2} \overrightarrow{E_{induced}}.\overrightarrow{dl}$ .

$\overrightarrow{E}$ is directed from 2 to 1 work has to be done in opposite direction from 1 to 2.

Now $\rho _{s}=\frac{Q}{S}$ . assume two conducting plates has equal surface area S

$\therefore$ to find out $\overrightarrow{E}$ at any point P in between the ‘2’ plates, use the concept of infinite sheet of charge distributions with densities $\rho _{s}$ and $-\rho _{s}$ .

$E_{\rho _{s}} = \frac{\rho _{s}}{2\epsilon _{o}}(-\overrightarrow{a_{z}})$ — from the positive charge distribution.

$E_{-\rho _{s}} = \frac{-\rho _{s}}{2\epsilon _{o}}(\overrightarrow{a_{z}})$ –with the negative charge distribution.

$\therefore$ Electric field intensity at P is the sum of electric field intensities due to two infinite charge distributions

$\overrightarrow{E_{total}} = \overrightarrow{E_{\rho _{s}}}+\overrightarrow{E_{-\rho _{s}}}$ .

$\overrightarrow{E_{total}} = \frac{\rho _{s}}{2\epsilon _{o}}(-\overrightarrow{a_{z}})+ \frac{-\rho _{s}}{2\epsilon _{o}}(\overrightarrow{a_{z}})$ .

$\overrightarrow{E_{total} }= \frac{\rho _{s}}{\epsilon _{o}}(-\overrightarrow{a_{z}})$ .

Now the potential difference between the two conductors is $\therefore V =-\int_{1}^{2} \overrightarrow{E_{induced}}.\overrightarrow{dl}$ .

$\therefore V =-\int_{1}^{2} \frac{\rho _{s}}{\epsilon _{o}}(-\overrightarrow{a_{z}}).\overrightarrow{dl}$ .

we know that $\overrightarrow{dl} = dx\ \overrightarrow{a_{x}}+ dy\ \overrightarrow{a_{y}}+ dz\ \overrightarrow{a_{z}}$ .

$\therefore V =\int_{1}^{2} \frac{\rho _{s}}{\epsilon _{o}}(\overrightarrow{a_{z}}).\ dz \ \overrightarrow{a_{z}}$ .

$V =\int_{0}^{d} \frac{\rho _{s}}{\epsilon _{o}}.\ dz$ .

$V = \frac{\rho _{s}d }{\epsilon _{o}}$ .

$V = \frac{Q\ d }{S\ \epsilon _{o}} \ \because \rho _{s}=\frac{Q}{S}$ .

$\therefore C= \frac{Q}{V}=\frac{S\ \epsilon _{o} }{d}$ .

If the medium between two parallel plates is air (or) free space (or) Vaccum use $\epsilon _{o}$ or else use $\epsilon$ .

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Infinite Line equivalent circuit

Consider the basic form of Transmission line with some impedance $Z_{o}$ at the Load end.

an infinite line can be approximated by an equivalent finite line with load impedance $Z_{o}$ as shown in the above figure, then the input impedance can be calculated from the voltage and current equations.

now at x= l , $V=V_{R} \ and \ I=I_{R}$ .

$V_{R}= V_{s}\cos h\gamma l-I_{s}Z_{o}\sin h\gamma l-----EQN(1)$ .

$I_{R}= I_{s}\cos h\gamma l-\frac{V_{s}}{Z_{o}}\sin h\gamma l----EQN(2)$ .

The load voltage is given by the equation $V_{R}=I_{R}Z_{o}$ .

$V_{s}\cos h\gamma l-I_{s}Z_{o}\sin h\gamma l = Z_{o}(I_{s}\cos h\gamma l-\frac{V_{s}}{Z_{o}}\sin h\gamma l)$

$V_{s}Z_{O}\cos h\gamma l-I_{s}Z_{o}^{2}\sin h\gamma l = Z_{o}(I_{s}Z_{o}\cos h\gamma l-V_{S}\sin h\gamma l)$

$V_{s}(Z_{O}\cos h\gamma l+Z_{R}\sin h\gamma l) = I_{s}Z_{o}(Z_{o}\sin h\gamma l+Z_{R}\cos h\gamma l)$

$Z_{S}=\frac{V_{S}}{I_{S}}=Z_{o} \frac{(Z_{o}\cos h\gamma l+Z_{o}\sin h\gamma l)}{(Z_{o}\cos h\gamma l+Z_{o}\sin h\gamma l)}$

$Z_{S} \ (or) \ Z_{in}=Z_{o}$ .

represents the source (or) input impedance of an infinite Transmission Line.

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E due to infinite line charge distribution

Consider an infinitely long straight line carrying uniform line charge with density $\rho _{L} C/m$ and lies on Z-axis from $-\infty$ to $+\infty$ .

Consider a point P at which Electric field intensity has to be determined which is produced by the line charge distribution.

from the figure let the co-ordinates of P are $(0,\rho ,0)$ ( a point on y-axis) and assume $dQ$ is a small differential charge confirmed to a point M $(0,0,Z)$ as co-ordinates.

$\therefore dQ$ produces a differential field $\overrightarrow{dE}$

$\overrightarrow{dE}=\frac{dQ}{4\pi \epsilon _{o}R^{2}}\widehat{a_{r}}$

the position vector $\overrightarrow{R}=-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}$ and the corresponding unit vector $\widehat{a_{r}} =\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{\sqrt{\rho ^{2}+Z^{2}}}$

$\therefore \overrightarrow{dE} =\frac{dQ}{4\pi \epsilon _{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{(\rho ^{2}+Z^{2})^{\frac{3}{2}}}})$

$therefore \overrightarrow{dE} =\frac{\rho _{L}dZ}{4\pi \epsilon _{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{(\rho ^{2}+Z^{2})^{\frac{3}{2}}}})$

then the Electric field strength $\overrightarrow{E}$ produced by the infinite line charge distribution $\rho _{L}$ is

$\overrightarrow{E} = \int \overrightarrow{dE}$

$\overrightarrow{E} = \int_{z=-\infty }^{\infty }\frac{\rho _{L}dZ}{4\pi \epsilon _{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{(\rho ^{2}+Z^{2})^{\frac{3}{2}}}})$

to solve this integral let $Z= \rho \tan \theta \Rightarrow dZ=\rho \sec ^{2}\theta d\theta$

as $Z\rightarrow -\infty \Rightarrow \theta \rightarrow \frac{-\pi }{2}$

$Z\rightarrow \infty \Rightarrow \theta \rightarrow \frac{\pi }{2}$

$\therefore \overrightarrow{E} = \int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} \frac{\rho _{L}}{4\pi\epsilon _{o}}(\frac{-\rho ^{2}\\sec ^{2}\theta \tan \theta d\theta \overrightarrow{a_{z}}+\rho ^{2}\sec ^{2}\theta d\theta \overrightarrow{a_{\rho }}}{(\rho ^{2}+\rho ^{2}\tan ^{2}\theta )^{\frac{3}{2}}})$

$\overrightarrow{E} = \int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} \frac{\rho _{L}}{4\pi\epsilon _{o}}(\frac{-\rho ^{2}\\sec ^{2}\theta \tan \theta d\theta \overrightarrow{a_{z}}+\rho ^{2}\sec ^{2}\theta d\theta \overrightarrow{a_{\rho }}}{\rho ^{3}\sec ^{3}\theta })$

$\overrightarrow{E} = \frac{\rho _{L}}{4\pi\epsilon _{o}\rho }$

$\overrightarrow{E}= \frac{\rho _{L}}{4\pi\epsilon _{o}\rho }$

$\therefore \overrightarrow{E}= \frac{\rho _{L}}{2\pi\epsilon _{o}\rho }\overrightarrow{a_{\rho }} Newtons/Coulomb$ .

$\overrightarrow{E}$ is a function of $\rho$ only, there is no $\overrightarrow{a_{z}}$ component and $\rho$ is the perpendicular distance from the point P to line charge distribution $\rho _{L}$ .

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Surface impedance

At high frequencies, the current is almost confined to a very thin sheet at the surface of the conductor which is used in many applications.

The surface impedance may be defined as the ratio of the tangential component of the electric field $\overrightarrow{E_{tan}}$ at the surface of the conductor to the current density (linear) $\overrightarrow{J_{s}}$ which flows due to this electric field.

given as $Z_{s}$ (or) $\eta _{s} = \frac{\overrightarrow{E_{tan}}}{\overrightarrow{J_{s}}}$ .

$\overrightarrow{E_{tan}}$ is the Electric field strength parallel to and at the surface of the conductor.

and $\overrightarrow{J}$ is the total linear current density which flows due to $\overrightarrow{E_{tan}}$ .

The $\overrightarrow{J_{s}}$ represents the total conduction per meter width flowing in this sheet.

Let us consider a conductor of the type plate, is placed at the surface y=0 and the current distribution in the y-direction is given by

Assume that the depth of penetration ( $\delta$ ) is very much less compared with the thickness of the conductor.

$J_{s}= \int_{0}^{\infty } \overrightarrow{J}.\overrightarrow{dy}$

$J_{s}= \int_{0}^{\infty } J_{o}e^{-\gamma y}dy$

$J_{s}= J_{o}(e^{-\gamma y})_{0}^{\infty }$

$J_{s}= \frac{J_{o}}{\gamma }$

from ohm’s law $\overrightarrow{J_{o}} = \sigma \overrightarrow{E_{tan}}$

$E = \frac{J_{o}}{\sigma }$ .

then $\eta _{s} = \frac{\gamma }{\sigma }$ .

$Z_{s}$ (or) $\eta _{s} = \frac{\gamma }{\sigma }$ .

we know that $\gamma = \sqrt{j\omega \mu (\sigma +j\omega \epsilon )}$

for good conductors .

then $\gamma \approx \sqrt{j\omega \mu \sigma }$

$\eta _{s} = \frac{\gamma }{\sigma } = \sqrt{\frac{j\omega \mu }{\sigma }}$ .

therefore the surface impedance of a plane good conductor which is very much thicker than the skin depth is equal to the characteristic impedance of the conductor.

This impedance is also known s input impedance of the conductor when viewed as transmission line conducting energy into the interior of metal.

when the thickness of the plane conductor is not greater compared to the depth of penetration , reflection of wave occurs at the back surface of the conductor.

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Application of Ampere’s circuit law-infinite sheet of current

consider an infinite sheet in the z=0 plane, which has uniform current density $\overrightarrow{k} = k_{y}\overrightarrow{a_{y}}$ 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

$\overrightarrow{H} = H_{x}\overrightarrow{a_{x}}+H_{y}\overrightarrow{a_{y}}+H_{z}\overrightarrow{a_{z}}$

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.

$\overrightarrow{H} = \left\{\begin{matrix} H_{o}\ \overrightarrow {a_{x}} \ for \ z>0\\ -H_{o}\ \overrightarrow {a_{x}} \ for \ z<0 \end{matrix}\right.$

from Ampere’s Circuit law $\oint \overrightarrow{H}. \overrightarrow{dl} = \int \left (\overrightarrow{H}_{1-2}. \overrightarrow{dl}_{1-2} +\overrightarrow{H}_{2-3}. \overrightarrow{dl}_{2-3}+\overrightarrow{H}_{3-4}. \overrightarrow{dl}_{3-4}+\overrightarrow{H}_{4-1}. \overrightarrow{dl}_{4-1} \right )$ .

the component $\oint \overrightarrow{H}. \overrightarrow{dl} = \int \left (\overrightarrow{H}_{1-2}. \overrightarrow{dl}_{1-2} +\overrightarrow{H}_{2-3}. \overrightarrow{dl}_{2-3}+\overrightarrow{H}_{3-4}. \overrightarrow{dl}_{3-4}+\overrightarrow{H}_{4-1}. \overrightarrow{dl}_{4-1} \right )$

$\overrightarrow{H}_{1-2}. \overrightarrow{dl}_{1-2}= \overrightarrow{H}_{1-0}. \overrightarrow{dl}_{1-0}+\overrightarrow{H}_{0-2}. \overrightarrow{dl}_{0-2}$

$\overrightarrow{H}_{1-2}. \overrightarrow{dl}_{1-2}= H_{o} \overrightarrow{a_{x}}.(\frac{a}{2})(-\overrightarrow{a_{z}})- H_{o} \overrightarrow{a_{x}}.(\frac{a}{2})(-\overrightarrow{a_{z}})$ .

$\overrightarrow{H}_{1-2}. \overrightarrow{dl}_{1-2}=0$ .

similarly $\overrightarrow{H}_{3-4}. \overrightarrow{dl}_{3-4} =0$ .

$\therefore \oint \overrightarrow{H}. \overrightarrow{dl} = \int \left (\overrightarrow{H}_{2-3}. \overrightarrow{dl}_{2-3}+\overrightarrow{H}_{4-1}. \overrightarrow{dl}_{4-1} \right)$ .

$\oint \overrightarrow{H}. \overrightarrow{dl} = -(H_{o}\ \overrightarrow{a_{x}}).(b \ -\overrightarrow{a_{x}})+(H_{o}\ \overrightarrow{a_{x}}).(b \ \overrightarrow{a_{x}})$ .

$\oint \overrightarrow{H}. \overrightarrow{dl} = 2H_{o}\ b$ .

$I= 2H_{o} \ b$ .

$k_{y}\ b = 2H_{o} \ b$ .

$H_{o} =\frac{1}{2} k_{y}$ .

as $\overrightarrow{H} = \left\{\begin{matrix} H_{o}\ \overrightarrow {a_{x}} \ for \ z>0 \\ -H_{o}\ \overrightarrow {a_{x}} \ for \ z<0 \end{matrix}\right.$

this will be changed to $\overrightarrow{H} = \left\{\begin{matrix} \frac{1}{2} k_{y}\ \overrightarrow {a_{x}} \ for \ z>0 \\ -\frac{1}{2} k_{y}\ \overrightarrow {a_{x}} \ for \ z<0 \end{matrix}\right.$

In general for a finite sheet of current density $\overrightarrow{k}$ A/m Magnetic field is generalised as $\overrightarrow{H} = \frac{1}{2} (\overrightarrow{k} X \overrightarrow{a_{n}})$ .