Inverse Z-Transform -Methods

Long Division  (or) Power Series ExpansionMethod:-

X(Z) = \sum_{n=-\infty }^{\infty }x[n]\ Z^{-n}

X(Z)can be expressed either in positive powers of Z or negative powers of Z.

if the sequence is causal X(Z) has negative powers of Z similarly the Non-causal sequence   X(Z)  negative powers of Z.

Let X(Z)=\frac{N(Z)}{D(Z)} =\frac{b_{0}+b_{1}Z^{-1}+b_{2}Z^{-2}+...........+b_{M}Z^{-M}}{a_{0}+a_{1}Z^{-1}+a_{2}Z^{-2}+...........+a_{N}Z^{-N}} .

X(Z) is causal it has ROC  \left | Z \right |> \left | r \right |  then X(Z) can be expressed as

X(Z) = x(0)+x(1)Z^{-1}+x(2)Z^{-2}+x(3)Z^{-3}+........

X(Z) is non-causal it has ROC \left | Z \right |<\left | r \right | then X(Z) can be expressed as

X(Z)=x(0)+x(-1)Z^{1}+x(-2)Z^{2}+x(-3)Z^{3}+...........

Partial fraction Method:-

Let X(Z)=\frac{N(Z)}{D(Z)} =\frac{b_{0}+b_{1}Z^{-1}+b_{2}Z^{-2}+...........+b_{M}Z^{-M}}{a_{0}+a_{1}Z^{-1}+a_{2}Z^{-2}+...........+a_{N}Z^{-N}}     and    a_{o}=1

if     M<N  ,  X(Z)   is a proper function .

if   M\geq N   ,  X(Z)   is improper function  so convert the improper function to proper function as

X(Z)=c_{o}+c_{1}Z^{-1}+c_{2}Z^{-2}+.........+c_{M-N}Z^{-(M-N)}+ \frac{N_{1}(Z)}{D(Z)}.

X(Z) = polynomial + rational \ proper \ function .

express    X(Z )  into powers of Z  as follows

X(Z ) = \frac{b_{o}Z^{N}+b_{1}Z^{N-1}+b_{2}Z^{N-2}+.......+b_{M}Z^{N-M}}{Z^{N}+a_{1}Z^{N-1}+a_{2}Z^{N-2}+.......+a_{N}}

then divide   X(Z)  by   Z

\frac{X(Z)}{Z} = \frac{b_{o}Z^{N-1}+b_{1}Z^{N-2}+b_{2}Z^{N-3}+.......+b_{M}Z^{N-M-1}}{Z^{N}+a_{1}Z^{N-1}+a_{2}Z^{N-2}+.......+a_{N}}  .

Now  express   \frac{X(Z)}{Z}  into partial fractions using different cases and find out the inverse Z-transform  for the function

X(Z ) = Z.(partial fraction \ expansion ) .

Convolution Method:-

express X(Z)   as a product of two functions X_{1}(Z)   and X_{2}(Z)  as follows X(Z) =X_{1}(Z) . X_{2}(Z)

then find the inverse Z- transforms of individual functions

x_{1}[n]\leftrightarrow X_{1}(Z)

x_{2}[n]\leftrightarrow X_{2}(Z)

by using convolution method find convolution of x_{1}[n]  and x_{2}[n]

i.e, x[n] = x_{1}[n] * x_{2}[n]

now x[n]   is the inverse Z-transform of X(Z) .

Cauchy Residue Theorem:-

f(Z) a function in Z if the derivative \frac{df(Z)}{dZ}   exists on and inside contour C and f(Z) has no poles at Z=Z_{o}   then.

\frac{1}{2\pi j}\oint_{c} \frac{f(Z)dZ}{Z-Z_{o}}=\left\{\begin{matrix} f(Z_{o}) \ if \ Z_{o} \ is\ inside \ C\\ 0 \ if \ Z_{o}\ is\ outside \ C \end{matrix}\right. .

if   (k+1)^{th} the derivative  of   f(Z)  exists on and has no poles at Z=Z_{o}   then.

\frac{1}{2\pi j}\oint_{c} \frac{f(Z)}{(Z-Z_{o})^k}dZ=\left\{\begin{matrix}\frac{1}{(k-1)!} \frac{d^{k-1}f(Z)} {dZ^{k-1}}\ if \ Z_{o} \ is\ inside \ C\\ 0 \ if \ Z_{o}\ is\ outside \ C \end{matrix}\right. .

the values on the right hand side are called Residue’s of the pole Z=Z_{o} .

if there are n no of poles inside C\frac{1}{2\pi j}\oint_{c} \frac{f(Z)dZ}{(Z-Z_{1})(Z-Z_{2})(Z-Z_{3})..(Z-Z_{n})}=\sum_{i=1}^{n}\lim_{Z\rightarrow Z_{i}} \left  .

 

 

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Complex Convolution Theorem of Z-Transforms

Complex Convolution Theorem :-

Let two signals x_{1}[n] , \ x_{2}[n]  which are complex signals then the product of these two signals be x_{3}[n] = x_{1}[n]\ x_{2}[n] .

x_{1}[n]\leftrightarrow X_{1}(Z) \ \ \ ROC :a_{1}< \left | Z \right | <b_{1}-R_{1}

x_{2}[n]\leftrightarrow X_{2}(Z) \ \ \ ROC :a_{2}< \left | Z \right | <b_{2}-R_{2}

Z\left \{ x_{1}[n]\ x_{2}[n] \right \}\leftrightarrow \ \ ?.

we know that  X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n} .

Z\left \{ x_{3}[n]\right \} = \sum_{n=-\infty }^{\infty }(x_{1}[n]\ x_{2}[n]) \ Z^{-n} .

Z\left \{ x_{3}[n]\right \} = \sum_{n=-\infty }^{\infty }(\ x_{2}[n]\ Z^{-n}) \ \frac{1}{2\pi j} \oint_{c}X_{1}(v)\ v^{n-1}\ dv .

Z\left \{ x_{3}[n]\right \} = \sum_{n=-\infty }^{\infty }(\ x_{2}[n]\ (\frac{Z}{v} )^{-n} \ \frac{1}{2\pi j} \oint_{c}X_{1}(v)\ v^{-1}\ dv .

Z\left \{ x_{3}[n]\right \} = \ \frac{1}{2\pi j} \oint_{c}X_{2}(\frac{Z}{v} ) X_{1}(v)\ v^{-1}\ dv .

Parseval’s Relation :-

The two complex valued signals x_{1}[n] , \ x_{2}[n]   then the Parseval’s relation states that

\sum_{n=-\infty }^{\infty } x_{1}[n] \ x_{2}^{*}[n] = \ \frac{1}{2\pi j} \oint_{c} X_{1}(v)\ X_{2}^{*}(\frac{1}{v^{*}} ) v^{-1}\ dv .

 

Proof:-

By using complex convolution theorem

\sum_{n=-\infty }^{\infty } x_{1}[n] \ x_{2}^{*}[n]\ Z ^{-n} = \ \frac{1}{2\pi j} \oint_{c} X_{1}(v)\ X_{2}^{*}(\frac{Z^{*}}{v^{*}} ) v^{-1}\ dv .

in the above equation substitute Z=1 , then

\sum_{n=-\infty }^{\infty } x_{1}[n] \ x_{2}^{*}[n] = \ \frac{1}{2\pi j} \oint_{c} X_{1}(v)\ X_{2}^{*}(\frac{1}{v^{*}} ) v^{-1}\ dv .

Hence Parseval’s relation is proved.

 

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Properties of Z-Transforms

1.Linearity Property:-

x_{1}[n]\leftrightarrow X_{1}(Z) \ \ \ ROC :a_{1}< \left | Z \right | <b_{1}-R_{1}

x_{2}[n]\leftrightarrow X_{2}(Z) \ \ \ ROC :a_{2}< \left | Z \right | <b_{2}-R_{2}

a\ x_{1}[n]+b\ x_{2}[n]\leftrightarrow \ \ ?.

we know that  Bi-lateral Z- Transform of a signal  x[n]  is X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n} .

Z\left \{a\ x_{1}[n]+b\ x_{2}[n]\right \} = \sum_{n=-\infty }^{\infty } \ \ \left \{ a\ x_{1}[n]+b\ x_{2}[n] \right \} Z^{-n}.

Z\left \{a\ x_{1}[n]+b\ x_{2}[n]\right \} = \sum_{n=-\infty }^{\infty } \ \ \left \{ a\ x_{1}[n]\ Z^{-n}+b\ x_{2}[n] \ Z^{-n}\right \}

Z\left \{a\ x_{1}[n]+b\ x_{2}[n]\right \} = \ a\ \sum_{n=-\infty }^{\infty } x_{1}[n]\ Z^{-n}+b \ \sum_{n=-\infty }^{\infty } x_{2}[n] \ Z^{-n}.

Z\left \{a\ x_{1}[n]+b\ x_{2}[n]\right \} = \ a \ X_{1}(Z)+b \ X_{2}(Z) .          ROC: R_{1} \cap R_{2}.

2.Time-shifting Property:-

x[n]\leftrightarrow X(Z) \ \ \ ROC : R

x[n-k]\leftrightarrow \ ?.

we know that  X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n} .

Z\left \{ x[n-k] \right \} = \sum_{n=-\infty }^{\infty } x[n-k] \ Z^{-n}.    Let     n-k=m \Rightarrow m= n+k

Here n is a variable and k is a constant.

Z\left \{ x[n-k] \right \} = \sum_{m\ =-\infty }^{\infty } x[m] \ Z^{-(m+k)} .

Z\left \{ x[n-k] \right \} = Z^{-k} \sum_{m\ =-\infty }^{\infty } x[m] \ Z^{-m}.

Z\left \{ x[n-k] \right \} = Z^{-k} X(Z)

x[n-k]\leftrightarrow \ Z^{-k} X(Z)\ , \ \ ROC:R.

from the above equation x[n-k]  forms Z Transform pair with Z^{-k} X(Z).

3. Scaling  in-Z-domain property:-

x[n]\leftrightarrow X(Z) \ \ \ ROC :r_{1}< \left | Z \right | <r_{2}-R_{1}

a^{n}x[n]\leftrightarrow \ \ \ ? ,

we know that  X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n} .

Z\left \{ a^{n}\ x[n] \right \} = \sum_{n=-\infty }^{\infty }a^{n}\ x[n] \ Z^{-n}.

Z\left \{ a^{n}\ x[n] \right \} = \sum_{n=-\infty }^{\infty }\ x[n] \ (a^{-1}Z)^{-n} .

Z\left \{ a^{n}\ x[n] \right \} = X(a^{-1}Z)

a^{n}x[n]\leftrightarrow \ X(\frac{Z}{a}), \ \ \ if \ a>0.   \ \ ROC \ of \ a^{n}x[n] \ is :\left | a \right |\ r_{1}< \left | Z \right | <\left | a \right |\ r_{2} .

4. Time-reversal property:-

x[n]\leftrightarrow X(Z) \ \ \ ROC :r_{1}< \left | Z \right | <r_{2}-R_{1}

x[-n]\leftrightarrow \ \ ?.

we know that  X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n} .

Z\left \{ x[-n] \right \} = \sum_{n=-\infty }^{\infty } x[-n] \ Z^{-n} .  Let  -n=\ m ,

Z\left \{ x[-n] \right \} = \sum_{m=-\infty }^{\infty } x[m] \ (Z^{-1})^{-m}\ \ \ ROC :r_{1}< \left | Z^{-1} \right | <r_{2}.

x[-n]\leftrightarrow \ X(Z^{-1})\ \ \ ROC :\frac{1}{\left | r_{2} \right |}< \left | Z \right | <\frac{1}{\left | r_{1} \right |}.

from the above equation x[-n]  forms Z Transform pair with X(\frac{1}{Z}).

5. Differentiation in Z-domain:-

x[n]\leftrightarrow X(Z) \ \ \ ROC :r_{1}< \left | Z \right | <r_{2}-R_{1}

n\ x[n]\leftrightarrow \ \ ?.

we know that  X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n} .

\frac{d(X(Z))}{dZ} = \sum_{n=-\infty }^{\infty } x[n] \frac{d (Z^{-n})}{dZ} .

\frac{d(X(Z))}{dZ} = \sum_{n=-\infty }^{\infty } \ -n \ x[n] Z^{(-n-1)} .

\frac{d(X(Z))}{dZ} = -\left  \ Z^{-n}\right ] Z^{-1} .

\frac{d(X(Z))}{dZ} = - Z^{-1} \ Z\left \{ n\ x[n] \right \} .

from the above equation \frac{d(X(Z))}{dZ}  forms Z-Transform pair with n\ x[n]  and the ROC is same as that of the original sequence x[n].

6. Convolution in Time-domain:-

x_{1}[n]\leftrightarrow X_{1}(Z) \ \ \ ROC :a_{1}< \left | Z \right | <b_{1}-R_{1}

x_{2}[n]\leftrightarrow X_{2}(Z) \ \ \ ROC :a_{2}< \left | Z \right | <b_{2}-R_{2}

x_{1}[n]* x_{2}[n]\leftrightarrow \ \ ?.

we know that  X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n} .

Z\left \{ x_{1}[n] *x_{2}[n]\right \} = \sum_{n=-\infty }^{\infty }(x_{1}[n] *x_{2}[n]) \ Z^{-n} .

Z\left \{ x_{1}[n] *x_{2}[n]\right \} = \sum_{k=-\infty }^{\infty }(x_{1}[k] x_{2}[n-k]) \sum_{n=-\infty }^{\infty }\ Z^{-n} .

Z\left \{ x_{1}[n] *x_{2}[n]\right \} = \sum_{k=-\infty }^{\infty }x_{1}[k] \sum_{n=-\infty }^{\infty }x_{2}[n-k] \ Z^{-n} .

Z\left \{ x_{1}[n] *x_{2}[n]\right \} = \sum_{k=-\infty }^{\infty }x_{1}[k] \ Z^{-k}\ X_{2}(Z) .

Z\left \{ x_{1}[n] *x_{2}[n]\right \} = \left  \ Z^{-k} \right ]\ X_{2}(Z) .

Z\left \{ x_{1}[n] *x_{2}[n]\right \} = X_{1}(Z) \ X_{2}(Z) .

 

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Properties of Fourier Transforms

1.Linearity Property:-

x_{1}(t)\leftrightarrow X_{1}(j\omega )

x_{2}(t)\leftrightarrow X_{2}(j\omega)

a\ x_{1}(t)+b\ x_{2}(t)\leftrightarrow \ \ ?.

we know that  Fourier Transform of a signal  x(t)  is X(j\omega) = \int_{-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ dt .

F\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \int_{-\infty }^{\infty }\ \left \{ a\ x_{1}(t)+b\ x_{2}(t) \right \} \ e^{-j\omega t}\ dt.

F\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \int_{-\infty }^{\infty }\ \left \{ a\ x_{1}(t)\ e^{-j\omega t}\ dt+b\ x_{2}(t) \ e^{-j\omega t}\ dt\right \}

F\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \ \ a \int_{-\infty }^{\infty }\ x_{1}(t)\ e^{-j\omega t}\ dt+b \int_{-\infty }^{\infty }\ x_{2}(t) \ e^{-j\omega t}\ dt.

F\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \ a \ X_{1}(j\omega) +b \ X_{2}(j\omega) .

2.Time-shifting Property:-

x(t)\leftrightarrow X(j\omega )

x(t-t_{o})\leftrightarrow \ ?.

we know that  X(j\omega) = \int_{-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ dt .

L\left \{ x(t-t_{o}) \right \} = \int_{-\infty }^{\infty }\ x(t-t_{o}) \ e^{- j\omega t}\ dt.    Lett-t_{o}=\lambda \Rightarrow dt= d\lambda

t \ limits \ : \ -\infty \ to \ \infty, \ \ \ \lambda \ \ limits \ : \ \infty \ to \ -\infty

L\left \{ x(t-t_{o}) \right \} = \int_{\lambda =-\infty }^{\infty }\ x(\lambda ) \ e^{- j\omega (\lambda +t_{o})}\ d\lambda .

L\left \{ x(t-t_{o}) \right \} =e^{-j\omega t_{o}} \int_{\lambda =-\infty }^{\infty }\ x(\lambda ) \ e^{-j\omega \lambda }\ d\lambda.

x(t-t_{o})\leftrightarrow \ e^{- j\omega t_{o}}\ X(j\omega).

from the above equation x(t-t_{o})  forms  Fourier Transform pair with e^{- j\omega t_{o}} \ X(j\omega).

3.Frequency-shifting Property:-

x(t)\leftrightarrow X(\omega )

?\ \leftrightarrow \ X(\omega -\omega _{o}).

we know that  X(\omega ) \ or X(j\omega ) = \int_{-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ dt .

L\left \{ e^{j\omega _{o}t}\ x(t) \right \} = \int_{t=-\infty }^{\infty }\ e^{j \omega _{o}t}\ x(t) \ e^{-j\omega t}\ dt.

L\left \{ e^{j\omega _{o}t}\ x(t) \right \} = \int_{t=-\infty }^{\infty }\ \ x(t) \ e^{-(\omega -\omega _{o})t}\ dt .

e^{j\omega _{o}t}x(t)\leftrightarrow \ X(\omega -\omega _{o}).

from the above equation e^{j\omega _{o}t}\ x(t)  forms  Fourier Transform pair with X(\omega -\omega _{o}).

4. Differentiation in time-domain:-

x(t)\leftrightarrow X(\omega )

\frac{dx(t)}{dt}\leftrightarrow \ ?.

we know that  inverse Fourier  Transform  x(t) =\frac{1}{2\pi } \int_{\omega =\infty }^{\infty }\ X( \omega ) \ e^{j\omega t} \ d\omega .

\frac{dx(t)}{dt} =\frac{1}{2\pi } \int_{\omega =\infty }^{\infty }\ X(\omega ) \ \frac{d(e^{j\omega t})}{dt} \ d\omega.

\frac{dx(t)}{dt} =\frac{1}{2\pi } \int_{\omega =\infty }^{\infty }\ X(\omega ) \ j \omega \ e^{j\omega t} \ d\omega .

\frac{dx(t)}{dt} =\frac{1}{2\pi } \int_{\omega =\infty }^{\infty }\ (\ j \omega \ X(\omega )) \ e^{j\omega t} \ d\omega.

\frac{dx(t)}{dt}\leftrightarrow \ j \omega \ X(\omega ).

from the above equation \frac{dx(t)}{dt}  forms Fourier Transform pair with \ j \omega \ X(\omega )

Similarly  \frac{d^{n}x(t)}{dt^{n}}\leftrightarrow \ \ (j \omega) ^{n}\ X(\omega ).

5.Differentiation in w-domain:-

x(t)\leftrightarrow X(\omega )

?\leftrightarrow \frac{dX(\omega )}{d\omega }.

we know that  X(\omega ) = \int_{t =-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ dt .

\frac{dX(\omega )}{d\omega } = \int_{t=-\infty }^{\infty }\ x(t) \ \frac{d(e^{-j\omega t}) }{d\omega } \ dt.

\frac{dX(\omega )}{d\omega } = \int_{t=-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ (-j t)\ dt .

\frac{dX(\omega )}{d\omega } = \int_{t=-\infty }^{\infty }\ (-jt \ x(t)) \ e^{-j\omega t} \ dt.

\frac{dX(\omega )}{d\omega }\leftrightarrow \ -jt\ x(t).

from the above equation \frac{dX(\omega )}{d\omega }  forms Fourier Transform pair with -jt\ x(t).

6. Conjugation property:-

x(t)\leftrightarrow X(\omega )

x^{*}(t)\leftrightarrow \ \ ?.

we know that  X(\omega ) = \int_{t=-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ dt .

F\left \{ x^{*}(t) \right \} = \int_{-\infty }^{\infty }\ x^{*}(t) \ e^{-j\omega t} \ dt.

F\left \{ x^{*}(t) \right \} = \int_{t =-\infty }^{\infty }( x(t ) \ e^{j \omega t} \ dt )^{*} .

x^{*}(t)\leftrightarrow \ X^{*}(-\omega ).

from the above equation x^{*}(t)  forms Fourier Transform pair with X^{*}(-\omega ).

7. Time-Scaling property:-

x(t)\leftrightarrow X(\omega )

x(at)\leftrightarrow \ \ ?.

we know that  X(\omega ) = \int_{-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ dt .

F\left \{ x(at) \right \} = \int_{-\infty }^{\infty }\ x(at) \ e^{-j\omega t} \ dt.          Let  at=\ \lambda ,      dt=\ \frac{d\lambda}{a}t \ limits \ : \ -\infty \ to \ \infty, \ \ \ \lambda \ \ limits \ : \ -\infty \ to \ \infty.

F\left \{ x(at) \right \} = \frac{1}{a}\int_{\lambda =-\infty }^{\infty }\ x(\lambda ) \ e^{-j(\frac{\omega }{a})\lambda } \ d\lambda .

x(at)\leftrightarrow \frac{1}{a} \ X(\frac{\omega }{a}), \ \ \ if \ a>0.

x(-at)\leftrightarrow \frac{1}{a} \ X(\frac{-\omega }{a}), \ \ \ if \ a<0 \ and \ (a\neq -1).

8. Convolution in Time-domain:-

x_{1}(t)\leftrightarrow X_{1}(\omega )

x_{2}(t)\leftrightarrow X_{2}(\omega )

x_{1}(t) * x_{2}(t)\leftrightarrow \ \ ?.

we know that  X(\omega ) = \int_{t=-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ dt .

F\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{t=-\infty }^{\infty }\ \left \{ x_{1}(t) * x_{2}(t) \right \}\ e^{-j\omega t} \ dt.

F\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{t=-\infty }^{\infty }\ \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau )\left \{ x_{2}(t-\tau ) \right \}\ e^{-j\omega t} \ dt \ d\tau.

F\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau )\left \{ \int_{t=-\infty }^{\infty } x_{2}(t-\tau )\ e^{-j\omega t} \ dt \right \} \ d\tau.

F\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau )\left \{ e^{-j\omega \tau } X_{2}(\omega ) \right \} \ d\tau.

F\left \{ x_{1}(t) * x_{2}(t) \right \} = \left \{ \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau ) e^{-j\omega \tau } \ d\tau \right \} \ X_{2}(\omega ).

F\left \{ x_{1}(t) * x_{2}(t) \right \} = \ X_{1}(\omega ) \ X_{2}(\omega )

x_{1}(t) * x_{2}(t)\leftrightarrow \ X_{1}(\omega ) \ X_{2}(\omega ).

 

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Derivation of Z and inverse Z-Transfomr

Derivation of Z-Transform:-

If x[n]  is the given sequence then it’s Discrete Time Fourier Transform  is  X(e^{j\omega })  .

i.e,    x[n]\leftrightarrow X(e^{j\omega })

x[n]\ r^{-n}\leftrightarrow X(r\ e^{j\omega }).

DTFT of x[n] = \sum_{n=-\infty }^{\infty } x[n] e^{-j\omega n} .

DTFT of  x[n]\ r^{-n} = \sum_{n=-\infty }^{\infty } x[n] \ r^{-n}e^{-j\omega n} .

X(r\ e^{j\omega }) = \sum_{n=-\infty }^{\infty } x[n] \ \left ( r\ e^{j\omega } \right )^{-n} .

Let    Z=r\ e^{j\omega }   a complex-variable.

X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z ^{-n}   . is the Z-Transform of a sequence  x[n] .

Inverse Z-Transform:-

The inverse DTFT of  X(e^{j\omega })  is    x[n] = \frac{1}{2\pi } \int X(e^{j\omega }) \ e^{j\omega n} \ d\omega.

x[n] \ r^{-n} = \frac{1}{2\pi } \int X(r\ e^{j\omega }) \ e^{j\omega n} \ d\omega .

x[n] = \frac{1}{2\pi } \int X(r\ e^{j\omega }) \ (r\ e^{j\omega })^{n} \ d\omega .

x[n] = \frac{1}{2\pi\ j Z } \int X(Z) \ Z^{n} \ dz .    Let  r\ e^{j\omega } = Z \Rightarrow \j \ r \ e^{j\omega }\ d\omega = dZ     and  \ j \ Z \ d\omega = dZ \Rightarrow d\omega =\frac{dz}{j \ Z} .

x[n] = \frac{1}{2\pi\ j } \int X(Z) \ Z^{n-1} \ dz – represents the inverse Z-Transform.

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Initial and Final Value Theorems

Initial-value Theorem:-

Use:- to find out the initial value of a signal x(t) without using inverse Laplace Transform.

x(0^{-})= \lim_{t\rightarrow 0^{-}}x(t)=\lim_{S\rightarrow \infty }s\ X(S).

Proof:-

we know that  L\left \{ \frac{dx(t)}{dt} \right \}\leftrightarrow S\ X(S)-x(0^{-}).

L\left \{ \frac{dx(t)}{dt} \right \}=\int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt.

i.e,       \int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt=S\ X(S)-x(0^{-}) .

\lim_{s\rightarrow \infty }\ \int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt= \lim_{s\rightarrow \infty }\ S\ X(S)- \lim_{s\rightarrow \infty }\ x(0^{-}) .

0= \lim_{s\rightarrow \infty }\ S\ X(S)- \lim_{s\rightarrow \infty }\ x(0^{-}) .

\lim_{s\rightarrow \infty }\ S\ X(S)= \lim_{s\rightarrow \infty }\ x(0^{-}) .

\ x(0^{-}) = \lim_{s\rightarrow \infty }\ S\ X(S) .

Hence proved.

final-value Theorem:-

Use:- to find out the final value of a signal x(t) without using inverse Laplace Transform.

x(\infty )= \lim_{t\rightarrow \infty }x(t)=\lim_{S\rightarrow 0 }s\ X(S).

Proof:-

we know that  L\left \{ \frac{dx(t)}{dt} \right \}\leftrightarrow S\ X(S)-x(0^{-}).

L\left \{ \frac{dx(t)}{dt} \right \}=\int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt.

i.e,       \int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt=S\ X(S)-x(0^{-}) .

\lim_{s\rightarrow 0 }\ \int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt= \lim_{s\rightarrow 0 }\ S\ X(S)- \lim_{s\rightarrow 0 }\ x(0^{-}) .

x(\infty )-x(0^{-})= \lim_{s\rightarrow 0 }\ S\ X(S)- \lim_{s\rightarrow 0 }\ x(0^{-}) .

x(\infty )-x(0^{-})= \lim_{s\rightarrow 0 }\ S\ X(S)- x(0^{-})

x(\infty )= \lim_{s\rightarrow 0 }\ S\ X(S) .

Hence proved.

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Properties of Laplace Transforms(Bi-lateral)

1.Linearity Property:-

x_{1}(t)\leftrightarrow X_{1}(S) \ \ \ ROC : R_{1}

x_{2}(t)\leftrightarrow X_{2}(S) \ \ \ ROC : R_{2}

a\ x_{1}(t)+b\ x_{2}(t)\leftrightarrow \ \ ?.

we know that  Laplace Transform of a signal  x(t)  is  X(S) = \int_{-\infty }^{\infty }\ x(t) \ e^{-St} \ dt .

L\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \int_{-\infty }^{\infty }\ \left \{ a\ x_{1}(t)+b\ x_{2}(t) \right \} \ e^{-St}\ dt.

L\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \int_{-\infty }^{\infty }\ \left \{ a\ x_{1}(t)\ e^{-St}\ dt+b\ x_{2}(t) \ e^{-St}\ dt\right \}

L\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \ \ a \int_{-\infty }^{\infty }\ x_{1}(t)\ e^{-St}\ dt+b \int_{-\infty }^{\infty }\ x_{2}(t) \ e^{-St}\ dt.

L\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \ \ a X_{1}(S) +b X_{2}(S) .          ROC: R_{1} \cap R_{2}.

2.Time-shifting Property:-

x(t)\leftrightarrow X(S) \ \ \ ROC : R

x(t-t_{o})\leftrightarrow \ ?.

we know that  X(S) = \int_{-\infty }^{\infty }\ x(t) \ e^{-St} \ dt .

L\left \{ x(t-t_{o}) \right \} = \int_{-\infty }^{\infty }\ x(t-t_{o}) \ e^{-St}\ dt.    Lett-t_{o}=\lambda \Rightarrow dt= d\lambda

t \ limits \ : \ -\infty \ to \ \infty, \ \ \ \lambda \ \ limits \ : \ \infty \ to \ -\infty

L\left \{ x(t-t_{o}) \right \} = \int_{\lambda =-\infty }^{\infty }\ x(\lambda ) \ e^{-S(\lambda +t_{o})}\ d\lambda .

L\left \{ x(t-t_{o}) \right \} =e^{-St_{o}} \int_{\lambda =-\infty }^{\infty }\ x(\lambda ) \ e^{-S\lambda }\ d\lambda.

x(t-t_{o})\leftrightarrow \ e^{-St_{o}}\ X(S) \ , \ \ ROC:R.

from the above equation x(t-t_{o})  forms a Laplace Transform pair with e^{-St_{o}} \ X(S).

3.Frequency-shifting Property:-

x(t)\leftrightarrow X(S) \ \ \ ROC : R

?\ \leftrightarrow \ X(S-S_{o}).

we know that  X(S) = \int_{-\infty }^{\infty }\ x(t) \ e^{-St} \ dt .

L\left \{ e^{S_{o}t}\ x(t) \right \} = \int_{t=-\infty }^{\infty }\ e^{S_{o}t}\ x(t) \ e^{-St}\ dt.

L\left \{ e^{S_{o}t}\ x(t) \right \} = \int_{t=-\infty }^{\infty }\ \ x(t) \ e^{-(S-S_{o})t}\ dt .

e^{S_{o}t}x(t)\leftrightarrow \ X(S-S_{o}) \ , \ \ ROC:R.

from the above equation e^{S_{o}t}x(t)  forms a Laplace Transform pair with X(S-S_{o}).

4. Differentiation in time-domain:-

x(t)\leftrightarrow X(S) \ \ \ ROC : R

\frac{dx(t)}{dt}\leftrightarrow \ ?.

we know that  inverse Laplace Transform  x(t) =\frac{1}{2\pi \ j} \int_{\sigma - j\infty }^{\sigma +j\infty }\ X(S) \ e^{St} \ dS .

\frac{dx(t)}{dt} =\frac{1}{2\pi \ j} \int_{\sigma - j\infty }^{\sigma +j\infty }\ X(S) \ \frac{d(e^{St})}{dt} \ dS.

\frac{dx(t)}{dt} =\frac{1}{2\pi \ j} \int_{\sigma - j\infty }^{\sigma +j\infty }\ X(S) \ S \ e^{St} \ dS .

\frac{dx(t)}{dt} =\frac{1}{2\pi \ j} \int_{\sigma - j\infty }^{\sigma +j\infty } (\ S\ X(S)) \ e^{St} \ dS.

\frac{dx(t)}{dt}\leftrightarrow \ S\ X(S).

from the above equation \frac{dx(t)}{dt}  forms a Laplace Transform pair with S\ X(S)

Similarly  \frac{d^{n}x(t)}{dt^{n}}\leftrightarrow \ S^{n}\ X(S).

5.Differentiation in S-domain:-

x(t)\leftrightarrow X(S) \ \ \ ROC : R

?\leftrightarrow \frac{dX(S)}{dS}.

we know that  X(S) = \int_{-\infty }^{\infty }\ x(t) \ e^{-St} \ dt .

\frac{dX(S)}{dS} = \int_{-\infty }^{\infty }\ x(t) \ \frac{de^{-St} }{dS}\ dt.

\frac{dX(S)}{dS} = \int_{-\infty }^{\infty }\ x(t) \ e^{-St} \ (-t)\ dt .

\frac{dX(S)}{dS} = \int_{-\infty }^{\infty }\ (-t \ x(t)) \ e^{-St} \ dt.

\frac{dX(S)}{dS}\leftrightarrow \ -t\ x(t) \ \ \ ROC:R.

from the above equation \frac{dX(S)}{dS}  forms a Laplace Transform pair with -t\ x(t).

6. Time-reversal property:-

x(t)\leftrightarrow X(S) \ \ \ ROC : R

x(-t)\leftrightarrow \ \ ?.

we know that  X(S) = \int_{-\infty }^{\infty }\ x(t) \ e^{-St} \ dt .

L\left \{ x(-t) \right \} = \int_{-\infty }^{\infty }\ x(-t) \ e^{-St} \ dt.          Let  -t=\ \lambda ,      -dt=\ d\lambdat \ limits \ : \ -\infty \ to \ \infty, \ \ \ \lambda \ \ limits \ : \ \infty \ to \ -\infty.

L\left \{ x(-t) \right \} = \int_{\lambda =\infty }^{-\infty }\ x(\lambda ) \ e^{S\lambda } \ (-d\lambda ) .

L\left \{ x(-t) \right \} = \int_{\lambda =-\infty }^{\infty }\ x(\lambda ) \ e^{-(-S) \lambda } \ d\lambda.

x(-t)\leftrightarrow \ X(-S).

from the above equation x(-t)  forms a Laplace Transform pair with X(-S).

7. Time-Scaling property:-

x(t)\leftrightarrow X(S) \ \ \ ROC : R

x(at)\leftrightarrow \ \ ?.

we know that  X(S) = \int_{-\infty }^{\infty }\ x(t) \ e^{-St} \ dt .

L\left \{ x(at) \right \} = \int_{-\infty }^{\infty }\ x(at) \ e^{-St} \ dt.          Let  at=\ \lambda ,      dt=\ \frac{d\lambda}{a}t \ limits \ : \ -\infty \ to \ \infty, \ \ \ \lambda \ \ limits \ : \ -\infty \ to \ \infty.

L\left \{ x(at) \right \} = \frac{1}{a}\int_{\lambda =-\infty }^{\infty }\ x(\lambda ) \ e^{(\frac{-S}{a})\lambda } \ d\lambda .

x(at)\leftrightarrow \frac{1}{a} \ X(\frac{S}{a}), \ \ \ if \ a>0.

x(-at)\leftrightarrow \frac{1}{a} \ X(\frac{-S}{a}), \ \ \ if \ a<0 \ and \ (a\neq -1).

 

8. Convolution in Time-domain:-

x_{1}(t)\leftrightarrow X_{1}(S) \ \ \ ROC : R_{1}

x_{2}(t)\leftrightarrow X_{2}(S) \ \ \ ROC : R_{2}

x_{1}(t) * x_{2}(t)\leftrightarrow \ \ ?.

we know that  X(S) = \int_{t=-\infty }^{\infty }\ x(t) \ e^{-St} \ dt .

L\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{t=-\infty }^{\infty }\ \left \{ x_{1}(t) * x_{2}(t) \right \}\ e^{-St} \ dt.

L\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{t=-\infty }^{\infty }\ \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau )\left \{ x_{2}(t-\tau ) \right \}\ e^{-St} \ dt \ d\tau.

L\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau )\left \{ \int_{t=-\infty }^{\infty } x_{2}(t-\tau )\ e^{-St} \ dt \right \} \ d\tau.

L\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau )\left \{ e^{-S\tau } X_{2}(S) \right \} \ d\tau.

L\left \{ x_{1}(t) * x_{2}(t) \right \} = \left \{ \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau ) e^{-S\tau } \ d\tau \right \} \ X_{2}(S).

L\left \{ x_{1}(t) * x_{2}(t) \right \} = \ X_{1}(S) \ X_{2}(S)

x_{1}(t) * x_{2}(t)\leftrightarrow \ X_{1}(S) \ X_{2}(S), ROC : R_{1} \cap R_{2}.

 

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Normal incidence on a Perfect Di-electric

Normal incidence on a Perfect Dielectric:-

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:-

\overrightarrow{E}_{i}=E_{io}\ e^{-\gamma _{1}z}\ \overrightarrow{a}_{x} .

\overrightarrow{H}_{i}=H_{io}\ e^{-\gamma _{1}z}\ \overrightarrow{a}_{y} .

\overrightarrow{H}_{i}=\frac{E_{io}}{\eta _{1}}\ e^{-\gamma _{1}z}\ \overrightarrow{a}_{y}.    the d.o.p of  H is  \overrightarrow{a}_{H}=\overrightarrow{a}_{k}X\overrightarrow{a}_{E}\Rightarrow \overrightarrow{a}_{z}X\overrightarrow{a}_{x}\Rightarrow \overrightarrow{a}_{y}.

Reflected waves:-

\overrightarrow{E}_{r}=E_{ro}\ e^{\gamma _{1}z}\ \overrightarrow{a}_{x} .

\overrightarrow{H}_{r}=-H_{ro}\ e^{\gamma _{1}z}\ \overrightarrow{a}_{y} .

\overrightarrow{H}_{r}=-\frac{E_{ro}}{\eta _{1}}\ e^{\gamma _{1}z}\ \overrightarrow{a}_{y}.  the d.o.p of  H is  \overrightarrow{a}_{H}=\overrightarrow{a}_{k}X\overrightarrow{a}_{E}\Rightarrow -\overrightarrow{a}_{z}X\overrightarrow{a}_{x}\Rightarrow -\overrightarrow{a}_{y}.

Transmitted waves:-

\overrightarrow{E}_{t}=E_{to}\ e^{-\gamma _{2}z}\ \overrightarrow{a}_{x} .

\overrightarrow{H}_{t}=H_{to}\ e^{\gamma _{2}z}\ \overrightarrow{a}_{y} .

\overrightarrow{H}_{t}=\frac{E_{to}}{\eta _{2}}\ e^{-\gamma _{2}z}\ \overrightarrow{a}_{y}.    the d.o.p of  H is  \overrightarrow{a}_{H}=\overrightarrow{a}_{k}X\overrightarrow{a}_{E}\Rightarrow \overrightarrow{a}_{z}X\overrightarrow{a}_{x}\Rightarrow \overrightarrow{a}_{y}.

Now the Transmission and reflection coefficients are defined as follows

reflection \ coefficient\ \rho (or)\ r= \frac{reflected\ electric\ field \ strength}{incident\ electric\ field \ strength} .

\rho _{E} = \frac{E_{ro}}{E_{io}} .

similarly the transmission coefficient is

transmission \ coefficient\ \tau (or)\ T= \frac{transmitted\ electric\ field \ strength}{incident\ electric\ field \ strength}.

\tau _{E} = \frac{E_{to}}{E_{io}} .

Derivation of coefficients:-

By using the boundary conditions,

the tangential components of E are continuous

i.e, \begin{vmatrix} E_{tan1} \end{vmatrix}=\begin{vmatrix} E_{tan2} \end{vmatrix} .

E_{io}+E_{ro}=E_{to} .

E_{io}+\rho \ E_{io}=\tau \ E_{io}.

1+\rho =\tau ------EQN(1).

the tangential components of H are discontinuous

i.e, \begin{vmatrix} H_{tan1} \end{vmatrix}-\begin{vmatrix} H_{tan2} \end{vmatrix} = J_{s} .  let us assume J_{s}=0 at the Boundary.

\begin{vmatrix} H_{tan1} \end{vmatrix}=\begin{vmatrix} H_{tan2} \end{vmatrix} .

H_{io}+H_{ro}=H_{to} .

\frac{E_{io}}{\eta _{1}}-\frac{E_{ro}}{\eta _{1}}=\frac{E_{to}}{\eta _{2}}\Rightarrow \frac{E_{io}}{\eta _{1}}-\rho \ \frac{E_{io}}{\eta _{1}}=\tau \ \frac{E_{io}}{\eta _{2}}.

\frac{1}{\eta _{1}}- \frac{\rho}{\eta _{1}}= \frac{\tau}{\eta _{2}}.

1-\rho =\frac{\eta _{1}}{\eta _{2}}\ \tau -------EQN(2).

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

\tau _{E}=\frac{2\eta _{2}}{\eta _{2}+\eta _{1}}     and  \rho _{E}=\frac{\eta _{2}-\eta _{1}}{\eta _{2}+\eta _{1}} .

similarly  the transmission and reflection coefficients  using magnetic field strength are

\tau _{H}=\frac{\eta _{1}-\eta _{2}}{\eta _{2}+\eta _{1}}    and  \rho _{H}=\frac{2\eta _{1}}{\eta _{2}+\eta _{1}} .

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Normal incidence on a perfect conductor

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 E_{t}=0 since the electric field intensity inside a perfect conductor is zero.

The incident  E_{i}(t) and reflected  E_{r}(t) 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 E_{i}=E_{i}e^{-\gamma z}

\because \alpha =0  in free space \beta =\beta _{1} for medium 1

\overrightarrow{E_{i}} = E_{i}e^{-j\beta z} (-\beta _{1}z a wave propagating in positive z-direction) and the reflected wave is \overrightarrow{E_{r}} = E_{r}e^{j\beta z} (\beta _{1}z a wave propagating in positive z-direction).

\overrightarrow{E}_{total} = E_{i}e^{-j\beta z}+ E_{r}e^{j\beta z} .

by using tangential components {E}_{tan1} = E_{tan2} .

{E}_{i} +{E}_{r} = 0

{E}_{i} =-{E}_{r}

The resultant wave is  \overrightarrow{E}_{total} = E_{i}e^{-j\beta_{1} z}+ E_{r}e^{j\beta_{1} z} .

\overrightarrow{E}_{total} = E_{i}e^{-j\beta_{1} z}- E_{i}e^{j\beta_{1} z} .

\overrightarrow{E}_{total}(z) = -2E_{i}\ j \sin \beta_{1} z

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

\widetilde{{E}_{total}}(z,t) = Re\left \{ -2E_{i}\ j \sin \beta_{1} z \ e^{j\omega t }\right \}

\widetilde{{E}_{total}}(z,t) = 2E_{i} \sin \beta_{1} z \ \sin \omega t .

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  \frac{\lambda }{4} and minimum amplitude at multiples of \frac{\lambda }{2} .

Similarly The resultant Magnetic field is

The resultant wave is  \overrightarrow{H}_{total} = H_{i}e^{-j\beta_{1} z}+ H_{r}e^{j\beta_{1} z} .

\overrightarrow{H}_{total} = H_{i}e^{-j\beta_{1} z}+ H_{i}e^{j\beta_{1} z} .

\overrightarrow{H}_{total}(z) = 2H_{i}\ \cos \beta_{1} z

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

\widetilde{{H}_{total}}(z,t) = Re\left \{ 2H_{i}\ \cos \beta_{1} z \ e^{j\omega t }\right \}

\widetilde{{H}_{total}}(z,t) = 2H_{i} \cos \beta_{1} z \ \cos \omega t .

this wave is  a stationary wave  it has minimum amplitude at odd multiples of  \frac{\lambda }{4} and maximum amplitude at multiples of \frac{\lambda }{2} .

 

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Nature of Magnetic materials

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.

\pm 9X 10^{-24} A-m^{2}  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 .

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Laplace’s and Poisson’s equations(Magnetostatics)

Laplace’s equation(Magneto statics):- 

From the equation  \oint_{s}\overrightarrow{B}.\overrightarrow{ds}=0\Rightarrow \overrightarrow{\bigtriangledown } .\overrightarrow{B}=0 .

since \overrightarrow{B}=\mu _{O}\overrightarrow{H}

\overrightarrow{\bigtriangledown }.(\mu _{O}\overrightarrow{H}) =0 .

\mu _{O}(\overrightarrow{\bigtriangledown }.\overrightarrow{H}) =0.

we know that for zero current density \overrightarrow{J}=0  , the Magnetic scalar potential  is \overrightarrow{H}=-\overrightarrow{\bigtriangledown }V_{m}

by replacing \overrightarrow{H}  in the  equation  \mu _{O}(\overrightarrow{\bigtriangledown }.\overrightarrow{H}) =0 ,

\mu _{O}(\overrightarrow{\bigtriangledown }.(-\overrightarrow{\bigtriangledown }V_{m})) =0 .

\mu _{O}(\bigtriangledown^{2}V_{m}) =0.

\bigtriangledown^{2}V_{m} =0 .   is known as Laplce’s equation in Magneto statics.

Poisson’s equation(Magneto statics):- 

In vector algebra a vector can be fully defined if it’s curl and divergence are defined.

\overrightarrow{\bigtriangledown } X\overrightarrow{H}=\overrightarrow{J}   -point form of Ampere’s law

\overrightarrow{\bigtriangledown } X\overrightarrow{B}=\mu _{o}\overrightarrow{J} because \overrightarrow{B}=\mu _{O}\overrightarrow{H}  .

from vector Magnetic potential  \overrightarrow{B}=\overrightarrow{\bigtriangledown } X \overrightarrow{A} .

\overrightarrow{\bigtriangledown } X(\overrightarrow{\bigtriangledown } X \overrightarrow{A})=\mu _{o}\overrightarrow{J}.

from the vector identity  \bigtriangledown ^{2}\overrightarrow{A}=\overrightarrow{\bigtriangledown }(\overrightarrow{\bigtriangledown }.\overrightarrow{A})-\overrightarrow{\bigtriangledown } X\overrightarrow{\bigtriangledown } X \overrightarrow{A}.

\overrightarrow{\bigtriangledown }(\overrightarrow{\bigtriangledown }.\overrightarrow{A})-\bigtriangledown ^{2}\overrightarrow{A}=\mu _{o}\overrightarrow{J} .

if \overrightarrow{\bigtriangledown }.\overrightarrow{A}=0 .

\bigtriangledown ^{2}\overrightarrow{A}=-\mu _{o}\overrightarrow{J} – this equation is known as Poisson’s equation for Magneto statics.

\overrightarrow{A}=A_{x}\overrightarrow{a_{x}}+A_{y}\overrightarrow{a_{y}}+A_{z}\overrightarrow{a_{z}}     and    \overrightarrow{J}=J_{x}\overrightarrow{a_{x}}+J_{y}\overrightarrow{a_{y}}+J_{z}\overrightarrow{a_{z}}

\bigtriangledown ^{2}(A_{x}\overrightarrow{a_{x}}+A_{y}\overrightarrow{a_{y}}+A_{z}\overrightarrow{a_{z}})=-\mu _{o}(J_{x}\overrightarrow{a_{x}}+J_{y}\overrightarrow{a_{y}}+J_{z}\overrightarrow{a_{z}}).

by equating the  respective components on each side

\bigtriangledown ^{2}A_{x}=-\mu _{o}J_{x} ,    \bigtriangledown ^{2}A_{y}=-\mu _{o}J_{y}    and  \bigtriangledown ^{2}A_{z}=-\mu _{o}J_{z}   are the scalar Poisson’s equations of Magneto statics.

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Relaxation time

Relaxation time (T_{r}) :-

from the equation \overrightarrow{J}=\sigma \overrightarrow{E}   and from Gauss’s law      \overrightarrow{\bigtriangledown }.\overrightarrow{E} = \frac{\rho _{v}}{\epsilon }   .

continuity equation  is      \overrightarrow{\bigtriangledown }.\overrightarrow{J} = -\frac{\partial \rho _{v}}{\partial t }.

\overrightarrow{\bigtriangledown }.(\sigma \overrightarrow{E}) = -\frac{\partial \rho _{v}}{\partial t } .

\sigma( \overrightarrow{\bigtriangledown }. \overrightarrow{E}) = -\frac{\partial \rho _{v}}{\partial t } .

\sigma(\frac{\rho _{v}}{\epsilon }) = -\frac{\partial \rho _{v}}{\partial t } .

\therefore \frac{\partial \rho _{v}}{\partial t }+(\frac{\sigma}{\epsilon })\rho _{v} = 0.

The solution to the above equation is of the form  \rho _{v} = \rho _{vo} \ e ^{-\frac{t}{T_{r}}}  .

where T_{r} = \frac{\epsilon }{\sigma }  is known as relaxation time and defined as the time it takes a charge placed in the interior of a material to drop to e^{-1}= 36.8 percent of it’s initial value.

\rho _{vo} is the initial charge density (i.e,  \rho _{v}  at t=0) the equation   \rho _{v} = \rho _{vo} \ e ^{-\frac{t}{T_{r}}}    shows that as a result of introducing  charge at some interior point of the material there is a decay of volume charge density  \rho _{v}  this decay is associated with the charge movement from the interior point at which it was introduced to the surface of the material.

T_{r}  –  is the time constant known as the relaxation time (or) rearrangement time.

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Boundary conditions static electric fields

Boundary conditions static Electric fields:-

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 (\epsilon _{1}) and Di-electric (\epsilon _{2}).
  • Conductor (\sigma ) and Di-electric(\epsilon ).
  • Conductor and free space.

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

  1. \oint_{s} \overrightarrow{D} .\overrightarrow{ds} =Q_{enclosed} –  Gauss’s law in Electrostatics.
  2. \oint_{l} \overrightarrow{E} .\overrightarrow{dl} = 0.

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

\overrightarrow{E}=\overrightarrow{E_{t}}+\overrightarrow{E_{n}}.

Di-electric (\epsilon _{1}) and Di-electric (\epsilon _{2}):-

In order to find out the \overrightarrow{E} , \overrightarrow{D}   at the interface between two different magnetic materials boundary conditions are required.

Consider two Di-electric materials having permeabilities  \epsilon _{1}   and  \epsilon _{2} as shown in the figure

To apply    \oint_{l} \overrightarrow{E} .\overrightarrow{dl} = 0  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 \oint_{l} \overrightarrow{E} .\overrightarrow{dl} = 0 .

E_{1t}\Delta w -E_{1n}\frac{\Delta h}{2}-E_{2n}\frac{\Delta h}{2}-E_{2t}\Delta w+E_{2n}\frac{\Delta h}{2}+E_{1n}\frac{\Delta h}{2}=0.

(since \Delta h=0   On the boundary)

E_{1t}\Delta w-E_{2t}\Delta w=0 .

E_{1t}=E_{2t}.

\frac{D_{1t}}{\epsilon _{1}}=\frac{D_{2t}}{\epsilon _{2}} , So the tangential components of  \overrightarrow{D}    are discontinuous where as \overrightarrow{E}   are continuous.

In order to apply \oint_{s} \overrightarrow{D} .\overrightarrow{ds} =Q , a surface is required which is nothing but Gaussian surface (or) a Pillbox enclosing some charge.

\oint_{s} \overrightarrow{D} .\overrightarrow{ds} = \oint_{side} \overrightarrow{D} .\overrightarrow{ds} + \oint_{top} \overrightarrow{D} .\overrightarrow{ds} + \oint_{bottom} \overrightarrow{D} .\overrightarrow{ds} .

(\oint_{s} \overrightarrow{D} .\overrightarrow{ds} =Q  -Because on the boundary  \Delta h=0, so there exists no side of surface).

D_{1n}\Delta S -D_{2n}\Delta S =\Delta Q.

D_{1n}-D_{2n}=\frac{\Delta Q}{\Delta S} .

D_{1n}-D_{2n}=\rho _{s}-----(1).

since \overrightarrow{D} = \epsilon \overrightarrow{E} ,

\epsilon _{1}E_{1n}-\epsilon _{2}E_{2n}=\rho _{s}-----(2)

From equations (1) and (2)  the normal components of \overrightarrow{D} and \overrightarrow{E}   are  discontinuous at the boundary .

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Magnetic Boundary conditions

Magnetic Boundary conditions:-

In order to find out the \overrightarrow{B} , \overrightarrow{H}  and  \overrightarrow{M}  at the interface between two different magnetic materials boundary conditions are required.

Consider two magnetic materials having permeabilities  \mu _{1}   and  \mu _{2} as shown in the figure

We will find out these boundary conditions by using

  1. \oint_{s} \overrightarrow{B} .\overrightarrow{ds} =0  –  Gauss’s law in Magneto statics.
  2. \oint_{l} \overrightarrow{H} .\overrightarrow{dl} = I–  Ampere’s law.

In order to apply \oint_{s} \overrightarrow{B} .\overrightarrow{ds} =0 , a surface is required which is nothing but Gaussian surface (or) a Pillbox.

\oint_{s} \overrightarrow{B} .\overrightarrow{ds} = \oint_{side} \overrightarrow{B} .\overrightarrow{ds} + \oint_{top} \overrightarrow{B} .\overrightarrow{ds} + \oint_{bottom} \overrightarrow{B} .\overrightarrow{ds} .

(\oint_{s} \overrightarrow{B} .\overrightarrow{ds} =0  -Because on the boundary  \Delta h=0, so there exists no side of surface).

B_{1n}\Delta S -B_{2n}\Delta S =0.

B_{1n}=B_{2n}-----------(1)

since \overrightarrow{B} = \mu \overrightarrow{H} ,      \mu _{1} H_{1n}= \mu _{2} H_{2n}.

H_{1n}= \frac{\mu _{2}}{\mu _{1}} H_{2n} ---------(2).

From equations (1) and (2)  the normal component of \overrightarrow{B}    is continuous at the boundary and the normal components of  \overrightarrow{H}   is discontinuous at the boundary.

we know that \overrightarrow{B} = \mu _{o}(1+ \chi _{m})\overrightarrow{H} .

\overrightarrow{B} = \mu _{o}\overrightarrow{H}+ \mu _{o}\overrightarrow{M} .

\overrightarrow{M} = \chi _{m}\overrightarrow{H} .

 M_{2n} = \chi _{m2} H_{2n} .

M_{1n} = \chi _{m1} H_{1n}.

M_{2n} = \chi _{m2} H_{2n}  since  H_{2n} = \frac{\mu _{1}}{\mu _{2}} H_{1n} .

M_{2n}=\chi _{m2} \frac{\mu _{1}}{\mu _{2}} H_{1n} .

M_{2n}=\frac{\chi _{m2}}{\chi _{m1}} \frac{\mu _{1}}{\mu _{2}} M_{1n}.

The magnetisation normal components are also discontinuous.

To apply    \oint_{l} \overrightarrow{H} .\overrightarrow{dl} = I  a path is required, which is a closed one in a plane normal to the boundary surface.

Here a closed path is abcda , which encloses a surface current density k on the surface of the boundary

Then \oint_{l} \overrightarrow{H} .\overrightarrow{dl} = I .

H_{1t}\Delta w -H_{1n}\frac{\Delta h}{2}-H_{2n}\frac{\Delta h}{2}-H_{2t}\Delta w+H_{2n}\frac{\Delta h}{2}+H_{1n}\frac{\Delta h}{2}=I.

(since \Delta h=0   On the boundary)

H_{1t}\Delta w-H_{2t}\Delta w=k \Delta w .

H_{1t}-H_{2t}=k.

\frac{B_{1t}}{\mu _{1}}-\frac{B_{2t}}{\mu _{2}}=k , So the tangential components of  \overrightarrow{B}   and  \overrightarrow{H}  are discontinuous.

(\overrightarrow{H_{1t}} -\overrightarrow{H_{2t}})X \overrightarrow{a_{n12}} = \overrightarrow{k}.

(or) (\overrightarrow{H_{1t}} -\overrightarrow{H_{2t}}) = \overrightarrow{k}X \overrightarrow{a_{n12}}.

When  \overrightarrow{k}=0\overrightarrow{H_{1t}}=\overrightarrow{H_{2t}}   and  \frac{\overrightarrow{B_{1t}}}{\mu _{1}} =\frac{\overrightarrow{B_{2t}}}{\mu _{2}} .

Similar to normal components of magnetisation the tangential components are

M_{2t} = \chi _{m2} H_{2t} .

M_{1t} = \chi _{m1} H_{1t}.

M_{2t} = \chi _{m2} H_{2t} since  H_{2t} = H_{1t}-k .

M_{2t}=\chi _{m2} (H_{1t}-k) .

M_{2n}=\frac{\chi _{m2}}{\chi _{m1}} M_{1t}-\chi _{m2} k.

Obtain the results for magnetisation by using the same procedure as that of  normal components.

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Differential form of Ampere’s Circuit law

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

\overrightarrow{H}  at  center point P  is \overrightarrow{H_{o}} =H_{xo} \ \overrightarrow{a_{x}}+H_{yo} \ \overrightarrow{a_{y}}+H_{zo} \ \overrightarrow{a_{z}} .

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.

from ampere’s circuit law 

\oint \overrightarrow{H} . \overrightarrow{dl}= \overrightarrow{H}_{1-2}.\overrightarrow{\Delta l}_{1-2}+\overrightarrow{H}_{2-3}.\overrightarrow{\Delta l}_{2-3}+\overrightarrow{H}_{3-4}.\overrightarrow{\Delta l}_{3-4}+\overrightarrow{H}_{4-1}.\overrightarrow{\Delta l}_{4-1}.

 

 

Biot-savart’s law

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

  1. is proportional to the product I dl and the \sin \alpha  the angle between the element and the line joining the point P to the element.
  2. and is inversely proportional to the square of the distance  R between P and the current element.

then the direction of \overrightarrow{dH} 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  \overrightarrow{dH} .

i.e,  dH \propto \frac{I \ dl \ \sin \alpha }{R^{2}} .

dH = \frac{ k\ I \ dl \ \sin \alpha }{R^{2}} .

where k is the constant of proportionality , k=\frac{1}{4\pi } .

\overrightarrow{dH} = \frac{ \ I \ dl \ \sin \alpha }{4\pi \ R^{2}}\ \overrightarrow{a_{R}}  A/m.

\overrightarrow{dH} = \frac{ \ I \ \overrightarrow{dl} X \ \overrightarrow{a_{R}} }{4\pi \ R^{2}} A/m.

\overrightarrow{dH}   is perpendicular to the plane that contains \overrightarrow{dl}   and \overrightarrow{a_{R}}.

\overrightarrow{dH} = \frac{ \ I \ \overrightarrow{dl} X \ \overrightarrow{R} }{4\pi \ R^{3}}  A/m.

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

\overrightarrow{H} = \oint \frac{ \ I \ \overrightarrow{dl} X \ \overrightarrow{a_{R}} }{4\pi \ R^{2}} 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.

\overrightarrow{H} = \int_{l} \frac{ \ I \ \overrightarrow{dl} X \ \overrightarrow{a_{R}} }{4\pi \ R^{2}} A/m.  —-for a line current element.

\overrightarrow{H} = \int_{s} \frac{ \overrightarrow{k} \ ds X \ \overrightarrow{a_{R}} }{4\pi \ R^{2}}  A/m. —-for a surface current element.

\overrightarrow{H} = \int_{v} \frac{ \overrightarrow{J} \ dv \ X \ \overrightarrow{a_{R}} }{4\pi \ R^{2}}  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) .

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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}}) .

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Magnetic Forces

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.

  1. The force can be due to a  moving charged particle in a  Magnetic field.
  2. on a current element in an external B  field.
  3. between two current elements.

Force on a charged particle:-

we know that \overrightarrow{E} = \frac{\overrightarrow{F}}{Q}  .

\overrightarrow{F_{e}} = Q \overrightarrow{E}-----EQN(1) .

where  \overrightarrow{F_{e}}  is the electric force on a stationary (or) moving electric charge in an electric field and is related to  \overrightarrow{E} .  where  \overrightarrow{F_{e}}  and   \overrightarrow{E} 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

\overrightarrow{F_{m}} = (Q\ \overrightarrow{u} \ X \overrightarrow{B})-----EQN(2) .

from the  equations    \overrightarrow{F_{e}}   is independent of velocity of the charge and performs work on the charge which changes its kinetic energy but  \overrightarrow{F_{m}}  depends on the charge velocity and is normal to it so work done \overrightarrow{F_{m}}.\overrightarrow{dl} =0 it does not cause increase in the kinetic energy of the charge.

 \overrightarrow{F_{m}}  is small compared to    \overrightarrow{F_{e}}  except at high velocities.

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

Then \overrightarrow{F} = \overrightarrow{F_{e}}+\overrightarrow{F_{m}} .

\overrightarrow{F} = Q(\overrightarrow{E}+\overrightarrow{u}X\overrightarrow{B}) .

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 \overrightarrow{F} = m\overrightarrow{a} .

\overrightarrow{F} = m\frac{d\overrightarrow{u}}{dt} .

\overrightarrow{F} = Q(\overrightarrow{u}X\overrightarrow{B}) .

m\frac{d\overrightarrow{u}}{dt}= Q(\overrightarrow{u}X\overrightarrow{B}) .

\frac{d\overrightarrow{u}}{dt}-\frac{1}{m} (\overrightarrow{u}X\overrightarrow{B})=Qm\overrightarrow{E} .

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 I\overrightarrow{dl} , in order to find out the force acting on the current carrying element by the  magnetic field \overrightarrow{B} .

I\overrightarrow{dl}\approx \overrightarrow{k}ds\approx \overrightarrow{J}dv .

I\overrightarrow{dl}=\frac{dQ}{dt}\overrightarrow{dl} .

I\overrightarrow{dl}=dQ\frac{\overrightarrow{dl}}{dt} .

I\overrightarrow{dl}=dQ\overrightarrow{u} .

I\overrightarrow{dl} is nothing but a elemental charge dQ moving with the velocity \overrightarrow{u} .

\overrightarrow{F}=Q\overrightarrow{u}X\overrightarrow{B} .

d\overrightarrow{F}=dQ\overrightarrow{u}X\overrightarrow{B}.

d\overrightarrow{F}=I\overrightarrow{dl}X\overrightarrow{B} .

\overrightarrow{F}=\oint_{l}I\overrightarrow{dl}X\overrightarrow{B} .

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

i.e,

The magnetic field produced by the current element  I\overrightarrow{dl}  does not exert force on the element itself just as a point charge does not exert force on itself.

So the  magnetic field  \overrightarrow{B}  that exerts force on I\overrightarrow{dl} must be from the another element in other words the magnetic field  \overrightarrow{B}is external to the current element I\overrightarrow{dl}.

Similarly we have  force equations for other current elements  \overrightarrow{k}ds  and \overrightarrow{J}dv as follows

\overrightarrow{F}=\oint_{s}\overrightarrow{k} dsX\overrightarrow{B}       and        \overrightarrow{F}=\oint_{v}\overrightarrow{J} dvX\overrightarrow{B} .

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

i.e,  \frac{d\overrightarrow{F}}{\overrightarrow{k}ds}=\overrightarrow{B}  .(or)    \overrightarrow{F_{m}} = Q\overrightarrow{u} X \overrightarrow{B}\Rightarrow \frac{\overrightarrow{F_{m}}}{Q}=\overrightarrow{u} X \overrightarrow{B}    similar to \overrightarrow{E} = \frac{\overrightarrow{F_{e}}}{Q} .

so the    \overrightarrow{B}  describes the force properties of a magnetic field.

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

Consider two current loops  I_{1}\overrightarrow{dl_{1}}  and  I_{2}\overrightarrow{dl_{2}} then by Biot- Savart’s law  both current elements produces respective magnetic fields so we may find the force on element sI_{1}\overrightarrow{dl_{1}} due to the field produced by I_{2}\overrightarrow{dl_{2}}.

Field produced by current element  is   I_{2}\overrightarrow{dl_{2}}  is  d\overrightarrow{B_{2}} .

So the force applied on  the element   I_{1}\overrightarrow{dl_{1}} is    d\overrightarrow{F_{1}}  by the field  d\overrightarrow{B_{2}} .

d(d\overrightarrow{F_{1}}) = I_{1}\overrightarrow{dl_{1}}Xd\overrightarrow{B_{2}} ..

d\overrightarrow{B_{2}} = \frac{\mu _{o}I_{2}\overrightarrow{dl_{2}}X \widehat{{a_{R21}}}}{4\pi R_{21}^{2}} .

d(d\overrightarrow{F_{1}}) = \frac{\mu _{o}I_{1}\overrightarrow{dl_{1}}XI_{2}\overrightarrow{dl_{2}}X \widehat{{a_{R21}}}}{4\pi R_{21}^{2}}.

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

\overrightarrow{F_{1}} =\frac{\mu _{o}I_{1}I_{2}}{4\pi}\oint_{l_{1}} \oint_{l_{2}}\frac{\ \overrightarrow{dl_{1}}X\ \overrightarrow{dl_{2}}X \ \widehat{{a_{R21}}}}{ R_{21}^{2}} .

Then the force  \overrightarrow{F_{2}}  acting on loop 2 due to the field produced by the current element  t I_{1}\overrightarrow{dl_{1}}   is nothing but \overrightarrow{F_{2}}=-\overrightarrow{F_{1}}

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

 

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H due to finite long straight conuctor

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 Z1 and Z2 above the origin with it’s upper and lower ends respectively subtending angles \alpha _{2} and \alpha _{1} at P.

P is the point at which  \overrightarrow{H}   is to be determined.

Consider a differential element \overrightarrow{dl}  along the Z-axis at a distance Z from the origin.

where \overrightarrow{dl} =dl \ \overrightarrow{a_{z}} .

\overrightarrow{R} = -Z \overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }} .

\widehat{a_{R}} = \frac{\overrightarrow{R} = -Z \overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{\sqrt{(Z^{2}+\rho ^{2})}} .

\overrightarrow{dl} X \ \widehat{a_{R}} = \frac{\rho dz \ \overrightarrow{a_{\phi }}}{\sqrt{(Z^{2}+\rho ^{2})}} .

\overrightarrow{H} = \oint \frac{ I \ \rho \ dz \ \overrightarrow{a_{\phi }}}{4\pi (Z^{2}+\rho ^{2})^{\frac{3}{2}}} .

\overrightarrow{H} = \oint_{Z_{1}}^{Z_{2}} \frac{ I \ \rho \ dz \ \overrightarrow{a_{\phi }}}{4\pi (Z^{2}+\rho ^{2})^{\frac{3}{2}}} .

as z=\rho \ cot\alpha    ,  Z_{2} = \rho \ \cot \alpha _{2}  and  dz = - \rho \ cosec^{2} \alpha \ d\alpha  .

\overrightarrow{H} =\frac{-I}{4\pi } \int_{\alpha _{1}}^{\alpha _{2}} \frac{\rho ^{2} \ cosec^{2}\alpha \ d\alpha \ \overrightarrow{a_{\phi }}}{ (\rho ^{2}+\rho ^{2} \cot ^{2}\alpha )^{\frac{3}{2}}} .

\overrightarrow{H} =\frac{-I}{4\pi\ \rho } \int_{\alpha _{1}}^{\alpha _{2}} \rho ^{2} \ \sin \alpha \ d\alpha \ \overrightarrow{a_{\phi }} .

\overrightarrow{H} =\frac{I}{4\pi\ \rho } \left \overrightarrow{a_{\phi }} .

Case 1 :- 

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

i.e,    Z_{1} =0 \ \Rightarrow \ \alpha _{1} = 90^{o}     and    Z_{2} =\infty \ \Rightarrow \ \alpha _{2} = 0^{o} .

then  \overrightarrow{H} = \frac{I}{4\pi \ \rho } \overrightarrow{a_{\phi }} .

Case 2:-

when conductor is infinite in length   A is at  -\infty  and B at \infty  implies  Z_{1} = -\infty \Rightarrow \alpha _{1} = 180 ^{o}  and  Z_{2} = \infty \Rightarrow \alpha _{2} = 0 ^{o} .

\overrightarrow{H} = \frac{I}{2\pi \ \rho } \overrightarrow{a_{\phi }} .

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application of Ampere’s circuit law to infinite line current element

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

In order to determine \overrightarrow{H} 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

  1. The field \overrightarrow{H} is either tangential (or) Normal to the path at each point of the closed path.
  2. The magnitude of \overrightarrow{H} must be same at all points of the path where  \overrightarrow{H} is tangential.

Now,   \overrightarrow{H}  is given by  \overrightarrow{H} =H _{\rho } \overrightarrow{a}_{\rho }+H _{\phi } \overrightarrow{a}_{\phi }+H _{z} \overrightarrow{a}_{z} .

The path we are assuming is in the direction of \phi  so  \overrightarrow{dl} = dl \overrightarrow{a}_{\phi } .

\overrightarrow{dl} = \rho \ d\phi \overrightarrow{a}_{\phi } .

Ampere’s law is used to find out \overrightarrow{H}  at P 

i.e, from Ampere’s circuit law  \oint \overrightarrow{H} . \overrightarrow{dl} = I_{enc}  .

\oint \overrightarrow{H} . \overrightarrow{dl} = I .

\oint \overrightarrow{H} . \overrightarrow{dl} =\oint(H _{\rho } \overrightarrow{a}_{\rho }+H _{\phi } \overrightarrow{a}_{\phi }+H _{z} \overrightarrow{a}_{z}) . \rho \ d\phi \overrightarrow{a}_{\phi } .

=\oint (H _{\phi } \overrightarrow{a}_{\phi }) . \rho \ d\phi \overrightarrow{a}_{\phi }

=\int_{\phi = 0}^{2\pi } H _{\phi } \rho \ d\phi

= H _{\phi } \ \rho \ 2\pi .

from Ampere’s law      H _{\phi } \ \rho \ 2\pi = I .

H _{\phi } =\frac{ I}{\ 2\pi\ \rho } .

\therefore \overrightarrow{H _{\phi } } = \frac{ I}{\ 2\pi\ \rho } \overrightarrow{a_{\phi }} .

Ampere’s law is applied to find the value of \overrightarrow{H}  at any point P in it’s field.

 

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Ampere’s Circuit law

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

 i.e, \oint \overrightarrow{H} .\overrightarrow{dl} = I_{enclosed} .

This is similar to Gauss’s law and can be applied to determine \overrightarrow{H} 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  \overrightarrow{H}at  any point A is given by

\overrightarrow{H} = \frac{I_{enc}}{2\pi R} \overrightarrow{a}_{\phi } .

\overrightarrow{H}.\overrightarrow{dl} = \frac{I_{enc}}{2\pi R} \overrightarrow{a}_{\phi }.dl \overrightarrow{a}_{\phi } .

\overrightarrow{H}.\overrightarrow{dl} = \frac{I_{enc}}{2\pi R}.dl.

\overrightarrow{H}.\overrightarrow{dl} = \frac{I_{enc}}{2\pi R}.R d\phi .

\overrightarrow{H}.\overrightarrow{dl} = \frac{I_{enc}}{2\pi }d\phi .

\oint \overrightarrow{H} . \overrightarrow{dl} = \int_{\phi =0}^{2\pi } \frac{I_{enc}}{2\pi }d\phi .

\oint \overrightarrow{H} .\overrightarrow{dl} = I_{enc} .

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

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Optimum filter

The function of a receiver in a binary Communication system is to distinguish between two transmitted signals x_{1}(t)\ and \ x_{2}(t)  (or) (s_{1}(t)\ and \ s_{2}(t)) in the presence of noise.

The performance of Receiver is usually measured in terms of the probability of error Pe 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 Pe .

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 \frac{x_{o1}(T)+x_{o2}(T)}{2} .

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  y(t) = x_{o1}(t)+n_{o}(t) .

The output of sampler is  y(T) = \left\{\begin{matrix} x_{o1}(T)+n_{o}(T) \ for \ binary \ i/p \ '1'\\ x_{o2}(T)+n_{o}(T) \ for \ binary \ i/p \ '0' \end{matrix}\right.

suppose if Binary ‘1’ is transmitted then the input is x(t) = x_{1}(t) , to find the probability of error this transmitted ‘1’ should be received as ‘0’.

this is possible  when the condition  \left | y(T) \right | <\frac{x_{o1}(T)+x_{o2}(T)}{2} is true.

1 will be received as 0    \Rightarrow x_{o1}(T)+n_{o}(T) <\frac{x_{o1}(T)+x_{o2}(T)}{2} .

n_{o}(T) <\frac{x_{o2}(T)-x_{o1}(T)}{2} .

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

  \left | y(T) \right | >\frac{x_{o1}(T)+x_{o2}(T)}{2} .

    \Rightarrow x_{o2}(T)+n_{o}(T) >\frac{x_{o1}(T)+x_{o2}(T)}{2} .

n_{o}(T) >\frac{x_{o1}(T)-x_{o2}(T)}{2} .

the conditions are  summarized in the table

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

f(n_{o}(T)) = \frac{1}{\sigma \sqrt{2\pi }} e^{-\frac{n_{o}^{2}(T)}{2}} .

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

\therefore P_{e} =  area under the curve n_{o}(T) >\frac{x_{o1}(T)-x_{o2}(T)}{2}   (or) area under the curve n_{o}(T) <\frac{x_{o2}(T)-x_{o1}(T)}{2} .

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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

  1. Pre-alias filter must be used to limit the band width of the signal to f_{m}  Hz.
  2. Sampling frequency must be  f_{s}>2f_{m} .

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

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Reconstruction filter(Low Pass Filter)

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  f_{s}>2f_{m})

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 fm Hz.

If we pass the sampled signal through a LPF having cut-off frequency of  fm  Hz.

From sampling theorem  

g(t) = x(t).\delta _{T_{s}}(t).

g(t)=\frac{1}{T_{s}}\left \{ 1+2\cos \omega _{s}t+2\cos 2\omega _{s}t+2\cos 3\omega _{s}t+..... \right \}.

g(t)     has a multiplication factor  \frac{1}{T_{s}}. To reconstruct  x(t)  (or)  X(f) , the sampled signal must be passed through an ideal LPF of Band Width of  f_{m}   Hz and gain  T_{s} .

\left | H(\omega ) \right |=T_{s} \ for \ -\omega _{m}\leq \omega \leq \omega _{m}.

h(t) = \frac{1}{2\pi } \int_{-\omega _{m}}^{\omega _{m}}T_{s}e^{j\omega t}\ d\omega.

h(t) = 2f_{m}T_{s} \ sinc(2\pi f_{m}t).

If sampling is done at Nyquist rate , then Nyquist interval is  T_{s} = \frac{1}{2f_{m}} .

 therefore  h(t) = \ sinc(2\pi f_{m}t) .

h(t) = 0.      at all Nyquist instants  t= \pm \frac{n}{2f_{m}}  , 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) .

g(t) = x(kT_{s})\delta (t-kT_{s}) .

x(t) =\sum_{k} x(kT_{s})\ h (t-kT_{s}) .

x(t) =\sum_{k} x(kT_{s})\ sinc(2\pi f_{m} (t-kT_{s})).

x(t) =\sum_{k} x(kT_{s})\ sinc(2\pi f_{m}t-k\pi ) .

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  fs  .

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Region of Convergence (ROC)

The range of values of the complex variable s for which Laplace Transform X(S)=\int_{-\infty }^{\infty }x(t)e^{-st}\ dt 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, x(t)= \frac{1}{2\pi j}\int_{\sigma -j\infty }^{\sigma +j\infty }X(S) e^{St} \ ds .

The path of integration is along S-plane S = \sigma +j\omega that is along \sigma +j\omega with \omega varying from -\infty \ to \ \infty  and moreover , the path of integration must lie in the ROC for X(S).

for example the signal e^{-at}u(t) , this is possible if \sigma >-a  so the path of integration is shown in the figure

Thus to obtain x(t) = e^{-at}u(t)   from X(S) = \frac{1}{s+a}   , the integration is performed through this path for the function  \frac{1}{s+a} . 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 j\omega axis in the S-plane.

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

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 Re\left \{ s \right \} =\sigma _{o} is in the ROC, then all values of s for which Re\left \{ s \right \} > \sigma _{o} will also be in the ROC.

i.e, if the signal is x(t) = e^{-at}u(t)  right-sided [0 \ to \ \infty ]  then X(S) = \frac{1}{s+a}  for ROC : Re\left \{ s \right \} > -a .

6. If x(t) is left-sided and if the line Re\left \{ s \right \} =\sigma _{o} is in the ROC, then all values of s for which Re\left \{ s \right \} < \sigma _{o} will also be in the ROC.

7. If x(t) is two-sided and if the line Re\left \{ s \right \} =\sigma _{o} is in the ROC, then the ROC consists of a strip  in the s-plane that includes the line Re\left \{ s \right \} = \sigma _{o} .

for the both sided signal , the ROC lies in the region \sigma _{1} < Re\left \{ s \right \}<\sigma _{2} . This ROC is the strip parallel to j\omega  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.

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Analogy between Vectors and Signals

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 \overrightarrow{A}.

Let us consider two vectors \overrightarrow{V_{1}}  and \overrightarrow{V_{2}} . It is possible to find out the component of one vector along the other vector.

In order to find out the component of vector \overrightarrow{V_{1}} along  \overrightarrow{V_{2}} . Let us assume it as C_{12}V_{2} ,  which is only the magnitude.

how do we represent physically the component of one vector \overrightarrow{V_{1}} along  \overrightarrow{V_{2}} ? This is possible by finding the projection of one vector on to the other.

i.e, by drawing a perpendicular from \overrightarrow{V_{1}}   to   \overrightarrow{V_{2}}

\overrightarrow{V_{1}} = C_{12}\overrightarrow{V_{2}} + \overrightarrow{V_{e}} .

There exists two other possibilities.

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

\overrightarrow{V_{1}}.\overrightarrow{V_{2}}=V_{1}V_{2}\cos \theta .

If \theta is the angle between two vectors \overrightarrow{V_{1}}  and \overrightarrow{V_{2}} , the component of \overrightarrow{V_{1}}  along \overrightarrow{V_{2}} is

\frac{\overrightarrow{V_{1}}.\overrightarrow{V_{2}}}{\left | V_{2} \right |}=V_{1}\cos \theta-----EQN(1).

The component  of \overrightarrow{V_{1}}  along \overrightarrow{V_{2}} is C_{12}V_{2}----EQN(2).

\therefore (1) = (2) .

\frac{\overrightarrow{V_{1}}.\overrightarrow{V_{2}}}{\left | V_{2} \right |} = C_{12}V_{2} .

C_{12}=\frac{\overrightarrow{V_{1}}.\overrightarrow{V_{2}}}{V_{2}^{2}} .

If two vectors are orthogonal  \overrightarrow{V_{1}}.\overrightarrow{V_{2}} =0 .

i.e, C_{12} =0.

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  f_{1}(t)   and f_{2}(t), now to approximate  f_{1}(t)  in terms of f_{2}(t)  over  t_{1}< t<t_{2} .

f_{1}(t) \approx C_{12}f_{2}(t) .

\therefore f_{1}(t) \approx C_{12}f_{2}(t)+f_{e}(t) .

f_{e}(t) = f_{1}(t)-C_{12}f_{2}(t) .

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  f_{e}(t) is to choose minimize the average value of f_{e}(t) .

i.e, as    \frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}}dt.

But the process of averaging gives a false indication.

i.e, for example while approximating a function  \sin t  with a null function  f(t)=0  is

  f_{1}(t) =C_{12}f_{2}(t).

\sin t =0. \ \ 0\leq t\leq 2\pi.

indicates that  \sin t =0  during    0   to  2\pi   without any error 

i.e,  f_{e}(t) = f_{1}(t)-C_{12}f_{2}(t) .

f_{e}(t) = \sin t .

Average value of error is = \frac{1}{2 \pi } \int_{0}^{2\pi } \sin t \ dt =0 .

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 \epsilon = \frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}}f_{e}^{2}(t) \ dt .

\epsilon = \frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}}(f_{1}(t)-C_{12}f_{2}(t))^{2} \ dt.

 

To find value which keeps error minimum  \frac{d\epsilon }{dC_{12}}=0 .

C_{12} = \frac{\int_{t_{1}}^{t_{2}}f_{1}(t)f_{2}(t) \ dt}{\int_{t_{1}}^{t_{2}}f_{2}^{2}(t) \ dt} .

C_{12}  Which is similar to C_{12}=\frac{\overrightarrow{V_{1}}.\overrightarrow{V_{2}}}{V_{2}^{2}} where \int_{t_{1}}^{t_{2}}f_{1}(t)f_{2}(t) \ dt   denotes the inner product between two  Real signals

\therefore  For the orthogonality of two signals C_{12} =0 

\Rightarrow \ \int_{t_{1}}^{t_{2}}f_{1}(t)f_{2}(t) \ dt =0.

 

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Relation between Laplace and Fourier Transform

The Fourier transform  of a signal x(t) is given as 

X(j\omega ) = \int_{-\infty }^{\infty } x(t) e^{-j\omega t}dt----EQN(I)

Fourier Transform exists only if \int_{-\infty }^{\infty } \left | x(t) \right |dt< \infty  

we know that s=\sigma + j\omega 

X(S) = \int_{-\infty }^{\infty } x(t) e^{-s t}dt

X(S) = \int_{-\infty }^{\infty } \left | x(t)e^{-\sigma t} \right | e^{-j\omega t}dt----EQN(II)

if we compare Equations (I) and (II) both are equal when  \sigma =0.

i.e, X(S) =X(j\omega)| \right |_{s=j\omega } .

This means that Laplace Transform is same as Fourier transform when s=j\omega.

Fourier Transform is nothing but the special case of Laplace transform where  s=j\omega indicates the imaginary axis in complex-s-plane.

Thus Laplace transform is basically Fourier Transform on imaginary axis in the s-plane.

 

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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 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.