Working /Operation of NPN and PNP Transistor

NPN Transistor Working:-

 

For Normal operation of NPN Transistor Emitter junction JE is Forward Biased and Collector junction JC is Reverse Biased.

The applied Forward Biased at Emitter-Base junction injects a large number of electrons into the N-region and these electrons have enough energy to overcome the JE junction and enter into the very thin lightly doped Base region.

Since Base is very lightly doped very few electrons recombine with the holes in the P-type Base region and constitutes a small Base current IB in μA.

The electrons in the Emitter region are more when compared to electrons in the Collector. Only 5% (or) 1% of injected electrons combines with the holes in Base to produce Iand remaining 95% (or0 99% of electrons diffuse into Collector region due to extremely small thickness of Base.

Since Collector junction is Reverse-Biased a strong Electro-static field develops between Base and Collector. The field immediately collects the diffused electrons which enters Collector junction and are collected by the Collector(Positive electrode).

Thus injected electrons from Emitter reaches Collector constituting a current known as I_{E}=I_{B}+I_{C} Thus Emitter current is sum of Base current and Collector current. I_{B} is very small in the Base region.

Current directions are  always from negative to positive and Majority carriers are electrons in NPN Transistor.

NPN Transistor is preferred over PNP since the mobility of electron is more than that of hole that is electron moves faster than holes.

PNP Transistor Working:-

For Normal operation of PNP Transistor Emitter junction JE is forward Biased and Collector junction JC is reverse biased.

The applied FB at Emitter-Base junction injects a large number of holes in the P-type emitter region and these holes have enough energy to enter into very thin lightly doped Base region. Base is very lightly doped N-type region. Therefore very few holes combines with the Base region and constitutes a small Base current IB (in Micro Amperes).

The holes in the Emitter region are more when compared to holes in the collector region.Only 5% or 1% of injected holes from Emitter combines with the electrons in the Base to produce IB and remaining 95% (or) 99% of holes diffuse into Collector region  due to extremely small thickness of Base.

Since Collector junction is Reverse-Biased a strong Electro-static field develops between Base and Collector. The field immediately collects the diffused holes which enters Collector junction and are collected by the Collector(negative electrode).

Thus injected holes from Emitter reaches Collector constituting a current known as I_{E}=I_{B}+I_{C}I_{B} is very small in the Base region.

Majority carriers are holes in PNP Transistor.

 

Conductivity of a Semi conductor

In a pure Semi conductor number of electrons = number of holes. Thermal agitation (increase in temperature) produces new electron-hole pairs and these electron-hole pair combines produces new charge particles.

 

one particle is of negative charge which is known as free electron with mobility \mu _{n} another in with positive charge known as free hole with mobility \mu _{p}.

two particles moves in opposite direction in an electric field \overrightarrow{E} and constitutes a current.

The total current density (J) with in the semi conductor.

\overrightarrow{J} = \overrightarrow{J_{n}} + \overrightarrow{J_{p}}

Total conduction current density = conduction current density due to electrons + conduction current density due to holes.

J_{n}= nq\mu _{n}E.

J_{p}= pq\mu _{p}E.

n- number of electrons/Unit-Volume.

p-number of holes/Unit-Volume.

E- applied Electric field strength V/m.

q-charge of electron/hole \approx 1.6X10^{-19}C.

J = nq\mu _{n}E +pq\mu _{p}E.

J = (n\mu _n +p\mu _{p})qE.

J=\sigma E.

where \sigma = (n\mu _{n}+p\mu _{p})q is the conductivity of semi conductor.

Intrinsic Semi conductor:-

In an  intrinsic semi conductor n=p=n_{i}

\therefore conductivity \sigma _{i}= (n_{i}\mu _{n}+ n_{i}\mu _{p})q

\sigma _{i}= n_{i}(\mu _{n}+ \mu _{p})q

where J_{i} is the current density in an intrinsic semi conductor J_{i} = \sigma _{i} E

Conductivity in N-type semi conductor:-

In N-type n> > p 

number of electrons > > number of holes

\therefore \sigma _{N}\simeq n\mu _{n}q

J _{N}= n\mu _{n}q.

Conductivity in P-type semi conductor:-

In P-type p> > n

number of holes > > number of electrons

\therefore \sigma _{p} \approx p\mu _{p}q.

J_{P}= p\mu _{p}q E.

 

 

Effect of negative feedback on Band width of an Amplifier

 Let the Band Width of an amplifier without feedback is = BW. Band width of an amplifier with negative feed back is BW_{f}= BW (1+A\beta ). Negative feedback increases Band width.

Proof:-  Consider an amplifier with gain ‘A’

Now the frequency response of the amplifier is as shown in the figure. Frequency response curve means gain (dB) Vs frequency (Hz)

the frequency response of an amplifier consists of three regions

  1. Low frequency region (< f_{1} -lower cut off frequency).
  2. Mid frequency region ( between f_{1} and f_{2}).
  3. High frequency region ( the region > f_{2} -upper cutoff frequency)

Gain in low- frequency region is given asA_{vl} = \frac{A_{v}}{1-j\frac{f_{1}}{f}}---------EQN(I),

A_{v} -open loop gain,

f– frequency,

f_{1}– lower cut off frequency, where Gain in constant region is A_{v}.

Gain in High-frequency region is A_{vh} = \frac{A_{v}}{1+j\frac{f}{f_{2}}}  .

In low-frequency region:-

since open loop gain  in low-frequency region is A_{vl} and gain with feedback is A_{vlf} = \frac{A_{vl}}{1+A_{vl}\beta }

From EQN(I) A_{vl} = \frac{A_{v}}{1-j\frac{f_{1}}{f}}   after substituting A_{vl} in the above equation 

A_{vlf} = \frac{\frac{A_{v}}{1-j\frac{f_{1}}{f}}}{1+\frac{A_{v}}{1-j\frac{f_{1}}{f}}\beta }    

        =\frac{A_{v}}{1-j\frac{f_{1}}{f}+A_{v}\beta }

  = \frac{A_{v}}{1+A_{v}\beta-j\frac{f_{1}}{f} }

Now by dividing the whole expression with (1+A_{v}\beta )

A_{vlf}= \frac{\frac{A_{v}}{(1+A_{v}\beta )}}{\frac{1+A_{v}\beta-j\frac{f_{1}}{f}}{(1+A_{v}\beta )} }

A_{vlf}= \frac{\frac{A_{v}}{(1+A_{v}\beta )}}{1-j\frac{f_{1}}{f}\frac{1}{(1+A_{v}\beta )} }

A_{vlf} = \frac{A_{vf}}{1-j\frac{f_{1}^{'}}{f}}, where A_{vf} = \frac{A_{v}}{1+A_{v}\beta }  and f_{1}^{'} = \frac{f_{1}}{1+A_{v}\beta }

for example lower cut-off frequency f_{1} =20 Hz  implies f_{1}^{'} = \frac{20}{1+A_{v}\beta }  is decreasing with negative feedback.

In High-frequency region:-

Gain with out feed back in High frequency region is A_{vh} = \frac{A_{v}}{1+j\frac{f}{f_{2}}}

Now Gain with negative feed back is A_{vhf} = \frac{A_{vh}}{1+A_{vh}\beta }

Substituting A_{vh} in the above equation 

A_{vhf} = \frac{\frac{A_{v}}{1+j\frac{f}{f_{2}}}}{1+\frac{A_{v}}{1+j\frac{f}{f_{2}}}\beta }

A_{vhf}=\frac{A_{v}}{1+j\frac{f}{f_{2}}+A_{v}\beta }

A_{vhf}= \frac{A_{v}}{1+A_{v}\beta+j\frac{f}{f_{2}} }

Now by dividing the whole expression with (1+A_{v}\beta )

A_{vhf}= \frac{\frac{A_{v}}{(1+A_{v}\beta )}}{\frac{1+A_{v}\beta+j\frac{f}{f_{2}}}{(1+A_{v}\beta )} }

A_{vhf}= \frac{\frac{A_{v}}{(1+A_{v}\beta )}}{1+j\frac{f}{f_{2}}\frac{1}{(1+A_{v}\beta )} }

A_{vhf} = \frac{A_{vf}}{1+j\frac{f}{f_{2}^{'}}}, where A_{vf} = \frac{A_{v}}{1+A_{v}\beta }  and f_{2}^{'} = f_{2}(1+A_{v}\beta )

for example lower cut-off frequency f_{2} =20K Hz  implies f_{1}^{'} = 20 K (1+A_{v}\beta )  is increasing with negative feedback.

In Mid-frequency region:-

Gain with out feed back is A_{v}

and the gain with negative feed back is A_{vf} = \frac{A_{v}}{1+A_{v}\beta }

With out feedback With feedback
lower cut-off frequency  is f_{1} lower cut-off frequency f_{1}^{'}= \frac{f_{1}}{1+A_{v}\beta }, increases
upper cut-off frequency is f_{2} upper cut-off frequency is f_{2}^{'} = f_{2}(1+A_{v}\beta )
BW = f_{2}-f_{1} BW_{f} = f_{2}^{'}-f_{1}^{'} increases

Thus negative feedback decreases lower cut-off frequency and increases upper cut-off frequency.

\therefore Over all gain decreases with  negative feedback and Band Width increases.

 

Classification (or) topologies of feedback Amplifiers

There are 4 different combinations possible with negative feedback in Amplifiers as given below

 

  1. Voltage-Series.
  2. Current-Series.
  3. Voltage-Shunt.
  4. Current-Shunt.

The first part represents the type of sampling at the output .

  • i.e ,  Voltage- Shunt connection.
  • Current-Series connection.

and the second part represents the type of Mixing at the input

  • Series- Voltage is applied at the input.
  • Shunt-Current is applied at the input.

For any Amplifier circuit we require 

  • High Gain
  • High Band Width
  • High Input Impedance
  • and Low Output Impedance.

Classification of feedback Amplifiers is also known as feedback Topologies.

Voltage-Series feedback Connection:-

at i/p side connection is Series and at o/p side connection used is Shunt  since o/p is collected is voltage.

Series connection increases i/p impedance and Voltage at the o/p indicates a decrease in o/p impedance.

i.e, Z_{if} = Z_{i}(1+A\beta )    and    Z_{of} = \frac{Z_{o}}{(1+A\beta )}.

Current-Series feedback Connection:-

Series connection increases i/p impedance and Current at the o/p indicates an increase in o/p impedance.

i.e, Z_{if} = Z_{i}(1+A\beta )    and    Z_{of} = Z_{o}(1+A\beta ).

Voltage-Shunt feedback Connection:-

In this connection, both i/p and o/p impedance decreases .

i.e, Z_{if} = \frac{Z_{i}}{(1+A\beta )}    and    Z_{of} = \frac{Z_{o}}{(1+A\beta )}.

Current-Shunt feedback Connection:-

Shunt connection decreases i/p impedance and Current at the o/p indicates an increase in o/p impedance.

i.e, Z_{if} = \frac{Z_{i}}{(1+A\beta )}    and    Z_{of} = Z_{o}(1+A\beta ).

Effect of negative feedback on different topologies:-

Type of f/b Voltage gain

 

Band Width with f/b i/p impedance o/p
impedance
Voltage-Series decreases increases increases decreases
Current-Series decreases increases increases increases
Voltage-Shunt decreases increases decreases decreases
Current-Shunt decreases increases decreases increases

Similarly negative feedback decreases noise and harmonic distortion for all the four topologies.

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p style=”text-align: justify;”>Note:-  for any of the characteristics in the above table, increase ‘s shown by multiplying the original value with (1+A\beta ) and decrease ‘s shown by dividing with (1+A\beta ).

Colpitts oscillator

Colpitt’s  Oscillator is an excellent circuit and is widely used in commercial signal generators upto 100MHz.

 

It consists of a single-stage inverting amplifier and an LC phase shift Network.

The two capacitors C_{1} and C_{2} provides potential divider used for providing V_{f}C_{1} is the feedback element and which provides positive feedback required for sustained Oscillations.

The amplifier circuit is a self-Bias Circuit with R_{1} , R_{2} and parallel combination of R_{E} with C_{E}.

V_{CC} is applied through a resistor  R_{C} (or) RFC choke some times. This RFC choke offers very high impedance to high frequency currents.

R_{C} value has chosen in such a way that it offers high impedance. Two coupling Capacitors C_{C1} and C_{C2} are used to block d.c currents, that means they do not permit d.c currents into tank circuit.

These capacitors C_{C1} and C_{C2} provides a path from Collector to Base through LC Network.

when V_{CC} is switched on , a transient current is produced in the tank circuit an consequently damped oscillations are setup in the circuit.

The oscillatory current in the tank circuit produces a.c voltages across C_{1} and C_{2} . If terminal 1 is more positive w.r. to 2 , then voltages across C_{1} and C_{2} are opposite thus providing a phase shift of 180^{o} between 1 and 2. 

as the transistor is operating in CE mode , it provides a phase shift of 180^{o}.

Therefore the over all phase shift provided by the circuit results 360^{o} which is an essential condition for developing oscillations.

If the feedback is adjusted so that the loop gain A\beta =1 then then the  circuit acts as an Oscillator.

The frequency of oscillation depends on the tank circuit and is varied by gang (or) group tuning of C_{1} and C_{2} means C_{1}=C_{2}.

working:-

The capacitors C_{1} and C_{2} are charged by V_{CC} and are discharged through the coil L setting up of oscillations with frequency 

f_{o}=\frac{1}{2\pi }\sqrt{\frac{1}{L}(\frac{1}{C_{1}}+\frac{1}{C_{2}})}.

these oscillations across C_{1} are applied to the Base-Emitter junction  and the amplified version of output is collected across Collector (the frequency of amplifier output is same as that of input of the amplifier) .

This amplified energy is given back to tank circuit to compensate losses.

therefore un damped oscillations results in the circuit.

Derivation for frequency of oscillations:-

chose \left | A\beta \right |\geq 1 for sustained oscillations.

Analysis(Qualitative):-

if Z_{1} , Z_{2}  and Z_{3}  are pure reactive elements  such that Z_{1}=\frac{1}{j\omega C_{1}} =\frac{-j}{\omega C_{1}} ,  Z_{2}=\frac{1}{j\omega C_{2}} =\frac{-j}{\omega C_{2}}   and  Z_{3}=j\omega L.

from the general condition for an Oscillator 

\left | A\beta \right | =1  \Rightarrow h_{ie}(Z_{1}+Z_{2}+Z_{3})+Z_{1}Z_{2}(1+h_{fe})+Z_{1}Z_{3}=0.

h_{ie}(-\frac{j}{\omega C_{1}}-\frac{j}{\omega C_{2}}+j\omega L)+\frac{j^{2}}{\omega ^{2}C_{1}C_{2}}(1+h_{fe})-\frac{j}{\omega C_{1}}.j\omega L=0

find the real and imaginary parts,

-j(\frac{1}{\omega C_{1}}+\frac{1}{\omega C_{2}}-\omega L)h_{ie}-\frac{1}{\omega ^{2}C_{1}C_{2}}(1+h_{fe})+\frac{L}{C_{1}}=0

equating imaginary part to zero  (\frac{1}{\omega C_{1}}+\frac{1}{\omega C_{2}}-\omega L)=0  ,  since h_{ie}\neq 0 .

\frac{\omega C_{1}+\omega C_{2}}{\omega^{2} C_{1}C_{2}}=\omega L.

after simplification 

\omega ^{2}=\sqrt{\frac{1}{L}(\frac{1}{C_{1}}+\frac{1}{C_{2}})}.

by substituting \omega =2\pi f    results f_{o}=\frac{1}{2\pi }\sqrt{\frac{1}{L}(\frac{1}{C_{1}}+\frac{1}{C_{2}})}.

substituting the value of \omega ^{2}  in the real part gives h_{fe}=\frac{C_{2}}{C_{1}}  . this is the condition for sustained oscillations.

 

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continuity equation

The continuity equation of the current is based on the principle of conservation of charge that is  the charge can neither be created not destroyed.

consider a closed surface S with a current density , then the total current I crossing the surface S is given by the volume V

The current coming out of the closed surface is 

since the direction of current is in the direction of positive charges, positive charges also move out of the surface because of the current I.

According to principle of conversation of charge, there must be decrease of an equal amount of positive charge inside the closed surface.

therefore the time rate of decrease of charge with in a given volume must be equal to the net outward current flow through the closed surface of the volume.

By Divergence theorem 

    implies  

for a constant surface the derivative becomes the partial derivative 

   -this is for the whole volume.

for a differential volume 

 , which is called as continuity of current equation (or) Point form (or) differential form of the continuity equation.

This equation is derived based on the principle of conservation charge states that there can be no accumulation of charge at any point.

for steady  (dc) currents      

from . The total charge leaving a volume is the same as total charge entering it. Kirchhoff’s law follows this equation.

This continuity equation states that the current (or) the charge per second, diverging from a small volume per unit volume is equal to the time rate of decrease of charge per unit volume at every point.

 

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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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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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\lambda,  t \ 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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Initial-value & 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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Derivation of Z-Transform/ inverse Z- Transform

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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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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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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Complex Convolution Theorem :-

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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