Unit Step

>>t=0:.001:10;

>>y=1*(t>=1);

>>plot(t,y);

>>title('Unit Step Function');

Unit Impulse

>>t=0:.001:5;

>>y=1*(t==1);

>>plot(t,y);

>>title('Unit Impulse');

Unit Ramp

>>t=0:.001:10;

>>t0=1;

>>y=(t-t0).*(t>1);

>>plot(t,y);

>>title('Unit Ramp');

Fig1.1:Unit Step

Fig2.1:Unit Impulse

Fig3.1:Unit Ramp

Unit Pulse

>>t=0:.001:10;

>>T=.01;

>>t=0:T/100:2*T;

>>t1=rem(t,T);

>>y=1.*(t1>0).*(t1<=T2)+(1).*(t1>T2).*(t1<=T);

>>plot(t,y);

>>title('Unit Pulse');

Exponentialy Decaying Sinosoydal Signal(Continuous)

>>t=0:.001:15;

>>y=exp(-.3*t).*sin(3*t);

>>plot(t,y);

>>title('Exponential Decaying Sinosoidal Signal');

Exponentialy Decaying Sinosoydal

Signal(Discrete)

>>t=0:.3:15;

>>y=exp(-.3*t).*sin(3*t);

>>stem(t,y);

>>title('Exponential Decaying Sinosoidal Signal In Discrete Manner');

Fig4.1:Unit Pulse

Fig5.1:Exponentialy Decaying Sinosoydal Signal(Continuous)

Fig6.1:Exponentialy Decaying Sinosoydal Signal(Discrete)

Triangular Signal (Continuous)

>>T=.1;

>>m=.5;

>>t=0:T/1000:3*T;

>>t1=rem(t,T);

>>y=m*t1.*(t1>0).*(t1<T/2)+m*(T- t1).*(t1>=T/2).*(t1<T);

>>plot(t,y);

>>title('Triangular Signal');

Triangular Signal (Discrete)

>>T=.1;

>>m=.5;

>>t=0:T/10:3*T;

>>t1=rem(t,T);

>>y=m*t1.*(t1>0).*(t1<T/2)+m*(T- t1).*(t1>=T/2).*(t1<T);

>>stem(t,y);

>>title('Triangular Signal In Discrete Manner');

Fig7.1:Triangular Signal(Continuous)

Fig8.1:Triangular Signal(Discrete)

Experiment No :2 Date : 22/08/2003

Name Of The Experiment :Transient Analysis Of R-L, R-C, R-L-C Series Circuits

Object: To study the transient response of R-L, R-C, R-L-C series circuit applying different sources like step, pulse & sine function to draw the respective response curve for each circuit.

Theory: In R-L-C circuit, both electromagnetic & electrostatic energies are involved, hence any sudden change in the condition of circuit involves the redistribution of these two forms of energy. The transient current produced due to this redistribution is known as double energy transients. But for only R-L or R-C circuit there are only one energy redistribution.

In an R-L-C circuit,the transient voltage accross the three circuit parameters are i_tR, Ldi_t/dt & q_t/C. Hence the equation of the transient voltage is i_t+Ldi_t/dt+q_t/C=0

Differenciating the above equation & putting i_t for dq_t/dt, we get

It's solution is given by i_t=k₁e^l1+k₂ e^l2

where l₁= -- l₂= -

Depending on the value of l₁ & l₂ from different condition of the circuit are distinguishable.

Apparatus used:

a)R-L-C series branch: It consists of resistor, inductor & capacitor in series with adjustable magnitude.

Power System Blockset > Elements>R-L-C series branch.

b)Step Function: It supplies signal for a defenit time of constant magnitude after or before that time period output is zero.

Simulink> Sources>Step function.

c)Sine Function: It supplies sinusoidal signal of defined frequency & magnitude.

Simulink> Sources>Sine Wave.

d)Controled Voltage Source: Source: It is a voltage source which is controlled by function generators.

Power System Blockset >Electrical Source>Controlled votage source.

e)Current Measurement: It is used to measure the current.

Power System Blockset >Measurement>Current Measurement.

f)Scope: It is a component which is used to picturise the output current or voltage with respect to time & plays a very important role in transient analysis.

Simulink>Sinks>Scope.

g)Signal Rms: This block measure the rms value of instantenious current or voltage.

Power System Blockset >Extra Library>measurement>Rms.

h)Display: It shows numerical values of input to it.

Simulink>Sinks>Display.

i)Ground: It is used for ground connection .

Power System Blockset >Connects>Ground

us

Fig1.2: R-L Series Circuit For Step Input

Fig2.2: Scope Output of R-L Series Circuit Fig3.2: R-C Scope Output Series Circuit

R=2W; L=6H(Step Input) R=4W; C=1F(Step Input)

Fig4.2: R-C Series Circuit For Step Input

Fig5.2: R-L-C Series Circuit For Step Input

Fig6.2:Scope Output of R-L-C Series Circuit Fig7.2: Scope Output of R-L-C Series Circuit

R=5W; L=1H; C=1F(x>1)(Step Input) R=2W; L=1H; C=1F(x=1)(Step Input)

Fig7.2: Scope Output of R-L-C Series Circuit

R=1W; L=2H; C=1F(x<1)(Step Input)

Fig8.2: Scope Output of R-L Series Circuit Fig9.2: Scope Output of R-C Series Circuit

R=2W; L=6H(Sine Input) R=4W; C=1F(x<1)(Sine Input)

Fig5.2: R-L-C Series Circuit For Sine Input

Fig11.2: Scope Output of R-L-C Series Circuit

R=2W; L=6H;C=1F(Sine Input)

Experiment No :3 Date : 22/08/2003

Name Of The Experiment :Transient Analysis Of R-L, R-C, R-L-C Parallel Circuits

Object: To study the transient response of R-L, R-C, R-L-C parallel circuit applying different sources like step, pulse & sine function to draw the respective response curve for each circuit.

Theory:

<A>For Step Excitation (V=V_m*U(t)):-

1)Parallel R-C Network: In this case the transient response is given by V(t)=IR(e^-t/RC).

2)Parallel R-L Network: In this case the transient response is given by V(t)=IR(1-e^-Rt/L).

3)Parallel R-L-C Network: In this case the transient response is given by

V(t)=I/wC(e^-xwnt)sin(w_n)

where x = & w_n=

<B>For Step Excitation (V=V_msin(wt)):-

1)Parallel R-C Network: In this case the transient response is given by

v(t) =

where z = & f = tan^-1

2)Parallel R-L Network: In this case the transient response is given by

v(t) =

whwre z = & f = tan^-1

3)Parallel R-L-C Network: In this case the transient response is given by

v(t) =

where z = & f = tan^-1

Apparatus used:

a)R-L-C parallel branch: It consists of resistor, inductor & capacitor in parallel with adjustable magnitude.

Power System Blockset > Elements>R-L-C parallel branch.

b)Step Function: It supplies signal for a defenit time of constant magnitude after or before that time period output is zero.

Simulink> Sources>Step function.

c)Sine Function: It supplies sinusoidal signal of defined frequency & magnitude.

Simulink> Sources>Sine Wave.

d)Controled Current Source: Source: It is a current source which is controlled by function generators.

Power System Blockset >Electrical Source>Controlled Current source.

e)Voltage Measurement: It is used to measure the voltage.

Power System Blockset >Measurement>Voltage Measurement.

f)Scope: It is a component which is used to picturise the output current or voltage with respect to time & plays a very important role in transient analysis.

Simulink>Sinks>Scope.

g)Signal Rms: This block measure the rms value of instantenious current or voltage.

Power System Blockset >Extra Library>measurement>Rms.

h)Display: It shows numerical values of input to it.

Simulink>Sinks>Display.

i)Ground: It is used for ground connection .

Power System Blockset >Connects>Ground.

Circuit Diagram & Outputs:

Fig1.3: R-L-C Parallel Circuit For Step Input

Fig2.3: Scope Output of R-L Parallel Circuit Fig3.3: Scope Output of R-C Parallel Circuit

R=1W; L=1H(Step Input) R=1W; C=1F(Step Input)

Fig4.3: Scope Output of R-L-C Parallel Circuit Fig3.3: Scope Output of R-L Parallel Circuit

R=1W; L=2H; C=1F(Step Input) R=1W; L=2H(Sine Input)

Fig2.3: Scope Output of R-C Parallel Circuit Fig3.3: Scope Output of R-C Parallel Circuit

R=5W; C=1F(Sine Input) R=5W; L=1H;C=0.01F(Sine Input)

Mithun Kundu ...................................

Teacher's Signature

Experiment No :4 Date : 29/08/2003

Name Of The Experiment :Determination of Laplace & Inverse Laplace Transform

Object: To determine laplase & inverse laplace transform using MATLAB.

Theory: Let f(t) be the function defined for all positive values of `t' , then F(s) =

Provided the integral exists. And F(s) is called the laplace transform of f(t). It is denoted as

L[ f(t) ] = F(s) =

Similarly, if F(s) is the laplace transform of f(t) & denoted as

L[ f(t) ] = F(s) or f(t) = L^-1 F(s)

Thus f(t) is called the inverse laplace transform of F(s).

Where s is the complex frequency in frequency domain. Actually laplace transform change the domain of function f(t) from time domain `t' to frequency domain `s'.

Source Codes & Outputs:

(1)Laplace transform of f(t)=e^atsin(wt)

» syms s t w a ;

»x=exp(a*t)*sin(w*t);

»disp(x);

exp(a*t)*sin(w*t)

»z=laplace(x);

»disp(z);

w/((s-a)^2+w^2)

(2)Laplace transform of f(t)=1/a²(1-e^-at-ate^-at )

» syms s t a ;

» x=(1/a^2)*(1-exp(-a*t)-a*t*exp(-a*t));

» disp(x);

1/a^2*(1-exp(-a*t)-a*t*exp(-a*t))

» z=laplace(x);

» disp(z);

1/a^2*(1/s-1/(s+a)-a/(s+a)^2)

(3)Inverselaplace transform of F(s)=1/(s+9)

» syms s t ;

» z=1/(s+1);

» disp(z);

1/(s+1)

» x=ilaplace(z);

» disp(x);

exp(-t)

» x1=0:.01:10;

» col=1;

» for t=0:.01:10

x1(col)=eval(x);

col=col+1;

end;

» t=0:.01:10;

» plot(t,x1,'k');

Fig4.1 : Plot of (3) Fig4.2 : Plot of (4)

(4)Inverselaplace transform of F(s)=5(s+2)/s²(s+1)(s+3)

» syms s t ;

» z=(5*(s+2))/(s^2*(s+1)*(s+3));

» disp(z);

(5*s+10)/s^2/(s+1)/(s+3)

» x=ilaplace(z);

» disp(x);

10/3*t-25/9+5/2*exp(-t)+5/18*exp(-3*t)

» x1=0:.01:10;

» col=1;

» for t=0:.01:10

x1(col)=eval(x);

col=col+1;

end;

» t=0:.01:10;

» plot(t,x1,'k');

Experiment No :5 Date :29/08/2003

Name Of The Experiment :Representation Of Poles & Zeros In Z-Plane.

Object: To represent the poles & zeros of a network's transfer function in Z-plane using MATLAB.

Theory: All network trannsform functions have the form of quotent of polynomials in s as following

N(s) = p(s)/q(s) = (a₀sⁿ+a₁s^n-1 +a₂s^n-2+......+a_n-1 s+a_n)/(b₀s^m+b ₁s^m-1+b₂s^m-2 +...+b_m-1s+b_m)

where a & b are the real & positive coefficients for passive network. now the equation p(s) = 0, has n roots, & the equation q(s) = 0, has m roots. Both p(s) & q(s) may be written as a product of linear factors involving this roots,

N(s) = H{(s-z₁)(s-z₂)(s-z₃ )......(s-z_n)}/{(s-p₁)(s-p ₂)(s-p₃)......(s-p_n)}

wher H=a₀/b₀ is a constant known as the scale factor & z₁, z₂, z₃,.....,z_n; p₁, p₂, p₃,....., p_m are complex frequencies.

Zeros: When the variables has the values z₁, z₂, z₃,....., z_n for which the network transform functin N(s) vanishes i.e. no output signal is obtained, because the numerator becomes `0'. Such complex frequencies are known as zeros of the network.

Poles: When the variables has the values p₁, p₂, p₃,....., p_m the network transform function becomes infinite i.e. output signal of magnitude infinity is obtained. Such complex frequencies which make the denominator of N(s) `0' are called the poles of the network.

In the equation (1) factors (s-z_j) are known as zero factors & factors (s-p_j) are known as pole factors. Poles & Zeros are useful in ndescribing network functions. Poles & Zeros are critical frequencies.

Fig 5.2 : Poles & zeros

Condition of stable & unstable system

(1)A network having real & complex poles & zeros will be stable if the real parts of the poles & zeros are negetive, i.e. poles & zeros are lying in the left half of the s-plane.

(2)A network having only real or complex (i.e. poles & zeros are lying on the axis of the s-plane) poles & zeros will be stable unless the input frequency function is equall to the networks transfer function.

(3)A network having real & complex poles & zeros will be unstable if the real part of the poles & zeros are

positive, i.e. poles & zeros are lying in the right half of the s-plain

Sourse Codes & Outputs:

(1)Calculation for poles & zeros in s-plane of F(s)=(5s³+3s²+5s+8)/(s³ +7s²+9s+5)

» B=[5 3 5 8];

» A=[1 7 9 5];

» sys1=tf(B,A)

Transfer function:

5 s^3 + 3 s^2 + 5 s + 8

-----------------------

s^3 + 7 s^2 + 9 s + 5

» p=pole(sys1)

p =

-5.5379

-0.7311 + 0.6070i

-0.7311 - 0.6070i

» z=zero(sys1)

z =

0.2337 + 1.2018i

0.2337 - 1.2018i

-1.0674

» pzmap(sys1);

(2)Calculation for system's transfer function having poles & zeros in s domain at p=-1,-2,3 & z=4,-8,9

» z=[-1 -2 3];

» p=[4 -8 9];

» k=[1];

» sys1=zpk(z,p,k)

Zero/pole/gain:

(s+1) (s+2) (s-3)

-----------------

(s-4) (s-9) (s+8)

» [B,A]=residue(z,p,k)

B =

1 -5 -29 192

A =

1 -5 -68 288

» sys2=tf(B,A)

Transfer function:

s^3 - 5 s^2 - 29 s + 192

------------------------

s^3 - 5 s^2 - 68 s + 288

»t=0:.01:1;

»impulse(sys2,t,'k');

coutput signal of input signal (using impulse & step input as well as input equal to system transfer function) on a system.

»r=[2 3 5];

»p=[1 2 4];

»k=[3];

»sys3=zpk(r,p,k)

Zero/pole/gain:

3 (s-2) (s-3) (s-5)

-------------------

(s-1) (s-2) (s-4)

»t=0:.01:1;

»impulse(sys3,t,'k');

»step(sys3,t,'k')

»impulse(sys3*sys3,t,'k');

Fig 5.5 : Pulse Response of sys3 (3) Fig 5.6 : Impulse Response of sys3*sys3 (3)

Fig 5.2 : Poles & zeros of (1)

Fig 5.4 : Impulse Response of sys3 (3)

Fig 5.3 : Impulse Response of sys2 (2)

Experiment No :6 Date : 05/09/2003

Name Of The Experiment :To Evaluate the Convolution Integral Of Two Specified Functions.

Object: To find the outpyt response y(t) of a network for the input waveform given by the function x(t) by performing the convolution integral using MATLAB.

Theory: Let two function f1(t) & f2(t) be laplace transformable,having the transform F1(s) &F2(s).The product of these two in the laplace transform of f(t) which result from the convolution of f1(t) &f2(t) as given by teh equation :

F(t) = L ^-1 [ F1(s) . F2(s) ] = =

where = t a dummy variable of t.

The in the above equation is known as convolution integral.

The convolution of f1(t) & f2(t) is denoted by a special notation , F(t)=f1(t)*f2(t).So we can write,

F(s) = L ^-1 [ f₁(t) * f₂(t) ] = L ^-1 [f₂(t)*f₁(t)] = F₁(s)*F₂(s)

Thus the inverse laplace transform of the product of F₁(s) & F₂(s).

APPLICATIONS: One important application of the convolution integral is to evaluate the output response of the network to an arbitrary input in terms of the impulse response of the network.So if the impulse of a given network be known to be h(t), the output v₂(t) for the input voltage v₁(t) is found by convolution integral to be same as given by the 1st equation.This convolution integral can be performed easily using MATLAB & hence the output response of any network for any given input function can be find out provided the impulse response h(t) for the network is known.

Sourse Codes & Outputs:

(1)Plot theoutput of two signals for a=10*sin(2*pi*20*t) & b=3*exp(-3*t);

»ts=.01;

»tp=.5;

»t=0:ts:tp;

»p=size(t);

»a=10*sin(2*pi*20*t);

»b=3*exp(-3*t);

»y=ts*conv(a,b);

»y1=y(:,1:p(2));

»plot(t,y1,'k');

(2)Plot theoutput of two signals for a = 3*exp(-2*t) & b=t .

»t=0:.02:2.98;

»a=ones(1,150);

»b=3*exp(-2*t);

»y=.02*conv(a,b);

»m=1:150;

»y1=y(1,m);

»plot(t,y1,'k');

Experiment No :8 Date : 19/09/2003

Name Of The Experiment :Ttransfer response of high pass, band pass and low pass filter.

Object: To find the transfer response of high pass, band pass and low pass filter.

Theory: The purpose of a filter network is to transmit or pass a desired frequency band without loss & stop or completely at attenuate al undesired frequencies. The propagation constant

a=a+ j* b

Where a is the attenuation constant & b is the phase constant. If a=0, then there is no attenuation in transmission. There is only a phase shift, & obviously mod (I1)=mod (I2). The operation is in the pass band of frequencies. When a > 0, mod (I2)<mod (I1), i.e. attenuation has occurred & the operation is in the stop band.

The mathematical relationship is: sinhg/2=(z1/4z2)^1/2

We can write

Sinh(a+ j*b/2) = sinh(a/2+j*b/2)

Sinh(g/2)=sinh(a/2)cos(b/2)+j*cosh( a/2)sin(b/2)=(z1/4z2)^1/2

Low Pass Filter:

Let Z₁=j*w*L & Z₂= -j/w*C. Then Z₁Z₂=L/C.

The pass band will be in range 0<Z₁/-4Z₂<1, i.e. the pass band starts at the point Z₁=-4Z₂. Obviously, the pass band starts at f=0 & continues to some higher frequency fc, the cut-off frequency. So it is a low pass filter. The cut-off frequency fc can be determined as Z₁= -4Z₂

jwA=-4/jwC=4j/wC

fc=1/p(LC)^0.5

sinh(g/2)=(-w^2LC/4)^0.5=jw((LC)^0.5)/2=j*f/fc.

So, in the pass band of frequency i.e. f<fc, so that in _1<Z₁/4Z₂<0 i.e. in 0<Z₁/-4Z₂<1we get f/fc<1 a=0, b=2sin Inv (f/fc) & in the attenuation or stop band of frequency i.e. f>fc, so that in -inf<Z₁/4Z₂<-1 or 1<Z₁/-4Z₂<inf we get f/fc>1, a=2cosh Inv (f/fc), b=p.

High pass Filter:

Let Z₁=-j/wC & Z₂=jwL. Then Z₁Z₂=k² & the filter design will be of const `k' type. The cut-off frequency fc at the point at which Z₁=-4Z₂, with a pass band from the cut-off frequency to infinity. All frequency below the cut-off frequency lies in an attenuation or stop band. The network is thus called a high pass filter. In the range -inf <_Z₁/4Z₂< -1, the stop band exit. Hence, the two cut-off frequency are given by the relation Z₁=4Z₂ =0 & Z₁/-4Z₂=1.

The limit Z₁/4Z₂=0 yields Z₁=0 or Z₂=inf. The limit Z₁/-4Z₂=1 yields fc=1/4p(LC)^0.5. Thus, with opposite type of reactance, the pass band exits fc to all higher frequencies up to infinity, & the attenuation or stop band exits from 0 to fc. Obviously the filter blocks low frequencies & passes high frequencies, which is the characteristic of a high-pass filter.

Now, in the range _1<Z₁/4Z₂<0, attenuation constant a=0, the pass band exits. The propagation const

sinh (g/2)=(Z1/4Z2)^0.5=(-1/4w^2LC)^0.5=-j/2w(LC)^0.5 =-j fc/f, as a=0 & cosh (a/2)=1. Therefore,

sin (b/2)=-fc/f & a=0.At f=fc, sin (b/2)=-1 >>b=(2n-1)p; n=0,1,2,3,4,5

At f=INF, b=0.

Thus, in the frequency range fc<f<INF, the pass band exits as the attenuation const a=0. The phase const b at f=fc is -p & gradually increases as f increases till it reaches 0 at f=INF.

Band pass Filter:

Band pass filter passes a band of frequencies & attenuate frequencies on both sides of the pass band. The series arm is made up of a series resonant frequency fc0. The shunt arm is an anti-resonant circuit having anti-resonant frequency fa. The reactance curves of series arm & shunt arm are shown in fig, where arbitrarily fa>fc. Depending on the value of (Z1/-4Z2), over the whole frequency range of the reactance curve, the pass & attenuation bands exits accordingly.

1. When Z₁ & Z₂ are of opposite type of reactance

For 1<Z₁/-4Z₂<INF, attenuation or stop band exits

For 0<Z₁/-4Z₂<INF, pass band exits

2.When Z₁& Z₂ are of same nature of reactance, stop band exits.

3.When Z₁ & Z₂ are of opposite type of reactance

For 0<Z₁/-4Z₂<1, pass band exits

For 1<Z₁/-Z₂<INF, stop band exits

The condition for the band pass is that of equal resonant frequencies, i.e. fa=fc0 for which w^2L₁C₁=1=w²L ₂C₂

The impedance of arms are Z₁=j (wL₁-1/ wC₁), Z₂=jwL₂ (-j/wC₂)/j (wL₂-1/wC₂)

For constant k type filter at cut-off frequency, Z₁=-4Z₂, Z₁^2=-4Z₁Z₂=-4Rk^2, Z₁=-j2Rk

Hence, at lower cut-off frequency f1 =-Z₁ at upper cut-off frequency f2 i.e.

1/ w₁-w₁L₁=w₂ L₁-1/w₂C₁

1-w₁²L₁C₁ =(w₁/w₂)( w₂²L₁C₁ -1)

Again as w₀²=1/L₁C₁

1-(f1/f0)²=(f1/f2)[(f2/f0)²-1]

f0=(f1f2)^0.5

The resonant frequency of individual arm should be the geometric mean of the two cut-off frequencies f1 & f2. The cut-off rate is slow and Z0 varies appreciably in the pass band. Hence, impedance is not possible.

Circuit diagram & Output Graphs:

(1)Desgine a high pass filter of resistance 600W & cut off frequency of 500 Hz.

Fig 8.1: High Pass Filter

(2)Desgine a low pass filter of resistance 500W & cut off ferquency 500Hz.

Fig 8.2: Low Pass Filter

(3)Desgine a Band pass filter For the range 10kHz to 30kHz.

Fig 8.3: Band Pass Filter

Fig 6.1

Fig 6.2

Experiment No :9 Date : 14/11/2003

Name Of The Experiment : To Determine The Impedence (Z) & Admittance (Y) Parameter Of Two Port Network

Object: To determine The Z (Z₁₁, Z₁₂, Z₂₁, Z₂₂) parameters & Y (Y₁₁, Y₁₂, Y₂₁, Y₂₂) parameters of a two port network using MATLAB.

Theory: A two port network is a special case of multiport network. Each part consists of two terminals. One is for input & other is for output. From teh defination of a port, the current at the input is equal to current at the output.

In any two port network, there are 4 variables, namely the current (I₁) & voltage (V₁) at the input and current (I₂) & voltage (V₂) at the output. There are however other voltage & currents that might be identified in the network, but they are available for measurement. Now only two among the 4 variables are independent & the specification of any two determines the remain two. The dependence of 2 of the 4 variables on the other 2 is described in a number of ways, depending on which of the variables are shown to be the independent variables.

OPEN CIRCUIT IMPEDENCE PARAMETERS:

In this case the voltage V₁ & V₂ are expressed in terms of I₁ & I₂ i.e. I₁ & I₂ are independent variables.

\ (V₁ , V₂) = f (I₁ , I₂)

i.e. [V] = [Z] [I] where Z is the impedence.

or, =

or, V₁ = Z₁₁ I₁+ Z₁₂ I₂ .............. (1)

V₂ = Z₂₁ I₁+ Z₂₂ I₂ ............... (2)

From equation (1) & (2) we get the Z parameters as follows :

Z₁₁= |_{= 0} The input dtiving point impedence with output part open circuited.

Z₂₁= |_{= 0} Forward transfer impedence with output port open circuited.

Z₁₂= |_{= 0} Reverse transfer impedence with input port open circuited.

Z₂₂= |_{= 0} The output point impedence with input port open circuited.

So we see that to find the Z parameters either the input on output port must be kept open circuited & so they are also called OPEN CIRCUIT PARAMETER.

The condition for reciprocity is: Z₁₂ = Z₂₁

& the condition for symmetry is: Z₁₁ = Z₂₂

Short circuit admittance Parameter

In case current I1 and I2 are expressed in terms of V₁ & V₂ i.e. V₁ & V₂ are the independent variable here,

(I₁ , I₂)=f ( V₁, V₂)

i.e. [I] = [Y] [V] where Y is the admittance.

or, =

or, I₁ = Y₁₁ V₁+ Y₁₂ V₂ ......... (3)

& I₂ = Y₂₁ V₁+ Y₂₂ V₂ ......... (4)

From equation (3) & (4) we get Y parameters :

Y₁₁= |_{= 0} Input driving point admittance with output shorted.

Y₁₂= |_{= 0} Reverse transfer admittance with input shorted .

Y₂₁= |_{= 0} Forward transfer admittance with output shorted .

Y₂₂= |_{= 0} Output driving point admittance with input shorted.

These parameters also called short circuit parameters.

Condition for reciprocity is Y₁₂ = Y_21.

Condition for symmetry is Y₁₁ = Y_22.

Circuit diagram & Output Graphs:

Fig 9.1: Circuit for calculating Z12 & Z22 (admittance parameter) of two-port network

Fig 9.2: Circuit for calculating Z11 & Z21 (admittance parameter) of two-port network

Fig 9.3: Circuit for calculating Y11 & Z21 (admittance parameter) of two-port network

Fig 9.4: Circuit for calculating Y12 & Y22 (admittance parameter) of two-port network

Experiment No :7 Date : 12/09/2003

Name Of The Experiment :Fourier Transform & Fourier Series.

Object: To find discrete Fourier transform for periodic & non-periodic signals.

Theory: If a periodic sequence of number {f_k}^N-1 of period N is given, then the discrete Fourier transform of the sequence Fn for n=0,1,....,N-1 defined by

Fn = e ^-2*p*n*k/N , where i =

But in matlab we use a zero or negetive index, so the sequecne is {fk}^N-1 & the discrete Fourier transform is comuted as

F_n= e^^{2*pi*i*(n-1)(k-1)/N} , where k=1,2,...,N.

It,s relation with Fourier transform is as follows:

This definition is related to the F.T. of a casual signal f(t) which is effectively zero for t>T as,

1>Sample time T_S is defined ie. T_S=T/N.

2>Sample points are defined as t_k=kT_S for k=0,1,,(N-1).

3>Then we defined f_k=f(t_k).

4>We define the frequency sampling points w_n=2*pi*n/T.

So, F(w_n)=, for n=0,,N-1.

For | x | >T, F(w_n)=_, for n=0,,N-1.

Sourse Codes & Outputs:

(1)FOURIER TRANSFORM:

syms t;syms w;

z1=int(0*exp(-i*w*t),t,-inf,-3)+int(1*exp(-i*w*t),t,-3,3)+int(0*exp(-i*w*t),t,3,inf);

fwarray=-10:.1:10;coll=1;

for w=-10:.1:10

fwarray(coll)=eval(abs(z1));

coll=coll+1;

end;

w=-10:.1:10;

plot(w,fwarray);

Fig 7.1: Discrete Fourier Transform

(2)FOURIER SERIES:

syms n;syms t;

T=4;m=2;w0=2*pi/T;

an=(2/T)*((int((m*t)*cos(n*w0*t),t,0,T/2)+int(m*(T-t)*cos(n*w0*t),t,T/2,T)));

bn=(2/T)*((int((m*t)*sin(n*w0*t),t,0,T/2)+int(m*(T-t)*sin(n*w0*t),t,T/2,T)));

a0=(1/T)*((int((m*t),t,0,T/2)+int(m*(T-t),t,T/2,T)));

y=0;

for n=1:30

y=y+eval(an)*eval(cos(n*w0*t))+eval(bn)*eval(sin(n*w0*t));

end;

y=y+a0;

yrow=0:.1:10;col=1;

for t=0:.1:10

yrow(col)=eval(y);

col=col+1;

end;

t=0:.1:10;

plot(t,yrow);

Fig 7.2: Fourier Series

_K=0

k=0