AC Analysis by MATLAB

8/2/2019 AC Analysis by MATLAB

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AC Circuit Analysis


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Compute Z and for the circuit shown with MATLAB

Let the given network be represented as shown in Figure below

where Z1 = j13-j8 = j5 , and Z2 = 10 +j5 and Z3 = 20 j16


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In the circuit of Figure , and the symbols V and A inside the circles denote an AC voltmeter * and

ammeter respectively. Assume that the ammeter has negligible internal resistance. The variable capacitor C is adjusted until

the voltmeter reads 25 V and the ammeter 5 A. Find the value of the capacitor.

Solution:

Since the instruments read absolute values, we are only need to be

concerned the magnitudes of the phasor voltage, phasor current,

and impedance. Thus,


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For the circuit of Figure 1, find vc(t) if vs1 = 15 V, vS2 (t) = 20 cos 1000t V , and is(t) = 4cos2000t A.

Plot vc(t) using MATLAB

This circuit is excited by a DC (constant) voltage source, an AC (sinusoidal) voltage source, and an AC

current source of different frequency. Therefore, we will apply the superposition principle. Let V'C be the

capacitor voltage due to vS1 acting alone, VC the capacitor voltage due to v S2(t) acting alone, and V'Cthe capacitor voltage due to iS(t) acting alone. Then, the capacitor voltage due to all three sources acting

simultaneously will be V = V + V' + V''.

With the DC voltage source acting alone, after steady-state conditions have been reached the inductors

behave like short circuits and the capacitor as an open circuit and thus the circuit is simplified as below.

Also Voltage source short circuited while current source open circuited.

By the voltage division expression

Fig 1


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Next, with the sinusoidal voltage source vs2(t) acting alone the reactance's are

By KCL

With MATLAB

5

-j 2


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Finally with the sinusoidal current source is(t) acting alone the reactance's are

By KCL

With MATLAB

Current source and its parallel resistance have beenreplaced with a voltage source with a series resistor

V = Is x R = 4 x 5 = 20


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These waveforms are plotted below using the following MATLAB code:

wt = linspace(0,2*2*pi);

Vc1=5;

Vc2=3.97*cos(wt-62.9*pi/180); % degree converted in radians

Vc3 = 3.43*cos(2*wt-114.8*pi/180);

plot(wt,Vc1,'*',wt,Vc2,'+',wt,Vc3,'x',wt,Vc1+Vc2+Vc3,'--')


8/27

RL Circuit For an RL circuit, the voltage v(t) and i(t) are given as;

v(t) = 10cos(377t)

i(t) = 5cos(377t+60o) Sketch v(t) and i(t) for t = 0 to 20 milliseconds

% RL circuit

% current i(t) and voltage v(t) are generated; t is time

t = 0:1E-3:20E-3;

v = 10*cos(377*t);

ang = (60*pi/180); % angle in radians (60 is degrees)

i= 5*cos(377*t+ang);

plot(t,v,'*',t,i,'o')

title('voltage and current of an RL circuit')xlabel('sec')

ylabel('voltage (V) and current(mA)')

text(0.003,1.5,'v(t)'); % text command place v(t) on x and y axis respectively.

text(0.009,2,'i(t)')


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0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02-10

-8

-6

-4

-2

0

2

4

6

8

10voltage and current of an RL circuit

sec

voltage(V)and

current(mA)

v(t)i(t)


10/27

Compute and sketch all phasor voltages for the circuit of Figure below. Then, use MATLAB to

plot these quantities in the t-domain

We will begin by selecting as our reference as shown on the phasor

diagram of Fig.

and

1


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below


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G=[1,0,0;-1/2,1/2-1/5j+1/3j,-1/3j;0,-1/3j,1/3j+1/5];

% Enter all values of the I matrix

I=[1 0 0]';

% Compute node voltages

V=G\I; % Got the values of V(1), V(2), V(3)

VR1=V(1)- V(2);

VL=V(2)- V(3);

% Compute magnitudes and phase angles of voltages

magV1=abs(V(1));

magV2=abs(V(2));

magV3=abs(V(3));

phaseV1=angle(V(1))*180/pi;

phaseV2=angle(V(2))*180/pi; phaseV3=angle(V(3))*180/pi;

magVR1=abs(VR1);

phaseVR1=angle(VR1)*180/pi;

magVL=abs(VL);

phaseVL=angle(VL)*180/pi;

% Denote radian frequency as w and plot wt for 0 to2*pi range

wt=linspace(0,2*pi);

V1=magV1*cos(wt+phaseV1);



VR1t=magVR1*cos(wt+phaseVR1);

VLt=magVL*cos(wt+phaseVL);

% Convert wt to degrees

deg=wt*180/pi;

% Print phasor voltages, magnitudes, and phase angles

fprintf(' \n');

% With fprintf only the real part of each parameter is processed

so we will use disp

disp('V1 = '); disp(V(1)); disp('V2 = '); disp(V(2));

disp('V3 = '); disp(V(3));

disp('VR1 = '); disp(VR1);

disp('VL = '); disp(VL);

fprintf('magV1 = %4.2f V \t', magV1);

fprintf('magV2 = %4.2f V \t', magV2);

fprintf('magV3 = %4.2f V', magV3);

fprintf(' \n'); fprintf(' \n'); % To create a gap between mag and phase

fprintf('phaseV1 = %4.2f deg \t', phaseV1);

fprintf('phaseV2 = %4.2f deg \t', phaseV2);

fprintf('phaseV3 = %4.2f deg', phaseV3);

fprintf(' \n'); fprintf(' \n');

fprintf('magVR1 = %4.2f V \t', magVR1); fprintf('phaseVR1 = %4.2f deg ', phaseVR1);

fprintf(' \n'); fprintf(' \n');

fprintf('magVL = %4.2f V \t', abs(VL));

fprintf('phaseVL = %4.2f deg ', phaseVL);

fprintf(' \n');

plot(deg,V1,deg,V2,deg,V3,deg,VR1t,deg,VLt)

fprintf(' \n');

1


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>> phasorvoltages

V1 =

1

V2 =

0.7503 - 0.1296i V3 =

0.4945 - 0.4263i

VR1 =

0.2497 + 0.1296i

VL =

0.2558 + 0.2967i

magV1 = 1.00 V magV2 = 0.76 V magV3 = 0.65 V

phaseV1 = 0.00 deg phaseV2 = -9.80 deg phaseV3 = -40.76 deg

magVR1 = 0.28 V phaseVR1 = 27.43 deg

magVL = 0.39 V phaseVL = 49.24 deg

With these values we have

vs(t) = v1(t) = coswt

V2(t) = 0.76cos(wt-9.80)

V3(t) = 0.65cos(wt-40.80)

vR1(t) = 0.28cos(wt+27.40)

VL(t) = 0.39cos(wt+49.20)

Note: With command legend(V1,V2,V3,VR1,VL) we can put thelegend on plot

0 50 100 150 200 250 300 350 400

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

V1

V2

V3

VR1

VL


14/27

For the circuit of Figure 1, vs(t) = 100cos1000t V.

Compute the average power delivered (or absorbed) by each device

jwL = j103 x 3 x 10-3 = j3 andj/wC= -j /103 x2 x104= -j5

The phasor equivalent circuit is shown below where

Fig 1

Fig 2

Z3

2


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MATLAB codes


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0


17/27

Steady state AC power Figure shows an impedance with voltage

across it given by v(t) and current throughit i(t).The instantaneous powerp(t) = v(t) i(t) ------------(1)

If v(t) and i(t) are periodic

with period T, the rms or

effective values of the

voltage and current are

Where Vrmsand Irms is the rms value of v(t) and i(t) respectively.

The average power dissipated by

The one port network is


18/27

The power factorpfis given as

If both the current i(t) and

voltage v(t) are both sinusoidal, i.e.

The rms value of the voltage v(t) and current is

The average power P is

The power factor, pf, is

The reactive power Q is

The complex power, S, is


19/27

Integration

In order to find the integral

The general forms of quad functions that can be used to find q

are

h fi if d i


20/27

For the fig if and . Determine

Find the average power, rms value of v(t) and the power factor using (a) analytical

solution and (b) numerical solution

Function files


21/27

Function files

(Lower codes are for function files)

function vsq=voltage1(t) % voltage 1 This function is used to define the voltage function

vsq=(10*cos(120*pi*t+30*pi/180)).^2; %Square of voltage and changing degree into radian also

(Save the above codes with name voltage1 as function file)

function isq=current1(t)

% current1This function is used to define the current function

isq=(6*cos(120*pi*t+60*pi/180)).^2; %Square of current and changing degree into radian also

(Save the above codes with name current1 as function file)

function pt = inst_pr(t)

%inst_pr This function is used to define % instantaneous power obtained by multiplying sinusoidal voltage and current

it = 6*cos(120*pi*t + 60*pi/180);

vt = 10*cos(120*pi*t + 30*pi/180); pt = it.*vt;

(Save the above codes with name inst_pr as function file)

T=2*pi/(120*pi); % 2*pi is the period of the sine wave 120 * pi is the time period of v(t) and i(t)


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T=2 pi/(120 pi); % 2 pi is the period of the sine wave , 120 pi is the time period of v(t) and i(t)

a = 0; % lower limit of integration

b = T; % upper limit of integration

x = 0:0.02:1; % Array

t=x.*b; %Array x is multiplied to T and get points on time period

v_int = quad('voltage1',a,b); %integration of instantaneous value

v_rms = sqrt(v_int/b); % rms of voltage

i_int = quad('currentl',a,b); % integration of instantaneous value

i_rms = sqrt(i_int/b); % rms of current

p_int =quad('inst_pr',a,b); % instantaneous value

p_ave = p_int/b; % average power

pf = p_ave/(i_rms*v_rms); % power factor

% analytical solution

p_ave_an = (60/2)*cos(30*pi/180); %Average Power is Multiplication of Vrms and Irms with difference of angle , an represent analytical

v_rms_an = 10.0/sqrt(2);

pf_an = cos(30*pi/180);

% results are printed

fprintf('Average power, analytical %f\n Average power, numerical: %f\n',p_ave_an,p_ave) fprintf('rms voltage, analytical: %f \n rms voltage, numerical: %f \n', v_rms_an, v_rms)

fprintf('power factor, analytical: %f \n power factor, numerical: %f \n', pf_an, pf)

Average power, analytical 25.980762

Average power, numerical: 25.980762

rms voltage, analytical: 7.071068

rms voltage, numerical: 7.071068

power factor, analytical: 0.866025

power factor, numerical: 0.866025


23/27

In figure, showing an unbalanced wye-wye system,

find the phase voltages VAN, VBN and VCN

Using KVL, we can solve for I1, I2, I3

- -------(1)

---(2)

---(3)

Simplifying Equations 1,2 and 3, we have

-----(4)

------(5)

-------(6)

Expressing the above equations in matrix form

We have

The phase voltages can be obtained as

The above matrix can be written as

Polar form exponential form


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Matlab Codes

% This program calculates the phasor voltage of

% an unbalanced three-phase system

% Z is impedance matrix, V is voltage vector

% I is current vectorZ = [6-13*j 0 0;

0 4+2*j 0;

0 0 6-12.5*j];

v2= 110*exp(j*pi*(-120/180)); % volt v2 In Radians

v3 = 110*exp(j*pi*(120/180)); % In Radians

V = [110; v2; v3]; % column voltage vector

I =inv(Z)*V; % solve for loop currents

%calculate the phase voltages

Van =(5+12*j)*I(1);

Vbn = (3+4*j)*I(2);

Vcn = (5-12*j)*I(3);

Van _abs = abs(Van);

Van_ang = angle(Van)*180/pi;

Vbn _abs = abs(Vbn);

Vbn_ang = angle(Vbn)*180/pi;

Vcn_abs = abs(Vcn);

Vcn_ang = angle(Vcn)*180/pi;

% print out results

fprintf('phasor voltage Van,magnitude: %f \n phasor

voltage Van, angle in degree: %f \n', Van_abs, Van_ang)

fprintf('phasor voltage Vbn,magnitude: %f \n phasor

voltage Vbn, angle in degree: %f \n', Vbn_abs,

Vbn_ang)

fprintf('phasor voltage Vcn,magnitude: %f \n phasor

voltage Vcn, angle in degree: %f \n', Vcn_abs, Vcn_ang)

The following results were obtained:

phasor voltage Van, magnitude: 99.875532

phasor voltage Van, angle in degree: 132.604994 phasor voltage Vbn, magnitude: 122.983739 phasor

voltage Vbn, angle in degree: -93.434949

phasor voltage Vcn, magnitude: 103.134238 phasor

voltage Vcn, angle in degree: 116.978859

110L-120= 110ej(-120)= 110ej(-120 x pi/180)

Polar form exponential form

Example of Isolation transformer


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Example of Isolation transformerFor the circuit below, find the voltage ratio V 2 / V 1

Solution:

The dots are given to us as shown. Now, we arbitrarily assign currents I1 &12 and

we write mesh equations for each mesh

With this current assignments I 2 leaves the dotted terminal of the right mesh and

therefore the mutual voltage has a negative sign. Then,

Mesh 1:

R1I1+ jwL1I1 jwMI2 = V in

( 0.5 + j18.85) I1 j18.85I 2 = 120I_0o --------------------------------------(1)


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Mesh 2:

jwMI 1 + jwL2 I 2 + RLOAD I 2 = 0 % w = 377 r/s

Or

j18.85I1 + (500 + j37.7)12 = 0 ---------------(2)

We will find the ratio V 2 / V 1 using the MATLAB code below where

V 1 = j w LI I1 = j18.85 I1 and V2 = I2 Rload

MATLAB codes

Z=[0.5+18.85j -18.85j; -18.85j 500+37.7j];

V=[120 0]';

I=Z\V; %It will solve I(1) and I(2)

fprintf(' \n');

fprintf('V1 = %7.3f V\t', abs(18.85j*I(1))); %abs of V1

fprintf('V2 = %7.3f V \t', abs(500*I(2))); %abs of V2

fprintf('Ratio V2/V1 = %7.3f \t', abs((500*I(2))/(18.85j*I(1))))

MATLAB Answer

V1 = 120.093 V V2 = 119.753 V Ratio V2/V1 = 0.997


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Thus the magnitude of Vload is practically same as the magnitude of Vin.

However we suspect that Vload will be out of phase with Vin. We can find the

phase of Vload by adding the following statement to the MATLAB code

fprintf('Phase V2 = %6.2fdeg',angle(500*I(2))*180/pi)

Answer

Phase V2 = -0.64 deg

This is a very small phase difference from the phase of Vin and thus we see that

both magnitude and phase of Vload are essentially the same as that of Vin.

If we increase the load resistance Rload to 1kwe will find that again themagnitude and phase of Vloadare essentially the same as that ofVin. Therefore, the

transformer of this example is an isolation transformer, that is , it isolates the load

from the source and the value of Vin appear across the load even though the load

changes. An isolation transformer is also referred to as a 1:1 transformer.

If in a transformer the secondary winding voltage is considerably higher than theinput voltage, the transformer is referred to as a step-up transformer. Conversely, if

the secondary winding voltage is considerably lower than the input voltage, the

transformer is referred to as a step-down transformer.

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AC Analysis by MATLAB