PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB
290 Practical MATLAB® Applications for Engineers XC = -j./(w*C); XL = j*w*L; ZLC = XC.*XL./(XC+XL); ZRLC = R.*ZLC./(R+ZLC); % pa ...
Alternating Current Analysis 291 ANALYTICAL Solution The circuit of Figure 3.71 is transformed into the phasor circuit diagram s ...
292 Practical MATLAB® Applications for Engineers Note that the aforementioned matrix equations can be expressed as [Z] * [I] = [ ...
Alternating Current Analysis 293 fprintf(‘The current i1(t)=%fcos(10t %f)amps\n’,I1mag,I1ang) fprintf(‘The current i2(t)=%fcos(1 ...
294 Practical MATLAB® Applications for Engineers ANALYTICAL Solution The two node equations are for node V 1 : (2 − j)V 1 + jV 2 ...
Alternating Current Analysis 295 % Solution for nodal voltages V1 and V2 V = inv(Y)*I; % Solution for the magnitude and phase an ...
296 Practical MATLAB® Applications for Engineers Example 3.16 The objective of this example is to verify the maximum power trans ...
Alternating Current Analysis 297 Example 3.17 The objective of this example is to verify the maximum power transfer theorem, whe ...
298 Practical MATLAB® Applications for Engineers MATLAB Solution % Script file: max _ power _ AC _ RL % Max. power delivered to ...
Alternating Current Analysis 299 Example 3.18 Create the script fi le I_versus V_ R L C that analyzes the effect of changing R o ...
300 Practical MATLAB® Applications for Engineers subplot(2,2,1) plot(W,Z1,’*’,W,Z2,’s’,W,Z3,’o’) grid on legend(‘R=540 Ohms’, ‘R ...
Alternating Current Analysis 301 C = 0.01 μF V= 5^0 ° L = 30 mH R 1 = 47 kΩ R 2 = 270 Ω Vc I IC IL FIGURE 3.80 Network of Exampl ...
302 Practical MATLAB® Applications for Engineers XL = j*w.*L;ZS=R2+XL; ZP = (ZS.*XC)./(ZS+XC); ZT = R1+ZP; YT =1./ZT; I = V./ZT; ...
Alternating Current Analysis 303 title(‘angle(VC) vs. w’,’FontSize’,9) xlabel(‘frequency in w (rad/sec)’), ylabel(‘angle (degree ...
304 Practical MATLAB® Applications for Engineers 11 10.5 10 9.5 9 8.5 0 5 10 15 0 5 10 15 0 0 5 10 15 0 0.2 0.4 0.6 0.8 1 51015 ...
Alternating Current Analysis 305 Example 3.20 Let the current in the series RLC circuit diagram shown in Figure 3.84 be given by ...
306 Practical MATLAB® Applications for Engineers title(‘[i(t)^2] vs. t’); ylabel (‘Amplitude’); grid on subplot(2,2,3) inti=int( ...
Alternating Current Analysis 307 subplot(2,2,3) Pl = iamps*Vl; ezplot(Pl) title(‘pl(t) vs t’) xlabel(‘ time ‘) ylabel(‘power (wa ...
308 Practical MATLAB® Applications for Engineers The voltage drop across the RL (in volts) is: 100009 2 599997 3 ------ t - ---- ...
Alternating Current Analysis 309 × 105 × 10 −^5 × 105 − 5 − 2 time 0 5 −^5 − 1 1 1.5 −0.5 time 0 5 0 0.5 − 5 t 0 5 0 1 2 3 − 5 t ...
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