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| author | priyanka | 2015-06-24 15:03:17 +0530 |
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| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /174/CH6/EX6.3/example6_3.sce | |
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diff --git a/174/CH6/EX6.3/example6_3.sce b/174/CH6/EX6.3/example6_3.sce new file mode 100755 index 000000000..750ee3a64 --- /dev/null +++ b/174/CH6/EX6.3/example6_3.sce @@ -0,0 +1,32 @@ +// To find the maximum time
+// Modern Electronic Instrumentation And Measurement Techniques
+// By Albert D. Helfrick, William D. Cooper
+// First Edition Second Impression, 2009
+// Dorling Kindersly Pvt. Ltd. India
+// Example 6-3 in Page 144
+
+
+clear; clc; close;
+
+// Given data
+R = 100*(10^3); // Value of resistance in ohm
+C = 0.1*(10^-6); // The value of integrating capacitor in F
+V_ref = 2; // The reference voltage in V
+V_out = 10; // The maximum limit of the output in V
+
+//Calculations
+T = R*C;
+printf("The integrator time constant = %0.3f s\n",T);
+V_s = V_ref/T; //Unit is V/s
+V = 1/V_s;
+printf("Therefore the integrator output = %0.3f s/V",V)
+disp('Therefore to integrate 10V');
+T_max = V*V_out; //The max time the ref voltage can be integrated
+printf("The time required = %0.4f s",T_max);
+
+//Result
+// The integrator time constant = 0.010 s
+// Therefore the integrator output = 0.005 s/V
+// Therefore to integrate 10V
+// The time required = 0.0500 s
+
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