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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /199/CH6/EX6.3/Example_6_3.sce | |
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1 files changed, 21 insertions, 0 deletions
diff --git a/199/CH6/EX6.3/Example_6_3.sce b/199/CH6/EX6.3/Example_6_3.sce new file mode 100755 index 000000000..2014dbb4f --- /dev/null +++ b/199/CH6/EX6.3/Example_6_3.sce @@ -0,0 +1,21 @@ +// Chapter6
+// Page.No-193, Figure.No-6.5(a)
+// Example_6_3
+// Components of peak amplifier
+// Given
+clear;clc;
+fp=16*10^3; // Peak frequency
+Af=10; // Gain at peak frequency
+C=0.01*10^-6; // Assume
+L=1/(((2*%pi*fp)^2)*10^-8); // Simplifying fp=1/(2*pi*sqrt(L*C))
+printf("\n Inductance is = %.4f H \n",L)
+L=10*10^-3; // Approximate
+R=30; // Assume the value of internal resistance of the inductor
+Xl=2*%pi*fp*L; // Inductive reactance
+Qcoil=Xl/R; // Figure of merit of the coil
+printf("\n Figure of merit of the coil is = %.1f \n",Qcoil)
+Rp=(Qcoil)^2*R; // Parallel resistance of the tank circuit
+printf("\n Parallel resistance of the tank circuit is = %.1f ohm \n",Rp)
+R1=100; // Assume the value of internal resisrance of the coil
+Rf=-Rp/(1-(Rp/(Af*R1))); // Simplifying Af=(Rf||Rp)/R1
+printf("\n Feedback resistance is = %.1f ohm \n",Rf)
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