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author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
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committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3554/CH15/EX15.3 | |
parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
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initial commit / add all books
Diffstat (limited to '3554/CH15/EX15.3')
-rw-r--r-- | 3554/CH15/EX15.3/Ex15_3.sce | 23 |
1 files changed, 23 insertions, 0 deletions
diff --git a/3554/CH15/EX15.3/Ex15_3.sce b/3554/CH15/EX15.3/Ex15_3.sce new file mode 100644 index 000000000..6f43e0acf --- /dev/null +++ b/3554/CH15/EX15.3/Ex15_3.sce @@ -0,0 +1,23 @@ +// Exa 15.3
+
+clc;
+clear all;
+
+// Given data
+
+Fh= 2*10^3;// Cutoff frequency in Hz
+
+// Solution
+
+disp(" Let C2=C3=0.0033 micro farads ");
+
+// Fh=1/(2*%pi*R*C); where R=R2=R3 and C2=C3=C;
+C=0.0033*10^-6; // farads
+// Therefore
+R=1/(2*%pi*Fh*C);
+printf(' Calculated value of R = %.1f K ohms. Let, R=R2=R3=22 k Ohms is selected\n',R/1000);
+// Since Rf/R1=0.586, therefore Rf=0.586*R1;
+// Let fix value of R1 as
+R1=10*10^3; // Ohms
+Rf=0.586*R1;
+printf(' The remaining components after calculation comes out to be as Rf= %.2f k Ohms and R1= %d k Ohms \n',Rf/1000,R1/1000);
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