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Diffstat (limited to '3682/CH7/EX7.2/Ex7_2.sce')
| -rw-r--r-- | 3682/CH7/EX7.2/Ex7_2.sce | 36 |
1 files changed, 36 insertions, 0 deletions
diff --git a/3682/CH7/EX7.2/Ex7_2.sce b/3682/CH7/EX7.2/Ex7_2.sce new file mode 100644 index 000000000..d1c1cd537 --- /dev/null +++ b/3682/CH7/EX7.2/Ex7_2.sce @@ -0,0 +1,36 @@ +// Exa 7.2
+
+clc;
+clear;
+
+// Given data
+
+n=4; // Fourth order Butterworth low-pass filter
+fH=1000; // Hz
+
+// Solution
+
+printf('Let C = 0.1 μF. \n');
+C=0.1*10^-6; // Farads
+// Since fH = 1/(2 * %pi * R*C);
+// Therefore;
+R = 1/(2*%pi*fH*C);
+printf(' The calculated value of R = %.1f kΩ. \n',R/1000);
+
+printf(' From Table 7.1, for n=4, we get two damping factors namely,\n alpha1 = 0.765 and alpha2 = 1.848.');
+alpha1=0.765;
+alpha2=1.848;
+A01 = 3-alpha1;
+A02 = 3-alpha2;
+printf('\n');
+printf('\n Then the pass band gain A01 = %.3f and A02 = %.3f. \n',A01,A02);
+printf('\n');
+printf(' The transfer function of the normalized second order low-pass Butterworth filter is 2.235 1.152 ');
+printf('\n ---------------- * ------------------');
+printf('\n Sn^2+0.765*Sn+1 Sn^2+1.848*Sn+1 ');
+
+// Since A01= 1 + Rf/Ri = 1 + 1.235;
+printf('\n Since A01= 2.235 so Let Rf1 = 12.35 kΩ and Ri1 = 10 kΩ to make A01 = 2.235.' );
+printf('\n Since A02= 1.152 so Let Rf2 = 15.20 kΩ and Ri1 = 100 kΩ to make A01 = 1.152.' );
+
+printf(' \n The circiuit realized is as shown in Fig. 7.7 with component value as mentioned above.');
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