From 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 Mon Sep 17 00:00:00 2001
From: prashantsinalkar
Date: Tue, 10 Oct 2017 12:27:19 +0530
Subject: initial commit / add all books

---
 3682/CH7/EX7.2/Ex7_2.sce | 36 ++++++++++++++++++++++++++++++++++++
 1 file changed, 36 insertions(+)
 create mode 100644 3682/CH7/EX7.2/Ex7_2.sce

(limited to '3682/CH7/EX7.2')

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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