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/CH4/EX4.3/Ex4_3.sce | 59 ++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 59 insertions(+)
 create mode 100644 3682/CH4/EX4.3/Ex4_3.sce

(limited to '3682/CH4/EX4.3')

diff --git a/3682/CH4/EX4.3/Ex4_3.sce b/3682/CH4/EX4.3/Ex4_3.sce
new file mode 100644
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+++ b/3682/CH4/EX4.3/Ex4_3.sce
@@ -0,0 +1,59 @@
+// Exa 4.3
+
+clc;
+clear;
+
+// Given data
+
+// An op-amp differentiator
+fa = 100; // Hz
+Vpp = 1; // Volts
+
+// Solution
+
+printf('Select, Fa = fmax = 100 Hz.\n');
+printf(' Let, C1 = 0.1 µF.\n');
+C1 = 0.1*10^-6; // Farads
+// since Fa = 1/(2*%pi*Rf*C1);
+// Therefore,
+Rf = 1/(2*%pi*fa*C1);
+printf(' Therefore, the calculated value of Rf = %.1f kΩ. \n',Rf/1000);
+
+printf(' Select, fb = 10*Fa = 1000 Hz.\n');
+fb = 1000; // Hz
+// Therefore
+R1 = 1/(2*%pi*fb*C1);
+printf(' The calculated value of R1 = %.2f kΩ. \n',R1/1000);
+// Since, RfCf = R1C1
+// Therefore we get,
+Cf = R1*C1/Rf;
+printf(' The calculated value of Cf = %.2f µF. \n',Cf*10^6);
+
+printf('\n\n For a sinusoidal input - \n\n');
+disp("since, vi = sin(2*%pi*100*t),  ");
+disp("From Eq. 4.69, vo = -Rf*C1* d/dt(vi), ");
+disp("Above equation yield following result once solved-  vo = -cos(2*%pi*100*t).");
+printf('\n The input and output waveforms are shown in Graphic window 0 ans 1 respectively. \n\n');
+// plotting wave forms
+
+t = [0:%pi:13*%pi];
+figure(0);
+
+a=gca(); // Handle on axes entity
+a.x_location = "origin"; 
+a.y_location = "origin"; 
+plot2d(t,sin(2*%pi*100*t));
+title('Sine-wave-input',"color","Red","fontsize",3);
+figure(1);
+
+a=gca(); // Handle on axes entity
+a.x_location = "origin"; 
+a.y_location = "origin"; 
+plot2d(t,-cos(2*%pi*100*t));
+title('Cosine-wave-output',"color","blue","fontsize",3);
+
+printf('\n For a square wave input -\n\n');
+printf('\n For a square wave input, say 1 V peak and 1 KHz,\n The output waveform will consist of positive and negative spikes of magnitude Vsat\n which is approximately 13 V for ± 15 V op-amp power supply.\n\n');
+printf(' During the timeperiods for which input is constant at ±  1V, the differentiated output will be zero. \n However, when input transits between ±1V levels, \n the slope of the input is infinite for an ideal square wave. \n\n The output, therefore, gets clipped to about ± 13V for a ± 15 V op-amp power supply.');
+
+printf('\n\n The output of a square wave input is a spike output as shown in Fig. 4.22(b). \n');
-- 
cgit