From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001
From: priyanka
Date: Wed, 24 Jun 2015 15:03:17 +0530
Subject: initial commit / add all books

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 991/CH20/EX20.9/Example20_9.sce | 16 ++++++++++++++++
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 create mode 100755 991/CH20/EX20.9/Example20_9.sce

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diff --git a/991/CH20/EX20.9/Example20_9.sce b/991/CH20/EX20.9/Example20_9.sce
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+//Example 20.9. refer fig.20.35(a).
+clc
+format(6)
+disp("(a) From Eq.(20.32), the time period, T = 2RC ln(R1+2R2 / R1)")
+disp("          T = 2RC ln(116*10^3 + 2*100*10^3/116*10^3)")
+disp("          T = 2RC ln(316*10^3/116*10^3)")
+disp("          T = 2RC        (since ln(316*10^3/116*10^3) = 1)")
+disp("Given     f = 1 kHz,  T = 1/f = 1 ms")
+disp("That is,   2RC = 1*10^-3 sec")
+disp("Therefore, the time constant RC = 0.5*10^-3 sec")
+r=(0.5)/0.01 // in k-ohm
+disp(r,"(b) With      C = 0.01 uF,  R(k-ohm) = 0.5*10^-3/0.01*10^-6 =")
+disp("(c) Maximum value of differential input voltage is")
+x=2*14*(100/(100+116))
+disp(x,"          2*Vsat*(R2 / R1+R2) = ")
+disp("Therefore, the peak values for the differential input voltage just exceed +-2 x 6.48 V")
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cgit