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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /991/CH20/EX20.9 | |
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initial commit / add all books
Diffstat (limited to '991/CH20/EX20.9')
| -rwxr-xr-x | 991/CH20/EX20.9/Example20_9.sce | 16 |
1 files changed, 16 insertions, 0 deletions
diff --git a/991/CH20/EX20.9/Example20_9.sce b/991/CH20/EX20.9/Example20_9.sce new file mode 100755 index 000000000..7a2d9aaeb --- /dev/null +++ b/991/CH20/EX20.9/Example20_9.sce @@ -0,0 +1,16 @@ +//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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