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author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
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committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3682/CH2/EX2.1 | |
parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
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initial commit / add all books
Diffstat (limited to '3682/CH2/EX2.1')
-rw-r--r-- | 3682/CH2/EX2.1/Ex2_1.sce | 20 |
1 files changed, 20 insertions, 0 deletions
diff --git a/3682/CH2/EX2.1/Ex2_1.sce b/3682/CH2/EX2.1/Ex2_1.sce new file mode 100644 index 000000000..9e450dd8e --- /dev/null +++ b/3682/CH2/EX2.1/Ex2_1.sce @@ -0,0 +1,20 @@ +// Exa 2.1
+
+clc;
+clear;
+
+// Given data
+
+// An amplifier( Refer Fig. 2.5(a) )
+Acl = -10; // Closed loop gain
+Ri = 10 * 10^3; // Input resistance of amplifier(Ω)
+
+// Solution
+
+// Since it is mentioned to design an amplifier, it means to calculate values for Rf(Feedback resistance) and R1.
+disp("Referring Fig. 2.5(a), we choose R1 as 10 kΩ i.e equal to input resistance of amplifier.");
+R1 = Ri;
+// Acl = -1 * Rf/R1;
+// Therefore;
+Rf= - Acl * Ri;
+printf(' The calculated value of Rf(Feedback resistane) is Rf = %d kΩ. \n',Rf/1000);
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