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4 files changed, 91 insertions, 0 deletions

diff --git a/3554/CH17/EX17.1/Ex17_1.sce b/3554/CH17/EX17.1/Ex17_1.sce
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+// Exa 17.1

+

+clc;

+clear all;

+

+// Given data

+// A 5 bit resistive divider

+

+n=5;// since 5 bit resistive divider

+Ip1=[1 1 0 1 1];// Digital input 1(1st element of array is MSB)

+Ip2=[1 0 1 1 0];//Digital input 2(1st element of array is MSB)

+V1=10;// Voltage corresponding to binary 1

+V0=0;// Voltage corresponding to binary 0

+

+// Solution

+

+LSB_weight=1/(2^n - 1);

+printf('The LSB weight = %.4f \n ',LSB_weight);

+LSB2_weight=2^(2-1)/(2^n-1);

+printf('The 2nd LSB weight = %.4f \n ',LSB2_weight');

+LSB3_weight=2^(3-1)/(2^n-1);

+printf('The 3rd LSB weight = %.4f \n ',LSB3_weight');

+LSB_op=V1*LSB_weight;// Change in output voltage due to change in LSB

+printf('The change in output voltage due to change in LSB = %.4f V \n ',LSB_op);

+LSB2_op=V1*LSB2_weight;

+printf('THe 2nd LSB causes a change in output voltage of %.4f V \n ',LSB2_op);

+LSB3_op=V1*LSB3_weight;

+printf('THe 3rd LSB causes a change in output voltage of %.4f V \n ',LSB3_op);

+Va=(V1*2^4+V1*2^3+V0*2^2+V1*2^1+V1*2^0)/(2^n-1);// output voltage for digital Ip1

+Vb=(V1*2^4+V0*2^3+V1*2^2+V1*2^1+V0*2^0)/(2^n-1);

+printf('The output voltage for a digital input 1 and 2 are %.2f V and %.3f V respectively \n ',Va,Vb);

diff --git a/3554/CH17/EX17.2/Ex17_2.sce b/3554/CH17/EX17.2/Ex17_2.sce
new file mode 100644
index 000000000..2141dcc59
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+++ b/3554/CH17/EX17.2/Ex17_2.sce
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+

+// Exa17.2

+

+clc;

+clear all;

+

+// Given data

+

+n=5;// 5 bit ladder

+V=10;// For binary 1

+

+// Solution

+// refering table 17.4(page no. 615)-Various Output voltage for corresponding MSB

+

+disp("The output voltage for each bit is as follows:");

+disp("");

+for i=1:n

+MSB(i)=V/2^i; //voltage corresponding to MSB i

+printf('  %d MSB          Va = V/2^%d = %.4f V \n ',i,i, MSB(i));

+end

diff --git a/3554/CH17/EX17.3/Ex17_3.sce b/3554/CH17/EX17.3/Ex17_3.sce
new file mode 100644
index 000000000..00703a724
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+++ b/3554/CH17/EX17.3/Ex17_3.sce
@@ -0,0 +1,18 @@
+// Exa 17.3

+

+clc;

+clear all;

+

+// Given data

+

+Vin=5;// Input voltage(Volts)

+Rin=2.5;// k Ohms

+Rf=1;//k Ohms

+

+// Solution

+

+Iin=Vin/Rin;//Input current(mA)

+If=Iin;

+Vout=-If*Rf;

+

+printf('The output voltage = %d Volts \n',Vout);

diff --git a/3554/CH17/EX17.4/Ex17_4.sce b/3554/CH17/EX17.4/Ex17_4.sce
new file mode 100644
index 000000000..db85b5236
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+++ b/3554/CH17/EX17.4/Ex17_4.sce
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+// Exa 17.4

+

+clc;

+clear all;

+

+// Given data

+Vref=5;//Reference voltage(V)

+R=5;// k Ohms

+

+// Solution

+

+disp("From fig. 17.18(c) , for a 4-bit D/A converter I=Vref/R* (D3+D2*2^-1+D1*2^-2+D0*^-3)");

+//16-input combinations are as follows

+Ip={[0 0 0 0];[0 0 0 1];[0 0 1 0];[0 0 1 1];[0 1 0 0];[0 1 0 1];[0 1 1 0];[0 1 1 1];[1 0 0 0];[1 0 0 1];

+[1 0 1 0];[1 0 1 1];[1 1 0 0];[1 1 0 1];[1 1 1 0];[1 1 1 1]};//[D3 D2 D1 D0 bits]

+

+disp("  Input Bits       Output Current(mA)    percent Fraction of maximum ");

+for i=1:16

+Iout(i)=Vref/R * (Ip(i,1)+Ip(i,2)*2^-1+Ip(i,3)*2^-2+Ip(i,4)*2^-3);

+

+printf('   %d  %d  %d  %d             %.3f            %.3f     \n',Ip(i,1),Ip(i,2),Ip(i,3),Ip(i,4),Iout(i),(Iout(i)/1.875)*100);//1.875(mA) is the highest output current

+end