11 files changed, 270 insertions, 0 deletions
diff --git a/3554/CH15/EX15.1/Ex15_1.sce b/3554/CH15/EX15.1/Ex15_1.sce
new file mode 100644
index 000000000..dcdb5d451
--- /dev/null
+++ b/3554/CH15/EX15.1/Ex15_1.sce
@@ -0,0 +1,19 @@
+//Exa 15.1
+
+clc;
+clear all;
+
+
+// Given data
+Fh=2;// kHz
+Af=2;// Pass band gain
+
+// Solution
+
+disp(" Let C1= 0.01 micro farads ");
+C=0.01;//micro farads
+R=1/(2*%pi*Fh*C);// k Ohms
+printf(' The calculated value of R is %.3f K ohms. Nearest practical value for R1 is 8.2 k Ohms\n',R);
+//Af=1+Rf/R1;
+// As Af=2. So, Rf=R1
+disp(" In this case , Rf=R1= 10 k Ohms is selected ");
diff --git a/3554/CH15/EX15.10/Ex15_10.sce b/3554/CH15/EX15.10/Ex15_10.sce
new file mode 100644
index 000000000..e395a330e
--- /dev/null
+++ b/3554/CH15/EX15.10/Ex15_10.sce
@@ -0,0 +1,35 @@
+// Exa 15.10
+
+clc;
+clear all;
+
+// Given data
+// Second order inverting Butterworth low pass filter
+// Refering Table 15.1 and 15.3 in page no 517 and 538 respectively
+
+Af=6;// DC gain
+Fc=1.5;// KHz
+Q=10;
+
+// Solution
+
+disp(" According to Table 15.1, the inverting configurations would normally be used to give an inverting low pass output. However, to obtain a gain of 6, an inverting uncommitted opamp has to br used, hence the non-inverting filter configuration must be used.");
+
+// From table 15.4 given on page no 538
+R2=316/Q;
+R3=100/(3.16*Q-1);
+// R1 treated as open circuit
+printf(' \n The R1 is open while R2 and R3 are %.1f ,k Ohms %.1f k Ohms respectively \n',R2,R3);
+// From equations 15.54 given on page no 538 we get R4 and R5
+R4=(5.03)*10^7/(Fc*10^3);//Ohms
+R5=R4;
+printf(' \n The calculated value of R4=R5=%.2f k Ohms \n',R4/1000);
+disp("  use R4=R5=33 k Ohms");
+
+disp(" Let R6=1.8 K ohms");
+R6=1.8; // K ohms
+R7=R6*Af;
+R8=(1/R6 + 1/R7)^-1;
+
+printf(' The values of R6 and R7 are %.1f k Ohms, %.3f K ohms respectively \n',R6,R7);
+printf(' The value of R8 = %.3f k Ohms \n ',R8);
diff --git a/3554/CH15/EX15.11/Ex15_11.sce b/3554/CH15/EX15.11/Ex15_11.sce
new file mode 100644
index 000000000..e8e877d74
--- /dev/null
+++ b/3554/CH15/EX15.11/Ex15_11.sce
@@ -0,0 +1,30 @@
+// Exa 15.11
+
+clc;
+clear all;
+
+// Given data
+
+Fc=4;// kHz
+Q=8;
+
+// Solution
+
+disp(" The FLT-U2 can be used as a notch filter by summing the inverted output of the bandpass filter designed with the input signal by means of the uncommitted opamp.");
+
+// From table 15.3 given on page no 538
+R2=100;// k Ohms
+R3=100/((3.40*Q)-1);
+// R1 treated as open circuit
+printf(' The R1 is open while R2 and R3 are %.1f , %.2f K ohms respectively \n',R2,R3);
+
+// From equations 15.54 given on page no 538 we get R4 and R5
+R4=(5.03)*10^7/(Fc*10^3);
+R5=R4;
+printf(' The calculated value of R4=R5=%.2f k Ohms(12 k Ohms) \n',R4/1000);
+disp(" Let R6=R7=R8=10 K ohms ");
+R=10000;//R=R6=R7=R8=10 k Ohms
+R9=(1/R+1/R+1/R)^-1;
+printf(' The value of R9 =%.2f K ohms \n',R9/1000);
+disp(" The complete circuit diagram is shown in fig. 15.26 on page no. 541.");
+// The value of R3 vary due to round off error.
diff --git a/3554/CH15/EX15.2/Ex15_2.sce b/3554/CH15/EX15.2/Ex15_2.sce
new file mode 100644
index 000000000..3dfba4a17
--- /dev/null
+++ b/3554/CH15/EX15.2/Ex15_2.sce
@@ -0,0 +1,16 @@
+// Exa 15.2
+
+clc;
+clear all;
+
+// Given data
+
+Wc=20*10^3; // Angular cutoff frequency in rad/s
+C=0.01*10^-6; //in  farads
+
+// Solution
+
+// As Wc=1/(R*C);
+R=1/(Wc*C);
+
+printf(' The value of resistance required = %d k Ohms \n',R/1000);
diff --git a/3554/CH15/EX15.3/Ex15_3.sce b/3554/CH15/EX15.3/Ex15_3.sce
new file mode 100644
index 000000000..6f43e0acf
--- /dev/null
+++ b/3554/CH15/EX15.3/Ex15_3.sce
@@ -0,0 +1,23 @@
+// Exa 15.3
+
+clc;
+clear all;
+
+// Given data
+
+Fh= 2*10^3;// Cutoff frequency in Hz
+	
+// Solution
+
+disp(" Let C2=C3=0.0033 micro farads ");
+
+// Fh=1/(2*%pi*R*C); where R=R2=R3 and C2=C3=C;
+C=0.0033*10^-6; // farads
+// Therefore
+R=1/(2*%pi*Fh*C);
+printf('  Calculated value of R = %.1f K ohms. Let, R=R2=R3=22 k Ohms is selected\n',R/1000);
+// Since Rf/R1=0.586, therefore Rf=0.586*R1;
+// Let fix value of R1 as
+R1=10*10^3; // Ohms
+Rf=0.586*R1;
+printf('  The remaining components after calculation comes out to be as Rf= %.2f k Ohms and R1= %d k Ohms \n',Rf/1000,R1/1000);
diff --git a/3554/CH15/EX15.4/Ex15_4.sce b/3554/CH15/EX15.4/Ex15_4.sce
new file mode 100644
index 000000000..211767976
--- /dev/null
+++ b/3554/CH15/EX15.4/Ex15_4.sce
@@ -0,0 +1,16 @@
+// Exa 15.4
+
+clc;
+clear all;
+
+// Given data
+// Second order filter
+
+R=47*10^3; // Ohms(R2=R3=R)
+C=0.0022*10^-6; // farads(C2=C3=C)
+
+// Solution
+
+Fl=1/(2*%pi*R*C); //low cutoff frequency(Hz)
+printf(' The low cutoff frequency for a high pass filter =%.2f kHz\n',Fl/1000);
+
diff --git a/3554/CH15/EX15.5/Ex15_5.sce b/3554/CH15/EX15.5/Ex15_5.sce
new file mode 100644
index 000000000..5e95c3259
--- /dev/null
+++ b/3554/CH15/EX15.5/Ex15_5.sce
@@ -0,0 +1,36 @@
+// Exa 15.5
+
+clc;
+clear all;
+
+// Given data
+
+Fl=100;// lower cutoff frequency in Hz
+Fh=1000;// higher cutoff frequency in Hz
+Af=4;// pass band gain
+
+// Solution
+
+// Wide bandpass filter design
+// 1. For a low pass filter Fh=1 KHz =1/(2*%pi*R*C);
+
+disp(" For a low pass filter section");
+disp(" Let C1=0.01 micro farads ");
+C1=0.01;// micro farads
+R1=1/(2*%pi*Fh*C1*10^-6);
+printf('The value of resistor = %.1f K ohms \n',R1/1000);
+
+// 2. For a high pass filter Fl=100 Hz=1/(2*%pi*R*C);
+disp(" For a high pass filter section");
+disp(" Let C2=0.01 micro farads ");
+C2=0.01;// micro farads
+R2=1/(2*%pi*Fl*C2*10^-6);
+printf(' The value of resistor = %d K ohms \n',R2/1000);
+
+disp(" Since the pass band gain is 4, the gain of the high pass and low pass filter sections are set each equal to 2. Therefore, R1=Rf=10 K ohms for both sections.");
+
+// Q for filter
+Fc=sqrt(Fl*Fh);
+
+Q=Fc/(Fh-Fl);
+printf(' The value of Q =%.2f which is less than 10, as expected for a wide band pass filter\n',Q);
diff --git a/3554/CH15/EX15.6/Ex15_6.sce b/3554/CH15/EX15.6/Ex15_6.sce
new file mode 100644
index 000000000..92ce4e67b
--- /dev/null
+++ b/3554/CH15/EX15.6/Ex15_6.sce
@@ -0,0 +1,26 @@
+// Exa 15.6
+
+clc;
+clear all;
+
+//Given data
+// Refering fig. 15.17- Narrow band pass filter
+
+Fc=1; // kHz
+Q=5; //Quality factor
+Avo=8; //Voltage gain
+Fc1=1.5;//New centre frequency(kHz)
+
+// Solution
+
+disp(" Let C1=C2=C3=C(say)=0.01 micro farads");
+C=0.01;//micro farads
+// But
+R1=Q/(2*%pi*Fc*10^3*C*10^-6*Avo); // From eqn. 15.45 on page no.530
+R2=Q/(2*%pi*Fc*10^3*C*10^-6*(2*Q^2-Avo));// From eqn. 15.47 on page no. 530
+R3=Q/(%pi*Fc*10^3*C*10^-6); // From eqn. 15.46 on agr no. 530
+printf(' The Values of R1, R2 and R3 are %.3f k Ohms(approx 10 k Ohms), %.3f k Ohms(approx 2 k Ohms and %.3f k Ohms(aprox 159 k Ohms) respectively\n',R1/1000,R2/1000,R3/1000); 
+// To change Fc to Fc1 we simply have to change R2 to R21 given as 
+R21=2000*(Fc/Fc1)^2;// since R2=2 k Ohms(approx)
+printf('  The calculated value of new R2 to change Fc from 1 kHz to 1.5 kHz keeping Avo(Voltage gain) and BW constant is = %.2f ohms (approx 820 Ohms) \n',R21);
+
diff --git a/3554/CH15/EX15.7/Ex15_7.sce b/3554/CH15/EX15.7/Ex15_7.sce
new file mode 100644
index 000000000..ef83d5ce0
--- /dev/null
+++ b/3554/CH15/EX15.7/Ex15_7.sce
@@ -0,0 +1,32 @@
+// Exa 15.7
+
+clc;
+clear all;
+
+// Given data
+// Refering fig. 15.20- Wide band reject filter
+
+Fl=100;// Hz
+Fh=1000;// Hz
+
+// Solution
+// 1. For a high pass filter Fh=1 KHz =1/(2*%pi*R*C);
+disp(" For a high pass filter section");
+disp(" Let C1=0.01 micro farads ");
+C1=0.01;// micro farads
+R1=1/(2*%pi*Fh*C1*10^-6);
+printf('  The value of resistor = %.1f k Ohms \n',R1/1000);
+
+// 2. For a low pass filter Fl=100 Hz=1/(2*%pi*R*C);
+disp(" For a low pass filter section");
+disp(" Let C2=0.01 micro farads ");
+C2=0.01;// micro farads
+R2=1/(2*%pi*Fl*C2*10^-6);
+printf('  The value of resistor = %.1f k Ohms \n',R2/1000);
+
+disp(" Since the pass band gain is 4, the gain of the high pass and low pass filter sections are set each equal to 2. Therefore, R1=Rf=10 k Ohmss for both section");
+disp(" Further, the gain of the summing amplifier is set to 1, therefore R2=R3=R4=10 k Ohms"); // K ohms
+R=10000;//Ohms(R=R2=R3=R4)
+Rom=(1/R+1/R+1/R)^-1;
+printf('  The value of Rom = %.1f k Ohms\n',Rom/1000);
+// There is a printing mistake as c=0.1 micro fard is printed instead of 0.01 micro farad.
diff --git a/3554/CH15/EX15.8/Ex15_8.sce b/3554/CH15/EX15.8/Ex15_8.sce
new file mode 100644
index 000000000..b7e84a363
--- /dev/null
+++ b/3554/CH15/EX15.8/Ex15_8.sce
@@ -0,0 +1,19 @@
+// Exa 15.8
+
+clc;
+clear all;
+
+// Given data
+// Active notch filter
+ 
+Fn=50; //Notch out frequency(Hz)
+
+// Solution
+
+disp(" Let C=0.047 micro farads");
+C=0.047; // micro farads
+R=1/(2*%pi*Fn*C*10^-6);
+
+printf('  The value of R is calculated as %d k Ohms \n',round(R/1000));
+disp("For R/2, two 68 k Ohms resistors connected in parallel are used and for the 2C components, two 0.047 micro farad capacitors connected in parallel are used.");
+
diff --git a/3554/CH15/EX15.9/Ex15_9.sce b/3554/CH15/EX15.9/Ex15_9.sce
new file mode 100644
index 000000000..fadd5f844
--- /dev/null
+++ b/3554/CH15/EX15.9/Ex15_9.sce
@@ -0,0 +1,18 @@
+// Exa 15.9
+
+clc;
+clear all;
+
+// Given data
+// Refering Fig. 15.2(a)- All pass filter
+f=2.5;// Input frequency in kHz
+
+// Solution
+
+disp(" Let C=0.01 micro farads and R= 15 k Ohms");
+C=0.01;// micro farads
+R=15;// k Ohms
+Phi=2*atan(2*%pi*f*C*R); // phase angle in radians
+
+printf(' This means that the output voltage Vo has the same frequency and amplitude as the input voltage but lags it by - %d degrees\n',Phi*180/%pi);
+