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diff --git a/Linear_Control_Systems_by_B_S_Manke/1-INTRODUCTION.ipynb b/Linear_Control_Systems_by_B_S_Manke/1-INTRODUCTION.ipynb new file mode 100644 index 0000000..737b769 --- /dev/null +++ b/Linear_Control_Systems_by_B_S_Manke/1-INTRODUCTION.ipynb @@ -0,0 +1,467 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1: INTRODUCTION" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6_10: final_value.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Caption:final_value\n", +"// example 1.6.10\n", +"//page 13\n", +"syms t s;\n", +"F=4/(s^2+2*s)\n", +"x=s*F\n", +"x=simple(x)\n", +"z=limit(x,s,0);//final value theorem\n", +"z=dbl(z);\n", +"disp(z,'f(0+)=')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6_1_iii: inverse_laplace_transform.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Caption:inverse_laplace_transform\n", +"// example 1.6.1.(iii)\n", +"//page 8\n", +"// F(s)=1/(s^2+4s+8)\n", +"s =%s ;\n", +"syms t ;\n", +"disp(1/(s^2+4*s+8),'F(s)=')\n", +"f=ilaplace(1/(s^2+4*s+8),s,t)\n", +"disp (f,' f(t)=')//result" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6_1_ii: inverse_laplace_transform.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Caption:inverse_laplace_transform\n", +"// example 1.6.1.(ii)\n", +"//page 7\n", +"// F(s)=s+6/(s(s^2+4s+3))\n", +"s =%s ;\n", +"syms t ;\n", +"[A]= pfss((s+6)/(s*(s^2+4*s+3))) // partial fraction of F(s)\n", +"A(1)=2/s;\n", +"F1 = ilaplace (A(1),s,t)\n", +"F2 = ilaplace (A(2),s,t)\n", +"F3 = ilaplace (A(3),s,t)\n", +"F=F1+F2+F3;\n", +"disp (F,' f (t)=')//result" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6_1_i: inverse_laplace_transform.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Caption:inverse_laplace_transform\n", +"// example 1.6.1.(i)\n", +"//page 7\n", +"// F(s)=1/(s*(s+1))\n", +"s =%s ;\n", +"syms t ;\n", +"[A]=pfss(1/((s)*(s+1))) // partial fraction of F(s)\n", +"F1 = ilaplace (A(1),s,t)\n", +"F2 = ilaplace (A(2),s,t)\n", +"F=F1+F2;\n", +"disp (F,' f(t)=')//result" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6_1_iv: inverse_laplace_transform.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Caption:inverse_laplace_transform\n", +"// example 1.6.1.(iv)\n", +"//page 8\n", +"// F(s)=s+2/(s^2+4s+6)\n", +"s =%s ;\n", +"syms t ;\n", +"disp((s+2)/(s^2+4*s+6),'F(s)=')\n", +"F=ilaplace((s+2)/(s^2+4*s+6),s,t)\n", +"disp (F,' f(t)=')//result" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6_1_v: inverse_laplace_transform.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Caption:inverse_laplace_transform\n", +"// example 1.6.1.(v)\n", +"//page 8\n", +"// F(s)=5/(s(s^2+4s+5))\n", +"s =%s ;\n", +"syms t ;\n", +"[A]= pfss (5/(s*(s^2+4*s+5))) // partial fraction of F(s)\n", +"F1= ilaplace (A(1),s,t)\n", +"F2= ilaplace (A(2),s,t)\n", +"F=F1+F2;\n", +"disp (F,'f(t)=')//result" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6_1_vi: program_laplace_transform.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Caption:program_laplace_transform\n", +"//example 1.6.1.(v)\n", +"//page 9\n", +"//this problem is solved in two parts because in this problem pfss function donot work.So, First we find partial fraction using method as we do in maths and then secondly we find inverse laplace transform as usual.\n", +"// partial fraction \n", +"s=%s\n", +"syms t;\n", +"num=(s^2+2*s+3);\n", +"den=(s+2)^3;\n", +"g=syslin('c',num/den);\n", +"rd=roots(den);\n", +"[n d k]=factors(g)\n", +"a(3)=horner(g*d(1)^3,rd(1))\n", +"a(2)=horner(derivat(g*d(1)^3),rd(1))\n", +"a(1)=horner(derivat(derivat(g*d(1)^3)),rd(1))\n", +"//inverse laplace\n", +"// partial fraction will be: a(1)/(s+1)+a(2)/((s+2)^2)+a(3)/((s+2)^3)\n", +"F1 = ilaplace (1/d(1),s,t)\n", +"F2 = ilaplace (-2/(d(1)^2),s,t)\n", +"F3 = ilaplace (2*1.5/(d(1)^3),s,t)\n", +"F=F1+F2+F3\n", +"disp (F,' f(t)=')//result" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6_2: solution_of_differential_equation.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Caption:solution_of_differential equation\n", +"// example 1.6.2\n", +"//page 9\n", +"//after taking laplace transform and applying given condition, we get :\n", +"//X(s)=2s+5/(s(s+4))\n", +"s=%s;\n", +"syms t\n", +"[A]=pfss((2*s+5)/(s*(s+4)))\n", +"A(1)=1.25/s\n", +"F1 =ilaplace(A(1),s,t)\n", +"F2 = ilaplace(A(2),s,t)\n", +"f=F1+F2;\n", +"disp (f,'f(t)=')//result" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6_3: solution_of_differential_equation.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Caption:solution_of_differential_equation\n", +"// example 1.6.3\n", +"//page 10\n", +"//after taking laplace transform and applying given condition, we get :\n", +"//X(s)=1/(s^2+2s+2)\n", +"s=%s;\n", +"syms t\n", +"f = ilaplace(1/(s^2+2*s+2),s,t);\n", +"disp (f,'f(t)=')//result" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6_4: solution_of_differential_equation.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Caption:solution_of_differential_equation\n", +"// example 1.6.4\n", +"//page 10\n", +"//after taking laplace transform and applying given condition, we get :\n", +"//Y(s)=(6*s+6)/((s-1)*(s+2)*(s+3))\n", +"s=%s;\n", +"syms t\n", +"[A]=pfss((6*s+6)/((s-1)*(s+2)*(s+3)))\n", +"F1 = ilaplace(A(1),s,t)\n", +"F2 = ilaplace(A(2),s,t)\n", +"F3 = ilaplace(A(3),s,t)\n", +"F=F1+F2+F3;\n", +"disp (F,'f(t)=')//result" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6_5: initial_value.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Caption:initial_value\n", +"// example 1.6.5\n", +"//page 11\n", +"//I(s)=(C*s/(RCs+1))*E(s)\n", +"//given: E(s)=100/s,R=2 megaohm ,C=1 uF\n", +"// so, I(s)=(((1*10^-6)*s)/(2*s+1))*(100/s)\n", +"syms t\n", +"p=poly([0 10^-6],'s','coeff');\n", +"q=poly([1 2],'s','coeff');\n", +"r=poly([0 1],'s','coeff');\n", +"F1=p/q;\n", +"F2=1/r;\n", +"F=F1*F2\n", +"f=ilaplace(F,s,t);\n", +"z=limit(f,t,0);//initial value theorem\n", +"z=dbl(z);\n", +"disp(z,'i(0+)=')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6_7: final_value.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Caption:final_value\n", +"// example 1.6.7\n", +"//page 12\n", +"//X(s)=100/(s*(s^2+2*s+50))\n", +"p=poly([100],'s','coeff');\n", +"q=poly([0 50 2 1],'s','coeff');\n", +"F=p/q;\n", +"syms s\n", +"x=s*F;\n", +"y=limit(x,s,0);//final value theorem\n", +"y=dbl(y)\n", +"disp(y,'x(inf)=')//result" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6_8: steady_state_value.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Caption:steady_state_value\n", +"// example 1.6.7\n", +"//page 12\n", +"//X(s)=s/(s^2*(s^2+6*s+25))\n", +"p=poly([0 1],'s','coeff');\n", +"q=poly([0 0 25 6 1],'s','coeff');\n", +"F=p/q;\n", +"syms s\n", +"x=s*F;\n", +"y=limit(x,s,0);//final value theorem\n", +"y=dbl(y)\n", +"disp(y,'x(inf)=')//result" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6_9: initial_values.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Caption:initial_values\n", +"// example 1.6.7\n", +"//page 13\n", +"//F(s)=(4*s+1)/(s^3+2*s)\n", +"s=%s;\n", +"syms t;\n", +"F=(4*s+1)/(s^3+2*s)\n", +"f = ilaplace (F,s,t);\n", +"y=limit(f,t,0);//initial value theorem\n", +"y=dbl(y);\n", +"disp(y,'f(0+)=')\n", +"// since F'(s)=sF(s)-f(0+) where L(f'(t))=F'(s)=F1\n", +"F1=(4*s+1)/(s^2+2)\n", +"f1= ilaplace(F1,s,t);\n", +"y1=limit(f1,t,0);//initial value theorem\n", +"y1=dbl(y1);\n", +"disp(y1,'f_prime(0+)=')\n", +"// since F''(s)=(s^2)*F(s)-s*f(0+)-f'(0+) where L(f''(t))=F''(s)=F2\n", +"F2=(s-8)/(s^2+2)\n", +"f2= ilaplace(F2,s,t);\n", +"y2=limit(f2,t,0);//initial value theorem\n", +"y2=dbl(y2);\n", +"disp(y2,'f_doubleprime(0+)=')" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |