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+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:9eef90d3b867f35b7e7b7ae71f5d96e96e47a495dc677c7492d34992d7905497"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 5:Large Signals Amplifiers"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.1,Page number 280"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "Rb=1*10**3 #base resistance(ohms)\n",
+ "Vcc=20 #supply voltage(V)\n",
+ "Rc=20 #collector resistance(ohms) \n",
+ "beeta=25 #current gain \n",
+ "Vbe=0.7 #base to emitter voltage(V) \n",
+ "ib=10*10**-3 #base current(ohms)\n",
+ "\n",
+ "#Calculations\n",
+ "Ibq=(Vcc-Vbe)/Rb #current(A)\n",
+ "Icq=beeta*Ibq #current(A)\n",
+ "Vceq=Vcc-(Icq*Rc) #collector voltage(V)\n",
+ "ic=beeta*ib #collector current(A)\n",
+ "Po=((ic/(math.sqrt(2)))**2)*Rc #output voltage(V)\n",
+ "Pi=Vcc*Icq #input power(W)\n",
+ "eta=(Po/Pi)*100 #efficiency \n",
+ "Pd=Pi-((Icq**2)*Rc)-Po #power dissipated(W) \n",
+ "\n",
+ "#Results\n",
+ "print\"input power is Pi\",Pi,\"W\"\n",
+ "print\"output power is Po\",Po,\"W\"\n",
+ "print\"power dissipated is\",round(Pd,1),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "input power is Pi 9.65 W\n",
+ "output power is Po 0.625 W\n",
+ "power dissipated is 4.4 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.2,Page number 283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration \n",
+ "Rl=500 #load resistance(ohms)\n",
+ "Vceq=50 #queinscent collector voltage(V)\n",
+ "beetamin=30 #current gain minimum(at Q)\n",
+ "Icq=0.4 #queinscent collector current(A)\n",
+ "Ibq=8 #queinscent base current(mA)\n",
+ "\n",
+ "#Calculations\n",
+ "Rac=Vceq/Icq #ac resistance(ohms)\n",
+ "beeta=(Icq*10**-3)/Ibq #current gain\n",
+ "Re=5/Icq #emitter resistance(ohms)\n",
+ "Rc=(512.5*Rac)/(512.5-Rac) #as Re+Rl=500+12.5=512.5\n",
+ "Vcc=5+Vceq+(Icq*Rc) #supply voltage(V) \n",
+ "Rb=(beetamin*Re)/10 #base resistance(ohms)\n",
+ "R1=39.5 #solving 125=Rc||(Rl+Re) and Vbb=Vcc*(R1/(R1+R2))\n",
+ "R2=750\n",
+ "Pi=120*Icq #Vcc chosen as 120\n",
+ "r=(Rc*Rl)/(Rc+Rl)\n",
+ "Poac=(100/(2*math.sqrt(2)))**2/r #output power(W)\n",
+ "etamax=Poac/Pi #efficiency\n",
+ "Poac1=(100/(2*math.sqrt(2)))**2/Rl #ac power absorbed by load(W)\n",
+ "eta=Poac1/Pi \n",
+ "Pc=(Icq**2)*Rc #power lost in Rc(W)\n",
+ "Pe=(Icq**2)*Re #power lost in Re(W)\n",
+ "Pd=Pi-Pc-Pe-Poac #power consumed(W)\n",
+ "\n",
+ "#Results\n",
+ "print\"input power is Pi\",Pi,\"W\"\n",
+ "print\"output power is Po\",round(Poac,2),\"W\"\n",
+ "print\"dissipated power is\",round(Pd,2),\"W\"\n",
+ "print\"values of R1,R2,Re and Rc are\",R1,\"ohms,\",R2,\"ohms,\",Re,\"ohms and\",round(Rc),\"ohms resp. (Calculated value of Rc is wrong in the book)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "input power is Pi 48.0 W\n",
+ "output power is Po 10.06 W\n",
+ "dissipated power is 9.49 W\n",
+ "values of R1,R2,Re and Rc are 39.5 ohms, 750 ohms, 12.5 ohms and 165.0 ohms resp. (Calculated value of Rc is wrong in the book)\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.3,Page number 285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "Pmax=10 #power maximum(W)\n",
+ "Ic=1 #collector current(A)\n",
+ "Vcemax=100 #max collector to emitter current(V)\n",
+ "Vcemin=2 #min collector to emitter current(V)\n",
+ "\n",
+ "#Calculations\n",
+ "#Part a\n",
+ "Vceq=46 #Vce at Q point \n",
+ "Icq=0.21 #Ic at Q point \n",
+ "Vcc=92 #supply voltage(V)\n",
+ "ic=0.42 #collector current(A) \n",
+ "\n",
+ "#Part b\n",
+ "Rl=Vceq/Icq #load resistance(ohms)\n",
+ "\n",
+ "#Part c\n",
+ "Pi=Vcc*Icq #input power(W)\n",
+ "Po=((ic/(2*math.sqrt(2)))**2)*Rl #output power(W)\n",
+ "eta=(Po/Pi)*100 #efficiency\n",
+ "\n",
+ "#Results\n",
+ "print\"Rl for maximum power input is\",round(Rl),\"ohms\"\n",
+ "print\"input power is is\",Pi,\"W\"\n",
+ "print\"Po is\",Po\n",
+ "print\"eta is\",eta,\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rl for maximum power input is 219.0 ohms\n",
+ "input power is is 19.32 W\n",
+ "Po is 4.83\n",
+ "eta is 25.0 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.4,Page number 286"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "Vcc=15 #supply voltage(V)\n",
+ "beeta=40. #current gain\n",
+ "Icq=5. #Ic at Q(mA)\n",
+ "Vceq=7.5 #Vce at Q(V) \n",
+ "icswing=10 #swing in ic(mA) \n",
+ "\n",
+ "#Calculations\n",
+ "#Part a\n",
+ "Rl=Vceq/Icq*10**-3 #load resistance(ohms) \n",
+ "\n",
+ "#Part b\n",
+ "Ibq=Icq/beeta #base current at Q(uA)\n",
+ "\n",
+ "#Part c\n",
+ "ibswing=icswing/beeta #swing in ib(mA)\n",
+ "Pac=Rl*(icswing/(2*math.sqrt(2)))**2 #ac power(W)\n",
+ "Pdc=Vcc*(Icq*10**-3) #dc power(W)\n",
+ "eta=(Pac/Pdc)*100 #efficiency\n",
+ "\n",
+ "#Results\n",
+ "print\"a)value of Rl is\",round(Rl/1E-6),\"ohms\"\n",
+ "print\"b)Ibq is\",round(Ibq/1E-3),\"uA\"\n",
+ "print\"c)ac power output is\",round((Pac/1E-3),2),\"mW\"\n",
+ "print\"efficiency is\",eta,\"%\"\n",
+ "print\"corresponding swing in ib is\",ibswing,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a)value of Rl is 1500.0 ohms\n",
+ "b)Ibq is 125.0 uA\n",
+ "c)ac power output is 18.75 mW\n",
+ "efficiency is 25.0 %\n",
+ "corresponding swing in ib is 0.25 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5,Page number 288"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "Vcc=Vce=10 #supply voltage(V) \n",
+ "Icq=140*10**-3 #Ic at Q point(A)\n",
+ "Rl=8 #load resistance(ohms)\n",
+ "vce=16 #instantaneous collector to emitter voltage(V)\n",
+ "ic=235*10**-3 #instantaneous collector current(A)\n",
+ "\n",
+ "#Calculations\n",
+ "RL=Vcc/Icq\n",
+ "r=math.sqrt(RL/Rl) #load resistance for max ac swing(ohms)\n",
+ "Po=(vce*ic)/(2*math.sqrt(2)*2*math.sqrt(2)) #output power(W)\n",
+ "Pi=Vcc*Icq #input power(W) \n",
+ "eta=Po/Pi #efficiency\n",
+ "Pd=Pi-Po #dissipated power(W) \n",
+ "\n",
+ "#Results\n",
+ "print\"a)transformation ratio is\",round(r)\n",
+ "print\"c)power output is\",Po,\"W\"\n",
+ "print\"efficiency is\",round(eta*100,2),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a)transformation ratio is 3.0\n",
+ "c)power output is 0.47 W\n",
+ "efficiency is 33.57 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.6,Page number 290"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "Rl=4.5 #load resistance(ohms)\n",
+ "Vceq=50 #Vc at point Q(V)\n",
+ "Icq=400*10**-3 #Ic at Q(A)\n",
+ "Re=12.5 #emitter resistance(ohms)\n",
+ "Vcemax=90 #from figure \n",
+ "Vcemin=10 #from figure\n",
+ "Icmax=730 #max Ic(mA)\n",
+ "Icmin=30 #min Ic(mA) \n",
+ "\n",
+ "#Calculations\n",
+ "#Part a\n",
+ "Rac=Vceq/Icq #ac resistance(ohms)\n",
+ "n=math.sqrt(Rac/Rl) #as n=N1/N2 and Rac=(N1/N2)^2*Rl\n",
+ "\n",
+ "#Part b\n",
+ "Vcc=Vceq+(Icq*Re) #supply voltage(V) \n",
+ "\n",
+ "#Part c\n",
+ "vce=Vcemax-Vcemin #instantaneous collector to emitter voltage(V)\n",
+ "ic=Icmax-Icmin #instantaneous collector current(mA)\n",
+ "Po=(vce*ic)/((2*math.sqrt(2))*(2*math.sqrt(2))) #output voltage(V)\n",
+ "Pi=Vcc*Icq #input voltage(V) \n",
+ "eta=(Po/Pi)*100 #efficiency\n",
+ "Pd=Pi-(Icq**2*Re)-Po*10**-3 #dissipated power(W)\n",
+ "\n",
+ "#Results\n",
+ "print\"a)transformation ratio is\",round(n,2)\n",
+ "print\"b)Vcc is\",Vcc,\"V\"\n",
+ "print\"c)power efficiency for the load is\",round((eta/1E+3),1),\"%\"\n",
+ "print\"power dissipated is\",Pd,\"W\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a)transformation ratio is 5.27\n",
+ "b)Vcc is 55.0 V\n",
+ "c)power efficiency for the load is 31.8 %\n",
+ "power dissipated is 13.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.7,Page number 295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Finding input power,output power,Pd,efficiency,\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "Vcc=30 #supply voltage(V)\n",
+ "Rl=16 #load resistance(ohms) \n",
+ "n=2 #transformation ratio\n",
+ "Im=1 #peak value of current(A)\n",
+ "etamax=78.54 #max efficiency(%)\n",
+ "\n",
+ "#Calculations\n",
+ "#Part a\n",
+ "Rl1=Rl*(n/2)**2 #load resistance(ohms)\n",
+ "Pi=(2*Vcc*Im)/math.pi #input power(W)\n",
+ "Pimax=(2*Vcc**2)/((math.pi)*Rl1) #input power max(W)\n",
+ "\n",
+ "#Part b\n",
+ "Po=((Im**2)*Rl1)/2 #output power(W)\n",
+ "Pomax=(Vcc**2)/(2*Rl1) #output power max(W)\n",
+ "\n",
+ "#Part c\n",
+ "eta=Po/Pi #efficiency\n",
+ " \n",
+ "\n",
+ "#Part d\n",
+ "P=((2*Vcc*Im)/math.pi)-((Im**2*Rl1)/2) #Power dissipated by transistors(W)\n",
+ "Pd=P/2 #power dissipated by each transistors\n",
+ "Pmax=(2*Vcc**2)/((math.pi)**2*Rl1) #max power dissipated by transistors\n",
+ "Pdmax=Pmax/2 #max power dissipated by each transistor\n",
+ "\n",
+ "#Results\n",
+ "print\"a)input power is\",round(Pi,1),\"W and max input power is\",round(Pimax,2),\"W\"\n",
+ "print\"b)output power \",Po,\"W and max output power is\",round(Pomax,2),\"W\"\n",
+ "print\"c)power efficiency for the load is\",round((eta/1E-2),2),\"% and its max value is\",etamax,\"%\"\n",
+ "print\"power dissipated by each transiator is\",round(Pd,1),\"W and max value is\",round(Pdmax,1),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a)input power is 19.1 W and max input power is 35.81 W\n",
+ "b)output power 8 W and max output power is 28.0 W\n",
+ "c)power efficiency for the load is 41.89 % and its max value is 78.54 %\n",
+ "power dissipated by each transiator is 5.5 W and max value is 5.7 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.8,Page number 296"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Variable declaration\n",
+ "Pd=10\n",
+ "\n",
+ "#Calculations\n",
+ "#Part a\n",
+ "Poacmax=10. #as Pd=Po(ac)max by class A\n",
+ "\n",
+ "#Part b\n",
+ "Pd=2*Poacmax #power dissipated(W)\n",
+ "Poacmax1=146/2 #max output power by class B\n",
+ "f=Poacmax1/Poacmax #factor by which power of class B is greater than class A\n",
+ " \n",
+ "#Results\n",
+ "print\"maximum signal output powerclass A produce is\",Poacmax,\"W\"\n",
+ "print\"maximum signal output powerclass produce is\",Poacmax1,\"W\"\n",
+ "print\"factor by which power in class b is larger than power in class A transformer is\",f"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "maximum signal output powerclass A produce is 10.0 W\n",
+ "maximum signal output powerclass produce is 73 W\n",
+ "factor by which power in class b is larger than power in class A transformer is 7.3\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.9,Page number 300"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "Vcc=30. #supply voltage(V)\n",
+ "Im=1 #peak value of current(A)\n",
+ "Rl=10. #load resistance(ohms)\n",
+ "\n",
+ "#Calculations\n",
+ "#Part a\n",
+ "Pi=(Vcc*Im)/math.pi #input power(W)\n",
+ "Pimax=(Vcc**2)/(math.pi*2*Rl) #max input power(W)\n",
+ "\n",
+ "#Part b\n",
+ "Po=((Im**2)*Rl)/2 #output power(W)\n",
+ "Pomax=(Vcc**2)/(8*Rl) #output power max(W)\n",
+ "\n",
+ "#Part c\n",
+ "eta=Po/Pi #efficiency\n",
+ "etamax=Pomax/Pimax #efficiency max \n",
+ "\n",
+ "#Part d\n",
+ "Pd=Pi-Po #Power dissipated by transistors(W) \n",
+ "Pmax=(Vcc**2)/(2*(math.pi)**2*Rl) #max power dissipated by transistors\n",
+ " \n",
+ "#Results\n",
+ "print\"a)input power is \",round(Pi,2),\"W and max input power is\",round(Pimax,2),\"W\"\n",
+ "print\"b)output power is \",Po,\"W and max output power is\",round(Pomax,2),\"W\"\n",
+ "print\"c)power efficiency for the load is\",round((eta/1E-2),2),\"% and its max value is\",round((etamax/1E-2),2),\"%\"\n",
+ "print\"power dissipated and its max value are\",round(Pd,2),\"W and\",round(Pmax,2),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a)input power is 9.55 W and max input power is 14.32 W\n",
+ "b)output power is 5.0 W and max output power is 11.25 W\n",
+ "c)power efficiency for the load is 52.36 % and its max value is 78.54 %\n",
+ "power dissipated and its max value are 4.55 W and 4.56 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.10,Page number 303"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "P1=2 #transistor power(W)\n",
+ "Rl=5*10**3. #load resistance()\n",
+ "Ic=35 #collector current(mA) \n",
+ "\n",
+ "#Calculations\n",
+ "Bo=40-Ic \n",
+ "B1=math.sqrt((2*P1)/Rl)\n",
+ "B2=Bo\n",
+ "D2=(B2/B1)*100 #second harmonic distortion(%)\n",
+ "\n",
+ "#Results\n",
+ "print\"second harmonic distortion is\",round((D2/1E+3),2),\"%\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "second harmonic distortion is 17.68 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.12,Page number 314"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "Vcc=15. #supply voltage(V)\n",
+ "Rl=10. #load resistance(ohms)\n",
+ "\n",
+ "#Calculations\n",
+ "#Part a\n",
+ "Immax=Vcc/Rl #max peak current(A)\n",
+ "Irmsmax=Immax/(math.sqrt(2)) #max rms current(A)\n",
+ "Pomax=Irmsmax**2*Rl #max output power(W)\n",
+ "Pi=(2*Vcc*Immax)/math.pi #max input power(W)\n",
+ "eta=Pomax/Pi #efficiency\n",
+ "\n",
+ "#Part b \n",
+ "Im=(2*Vcc)/(math.pi*Rl) #peak current(A)\n",
+ "Pdmax=((2*Vcc*Im)/(math.pi))-((Im**2*Rl)/2) #max power dissipated(W)\n",
+ "eta1=((Im**2)*Rl*math.pi)/(2*2*Vcc*Im) #efficiency\n",
+ "\n",
+ "#Results\n",
+ "print\"a)max signal output power,collector dissipation are\",Pomax,\"W,\",round(Pi,2),\"W and efficiency is\",round((eta/1E-2),2),\"%\"\n",
+ "print\"b)max dissipation of each transistor and corresponding efficiency is\",round(Pdmax,2),\"W and\",eta1,\"resp.\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a)max signal output power,collector dissipation are 11.25 W, 14.32 W and efficiency is 78.54 %\n",
+ "b)max dissipation of each transistor and corresponding efficiency is 4.56 W and 0.5 resp.\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.13,Page number 315"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculations\n",
+ "eta=0.5 #As Po(ac)=Vcc^2/2*pi^2*Rl and Pi(dc)=Vcc^2/pi^2*Rl\n",
+ " #put these in eta=Po(ac)/Pi(dc) which is 1/2=0.5 \n",
+ " \n",
+ "#Results\n",
+ "print\"push pull amplifier efficiency is\",round(eta/1E-2),\"%\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "push pull amplifier efficiency is 50.0 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file