Class Assignment “
REFLECTION COEFFICIENT SOLUTION
FOR L MATCHING NETWORK USING MATLAB
”
Subject: Microwave Engineering
By: Enggar Fransiska (0103513001)
ELECTRICAL ENGINEERING DEPARTMENT FACULTY OF SCIENCE AND TECHNOLOGY UNIVERSITY OF AL AZHAR INDONESIA IN DONESIA 2015
REFLECTION COEFFICIENT SOLUTION FOR L MATCHING NETWORK USING MATLAB Enggar Fransiska (0103513001) Electrical Engineering Department – Department – Undergraduate, Undergraduate, Faculty of Science and Technology, University of Al Azhar Indonesia, Jl. Sisingamangaraja, Sisingamangaraja, Kebayoran Baru, Jakarta Selatan, 12110 Email :
[email protected]
Impedance matching in transmission line is very important because of some reasons. First, maximum power is delivered to load when Transmission Line (TL) is matched at both the load and source ends. This configuration satisfies the conjugate match condition. Second, with a property matched TL, more signal power is transferred to the load, which increases the sensitivity of the device. Third, some equipment (such as contain amplifiers) can be damaged when too much power is reflected back to the source. The method to make impedance matching is by adding circuit between transmission line and the load, see Figure 1.1. The matching network consist of conductor and inductor which has eight possible combinations, see Figure 1.2.
Figure 1.1
Figure 1.2 The L and C values of these elements can be identified b y the sign of reactance or susceptance. The type of the element can be seen in table tabl e 1.1.
Table 1.1 There are two ways to compute L matching problems, first analytically then using the Smith chart. As in the text, we’ll solve the problem of L matching to plot reflection coefficient using Matlab. Assignments given bellow will show the detail. Assignment 1. Show that these two solution will make the TL match at frequency 500 Mhz using Matlab.
Figure1.3 Answer: Solution1 C1=0.92 pF = 0.92 x 10 -12 F, ZC1=
(see table 1.1)
(0.90 ) = 5.770 = 5.7700 − = 5.770 −73.3 Ω = =
L1 = 38.8 nH = 38.8 x 10 -9 H, ZL1 = jωL1 (see table 1.1)
= j2πf (38.8 x 10 -9) = j2πf x 109 (38.8 x 10 -9) = j2πf (38.8) = j243.66f Ω
+ . (00−00)( ) = 00−00+.
Zparalel 1 =
=
4
00−(00+73.3)
Ω
Zin1 = Ztotal1 = ZL1+ (ZL1//ZC1) = j243.66f +
Reflection Coefficient 1=
4
00−(00+73.3)
Ω
−0 +0
Solution 2 C2 = 2.61 pF = 2.61 x 10 -12 F,
ZC2 =
(see table 1.1)
L2 = 46.1 nH = 46.1 x 10 -9 H, ZL2 = jωL2 (see jωL2 (see table 1.1)
(.60 ) = 6.39080 = 6.39080 0 − = 6.39080 −6 Ω =
= j2πf (46.1 x 10 -9)
=
= j2πf x 109 (46.1 x 10 -9) = j2πf (46.1) = j289.508f Ω
+ (00−00)(89.508) = 00−00+89.508 5790.6+8950.8f Ω = 00+(−00+89.508)
Zparalel2 =
Zin2 = Ztotal2 = ZC2+(ZL//ZL2) =
−6 + 5790.6+8950.8f 00+(−00+89.508)
Reflection Coefficient 2=
Ω
−0 +0
Solution 1 and solution 2 has identic load impedance (ZL) and Z0. ZL = (200-J100) Ω Z0 = 100 Ω
MATLAB SOLUTION % ENGGAR FRANSISKA % L MATCHING CODE FOR REFLECTION COEFFICIENT WITH FREQUENCY clear all all; ; clc; ZL=200-j*100; Z0=100; f=0:0.1:1; w=2*pi.*f*1e9; %Solution 1 C1=0.92e-12; L1=38.8e-9; Zc1=1./(j*w*C1); Zl1=j*w*L1; parallel1=(ZL*Zc1)./(ZL+Zc1); Zin1=Zl1+parallel1; Ref1=(Zin1-Z0)./(Z0+Zin1); %Solution 2 C2=2.61e-12; L2=46.1e-9; Zc2=1./(j.*w.*C2); Zl1=j.*w.*L2; parallel2=(ZL.*Zl1)./(ZL+Zl1); Zin2=Zc2+parallel2; Ref2=(Zin2-Z0)./(Z0+Zin2);
%Load Impedance %Z0 %Frequency Range %Omega
%Value of C1 %Value of L1 %Impedance of Capacitor 1 %Impedance of Inductor 1 %Parallel Impedance 1 %Ztotal1=Zin1 %Refflection Coefficient 1
%Value of C2 %Value of L2 %Impedance of Capacitor 2 %Impedance of Inductor 2 %Parallel Impedance 2 %Ztotal2=Zin2 %Refflection Coefficient 2
%Result plot(f,abs(Ref1),f,abs(Ref2)); legend('Solution legend('Solution #1', #1' ,'Solution #2', #2' ,'Location' ,'NorthEast' 'NorthEast'); ); title('Reflection title( 'Reflection Coefficient Vs Frequency (GHz)' ); xlabel('f xlabel('f (GHz)'); (GHz)' ); ylabel(' ylabel(' Reflection Coefficient Magnitude' );
Figure 1.4 This figure shows that solution 1 and solution 2 will match at 0.5Ghz or 500 Mhz (reflection coefficient equal to zero). At other frequency it isn’t match (reflection coefficient isn’t zero).
Assignment 2 Show that this circuit in TL will match at 1 Ghz.
Figure 1.5 Answer:
ZL = 25+j30 Ω Z0 = 50 Ω C1
= 3.18 pF = 3.18 x 10 -12 F,
ZC1 =
1 1
(see table 1.1)
(3.80 ) = (9.970 ) = (0)(9.970) − = 9.970 −50.07 Ω = =
Zseri = ZL+ZC2
−5) 5 = 25 + j (30 - ) Ω
= 25+j30 + (
Zin = Ztotal = ZC1 // Zseri
+ . [5 + j 30 − ] = . Ω +[ 5 + j 30 − ] =
Reflection Coefficient =
−0 +0
C2 = 31.8 pF = 31.8 x 10 -12 F, ZC2 =
1 2
(see table 1.1)
(3.80 ) = (99.70 ) = (0)(99.70 ) − = 99.70 −5 Ω = =
MATLAB SOLUTION % ENGGAR FRANSISKA % L MATCHING CODE FOR REFLECTION COEFFICIENT WITH FREQUENCY % EXAMPLE N7.1 (SLIDE) clear all all; ; clc; ZL=25+j*30; Z0=50; f=0:0.1:5; w=2*pi.*f*1e9;
%Load Impedance %Z0 %Frequency %Omega
%Solution C1=3.18e-12; C2=31.8e-12; Zc1=1./(j.*w.*C1); Zc2=1./(j.*w.*C2); Zseri=ZL+Zc2; Zin=(Zc1.*Zseri)./(Zc1+Zseri); Ref=(Zin-Z0)./(Z0+Zin);
%Value of C1 %Value of C2 %Impedance of Capacitor 1 %Impedance of Capacitor 2 %Impedance of serial circuit %Zin = Total of Parallel Impedance %Refflection Coefficient
%Result plot(f,abs(Ref)); title('Reflection title( 'Reflection Coefficient Vs Frequency (GHz)' ); xlabel('f xlabel('f (GHz)'); (GHz)' ); ylabel('Reflection ylabel('Reflection Coefficient Magnitude' );
Figure 1.5 This figure shows that The circuit will match at 1G (reflection coefficient equal to zero). At other frequency it isn’t match (reflection coefficient isn’t zero).
To make easy code in Matlab, input the t he equation inside orange box. That is the simple equation.
Reference : D. M. Pozar, Microwave Pozar, Microwave Engineering , 4th ed. New York: Wiley, Wile y, 2012.
