Chapter 7 POWER DIVIDERS DIRECTIONAL COUPLERS AND HYBRIDS¶

exa 7.1 page.no:360¶

In [1]:
#function:
def parallel_impedence(Z1,Z2):
return (Z1*Z2)/(Z1+Z2);

# program to compute the reflection coefficients seen looking in to the output port.
# as the power is divided in to 2:1 ratio. and Pin=(1/2)⇤Voˆ2/Zo;
# so,P1=(1/3)⇤Pin;and P2=(2/3)⇤Pin ............( i)
Zo =50.;
Z1=3.*Zo; # from above condition .............( i)
Z2=(3/2)*Zo;
Zin=parallel_impedence(Z1,Z2); # input impedence to the junction .
if Zin==Zo:
print "input is matched to the 50 ohm sources"
else:
print "not matched"
Zin1=parallel_impedence(Zo,Z2); # looking in to the 150 ohm source.
Zin2=parallel_impedence(Zo,Z1); # looking in to the 75 ohm source.
tao1=(Zin1-Z1)/(Zin1+Z1);
tao2=(Zin2-Z2)/(Zin2+Z2);
print "reflection coefficient looking at 150 ohm  line = %.2f"%tao1
print "reflection coefficient looking at 75 ohm line = %.2f"%tao2

not matched
reflection coefficient looking at 150 ohm  line = -0.71
reflection coefficient looking at 75 ohm line = -0.14


exa 7.2 page.no:365¶

In [2]:
# program to design an equisplit wilkinson power divider for 50 ohm system impedence .
from math import sqrt

Zo =50;
Z=sqrt(2)*Zo; # impedence of quarter wave transmission line .
R=2*Zo; # shunt resistor .
print "the shunt resistance value should be in ohm= ",R
print "the quarter wave transmission line in the divide should have a characteristic impedence in ohm = ",Z

the shunt resistance value should be in ohm=  100
the quarter wave transmission line in the divide should have a characteristic impedence in ohm =  70.7106781187


exa 7.3 page.no:372¶

In [3]:
# program to design bethe-hole coupler for x-band wave guide .
from math import pi,asin

f=9*10**9;C=20;a=0.02286;b=0.01016;
Ko=188.5;B=129;Z10 =550.9;
P10=4.22*10**-7;lamdao=0.0333;
uo=4*pi*10**-7;eo=8.854*10**-12;w=2*pi*f;
s=(a/pi)*asin(lamdao/sqrt(2*(lamdao**2-a**2)))*10**3;
# a=10⇤b;# as C=20db; # take x=a/b; so x=10;
ro=(P10/((10*w)*((((2*eo/3)+(4*uo)/(3*Z10**2))*0.944)-((4*pi**2*uo*0.056)/(3*B**2*a**2*Z10**2)))))**(1/3)*10**3;
print "the aperture position in mm = %.2f"%s
print "the aperture size in mm = ",ro
print "NOTE:the above shown results completes the design of the betha hole coupler "

the aperture position in mm = 9.72
the aperture size in mm =  1000.0
NOTE:the above shown results completes the design of the betha hole coupler


exa 7.4 page.no:378¶

In [4]:
# program to design a four hole chebysev coupler in xband wave guide using round aperture located at s=a/4.
from math import pi
from mpmath import cosh,acosh,sin,cos,asec

a=0.02286;b=0.01016;lamdao=0.0333;
ko=188.5;bta=129; Z10=550.9;
P10=4.22*10**-7;f=9*10**9;no=377;N=3;
s=a/4;
kf=((2*ko)/(3*no*P10))*((sin(pi*s/a)**2) -(2*(bta**2)/(ko**2))*((sin(pi*s/a)**2)+((pi**2)/((bta**2)*(a**2)))*(cos(pi*s/a)**2)));
kf=abs(kf)
kb=((2*ko)/(3*no*P10))*((sin(pi*s/a)**2)+(2*(bta**2) /(ko**2))*((sin(pi*s/a)**2)-((pi**2)/((bta**2)*(a **2)))*(cos(pi*s/a)**2)));
kb=abs(kb)
x=cosh(acosh(100)/3); # x=sec(thetam).
thetam=asec(x)*180/pi; # so , thetam=70.6 and at the band edge .
k=10**(-171.94/20);
ro=(((k/2)**(1/3))*x)*1000;
r1=(1.5*k*((x**3)-x))**(1/3)*1000;
print "kf = %.2f"%kf
print "kb = %.2f"%kb
print "thetam in degree = %.2f"%thetam
print "ro in mm= %.2f"%ro
print "r1 in mm= %.2f"%r1

kf = 394835.06
kb = 345030.82
thetam in degree = 70.59
ro in mm= 3009.49
r1 in mm= 1000.00


exa 7.5 page.no:382¶

In [5]:
#program to design a 50 ohm branchline quadrature hybrid junction
from math import sqrt

Zo =50.;
Z=Zo/sqrt(2)
print "the branch line impedence in ohm will be = %.2f"%Z

the branch line impedence in ohm will be = 35.36


exa 7.6 page.no:387¶

In [6]:
# program to determine the even and odd mode characteristic impedence.
from sympy import symbols,sqrt

C,A,d,W,C11,C12,Ce,Co,v,eo,er,s,b,uo,Zoe,Zoo,eipsila=symbols('C,A,d,W,C11,C12,Ce,Co,v,eo,er,s,b,uo,Zoe,Zoo,eipsila');
C=A*eipsila/d;
C11=(eo*er*W)/((b-s)/2)+(eo*er*W)/((b+s)/2);
C12=er*eo*W/s;
Ce=C11;
Co=C11+2*C12
v=1/sqrt(er*eo*uo);
Zoe=1/(v*C11); # as Ce=C11;
Zoo=1/(v*Co);
print "Zoe = ",Zoe," zoo = ",Zoo

Zoe =  sqrt(eo*er*uo)/(W*eo*er/(b/2 + s/2) + W*eo*er/(b/2 - s/2))  zoo =  sqrt(eo*er*uo)/(W*eo*er/(b/2 + s/2) + W*eo*er/(b/2 - s/2) + 2*W*eo*er/s)


exa 7.7 page.no:394¶

In [7]:
#design a 20 db single section coupled line coupler in stripline .
from math import sqrt

C=10**(-20/20);f=3*10**9;eipsila=2.56;Zo=50;b=0.00158;
Zoe=Zo*sqrt((1+C)/(1-C));
Zoo=Zo*sqrt((1-C)/(1+C));
Zoe=eipsila*Zoe;
Zoo=eipsila*Zoo;
x=0.72; #x=w/b.
y=0.34; # y=s/b.
w=0.72*b*100;
s=0.34*b*100;
print "conductor width in cm = %.3f"%w
print "conductor seperation in cm = %.3f"%s

conductor width in cm = 0.114
conductor seperation in cm = 0.054


exa 7.8 page.no:396¶

In [8]:
# design a three section 20 db coupler with a binomial response .
from sympy import symbols,sqrt

Zo=50;f=3*10**9;N=3;
C,C1,C2,theta=symbols('C,C1,C2,theta')
C=10**(-20/20);
print "for a maximally flat response for a three= section coupler doupble derivative of C will be zero"
C1=0.0125;C2=0.125;C3=0.0125;
Zoe1=Zo*sqrt((1+C1)/(1-C1));
Zoe3=Zo*sqrt((1+C3)/(1-C3));
Zoo1=Zo*sqrt((1-C1)/(1+C1));
Zoo3=Zo*sqrt((1-C1)/(1+C1));
Zoe2=Zo*sqrt((1+C2)/(1-C2));
Zoo2=Zo*sqrt((1+C2)/(1-C2));
print "the even and odd mode characteristic impedences for each section are :"
print "Zoe1 = %.2f"%Zoe1,"\nZoo1= %.2f"%Zoo1,"\nZoe2= %.2f"%Zoe2,"\nZoo2= %.2f"%Zoo2,"\nZoe3= %.2f"%Zoe3,"\nZoo3= %.2f"%Zoo3

for a maximally flat response for a three= section coupler doupble derivative of C will be zero
the even and odd mode characteristic impedences for each section are :
Zoe1 = 50.63
Zoo1= 49.38
Zoe2= 56.69
Zoo2= 56.69
Zoe3= 50.63
Zoo3= 49.38


exa 7.9 page.no:407¶

In [9]:
#design a 180 deg. ring hybrid for a 50 ohm system impedence .
from math import sqrt

Zo =50;
Z=sqrt(2)*Zo;
print "the characteristic impedence of the ring transmission line in ohm is = %.2f"%Z

the characteristic impedence of the ring transmission line in ohm is = 70.71