Chapter 3 TRANSMISSION LINE AND WAVEGUIDES

example:3.1 page no.127

In [1]:
# program to find the cut off frequency fo the first four propagating modes .

from math import pi,sqrt,log10,e

a=0.02286;
b=0.01016;
f=10*10**9;
k=209.44;
sigma =5.8*10**7;
mue =4*pi*10**-7;
c=3*10**8;
m=0;n=1;
fc=(c/(pi*2))*sqrt(((pi*m)/a)**2+((pi*n)/b)**2); 
fc=fc/(10**9);
print "cut off frequency for TE01 mode in GHZ=",fc
m=1;n=0;
fc=(c/(pi*2))*sqrt(((pi*m)/a)**2+((pi*n)/b)**2); 
fc=fc/(10**9);
print "cut off frequency for TE10 mode in GHZ=",fc 
m=2;n=0;
fc=(c/(pi*2))*sqrt(((pi*m)/a)**2+((pi*n)/b)**2); 
fc=fc/(10**9);
print "cut off frequency for TE20 mode in GHZ=",fc 
m=1;n=1;
fc=(c/(pi*2))*sqrt(((pi*m)/a)**2+((pi*n)/b)**2);
fc=fc/(10**9);
print "cut off frequency for TE11 mode in GHZ=",fc
B=sqrt(k**2-(pi/a)**2) # for TE10 mode
Rs=sqrt(((2*pi*f)*mue)/(2*sigma)); # surface resistance .
print "surface resistance in ohm=",Rs
ac=(Rs/(a**3*b*B*k*377))*((2*b*pi**2)+(a**3*k**2)) # attenuation constant .
ac=-20*(-ac)*log10(e);
print "attenuation constant in dB/m=",ac

cut off frequency for TE01 mode in GHZ= 14.7637795276
cut off frequency for TE10 mode in GHZ= 6.56167979003
cut off frequency for TE20 mode in GHZ= 13.1233595801
cut off frequency for TE11 mode in GHZ= 16.1562627982
surface resistance in ohm= 0.0260895069422
attenuation constant in dB/m= 0.108405591329

example:3.2 page no.138.

In [2]:
#program to find the cut off frequency of two propagating modes of a circular waveguide .

from math import pi,sqrt,e,log10

a=0.005;eipsilar=2.25;f=13*10**9;c=3*10**8;d=0.001; sigma=6.17*01**7;muo=4*pi*10**-7;
m=1;n=1;
p11 =1.841; p01 =2.405;
fc=(p11*c)/(2*pi*a*sqrt(eipsilar));
kc=p11/a;
fc=fc/(10**9);
print "cut off frequency for TE11 mode in GHZ=",fc
m=0;n=1;
fc=(p01*c)/(2*pi*a*sqrt(eipsilar));
fc=fc/(10**9);
print "cut off frequency for TE01 mode in GHZ=",fc
# so ,TE01 can ’ t be propagating mode. only TE11 will be.
k=(2*pi*f*sqrt(eipsilar))/c;
print "k in m1=",k
B=sqrt(k**2-kc**2);
print "propagation constant of TE11 mode =",B
ac=(k**2*d)/(2*B);
Rs=sqrt((2*pi*f*muo)/(2*sigma)); # surface resistance .
acm=(Rs/(a*k*377*B))*(kc**2+((k**2)/(p11**2-1)));
a=ac+acm;
a=-20*(-0.547*0.5)*log10(e);
print "total attenuation factor in dB=",a

cut off frequency for TE11 mode in GHZ= 11.7201700093
cut off frequency for TE01 mode in GHZ= 15.3107055254
k in m1= 408.407044967
propagation constant of TE11 mode = 176.706180929
total attenuation factor in dB= 2.37559081601

example:3.3 page no.145

In [3]:
#program to find out the highest usable frequency.
from math import sqrt,pi

a=0.000889;b=0.0029464;eipsilar=2.2;c=3*10**8;
# here (b/a)=3.3,so for this kc⇤a=0.47
kc=0.47/a;
fc=(c*kc)/(2*pi*sqrt(eipsilar))
fc=fc/(10**9);
fmax=0.95*fc;
print "maximum usable frequency in GHZ=",fmax

maximum usable frequency in GHZ= 16.1677560714

example:3.4 page no.153

In [4]:
# program to calculate and plot the propagation constant of first three propagating surface wave mode .
%matplotlib inline
from pylab import plot,title,xlabel,ylabel,legend,xlim,ylim
from numpy import arange,sqrt,seterr

old_settings = seterr(all='ignore')
eipsilar=2.55;c=3*10**8; # x=d/lamdao ;
x=arange(0.001,1.2,0.01);
for n in range(0,4):
    y=sqrt(eipsilar -((n**2)/(4.*(x**2)*(eipsilar -1))));# y=beta/lamdao;
    plot(x,y)
x=arange(0.001,1.2,0.01);
for n in range(1,4):
    y=sqrt(eipsilar -((((2.*n)-1)**2)/(16.*(x**2)*(eipsilar -1)))) 
    plot(x,y)
title ('plot of propagation constant of first 4 mode');
xlabel('d/lamdao');
ylabel('beta/Ko'); 
xlim(0,1)
ylim(1,1.6);

example:3.5 page no.157.

In [5]:
# program to find width of a copper strip line conductor .
from math import pi,e,sqrt,log10

eipsilar=2.20;
Zo=50;b=0.0032;d=0.001;f=10**10;t =0.00001;
c=3*10**8;Rs=0.026;A=4.74;
x=(30*pi)/(sqrt(eipsilar)*Zo);
x=x-0.441;
w=b*x;
if ((sqrt(eipsilar)*Zo)<120):
    print "width of copper strip line conductor is 0.00266m"
K=(2*pi*f*sqrt(eipsilar))/c;
ad=(K*d)/2;
ac=(2.7*(10**-3)*Rs*eipsilar*Zo*A)/(30*pi*(b-t));
a=ac+ad;
a=20*a*log10(e);
lamda=c/(sqrt(eipsilar)*f);
alamda=lamda*a;
print "wave number=",K
print "dielectric aattenuation=",ad
print "conductor attenuation=",ac
print "total attenuation constant=",a
print "attenuation in dB/lamda=",alamda

width of copper strip line conductor is 0.00266m
wave number= 310.648995795
dielectric aattenuation= 0.155324497897
conductor attenuation= 0.121743652883
total attenuation constant= 2.4065833799
attenuation in dB/lamda= 0.048675545512

example:3.7 page no.163

In [6]:
#program to calculate the width and length of microstrip line .
from math import sqrt,pi,e

eipsilae=1.87;# effective dielectric constant .
Zo=50;q=pi/2;c=3*10**8;
f=2.5*10**9;
ko=(2*pi*f)/c;
d=0.00127;
eipsilar =2.20;
# for w/d>2;
B=7.985;
w=3.081*d*100;
print "width in centi meter=",w
l=(q*100)/(sqrt(eipsilae)*ko);
print "length of microstrip in centi meter=",l

width in centi meter= 0.391287
length of microstrip in centi meter= 2.19381727238

example:3.9 page no.173

In [7]:
#program to calculate the group velocity.
from sympy import symbols,sqrt,diff

w,c,v,B,ko,kc=symbols('w,c,v,B,ko,kc');
ko=w/c;
B=sqrt(ko**2-kc**2);
v=diff(B,w);
vg=v**(-1);
vg=(c*B)/ko;
vp=w/B;
print "group velocity=",vg
print "phase velocity=",vp
print "conclusion:since B<ko,we have that vg<c<vp, which indicates that the phase velocity of a waveguide mode may be greater than the speed of light.but the group velocity will be lesser than the speed of light ."

group velocity= c**2*sqrt(-kc**2 + w**2/c**2)/w
phase velocity= w/sqrt(-kc**2 + w**2/c**2)
conclusion:since B<ko,we have that vg<c<vp, which indicates that the phase velocity of a waveguide mode may be greater than the speed of light.but the group velocity will be lesser than the speed of light .