CHAPTER-2 - Best Shop for Test Banks and Solution Manuals · ANSWER: RC =: (a) 0.3888 (b) -0.2710...
Transcript of CHAPTER-2 - Best Shop for Test Banks and Solution Manuals · ANSWER: RC =: (a) 0.3888 (b) -0.2710...
CHAPTER-2
Q-2.1 A wave propagates in a non-magnetic media having relative dielectric constant εr.
Find its value if (a) η = 180Ω, (b) the wavelength at 10GHz is 2cm and (c) β = 0.001, f =
25000Hz
MATLAB CODE
% For nonmagnetic media mur = 1 or mu = mu0
mu0 = 4*pi*10^-7; aps0 = 8.854*10^-12;
% Required relation v = 1/sqrt(mu*aps), eta = sqrt(mu/aps), beta =2*pi/lemda
%and lemda = v/f =v0/(f*sqrt(mur*apsr)
muap = mu0*aps0; apmu = mu0/aps0; v0 = 1/sqrt(muap); eta0 = sqrt(apmu);
% (a) Given
eta = 180;
% From relation eta = sqrt(mu/aps) the obtained value of apsr is:
apsr1 = (eta0/eta)^2
% (b) Given
lemda = 2*10^-2; f = 10^10; v = f*lemda;
% From the relations lemda = v/f, v = v0/sqrt(apsr)
apsr2 = (v0/v)^2
% (c ) Given
beta = 0.001; f = 25000;
% From the relation β = 2π/λ=2πf/υ=2πf/(υ0/√εr)
apsr3 = ((beta*v0)/(2*pi*f))^2
ANSWER: (a) εr = 4.3805 (b) εr = 2.2469 (c) εr = 3.6426011
Q-2.2 A wave propagates at 100MHz in a dielectric media having some component of
conductivity (σ) of the order of 10-5 ℧/m. and μ = μ0. Find the values of α, β, υ and η
MATLAB CODE
% Given
pi = 3.1415; f0 = 10^8; sigma=10^-5; i = sqrt(-1);
mu0 = 4*pi*10^-7; aps0 = 8.854*10^-12; mur =1; apsr =1;
mu = mu0*mur; aps = aps0*apsr; muap = mu0*aps0;
muapt = sqrt(muap); apmu = mu0/aps0;
v0 = 1/muapt
eta0 = sqrt(apmu)
w = 2*pi*f0; wv = w/v0;
alfa = sigma*eta0/2
ws = w^2; apss = aps^2; sigs = sigma^2;
A1 = 8*ws*apss;
A11 = sigs/A1
A2 = 1+A11;
beta = wv*A2
% v =v0*(1-A11)
A4 = sigma/(2*w*aps);
eta=eta0*(1+i*A4)
ANSWER:
v0 = 2.9980e+008 eta0 = 376.7288 alfa = 0.0019 beta = 2.0957 υ ≈ 2.9980e+008 (1 - 4.0392e-007) eta = 3.7673e+002 +3.3860e-001i
Q-2.3 Find the depth of penetration at 1000Hz in (a) silver (σ = 6.17 x 107 ℧/m), (b)
Aluminium (σ = 3.72 x 107 ℧/m), (c) Brass (1.5 x 107℧/m) and (d) fresh water (σ = 10-3 ℧/m)
MATLAB CODE
% Relation for Rs is given by Eqn.(2.17)
% Also given
f = 1000; pi = 3.1416; mu = 4*pi*10^-7; k = pi*mu*f; k1 = sqrt(1/k);
% (a) for silver
sigma= 6.17*10^7; sig = sqrt(sigma);
delta = k1/sig
% (b) for Aluminium
sigma= 3.72*10^7; sig = sqrt(sigma);
delta = k1/sig
% (c) for brass
sigma= 1.5*10^7; sig = sqrt(sigma);
delta = k1/sig
% (a) for fresh water
sigma= 10^-3; sig = sqrt(sigma);
delta = k1/sig
ANSWER:
delta =: (a) 0.002m (b) 0.0026m (c) 0.0041m (d) 503.2909m
Q-2.4 The conductivity (σ) and depth of penetration (δ) at 1000Hz are given for slabs made
of: (a) silver σ = 6.17 x 107 ℧/m, δ = 0.00202m (b) copper σ = 5.8 x 107 ℧/m, δ = 0.00209m
(c) aluminium σ = 3.72 x 107 ℧/m, δ = 0.00261m and (d) brass σ = 1.5 x 107 ℧/m, δ =
0.00411m. Find the surface resistance of these slabs at 1000Hz. Also find the change if this
frequency is raised to 10MHz.
MATLAB CODE
% In view of Eqn. (2.19b) Rs = 1/ (sigma*delta)
% Thus Rs at 1000Hz for:
%(a) Silver:
sigma = 6.17*10^7; delta = 0.00202; A1 = sigma*delta;
Rssi = 1/A1
%(b) Copper:
sigma = 5.8*10^7; delta = 0.00209; A2 = sigma*delta;
Rscu = 1/A2
%(c) Aluminium:
sigma = 3.72*10^7; delta = 0.00261; A3 = sigma*delta;
Rsal = 1/A3
%(d) Brass:
sigma = 1.5*10^7; delta = 0.00411; A4 = sigma*delta;
Rsbr = 1/A4
% In view of the relation Rs = (1/σ) (1/δ) = (1/σ) √(πμσ) √f , Rs is proportional
% to √f. If all other parameters are constant Rs = k √f. Thus at f1 Rs1 = k %√f1. and at f2
Rs2 = k √f2. The change in Rs will correspond to
% Rs2 = Rs1 x sqrt(f2/f1)
f1 = 10^3; f2 = 10^7; fr = sqrt(f2/f1);
% New values of Rs are
%(a) Silver:
Rssin = Rssi*fr
%(b) Copper:
Rscun = Rscu*fr
%(c) Aluminium:
Rsaln = Rsal*fr
%(d) Brass:
Rsbrn = Rsbr*fr
ANSWER:
Values of Rs at 1000Hz for
Silver - Rssi = 8.0235e-006 Copper - Rscu = 8.2495e-006
Aluminium - Rsal = 1.0300e-005 Brass - Rsbr = 1.6221e-005
Values of Rs at 10MHz for
Silver - Rssin = 8.0235e-004 Copper - Rscun = 8.2495e-004
Aluminium - Rsaln = 0.0010 Brass - Rsbrn = 0.0016
Q-2.5 Find the time average power density at x = 1 if for a uniform plane wave is given by E
= 30 e-αx cos (108t – βx) az V/m and the dielectric constant of the propagating media is: (a) ε
= ε0, μ = μ0 (b) εr = 2.26, σ = 0 and (c) εr = 3.4, σ/ωε = 0.2
MATLAB CODE
% Given E = 30 e-αx cos (108t – βx) az V/m
% From the expression of E
E0z = 30; w = 10^8; x = 1; mu0 = 4*pi*10^-7; aps0 = 8.854*10^-12;
eta0 = sqrt(mu0/aps0)
% (a)
mur = 1; apsr = 1; mu = mur*mu0; aps = apsr*aps0;
eta1 = sqrt(mu/aps)
Pxav= (E0z^2)/(2*eta1)
% (b)
mur = 1; apsr = 2.26; mu = mur*mu0; aps = apsr*aps0;
eta2 = sqrt(mu/aps)
Pxav= (E0z^2)/(2*eta2)
% (c)
mur = 1; apsr = 3.4; mu = mur*mu0; aps = apsr*aps0;
eta3 = sqrt(mu/aps)
Px= (E0z^2)/(2*eta3);
% Given loss tangent tan(theta) = sigma/w*aps = 0.2
theta = atan (0.2); cs = cos(theta); sigma = 0.2*w*aps;
alfa = (sigma*eta3)/2; Pxav = Px*exp(-2*alfa*x)*cs
ANSWER: (a) eta1 = 376.7347 Pxav = 1.1945
(b) eta2 = 250.6002 Pxav = 1.7957
(c) eta3 = 204.3131 Pxav = 1.9098
Q-2.6 Find the reflection and transmission coefficients for E and H of a uniform plane wave
traveling in region-1 normally incident at the surface of region-2. The relative dielectric
constants are: (a) ε1 = 2.53 and ε2 = 1 (b) ε1 = 1 and ε2 = 2.53 (c) ε1 = 2.53 and ε2 = 2.26.
MATLAB CODE
%(a)
aps1 = 2.53; aps2 = 1; a = sqrt(aps1); b = sqrt(aps2);
c = a - b; d = a + b; e = b - a;
% Reflection coefficient for E
Taue = c/d
%Transmission coefficient for E
Touh=2*a/d
%Reflection coefficient for H
Te = e/d
%Transmission coefficient for H
Th=2*b/d
%(b)
aps1 = 1; aps2 = 2.53; a = sqrt(aps1); b = sqrt(aps2);
c = a - b; d = a + b; e = b - a;
% Reflection coefficient for E
Taue = c/d
%Transmission coefficient for E
Touh=2*a/d
%Reflection coefficient for H
Te = e/d
%Transmission coefficient for H
Th=2*b/d
%(c)
aps1 = 2.53; aps2 = 2.26; a = sqrt(aps1); b = sqrt(aps2);
c = a - b; d = a + b; e = b - a;
% Reflection coefficient for E
Taue = c/d
%Transmission coefficient for E
Touh=2*a/d
%Reflection coefficient for H
Te = e/d
%Transmission coefficient for H
Th=2*b/d
ANSWER:
(a) Taue = 0.2280 Touh = 1.2280 Te = -0.2280 Th = 0.7720 (b) Taue = -0.2280 Touh = 0.7720 Te = 0.2280 Th = 1.2280 (c) Taue = 0.0282 Touh = 1.0282 Te = -0.0282 Th = 0.9718
Q-2.7 Calculate the percentage of reflected and transmitted powers for a uniform plane wave
traveling between two regions. The transmission and reflection coefficients for E and H are:
Reflection coefficient for: Transmission coefficient for:
% E =RE H = RH E = TE H = TH
%(a) 0.227977 - 0.227977 1.228 0.772
%(b) -0.227977 0.227977 0.772 1.228
%(c) 0.02817 -0.02817 1.0282 0.97179
MATLAB CODE
% Let Ei is the incident E and Hi is the incident H
% Er is the reflected E and Hr is the reflected H
% Et is the transmitted E and Ht is the transmitted H.
% Pin, Pr and Pt are the incident, reflected and transmitted powers
% RE is the reflection coefficient for E
% RH is the reflection coefficient for H
% TE is the transmission coefficient for E
% TH is the transmission coefficient for H
% Assuming
Ei =1; Hi = 1; Pin= Ei*Hi;
%(a) Given
RE = 0.227977; RH = - 0.227977; TE =1.228; TH = 0.772;
Er = RE*Ei; Hr = RH*Hi;
Pr = Er*Hr
Et = TE*Ei; Ht = TH*Hi;
Pt= Et * Ht
%(b) Given
RE =-0.227977; RH =0.227977; TE =0.772; TH =1.228;
Er = RE*Ei; Hr = RH*Hi;
Pr = Er*Hr
Et = TE*Ei; Ht = TH*Hi;
Pt= Et * Ht
%(c) Given
RE = 0.02817; RH = -0.02817; TE =1.0282; TH = 0.97179;
Er = RE*Ei; Hr = RH*Hi;
Pr = Er*Hr
Et = TE*Ei; Ht = TH*Hi;
Pt= Et * Ht
ANSWER:
(a) Pr = -0.0520 or 5.2% Pt = 0.9480 or 94.8%
(b) Pr = -0.0520 or 5.2% Pt = 0.9480 or 94.8%
(c) Pr = -7.9355e-004 or o.o79355% Pt = 0.9992 or 99.92%
Q-2.8 Find the reflection coefficients for a vertically polarized wave obliquely incident on
the interface between two regions making 30o angle with the perpendicular drawn at the
boundary surface. The relative dielectric constants of regions- 1 and 2 are: (a) ε1 = 2.53 and
ε2 = 1 (b) ε1 = 1 and ε2 = 2.53 (c) ε1 = 2.53 and ε2 = 2.26.
MATLAB CODE
% For a vertically polarized wave the reflection coefficient is given by
% Er/Ei = [cosθ1 - √{(ε2/ε1) - sin2θ1}]/[cosθ1 +√{(ε2/ε1)- sin2θ1}]
% Given
pi = 3.1415; theta = pi/6; Cs = cos(theta); Ss = sin(theta); Sst = Ss^2;
%(a)
aps1 = 2.53; aps2 = 1; aps21= (aps2/aps1);
A1 = (aps21-Sst); A2 = sqrt(A1);
RC= (Cs-A2)/(Cs+A2)
%(b)
aps1 = 1; aps2 = 2.53; aps21= (aps2/aps1);
A1 = (aps21-Sst); A2 = sqrt(A1);
RC= (Cs-A2)/(Cs+A2)
%(c)
aps1 = 2.53; aps2 = 2.26; aps21= (aps2/aps1);
A1 = (aps21-Sst); A2 = sqrt(A1);
RC= (Cs-A2)/(Cs+A2)
ANSWER: RC =: (a) 0.3888 (b) -0.2710 (c) 0.0384
Q-2.9 Find the reflection coefficients for a parallel polarized wave obliquely incident on the
interface between two regions making an angle of 30o with the perpendicular drawn at the
boundary surface. The relative dielectric constants of the two regions are: (a) ε1 = 2.53 and ε2
= 1, (b) ε1 = 1 and ε2 = 2.53 (c) ε1 = 2.53 and ε2 = 2.26.
Er/Ei = [(ε2/ε1) cosθ1 - √{(ε2/ε1) – sin2θ1}] / [(ε2/ε1) cosθ1 - √{(ε2/ε1) – sin2θ1}]
MATLAB CODE
% For a parallel polarized wave the reflection coefficient is given by
% Er/Ei=[(ε2/ε1)cosθ1 -√{(ε2/ε1)–sin2θ1}] / [(ε2/ε1)cosθ1-√{(ε2/ε1)– sin2θ1}]
% Given
pi = 3.1415; theta = pi/6; Cs = cos(theta); Ss = sin(theta); Sst = Ss^2;
%(a)
aps1 = 2.53; aps2 = 1; aps21= (aps2/aps1);
A1 = (aps21-Sst); A2 = sqrt(A1); A3 = aps21*Cs;
RC= (A3-A2)/(A3+A2)
%(b)
aps1 = 1; aps2 = 2.53; aps21= (aps2/aps1);
A1 = (aps21-Sst); A2 = sqrt(A1); A3 = aps21*Cs;
RC= (A3-A2)/(A3+A2)
%(c)
aps1 = 2.53; aps2 = 2.26; aps21= (aps2/aps1);
A1 = (aps21-Sst); A2 = sqrt(A1); A3 = aps21*Cs;
RC= (A3-A2)/(A3+A2)
ANSWER: RC =: (a) -0.0537 (b) 0.1840 (c) -0.0181
Q-2.10 Find the Brewster’s angle if the relative dielectric constants for the two regions are:
(a) ε1 = 2.53 and ε2 = 1 (b) ε1 = 1 and ε2 = 2.53 (c) ε1 = 2.53 and ε2 = 2.26.
MATLAB CODE
% The Brewster’s angle is given as θ1 = tan-1 √(ε2/ε1)
%(a)
aps1 = 2.53; aps2 = 1; aps21= (aps2/aps1);
saps = sqrt(aps21); thetar = atan(saps);
thetad = thetar*180/pi
%(b)
aps1 = 1; aps2 = 2.53; aps21= (aps2/aps1);
saps = sqrt(aps21); thetar = atan(saps);
thetad = thetar*180/pi
%(c)
aps1 = 2.53; aps2 = 2.26; aps21= (aps2/aps1);
saps = sqrt(aps21); thetar = atan(saps);
thetad = thetar*180/pi
ANSWER: θ1 =: (a) 32.1583o (b) 57.8444o (c) 43.3856o
Q-2.11 A wave propagates between two parallel planes separated by 3cm. The space
between planes is filled by a non-magnetic material having relative dielectric constant εr
where (a) εr = 1 (b) εr = 2.26 and (c) εr = 2.53. Find the cutoff frequencies for m = 1, 2 and 3.
MATLAB CODE
% Given
mu0 = 4*pi*10^-7; aps0 = 8.854*10^-12; pi = 3.1416;
v0 = 3*10^8; a = 3*10^-2; v = v0/(2*a);
% In view of Eqn.(2.) fc = v0(m/2a)/sqrt(apsr)
% (a)
apsr =1; aps = 1/sqrt(apsr);
for m = 1:1:3
fc = m*(v/aps)
end
% (b)
apsr =2.26; aps = 1/sqrt(apsr);
for m = 1:1:3
fc = m*(v/aps)
end
% (c)
apsr =2.53; aps = 1/sqrt(apsr);
for m = 1:1:3
fc = m*(v/aps)
end
ANSWER: Cutoff frequency (fc) in Hz
m =: 1 2 3
(a) 5.0000e+009 1.0000e+010 1.5000e+010 (a) 7.5166e+009 1.5033e+010 2.2550e+010 (b) 7.9530e+009 1.5906e+010 2.3859e+010
Q-2.12 A 10GHz, TE10 wave propagates between two parallel planes separated by 5cm. The
space between planes is filled by a non-magnetic material with εr = 2.26. Find β, λand υ.
MATLAB CODE
% The required relations are: β = √[ω2με - (mπ/a)2] λ = (2π/β) υ = λf
% Given TE10 mode
pi = 3.1415; a = 5*10^-2; muo = 4*pi*10^-7; aps0 = 8.854*10^-12; apsr = 2.26; mur
= 1; f0 = 10^10; m = 1; w = 2*pi*f0; wt = w^2;
muap = mu0*aps0; muapr = mur*apsr*muap; A1 = wt*muapr;
mpa = m*pi/a; mpas = mpa^2; A2 = A1-mpas;
beta = sqrt(A2)
lemda = (2*pi/beta)
v = lemda*f0
ANSWER: beta = 308.7288rad./m lemda = 0.0204cm
v = 2.0351e+008 m/sec
Q-2.13 The cutoff frequency of a wave is 80% of its operating frequency. Find the wave
impedances for TE and TM waves if its characteristic impedance is 120π.
MATLAB CODE
% The required relations are: For TE wave: Zyx + = η/√{1-( / ) }
% For TM wave: Zxy+ = η √{1-( / ) }
% Given
fc = 0.8*f0; pi=3.14; fc0 = 0.8; fcs = fc0^2;
A1 = 1 - fcs; A2 = sqrt(A1); eta=120*pi;
%For TE wave:
Zyx = eta/A2
%For TM wave:
Zxy = eta*A2
ANSWER: Zyx + = 628.0000Ω Zxy
+ = 226.0800Ω
Q-2.14 A 20GHz, TE10 wave propagates between two parallel planes 3cm apart. Find wave
impedance if the space between planes is occupied by a material having μr = 2 and εr = 2.5.
MATLAB CODE
% The required relation is: =η/√{1-( / ) }
% Given
pi = 3.1415; a = 3*10^-2; muo = 4*pi*10^-7; aps0 = 8.854*10^-12; apsr = 2.5; mur
= 2; f = 2*10^10; m = 1; muap = mu0*aps0;
muapr = mur*apsr*muap; v = 1/sqrt(muapr); lemda = v/f;
A1 = m*lemda/(2*a); A2 = A1^2; A3 = 1 - A2;
A4 = sqrt(A3); eta0 = 120*pi;
Zyx = eta0/A4
ANSWER: Zyx = 379.3552Ω
Q-2.15 A 10GHz wave propagates between two parallel planes separated by 5cm. Find
attenuation constants (α) for TE10 wave if the space between planes is filled with a material
having μr = 1.2 and σm = 5.8 x 107 ℧/m and εr = 2.
MATLAB CODE
% The required relations are given below:
% Given TE10 mode
pi = 3.1415; a = 5*10^-2; muo = 4*pi*10^-7; aps0 = 8.854*10^-12; apsr = 2; mur =
1.2; f = 10^10; m = 1; sigma = 5*10^7; ac = a^3;
mps = (m*pi)^2; mpa=m*pi/a; mpas=mpa^2; w = 2*pi*f;
ws=w^2; mu = mur*mu0; aps = apsr*aps0; muap = mu*aps;
musig = mu/(2*sigma); wma = ws*muap; A1 = wma-mpas;
beta = sqrt(A1)
A2 = (w*mu)/(2*sigma);
Rs = sqrt(A2)
A4 = beta*w*mu*ac;
Alpha =( 2*mps*Rs)/A4
ANSWER: β = 318.5317rad/m Rs = 0.0308Ω α =1.6105e-004nepers/m
Q-2.16 Two strip shaped conductors of width b (= 5cm) and thickness t (= 1cm) in a two
parallel wire transmission line are separated by a distance a (= 10cm). Calculate the
inductance, capacitance and conductance, velocity of propagation and the characteristic
impedance if μ = μ0, ε = ε0 and σ = 3.72 x 107℧/m.
MATLAB CODE
% Given
a=10*10^-2; b=5*10^-2; t= 10^-2; pi=3.14; mu0 = 4*pi*10^-7; aps0 =
8.854*10^-12; sigma =3.72*10^7; mu = mu0; aps = aps0;
L = mu*(a/b)
C = aps*(b/a)
G = b*(sigma/a)
v = 1/sqrt(L*C)
Z = sqrt(L/C)
ANSWER: L = 2.5120e-006H/m C = 4.4270e-012F/m G = 18600000℧/m v =2.9987e+008 m/sec Z = 753.2776Ω
Q-2.17 A transmission line carries a sinusoidal signal of 109rad/sec. Find the velocity of
propagation if its parameters are: inductance L (= 0.4μH/m) and capacitance C (= 40pF/m)
and (a) R = 0, G = 0 (b) R = 0.1Ω/m, G = 10-5℧/m and (c) R = 300Ω/m, G = 0.
MATLAB CODE
% Given
L = 0.4*10^-6; C = 40*10^-12; w = 10^9; i = sqrt(-1);
WL =w*L; WC =w*C ;
% (a)
R = 0; G = 0; A1 = R + i*WL; A2 = G + i*WC; gama = sqrt(A1*A2);
alpha = real(gama); beta = imag(gama);
v = w/beta
%(b)
R = 0.1; G = 10^-5; A1 = R + i*WL; A2 = G + i*WC;
gama = sqrt(A1*A2); alpha = real(gama); beta = imag(gama);
v = w/beta
%(c)
R = 300; G = 0; A1 = R + i*WL; A2 = G + i*WC;
gama = sqrt(A1*A2); alpha = real(gama); beta = imag(gama);
v = w/beta
ANSWER: v =: (a) 250000000m/s (b) 250000000m/s (c) 2.3570e+008
Q-2.18 A 10cm long lossless transmission line has a characteristic impedance of 50Ω . Find
its input impedance at 50MHz if it is terminated in (a) open circuit (b) short circuit and (c)
10pF capacitor.
MATLAB CODE
% Given
f = 50*10^6; Z0 = 50; L = 0.1; w = 2*pi*f; v = 3*10^8;
beta = w/v; bet = beta*L; SN = sin(bet);
CS = cos(bet); TN = tan(bet); CT = cot(bet) ;
%(a) Line is open circuit or ZR is infinite
Zin = -i*Z0*CT
%(b) Line is short circuited or ZR = 0
Zin = i*Z0*TN
%(c) Line is terminated in 10pF capacitor
C = 10^-11; WC = w*C; ZR = -i*(1/WC);
A1 = ZR*CS + i*Z0*SN; A2 = Z0*CS + i*ZR*SN;
Zin = Z0*(A1/A2)
ANSWER : Zin =: (a) 0 -4.7596e+002i (b) 0 + 5.2525i
(c) 0 -1.8766e+002i
Q-2.19 A 100MHz transmission line has inductance L (= 0.4μH/m), capacitance C (=
40pF/m), resistance R (= 0.1Ω/m), and conductance G (= 10-5℧/m) Find propagation
velocity, phase shift constant (β), characteristic impedance (Z0), and attenuation constant (α).
MATLAB CODE
% Given
L = 0.4*10^-6; C = 40 *10^-12; R = 0.1; G = 10^-5; f = 10^8; pi=3.14;
i=sqrt(-1); w = 2*pi*f;
v = 1/sqrt(L*C)
beta = w/v
XL = w*L; XC = w*C; Z = R + i*XL; Y = G + i*XC;
gama = sqrt(Z*Y);
Z0 = sqrt(Z/Y)
alfa = (R/2)*sqrt(C/L) + (G/2)*sqrt(L/C)
ANSWER: v =2.5000e+008 m/s beta = 2.5120 rad/m
Z0 = 1.0000e+002 -2.7623e-018i ≈ 100 Ω alfa = 1.0000e-003 nepers/m
Q-2.20 A 50MHz transmission line with characteristic impedance of 50Ω is terminated in
200Ω. Find its reflection coefficient (Γ) and VSWR (s). Also find its quality factor (Q) if this
line is a resonant section and its inductance is L (= 0.4μH), capacitance C (= 40pF/m),
resistance is R (= 0.1Ω) and conductance G (= 10-5 ℧/m).
MATLAB CODE
% Given
ZR = 200; Z0 = 50; L = 0.4*10^-6; R = 0.1; C = 40*10^-12; G = 10^-5;
pi=3.14; f=50*10^6; alfa= 0.001; f = 50*10^6; w = 2*pi*f ;
% Reflection coefficient
RC = (ZR -Z0)/(ZR + Z0)
% Voltage standing wave ratio
SWR = (1 + RC)/(1 - RC)
% Quality factor
Q = (w*L)/R
ANSWER: Γ = 0.6000 ρ = 4 Q = 1.2560e+003
Q-2.21 A 300MHz transmission line of characteristic impedance 50Ω is terminated in an
unknown impedance R + jX. The voltage standing wave ratio on the line is 3. Find R and X
if the distance (L2) between the first minima and the terminating end is (a) 70cm (b) 30cm.
MATLAB CODE
% Given
f = 300*10^6; R0 = 50; ro = 3; v = 3*10^8; pi = 3.1416;
w = 2*pi*f; beta = w/v;
%(a)
L2 = 0.7; va = beta*L2; vs=sin(va); vss = vs^2;
vc = cos(va); vcs = vc^2; ros = ro^2; A1 = ro*R0;
A2 = ros*vcs + vss;
R = A1/A2
A3 = R0*(ros - 1)*vs*vc;
X = -A3 /A2
%(b)
L2 = 0.3; va = beta*L2; vs=sin(va); vss = vs^2;
vc = cos(va); vcs = vc^2; ros = ro^2;
A1 = ro*R0; A2 = ros*vcs + vss;
R = A1/A2
A3 = R0*(ros - 1)*vs*vc;
X = -A3 /A2
ANSWER: R = 85.0396 Ω X = -66.6448 Ω
R = 85.0363 Ω X = 66.6449 Ω
Q-2.22 Calculate the inductance (L) and capacitance (C), velocity of propagation (υ),
wavelength (λ) and the characteristic impedance (Z0) for a lossless coaxial line operating at
3GHz and having:
(a) a = 1mm, b = 3mm, μr = εr = 1
(b) a = 1mm, b = 3mm, μr = 1 and εr = 3
(c ) a = 1mm, b = 5mm, μr = 1 and εr = 3
MATLAB CODE
% Given
f = 3*10^9; mu0 = 4*pi*10^-7; aps0 = 8.854*10^-12;
muap0 = mu0*aps0; v0 = 1/sqrt(muap0); lemda0 = v0/f
%(a)
a = 10^-3; b = 3*10^-3; mur = 1; apsr = 1; mu = mu0*mur;
aps = aps0*apsr; muapr = mur*apsr; muaprs = sqrt(muapr);
v = v0/muaprs
lemda = lemda0/muaprs
Lba = log (b/a);
L = (mu/(2*pi))*Lba
C = (2*pi*aps)/Lba
Z0 = 60*muaprs*Lba
%(b)
a = 10^-3; b = 3*10^-3; mur = 1; apsr = 3; mu = mu0*mur
aps = aps0*apsr; muapr = mur*apsr; muaprs = sqrt(muapr);
v = v0/muaprs
lemda = lemda0/muaprs
Lba = log (b/a);
L = (mu/(2*pi))*Lba
C = (2*pi*aps)/Lba
Z0 = 60*muaprs*Lba
%(c)
a = 10^-3; b = 5*10^-3; mur = 1; apsr = 3; mu = mu0*mur;
aps = aps0*apsr; muapr = mur*apsr; muaprs = sqrt(muapr);
v = v0/muaprs
lemda = lemda0/muaprs
Lba = log (b/a);
L = (mu/(2*pi))*Lba
C = (2*pi*aps)/Lba
Z0 = 60*muaprs*Lba
ANSWER:
(a) L = 2.1972e-007 H/m C = 5.0638e-011F/m v =2.9980e+008 m/s
lemda = 0.0999m Z0 = 65.9167 Ω
(b) L = 2.1972e-007 H/m C = 1.5191e-010 F/m v = 1.7309e+008 m/s
lemda = 0.0577m Z0 = 114.1711 Ω
(c) L = 3.2189e-007 H/m C = 1.0370e-010 F/m v = 1.7309e+008 m/s
lemda = 0.0577m Z0 = 167.2577 Ω
Q-2.23 The real component of a wave traveling in x-direction in a lossless media is
expressed by the relation )cos()],(Re[ 1 xtCtxEy βω −= . Plots Ey for ωt = 0, π/4, π/2, 3π/4
and π if C1 = 1 and β = 1 rad./m. Take x = 0, 0.1, 0.2, …10
MATLAB CODE
%Given
C1 = 1; bta = 1; x = 0:0.1:10; t = bta*x; y = C1*cos(t); y1 = C1*cos(t- pi/4);
y2 = C1*cos(t- pi/2); y3 = C1* cos(t-3 *pi/4); y4 = C1*cos(t- pi);
plot(t,y,'r',t,y1,'g',t,y2,'b',t,y3,'o',t,y4,'p ')
RESULT: Traveling wave pattern
Q-2.24 The real component of a 10GHz wave traveling in a lossless media in x-direction and
having forward and reflected components is expressed by the relation
)cos()cos()],(Re[ 21 xtCxtCtxEy βωβω ++−= . Plots Ey for time instants t = 0, T/8, T/4,
3T/8 and T/2 if C1 = 1 and C2 = 0.5. Take x = 0, 0.1, 0.2, …1
MATLAB CODE
%Given
C1 = 1; C2 = 1; bta = 1; x = 0:0.2:10; t = bta*x;
y = C1*cos(t) + C2* cos(t); y1 = C1*cos(t- pi/4) + C2*cos(t+ pi/4);
y2 = C1*cos(t- pi/2) + C2*cos(t+ pi/2); y3 = C1* cos(t-3*pi/4) + C2* cos(t+3*pi/4);
y4 = C1*cos(t- pi) + C2*cos(t+ pi);
plot(t,y,'r',t,y1,'g',t,y2,'b',t,y3,'o',t,y4,'p ')
RRESULT: Standing wave pattern
0 1 2 3 4 5 6 7 8 9 10-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Q-2.25 The real component of a wave traveling in a lossy media in x-direction is expressed
by the following relations )cos()],(Re[ 1 xteCtxE xy βωα −= − . Plots Ey for ωt = 0, π/4,
π/2, 3π/4 and π if C1 = 1, α = 1.885 x 10-3 Nep./m and β = 0.001 rad./m. Take x = 0, 0.1,
0.2, …1
MATLAB CODE
%Given
C1 = 1; bta = 1; alfa = 0.1; x = 1:2:101; x1 = 0.2*x; t = bta*x1;
C = C1*exp(-alfa*x1); D = cos (t); D1 = cos (t-pi/2); D2= cos(t-pi);
y = C.* D; y1 = C.* D1; y2 = C.* D2
plot(t,y,'r',t,y1,'g',t,y2,'b')
RESULT: Decaying traveling wave pattern
0 1 2 3 4 5 6 7 8 9 10-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Q.2.26 Illustrate the orientation of E field vectors at different time instants for linearly
polarized wave when the phase difference between Ex and Ey is (a) 0o (b) 180o. Assume E0x
& E0y to be the max. amplitudes of Ex and Ey respectively.
MATLAB CODE RESULT:
% (a) Given phase difference = 0 rad.
E0x = 0.8; E0y = 0.6;
phi = 0; f = 1000;
T = 1/f; w = 2*pi*f;
t = 0:0.1*T: T;
E = E0x.*sin(w.*t) + i*E0y.*sin(w.*t - phi);
compass(E);
0 5 10 15 20 25-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
% (b) Given phase difference = pi rad.
E0x = 0.8; E0y = 0.6;
phi = pi; f = 1000;
T = 1/f; w = 2*pi*f;
t = 0:0.1*T: T;
E = E0x.*sin(w.*t) + i*E0y.*sin(w.*t - phi);
compass(E);
Q.2.27 Illustrate the orientation of E field vectors at different time instants for circularly
polarized wave when the phase difference between Ex and Ey is (a) 90o (b) 270o. Assume
E0x & E0y to be the maximum amplitudes of Ex and Ey respectively.
MATLAB CODE RESULT:
% (a) Given phase difference = pi/2 rad.
E0x = 1; E0y = 1;
phi = pi/2; f = 1000;
T = 1/f; w = 2*pi*f;
t = 0:0.1*T:T;
E = E0x.*sin(w.*t) + i*E0y.*sin(w.*t - phi);
compass(E);
% (b) Given phase difference = 3pi/2 rad.
E0x = 1; E0y = 1;
phi = 3*pi/2; f = 1000;
T = 1/f; w = 2*pi*f;
t = 0:0.1*T:T;
E = E0x.*sin(w.*t) + i*E0y.*sin(w.*t - phi);
compass(E);
Q.2.28 Illustrate the orientation of E field vectors at different time instants for elliptically
polarized wave when the phase difference between Ex and Ey is (a) 45o (b) 135o. Assume
E0x & E0y to be the maximum amplitudes of Ex and Ey respectively.
MATLAB CODE RESULT:
% (a) Given phase difference = pi/4 rad.
E0x = 0.8; E0y = 0.6;
phi = pi/4; f = 1000;
T = 1/f; w = 2*pi*f;
t = 0:0.1*T:T;
E = E0x.*sin(w.*t) + i*E0y.*sin(w.*t - phi);
compass(E);
Join the tips of arrows to get an ellipse
% (b) Given phase difference = 3pi/4 rad.
E0x = 0.8; E0y = 0.6;
phi = 3*pi/4; f = 1000;
T = 1/f; w = 2*pi*f;
t = 0:0.1*T:T;
E = E0x.*sin(w.*t) + i*E0y.*sin(w.*t - phi);
compass(E);
Join the tips of arrows to get an ellipse
INTRODUCTION
MICROWAVES (MWs)
• extremely short waves
• 1/3m to mm or even less
• f 1GHz but < optical freq.
• Include UHF, SHF & EHF bands
At MWs
Lumped element circuit theory becomes inaccurate
Distributed circuit elements become more relevant
Tx-line theory becomes more
useful for design and analysis of
systems
Open-wire and coaxial lines are
replaced by waveguides and
planar lines
Lumped-element tuned
circuits are replaced by
cavity resonators
Effects of reflection,
refraction, polarization and
absorption become more
significant
Frequency bands are segments of frequency spectrum
Help in identification of applications and technologies used
in different frequency ranges
EXTREMELY LOW FREQUENCY (ELF)
SUPPER LOW FREQUENCY (SLF)
ULTRA LOW FREQUENCY (ULF)
VERY LOW FREQUENCY (VLF)
LOW FREQUENCY (LF)
MEDIUM FREQUENCY (MF)
HIGH FREQUENCY (HF),
VERY HIGH FREQUENCY (VHF),
ULTRA HIGH FREQUENCY (UHF),
SUPPER HIGH FREQUENCY (SHF),
EXTREMELY HIGH FREQUENCY (EHF)
TERA HZ FREQUENCY (THF)
GENERAL RULE
Frequency (f) of a band lies
between
3 10n-1 to 3 10n Hz
where n is the number of band
Example
For HF (n = 7)
f lies between 3 to 30 MHz
Symbols assigned to different bands
by NATO
A to M in continuity
by IEEE
HF, VHF, UHF, L, S, C, X, Ku. K, Ka, Q, V and W
Frequency ranges
cm wave: 3 to 30GHz
mm wave: 30 to 300GHz
sub mm wave: 300 to 3000GHz
ADVANTAGES
In Communication
Very reliable for signal’s
transmission & reception
Consumes lesser BW of
information channel
Permits more channels than at
lower carrier frequencies
IN RADARS
MW antennas - extremely directive
A pencil beam of 1⁰ beamwidth
can be obtained from an antenna of:
6900feet diameter at 10 MHz
6.9ft diameter At 10 GHz
Neither reflected nor absorbed by
ionosphere.
Suitable for space communication,
including radio-astronomy and satellites
Widely used for telephone network, TV
systems, railway and military applications
through LOS and Tropo links & Satellites.
Provides insight into the nature of
molecules and their interaction
Provides valuable information
about relaxation processes and
electron spin resonance
phenomena
Due to existence of molecular resonances
at MWs in certain crystal materials
energy can be generated or converted
by means of atomic oscillations
in and around avalanche conditions
Gunn, Read, Impatt diodes, etc. are based
on this property.
More easily directed, controlled and
concentrated
Extensively used in industry
for heating, drying, testing, analysis
and processing
Also used for cocking and physical
diathermy
Microwave techniques
may also be used in
extremely fast computer
applications
Pulses with very small widths are
used in high speed logic circuits
COMPONENTS OF MW SYSTEM
Microwave generation
Microwave processing
Microwave transmission
Microwave measurements
Microwave antennas
APPLICATIONS
Application Freq.
band(s) Frequency
range(s) in GHz Air borne whether avoidance
radar C 5.4
Anti aircraft weapon system MMW 94 & 140 Astronomical radars P,L, S &
C 0.408, 0.43, 0.7,
1.295, 2.388, 8 Automatic Toll Collection P & C 0.905 & 5 – 6 Battle field surveillance radar MMW 70 Cellular Phone P 0.824 – 0.849 &
0.869 – 0.895 Cellular Video Ka 28 Collision Avoidance Radar MMW 60, 77 & 94 Direct Broadcast Satellite X 11.7 – 12.5 Global Digital Satellite C 4-8
Global Positioning Satellite L 1.575 & 1.227 Ground Based Radar,
Navigation X 8-12
GSM P 0.890 – 0.915 &
0.935 – 0.960 Instrumentation radars S & C 2.9, 5.6, 1 – 4 Missile guidance seeker
system with radiometric
sensors
MMW 35 & 94
Paging P 0.931-0.932 Personal Communication
System L & S 1.85 – 1.99 & 2.18
–2.20
Radio altimeters C 4.2 – 4.4
Ship borne navigation radars X 9.345 – 9.405
Special Telecom Services Ka 27-40
Surveillance and acquisition
radar
MMW 94
Synthetic Aperture Radar P 0.23 - 1
VSAT Networks Ku 12-18
Wide Area Computer Networks MMW 60
Wireless Local Area Network S 2-4
Wireless Local Area Networks S & C 2.4-2.48 & 5.4
Courtesy of CRC Press/Taylor & Francis Group
002x001.eps
f1(x – υ0t1) f2(x+ υ0t1)
f2(x + υ0t2)
t = t1
(a) (b)
t = –t1
t = –t2
xx
x
t1 – t2
t = t2
t2 – t1
f1(x – υ0t2)
Figure 2.1Two forms of wave. (a) Forward wave. (b) Backward wave.
Courtesy of CRC Press/Taylor & Francis Group
002x002.eps
Ey(x, t) Ey(x, t)
0 0
(a) (b)
x x
ωt = π/4t = T/4
t = T/8
t = 0
t = 3T/8ωt = π/2 υ = ω/β
υ = ω/β
ωt = 0
Figure 2.2Wave in lossless media. (a) Travelling wave. (b) Standing wave.
Courtesy of CRC Press/Taylor & Francis Group
002x003.eps
Ey(x, t)ωt=0 ωt=π/2
ωt=π/4
0
e–αx
x
Figure 2.3Travelling waves in lossy media.
Courtesy of CRC Press/Taylor & Francis Group
002x004.eps
Perfect
0.1 Dielectrics Conductors1 ∞
Good Average Poor Poor Average Good Perfectσ/ωε
Figure 2.4Dielectrics and conductors.
Courtesy of CRC Press/Taylor & Francis Group
002x005.eps
x
x
yϕ = 0°
t =
t =
0
0
8T
ϕ = 180°y
0
0
Ex
Ex
Ey
Ey
z
z
8T
Magnitude and angular orientation of vector E at different time instantsϕ Phase shift between Ex and Ey
4T
4T
2T
2T
83T
83T
85T
85T
23T
23T
87T T
T87T
|Ex| = |Ey| E = Ex2 + Ex2
Figure 2.5Linear polarisation.
Courtesy of CRC Press/Taylor & Francis Group
002x006.eps
Ex
Ex
x
x
z
z
0
y
y
0
Ey
Ey
φ = 270°
φ = 90°
t = 0 8T
4T
2T
83T
85T
23T
87T T
t = 0 8T
4T
2T
83T
85T
23T T8
7T
|Ex| = |Ey| E = Ex2 + Ey
2 φ Phase shift between Ex and EyMagnitude and angular orientation of vector E, at different time instants
If an envelope is traced by joining the tips of rotating vectors the resultant will appear to be a circle
Figure 2.6Circular polarisation.
Courtesy of CRC Press/Taylor & Francis Group
002x007.eps
Ex
Ex
x
x
Ey
T/4T/8 3T/8 5T/8 7T/8
T/2 T t3T/2
Ey
y
y
0
0
0
φ = 45°
φ = 225°
z
z
t = 0 T/8 T/4 T/23T/4 5T/8 3T/2 7T/8 T
t = 0 T/8 T/4 T/23T/4 5T/8 3T/2 7T/8 T
|Ex| = |Ey| E = Ex2 + Ey
2
Magnitude and angular orientation of vector E,at different time instants
If an envelope is traced joining the tips of these rotating vectors theresultant will appear to be an ellipse (shown by dotted lines)
Figure 2.7Elliptical polarisation.
Courtesy of CRC Press/Taylor & Francis Group
002x008.eps
Represents I2Rlosses in the
volume
P1 = –�v(E · J)dv
P2 = – �v( H2)dv
Volume V enclosedby surface S
Represents rate ofdecrease of electric
energy in the volume
Represents rate ofdecrease of magneticenergy in the volume
��t
µ2
P2 = – �v( E2)dv��t
ε2
Figure 2.8Components of Poynting vector.
Courtesy of CRC Press/Taylor & Francis Group
002x009.eps
F
0
90°ñ
B řC
A
M
SWave
normal
(a) (b)
Radiusvector
F
P
λ = 2π/β
λυ = ω/β
υ0 or c υz λz
υyθ
λy
P
Strikingwave
Q
P′
R′
Q′
R
zz
x
y y
Figure 2.9Direction cosines, wavelength and phase velocity. (a) Direction cosines. (b) Wavelength and velocity components.
Courtesy of CRC Press/Taylor & Francis Group
002x010.eps
Reflection in case of
Conductors
Normalincidence
Polarisation Polarisation Polarisation
Obliqueincidence
Normalincidence
Obliqueincidence
Dielectrics
Polarisation
Figure 2.10Classification of cases of reflection.
Courtesy of CRC Press/Taylor & Francis Group
002x011.eps
E perpendicular toplane of incidence
E parallel toplane of incidence
E coming out
E going in
Medium-2
Medium-1Plane of incidence
Boundary surface
Figure 2.11Illustration of E-field on the plane of incidence.
Courtesy of CRC Press/Taylor & Francis Group
002x012.eps
Ei
z
0
σ = 0 σ = ∞
x
Er
Figure 2.12Normal incidence—perfect conductor.
Courtesy of CRC Press/Taylor & Francis Group
002x013.eps
Et Htt = 0
σ = ∞ σ = ∞
T/8T/4
3T/8
(a) (b)
t = T/4
5T/8T/23T/8
Figure 2.13Standing waves for (a) E and (b) H fields.
Courtesy of CRC Press/Taylor & Francis Group
002x014.eps
z
x
Hi
Hr
0
Ht
σ = 0
η1= µ1/ε1
η2= µ2/ε2
Figure 2.14Normal incidence—perfect dielectric.
Courtesy of CRC Press/Taylor & Francis Group
002x015.eps
Ei Ei
θ1 θ1θ1 θ1
θ2 θ2
ErEr
HiHi
HtHt
Et Et
Medium-1 Medium-1
Medium-2
(a) (b)
Medium-2
HrHr
z z
y y0 0
Figure 2.15Illustration of two forms of polarisation. (a) Perpendicular or horizontal polarisation. (b) Parallel or vertical polarisation.
Courtesy of CRC Press/Taylor & Francis Group
002x016.eps
Minima
0
z
yn n′
Ei HrEr
Hi
z
y
Maxima
Maxima
Maxima
Incidentray
Reflectedray
Minima
Minima
Minima
(a) (b)
σ = ∞
θ θ
σ = ∞
Figure 2.16Perpendicular polarisation case. (a) Orientation of E and H field. (b) Resulting standing wave pattern along z-axis.
Courtesy of CRC Press/Taylor & Francis Group
002x017.eps
Ei Er
Hr
z
0
θθ
σ = ∞ y
Ez
EyHi
Figure 2.17E in parallel polarisation.
Courtesy of CRC Press/Taylor & Francis Group
002x018.eps
Incident rays
θ1
η1 ε1
η2 ε2
θ3
θ2
Boundarysurface
Medium-1
E
A
D
y
z
B
C
Medium-2
Reflected rays
Transmitted rays
Figure 2.18Incident, reflected and transmitted rays in case of oblique incident.
Courtesy of CRC Press/Taylor & Francis Group
002x019.eps
x
y
0 Pa
z
Figure 2.19Configuration of two parallel planes.
Courtesy of CRC Press/Taylor & Francis Group
002x020.eps
x x x x
z z y y
Hx
Hx and Hz Ey aloneResultant H-field Resultant E-field
Hz Ey EH
(a) (b)
Figure 2.20Field configuration for TE1,0 mode. (a) Field distribution in x–z plane. (b) Field distribution in x–y plane.
Courtesy of CRC Press/Taylor & Francis Group
002x021.eps
x x
z
x x
z y y
EzHy HEx E
Ex and Ez Resultant E-field
(a) (b)
Resultant H-fieldHy alone
Figure 2.21Field configuration for TM1,0 mode. (a) Field distribution in x–z plane. (b) Field distribution in x–y plane.
Courtesy of CRC Press/Taylor & Francis Group
002x022.eps
E
x
a
x
b
b0
H
(a) (b) (c)
EE
H
Ha
z
z y
y
Figure 2.22Field distribution for TEM wave. (a) In x–z plane. (b) In y–z plane. (c) In x–y plane.
Courtesy of CRC Press/Taylor & Francis Group
002x023.eps
X
Hx
(a) (b)
a a
θ θvg vgEx
Directionof wave
propagation
Directionof wave
propagation
X
YZ
YZ
Figure 2.23Zig-zag paths and field components of (a) TE and (b) TM waves.
Courtesy of CRC Press/Taylor & Francis Group
002x024.eps
Region ofno wave
propagationTE wave
TM wave
z
η
ffc
Figure 2.24Variation of impedances with frequency.
Courtesy of CRC Press/Taylor & Francis Group
002x025.eps
TM wave
TE wavefc 3fc f
TEM waveAtte
nuat
ion
α
Figure 2.25Variation of attenuation with frequency.
Courtesy of CRC Press/Taylor & Francis Group
002x026.eps
x
yx
a
A
BG H
EE
F K H
y
bz z
C
D
y
b
z
Δz
Δz
a
a
xz
b
EH
t
0
0
Pb
TEM Wave
(a)
(d)(e)
(c)Fig (d) seenfrom front
Fig (c) seenfrom side
(b) Two strips removed
from planes
Fig (e) seendownward
Figure 2.26Configuration of transmission line and field distribution.
Courtesy of CRC Press/Taylor & Francis Group
002x027.eps
IS
ZS ZR VR
IR
VS
Z0 γ
z = –l z = 0
Figure 2.27Circuit representation of transmission line.
Courtesy of CRC Press/Taylor & Francis Group
002x028.eps
Standing wavesV
V V
VI
I I
IIm
Im
Im
Z0 = R0
R = Z0 R ≠ R0Z0 = R0 Z0 = R0ZR = Z0
Z0 = R0
x = 0
x = 0 x = 0
x=0
ZR = 0 ZR = ∞
ZR ≠ Z0
Vm
Vm
Vm
No standing waves
(b)(a)
(d)(c)
Standing waves
Standing waves
Figure 2.28Voltage and current distribution for different terminations. (a) Line short circuited. (b) Line open circuited. (c) Line terminated in characteristic impedance. (d) Terminated impedance is not equal to characteristic impedance.
Courtesy of CRC Press/Taylor & Francis Group
002x029.eps
R1 < R0
x = 0
R ± jX
R1
L2
L2
λ/2
λ/2 – L1
λ/2L2L1
L1
L1
ZR
ZR
a
(a)
Wave sloping downat termination
Wave sloping upat termination
(b) (d)
(c)
b c
R1 < R0
Figure 2.29Voltage and current distributions for termination in complex impedance Z = R ± X. (a) Z is located at x = 0, (b) Z is replaced by R at a distance L1 from termination, (c) X is inductive and (d) X is capacitive.
Courtesy of CRC Press/Taylor & Francis Group
002x030.eps
Reflection withoutphase reversalfrom OC end
OCend
SCend
(a) (b)Quarter
wavesection
b
a
11
11
11
2
2
2 2
2
4 4
4
35
6
3
3
35
c
Feedpoint
Reflection withphase reversalfrom SC end
Figure 2.30Resonance phenomenon. (a) Quarter-wave section and (b) voltage buildup.
Courtesy of CRC Press/Taylor & Francis Group
002x031.eps
SC1
SC
SC
j Z0 tan βl
–j Z0 tan βl
–j Z0 cot βl
2 Z02/RI
2 Z02/RI
λ/4λ/2
RI/2
RI/2
j Z0 cot βl
SC
OC
OC
OC
OC
2
3
4
5
6
7
8
Figure 2.31Line sections with equivalent elements for different lengths and terminations.
Courtesy of CRC Press/Taylor & Francis Group
002x032.eps
0
(a) (b)
0π/2 π/23π/2 3π/22π 2ππ π
tan βl
βl βl
cot βl
jZ0 tan βl – jZ0 tanβl –jZ0 cot βl jZ0 cot βl – jZ0 cot βl jZ0 cot βl
2Z02/Rl 2Z0
2/Rl2Z02/Rl 2Z0
2/RlRl/2 Rl/2Rl/2Rl/2
jZ0 tanβl – jZ0 tanβl
Figure 2.32Circuit elements for different electrical lengths and terminations. (a) Short-circuited lines. (b) Open-circuited lines.
Courtesy of CRC Press/Taylor & Francis Group
002x033.eps
SCx
– xλ/4
OCVx
XLXCVs
Rx
Rs
λ4
Figure 2.33Transmission line as a tuned circuit.
Courtesy of CRC Press/Taylor & Francis Group
002x034.eps
Line 1
ZS1 ZR1 = ZS2
ZS1 = Z021/ZR1 ZS2 = Z02
2/ZR2
Z01 Z02ZR2
Line 2
Figure 2.34Impedance transformation.
Courtesy of CRC Press/Taylor & Francis Group
002x035.eps
Generator
(a) (b)
GeneratorAntennaZOCZSCZOC
SC
SCOC
λ/12λ/4
λ/6
Antenna
Figure 2.35Suppression of (a) third harmonic and (b) even harmonics.
Courtesy of CRC Press/Taylor & Francis Group
002x036.eps
YC YA
YS
YSCYR
YC
YS
YA
LoadYR
S
d
βS
βSβd
βd
Figure 2.36Single-stub matching.
Courtesy of CRC Press/Taylor & Francis Group
002x037.eps
YSCβS2
βd βll
βS1
βd
βl
YS1
YSC
YCYR
YS2
YS1
Y11
Yd
Y22
YR
Load
YS2
Y11Y22YdYC
d2
2 1
1
Figure 2.37Double stub matching.
Courtesy of CRC Press/Taylor & Francis Group
002x038.eps
V = 0
Innerconductor
E lines
H lines
Dielectricor air
Outerconductor
(a) (b) (c) (d)
V = V0z
y
x
a
2a
2bb
pφ
Figure 2.38Coaxial cable. (a) Geometry, (b) cross section, (c) TEM mode and (d) TE11 mode.
Courtesy of CRC Press/Taylor & Francis Group
002x039.eps
Unconfined fieldInner conductor
E lines
Outer conductor
Confinementregion
(a) (b)
Field confinedbetween conductors
Figure 2.39End views and field distributions. (a) Two-wire Tx line. (b) Coaxial cable.
Courtesy of CRC Press/Taylor & Francis Group
002x040.eps
Conductors(a)
S1 S2
(b)
Figure 2.40Relative spacing in coaxial cable and circular waveguide. (a) Coaxial cable. (b) Circular waveguide.