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

delta = k1/sig

% (c) for brass

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:

Rscu = 1/A2

%(c) Aluminium:

Rsal = 1/A3

%(d) Brass:

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)

mur = 1; apsr = 1; mu = mur*mu0; aps = apsr*aps0;

eta1 = sqrt(mu/aps)

Pxav= (E0z^2)/(2*eta1)

mur = 1; apsr = 2.26; mu = mur*mu0; aps = apsr*aps0;

eta2 = sqrt(mu/aps)

Pxav= (E0z^2)/(2*eta2)

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

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

aps1 = 1; aps2 = 2.53; a = sqrt(aps1); b = sqrt(aps2);

c = a - b; d = a + b; e = b - a;

Taue = c/d

Touh=2*a/d

Te = e/d

Th=2*b/d

aps1 = 2.53; aps2 = 2.26; a = sqrt(aps1); b = sqrt(aps2);

c = a - b; d = a + b; e = b - a;

Taue = c/d

Touh=2*a/d

Te = e/d

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;

Pr = Er*Hr

Pt= Et * Ht

%(c) Given

RE = 0.02817; RH = -0.02817; TE =1.0282; TH = 0.97179;

Pr = Er*Hr

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;

aps1 = 2.53; aps2 = 1; aps21= (aps2/aps1);

A1 = (aps21-Sst); A2 = sqrt(A1);

RC= (Cs-A2)/(Cs+A2)

aps1 = 1; aps2 = 2.53; aps21= (aps2/aps1);

RC= (Cs-A2)/(Cs+A2)

aps1 = 2.53; aps2 = 2.26; aps21= (aps2/aps1);

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;

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)

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)

apsr =1; aps = 1/sqrt(apsr);

for m = 1:1:3

fc = m*(v/aps)

apsr =2.26; aps = 1/sqrt(apsr);

for m = 1:1:3

fc = m*(v/aps)

apsr =2.53; aps = 1/sqrt(apsr);

for m = 1:1:3

fc = m*(v/aps)

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 ;

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

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

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;

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

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 = 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

a = 10^-3; b = 3*10^-3; mur = 1; apsr = 3; mu = mu0*mur

v = v0/muaprs

Lba = log (b/a);

L = (mu/(2*pi))*Lba

C = (2*pi*aps)/Lba

Z0 = 60*muaprs*Lba

a = 10^-3; b = 5*10^-3; mur = 1; apsr = 3; mu = mu0*mur;

v = v0/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

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

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

% (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;

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;

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;

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;

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;

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

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

t1 – t2

t = t2

t2 – t1

f1(x – υ0t2)

Figure 2.1Two forms of wave. (a) Forward wave. (b) Backward wave.

002x002.eps

Ey(x, t) Ey(x, t)

(a) (b)

ωt = π/4t = T/4

t = T/8

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.

002x003.eps

Ey(x, t)ωt=0 ωt=π/2

ωt=π/4

e–αx

Figure 2.3Travelling waves in lossy media.

002x004.eps

Perfect

0.1 Dielectrics Conductors1 ∞

Good Average Poor Poor Average Good Perfectσ/ωε

Figure 2.4Dielectrics and conductors.

002x005.eps

yϕ = 0°

ϕ = 180°y

Magnitude and angular orientation of vector E at different time instantsϕ Phase shift between Ex and Ey

|Ex| = |Ey| E = Ex2 + Ex2

Figure 2.5Linear polarisation.

002x006.eps

φ = 270°

φ = 90°

t = 0 8T

23T T8

|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.

002x007.eps

T/4T/8 3T/8 5T/8 7T/8

T/2 T t3T/2

φ = 45°

φ = 225°

t = 0 T/8 T/4 T/23T/4 5T/8 3T/2 7T/8 T

|Ex| = |Ey| E = Ex2 + Ey

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.

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

P2 = – �v( E2)dv��t

Figure 2.8Components of Poynting vector.

002x009.eps

90°ñ

normal

(a) (b)

Radiusvector

λ = 2π/β

λυ = ω/β

υ0 or c υz λz

Strikingwave

Figure 2.9Direction cosines, wavelength and phase velocity. (a) Direction cosines. (b) Wavelength and velocity components.

002x010.eps

Reflection in case of

Conductors

Normalincidence

Polarisation Polarisation Polarisation

Obliqueincidence

Normalincidence

Obliqueincidence

Dielectrics

Polarisation

Figure 2.10Classification of cases of reflection.

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.

002x012.eps

σ = 0 σ = ∞

Figure 2.12Normal incidence—perfect conductor.

002x013.eps

Et Htt = 0

σ = ∞ σ = ∞

T/8T/4

(a) (b)

t = T/4

5T/8T/23T/8

Figure 2.13Standing waves for (a) E and (b) H fields.

002x014.eps

σ = 0

η1= µ1/ε1

η2= µ2/ε2

Figure 2.14Normal incidence—perfect dielectric.

002x015.eps

θ1 θ1θ1 θ1

θ2 θ2

Medium-1 Medium-1

Medium-2

(a) (b)

Medium-2

y y0 0

Figure 2.15Illustration of two forms of polarisation. (a) Perpendicular or horizontal polarisation. (b) Parallel or vertical polarisation.

002x016.eps

Minima

yn n′

Ei HrEr

Maxima

Incidentray

Reflectedray

Minima

(a) (b)

σ = ∞

Figure 2.16Perpendicular polarisation case. (a) Orientation of E and H field. (b) Resulting standing wave pattern along z-axis.

002x017.eps

σ = ∞ y

Figure 2.17E in parallel polarisation.

002x018.eps

Incident rays

η1 ε1

η2 ε2

Boundarysurface

Medium-1

Medium-2

Reflected rays

Transmitted rays

Figure 2.18Incident, reflected and transmitted rays in case of oblique incident.

002x019.eps

Figure 2.19Configuration of two parallel planes.

002x020.eps

x x x x

z z y y

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.

002x021.eps

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.

002x022.eps

(a) (b) (c)

Figure 2.22Field distribution for TEM wave. (a) In x–z plane. (b) In y–z plane. (c) In x–y plane.

002x023.eps

(a) (b)

θ θvg vgEx

Directionof wave

propagation

Directionof wave

propagation

Figure 2.23Zig-zag paths and field components of (a) TE and (b) TM waves.

002x024.eps

Region ofno wave

propagationTE wave

TM wave

Figure 2.24Variation of impedances with frequency.

002x025.eps

TM wave

TE wavefc 3fc f

TEM waveAtte

Figure 2.25Variation of attenuation with frequency.

002x026.eps

TEM Wave

(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.

002x027.eps

ZS ZR VR

z = –l z = 0

Figure 2.27Circuit representation of transmission line.

002x028.eps

Standing wavesV

Z0 = R0

R = Z0 R ≠ R0Z0 = R0 Z0 = R0ZR = Z0

Z0 = R0

x = 0 x = 0

ZR = 0 ZR = ∞

ZR ≠ Z0

No standing waves

(b)(a)

(d)(c)

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.

002x029.eps

R1 < R0

R ± jX

λ/2 – L1

λ/2L2L1

Wave sloping downat termination

Wave sloping upat termination

(b) (d)

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.

002x030.eps

Reflection withoutphase reversalfrom OC end

(a) (b)Quarter

wavesection

Feedpoint

Reflection withphase reversalfrom SC end

Figure 2.30Resonance phenomenon. (a) Quarter-wave section and (b) voltage buildup.

002x031.eps

j Z0 tan βl

–j Z0 tan βl

–j Z0 cot βl

2 Z02/RI

λ/4λ/2

j Z0 cot βl

Figure 2.31Line sections with equivalent elements for different lengths and terminations.

002x032.eps

(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.

002x033.eps

– xλ/4

XLXCVs

Figure 2.33Transmission line as a tuned circuit.

002x034.eps

Line 1

ZS1 ZR1 = ZS2

ZS1 = Z021/ZR1 ZS2 = Z02

Z01 Z02ZR2

Line 2

Figure 2.34Impedance transformation.

002x035.eps

Generator

(a) (b)

GeneratorAntennaZOCZSCZOC

λ/12λ/4

Antenna

Figure 2.35Suppression of (a) third harmonic and (b) even harmonics.

002x036.eps

LoadYR

βSβd

Figure 2.36Single-stub matching.

002x037.eps

YSCβS2

βd βll

Y11Y22YdYC

Figure 2.37Double stub matching.

002x038.eps

Innerconductor

E lines

H lines

Dielectricor air

Outerconductor

(a) (b) (c) (d)

V = V0z

Figure 2.38Coaxial cable. (a) Geometry, (b) cross section, (c) TEM mode and (d) TE11 mode.

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.

002x040.eps

Conductors(a)

Figure 2.40Relative spacing in coaxial cable and circular waveguide. (a) Coaxial cable. (b) Circular waveguide.

CHAPTER-2 - Best Shop for Test Banks and Solution Manuals · ANSWER: RC =: (a) 0.3888 (b) -0.2710...

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