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Page 1: Microwaves and Radar - Navodaya Institute of … and Radar Time: 3 hrs. Max. Marks: ... where ZY ZY2 ZY j γ= ⇒γ = ⇒γ ... on smith chart & name that point as 'A'. b) ...

December 2012Fifth Semester B.E. Degree Examination (10EC54)

Microwaves and RadarTime: 3 hrs. Max. Marks: 100

Note: 1. Answer any FIVE full questions, selecting at least two questions from each part.

PART - A

1. a. Derive transmission line equations by the method of distributed circuit theory. (09 Marks)Ans:

v(z, t)

i(z, t)

i(z, t) z�z = (R +j�L)�z

i(z,+�z, t)

G�z C�z

i(z+�z, t)

v(z + �z, t)

�z

The change in voltage w.r.t the distance �z along the line is given by

iv Ri z L z

t

v iRi L

z t

∂Δ = Δ + Δ∂

Δ ∂∴ = +Δ ∂

v vAs z 0,

z z

Δ ∂Δ → →Δ ∂

v iRi L

z t

∂ ∂∴ = +∂ ∂ (1)

Similarly change in current w.r.t �z is given byv

i G zv C zt

i vGV C

z t

∂Δ = Δ + Δ∂

Δ ∂∴ = +Δ ∂

i iAgain as z 0,

z z

Δ ∂Δ → →Δ ∂

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Page 2: Microwaves and Radar - Navodaya Institute of … and Radar Time: 3 hrs. Max. Marks: ... where ZY ZY2 ZY j γ= ⇒γ = ⇒γ ... on smith chart & name that point as 'A'. b) ...

December 2012 Microwave and RadarSet 2 - ��

i VGV C

z t

∂ ∂∴ = +∂ ∂

(2)

Differentiating eqn (1) w.r.t z partially, we get2

2

v i iR L

z z t z

∂ ∂ ∂ ∂⎛ ⎞= + ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ (3)

Using (2) in (3)2

2

v v VR GV C L GV C

z t t t

∂ ∂ ∂ ∂⎡ ⎤ ⎛ ⎞= + + +⎜ ⎟⎢ ⎥ ⎝ ⎠∂ ∂ ∂ ∂⎣ ⎦

( )2 2

2 2

v v vRGV RC GL LC

z t z

∂ ∂ ∂∴ = + + +∂ ∂ ∂

(4)

∂ ∂ ∂= + + +∂ ∂ ∂

2 2

2 2

i i iSimilarly, RGi (RC GL) LCz t t

(5)

Redenote the V & I in sinusodial formV = V (z, t) = V(z) ej�t (6)i = i (z, t) = I (z) ej�t (7)differentiating eqn (6) twice w.r.t z we get

2 2j t

2 2

V d v(z)e

z dzω∂ =

∂ (8)

differentiating eqn (6) w.r.t. t we get

( )j tvv(z) e j

tω∂ = ω

∂ (9)

differentiating again w.r.t. t

( )2

2j t2

vV(z) e j

tω∂ = ω

( )2

2 j t2

vv z e

tω∂∴ = −ω

∂ (10)

Using eqn (8) to eqn (10) in eqn (4) we get

( ) ( ) ( )( )2

j t j t j t 2 j t2

d V(z)e RGV(z)e RC GL V z j e LC V z e

dzω ω ω ω⎡ ⎤= + + ω + −ω⎣ ⎦

( ) ( )

( ) ( )

⎡ ⎤= + ω + − ω⎣ ⎦

= ⎡ + ω + ω + ω ⎤⎣ ⎦

22

2

d v(z) RG j RC GL LC V zdz

R G j C j L G j C V(z)

( )( )2

2

d V(z)R j L G j C V(z)

dz∴ = + ω + ω (11)

Let Z = R + j�L, Y = G + j�C (12)Z � series impedance/unit lengthy � series admittance/unit length

2

2

d V(z)ZY V(z)

dZ= (13)

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Page 3: Microwaves and Radar - Navodaya Institute of … and Radar Time: 3 hrs. Max. Marks: ... where ZY ZY2 ZY j γ= ⇒γ = ⇒γ ... on smith chart & name that point as 'A'. b) ...

Microwave and Radar December 2012 Set 2 - ��

or 2

2

d V(z)ZY V(z) 0

dZ= = (14)

Similarly, 2

2

d I(z)ZY I(z) 0

dZ− = (15)

� Transmission line equations are

= − = −dv dIZI, YVdz dz

(16)2 2

2 22 2

d v d Iv, I

dz dz= γ = γ (17)

2where ZY ZY

ZY j

γ = ⇒ γ =

⇒ γ = = α + β

1. b. A single stand turner is to match a lossless line of 100� to a load of (800 + j300) �. The frequency is 3GHz.

i) Find the distance in meters from the load to the tuning stub ii) Determine the length in meters of the short-circuited stub (06 Marks)Ans: Given Data,

Zo = 400�ZL = 800 + j 300 �l = ?d = ?Normalization:

L

o

1

Z 800 j3002 j0.75

Z 400

z 2 j 0.75

+= = +

⇒ = +

Procedure

a) Plot the normalized zL on smith chart & name that point as 'A'.

b) Draw a line from centre to zL point, i.e 0 to A

c) With OA as radius & 'O' as centre draw a circle & name it as VSWR circle

d) Draw a line diametrically opposite to A (zL) point to meet the VSWR circle point A� (y point)

e) Extend A to periphery, let that point be B

f) Locate the intersection point of VSWR circle & G = 1 circle and name this point as C. Draw a line from O to C as OC & extend it to periphery. Let this point be C.

To � nd 'd':

g) Move from point B to C in the clockwise direction (towards generator), so,

8

9

dBC 0.1853

c 3 10where 0.1m

f 3 10

= =λ

×λ = = =×

´

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Page 4: Microwaves and Radar - Navodaya Institute of … and Radar Time: 3 hrs. Max. Marks: ... where ZY ZY2 ZY j γ= ⇒γ = ⇒γ ... on smith chart & name that point as 'A'. b) ...

December 2012 Microwave and RadarSet 2 - ��

� d = 0.1853

� d = 0.1853 × 0.1

��d = 0.01853mTo � nd l:

h) Find the value of susceptance at point C. i.e + 0.85

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Page 5: Microwaves and Radar - Navodaya Institute of … and Radar Time: 3 hrs. Max. Marks: ... where ZY ZY2 ZY j γ= ⇒γ = ⇒γ ... on smith chart & name that point as 'A'. b) ...

Microwave and Radar December 2012 Set 2 - ��

To compensate this, stub is to be inserted at the point of – 0.85 susceptance, mark this point as D, on the periphery. From the short circuit end move towards D.

i.e PD 0.1308 0.1308 0.1m

0.01308m

= = ⇒ = ×λ

⇒ =

ll

l

1. c. De� ne re� ection co-ef� cient & derive an expression for re� ection co-ef� cient at load in terms of load impedance. (05 Marks)

Ans: It is de� ned as the ratio of amplitudes of re� ected voltage to incident voltage at receiving end & is denoted by K.V(Z) = V+ e

�z + V– e–�z (1)

From de� nition z

z

V eK

V e

−γ−

+γ+

=

We know that

zV V(Z) V e V eγ −γ

+ −== = +l ll l (2)

L z lo o

V e V eI I(z) |

Z Z

−γ −γ+ −

== = −l l

(3)

l

l

eL o

V V e VZ Z

I V e V e

γ −γ−

γ −γ+ −

⎡ ⎤+ += = ⎢ ⎥−⎣ ⎦

l

l l

l

(4)

Using componendo, dividendo method we get

L o

L o

L O

L O

L O

L O

Z Z V e V e V e V e

Z Z V e V e V e V e

Z ZV eK

V e Z Z

Z ZK

Z Z

γ −γ γ −γ+ − + −

γ −γ γ −γ+ − + −

−γ−

+γ+

⎡ ⎤− + − += ⎢ ⎥+ − + +⎣ ⎦

−⇒ = =

+

−⇒ =

+

l l l l

l l l l

l

l

2. a. Using the Helmhotz equation, derive the � eld equations or TE modes in rectangular waveguides. (09 Marks)

Ans: The TEmn modes in a rectangular guide are characterized by EZ = 0From a given helmholtz equation �2Hz = �2Hz

Z

b

x

y

a

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Page 6: Microwaves and Radar - Navodaya Institute of … and Radar Time: 3 hrs. Max. Marks: ... where ZY ZY2 ZY j γ= ⇒γ = ⇒γ ... on smith chart & name that point as 'A'. b) ...

December 2012 Microwave and RadarSet 2 - ��

A solution in the form of

j z2 m m n n

m x m x n y n yH A sin B cos C sin D cos e

a a b b− β⎡ ⎤⎡ π π ⎤ ⎡ π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + × +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣⎣ ⎦

(1)

where Kx = m�/a & ky = n�/b are replaced.For a lossless dielectric, Maxwell's curl equations in frequency domain are� × E = –j��H (2)� × H = j�� � !�" In rectangular coordinates, their components are

yzx

EEj H

y z

∂∂− = − ωμ

∂ ∂ (4)

x zy

E Ej H

z x

∂ ∂− = − ωμ

∂ ∂ (5)

y xz

E Ej H

x y

∂ ∂− = − ωμ

∂ ∂ (6)

yzx

HHj E

y z

∂∂− = ω ∈

∂ ∂ (7)

x zy

H Hj E

z x

∂ ∂− = ω ∈

∂ ∂ (8)

y xz

H Hj E

x y

∂ ∂− = ω ∈

∂ ∂ (9)

with substitution g zj & E 0,equationssimplified toz

∂ = − β =∂

�g Ey = –��Hx (10)

�g Ex = �� Hy (11)

y y

z

E Ej H

x y

∂ ∂− = − ωμ

∂ ∂ (12)

zg y x

Hj H j E

y

∂+ β = ωμ

∂ (13)

zg x y

Hj H j E

x

∂− β − = ω ∈

∂ (14)

y xH H

0x y

∂ ∂− =

∂ ∂ (15)

Solving these equations from (10) to (15) for Ex, Ey, Hx & Hy in terms of Hz will give the TE mode � eld equations in rectangular waveguides as

zx 2

c

HjE

K y

∂− ωμ=∂

(16)

zy 2

c

HjE

K x

∂+ ωμ=∂

(17)

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Page 7: Microwaves and Radar - Navodaya Institute of … and Radar Time: 3 hrs. Max. Marks: ... where ZY ZY2 ZY j γ= ⇒γ = ⇒γ ... on smith chart & name that point as 'A'. b) ...

Microwave and Radar December 2012 Set 2 - ��

Ez = 0

g zx 2

c

j HH

K x

− β ∂=

∂ (18)

g zy 2

c

j HH

K y

− β ∂=

∂ (19)

j gZz m m n n

m x n x n y n yH A sin B cos C sin D cos e

a a b b− β⎡ π π ⎤ ⎡ π π ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + × +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

(20)

2 2 2c gwhere k = ω μ ∈−β

Differentiating equation (20) w.r.t x & y then substituting results in (16) to (19) yield set of equations.

zx n

zy m

z

HSince E 0, 0 at y 0, Hence C 0

y

HSince E 0, then 0,at x 0 HenceA 0

x

HIt is concluded that, 0

x

∂= = = =

∂= = = =

∂∂

=∂

� Magnetic � eld in the +ve z direction is given by

gj Z

z oz

m x n yH H cos cos e

a b− βπ π⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(21)

substituting equation (21) in (16) to (19) gives TEMN � eld equations

gj Z

x ox

m x n yE E cos sin e

a b− βπ π⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(22)

gj Z

y oy

m x n yE E sin cos e

a b− βπ π⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(23)

Ez = 0

gj Z

x ox

m x n yH H sin cos e

a b− βπ π⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ (24)

gj Z

y oy

m x n yH H cos sin e

a b− βπ π⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(25)

2. b. With a neat sketch, explain the 4 port microwave circulator & also obtain the s-matrix. (10 Marks)Ans:

Port 4

Port 2

Port 3 Port 1

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Page 8: Microwaves and Radar - Navodaya Institute of … and Radar Time: 3 hrs. Max. Marks: ... where ZY ZY2 ZY j γ= ⇒γ = ⇒γ ... on smith chart & name that point as 'A'. b) ...

December 2012 Microwave and RadarSet 2 - ��

Microwave circulator is a multiport waveguide junction in which the wave can � ow only from the nth port to the (n + 1)th port in one direction.

Four port CirculatorOne type of 4-port circulator is a combination of two 3dB side hole directional coupler & a rectangular waveguide with two non reciprocal phase shifters as shown below.

Port 43 = 00

1 = 1800 2 = 1800

4 = 900

900

900900

Phase Shifter

Phase Shifter

Primary guide

Secondary guide

2700

1800

1800Port 2Port 1

Port 3

Coupler 1 Coupler 2

Schematic diagram of 4 port circulator

Each of the two 3dB couplers in the circulator introduces a phase shift of 90° and each of the two phase shifters produces certain phase shift.When the wave is incident at port-1, it splits into two components by coupler 1. The wave in primary guide arrives at port - 2 with relative phase change of 180°. The second wave propagates through 2 couplers & secondary guide and arrives at port 2 with phase shift of 180°.Since the two waves reaching port 2 are in phase, the power transmission is obtained from port 1 to port 2. The wave propagates through the primary guide, phase shifter & coupler 2 and arrives at port - 4 with a phase shift of 270°. The wave travels through coupler 1 & the secondary guide & it arrives at port 4 with a phase shift of 90°.Since the 2 waves at port - 4 are out of phase by 180° � net power at port 4 is zero.A perfectly matched, lossless & nonreciprocal 4 port-circulator has a S matrix of the form

12 13 14

21 23 24

31 32 34

41 42 43

0 S S S

S 0 S SS

S S 0 S

S S S 0

⎡ ⎤⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

It can be simpli� ed as

0 0 0 1

1 0 0 0S

0 1 0 0

0 0 1 0

⎡ ⎤⎢ ⎥⎢ ⎥= ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

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Page 9: Microwaves and Radar - Navodaya Institute of … and Radar Time: 3 hrs. Max. Marks: ... where ZY ZY2 ZY j γ= ⇒γ = ⇒γ ... on smith chart & name that point as 'A'. b) ...

Microwave and Radar December 2012 Set 2 - ��

2. .c. An air-� lled rectangular waveguide of inside dimensions 7 × 3.5 cm operates in the dominant TE10 mode. Find

i) The cutoff frequency ii) The phase velocity of the wave in the guide at a frequency of 3.5 MHZ iii) The guided wavelength at the same frequency (03 Marks)

Ans:

i) 2 2 8

c 02 2 9

8

c 10 2

1 m n c 3 10f ,

a b f 3.5 102

c 3 10f for TE mode 2.14 GHz

2a 2 7 10−

×= + λ = =×με

×⇒ = = =× ×

ii) ( ) ( )

88

g 2 2

c 3 10V 3.78 10 m / s

1 fc f 1 2.14 3.5

×= = = ×− −

iii) ( )

( )( )

8 9

0g 2 2

3 10 3.5 1010.8cm

1 fc f 1 2.14 3.5

× ×λλ = = =

− −

3. a. With neat sketches, explain the IMPATT diode & draw the negative resistance curve (10 Marks)

Ans: The IMPATT diode consists of 4 layers namely n+ – P -i – p+ layers, where +� refers to high doping pro� le.v(or)i � intrinsic layerThe device consists of 2 regions, one is P region at which avalanche multiplication occurs. This region is also called avalanche region. The other region is i (or ) v region through which the generated holes must drift in moving to p+ contact.The space between n+ – P & i – p+ junction is called the space-charge region.When the applied reverse bias voltage is greater than breakdown voltage, the space charge region extends from n+p junction through the P & i regions, to the i – p+ junction. Maximum � eld occurs at n+p junction as shown in � g. b. Due to this high � eld electrons move into n+ region & holes drift through the space charge region to the p+ region. The holes near n+p junction aquire energy to knock valence electrons into the conduction band thus producing electron-hole pair, this is called avalanche multiplication.When the IMPATT diode is mounted in a microwave resonant circuit, an AC voltage can be maintained at a given frequency in a circuit & the total � eld across the diode is the sum of ac & dc � elds. This � eld causes breakdown at the n+p junction during the +ve half cycle of ac voltage, if the � eld is above breakdown voltage and carrier current. Io(t) generated at n+p junction by the avalanche multiplication grows exponentially with time. During negative half cycle, when the � eld is below breakdown voltage the Io(+) decays exponentially.As the injected holes traverse the drift space, they induce a Ie(t) in the external circuit as shown in � g. (e).

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Page 10: Microwaves and Radar - Navodaya Institute of … and Radar Time: 3 hrs. Max. Marks: ... where ZY ZY2 ZY j γ= ⇒γ = ⇒γ ... on smith chart & name that point as 'A'. b) ...

December 2012 Microwave and RadarSet 2 - �

Vdc

n+

x

0

0

1 2 31013

5 × 106

1020

E(x)

+P P+i (or) u (a) Silicon Structure

(b) Field distribution

(c) Doping pro� le

Concentration

Distance

Electric � eld

Space charge region

+ --

---

-+++

3�/22� Wt

va

�/2 �

(d) Applied ac voltage

Id

(e) Io(t) & Ie(t)Io(t)

Ie(t)

#$� #$�

L

Negative Resistance curve:

1 cos� �� ��� �� ��� �Negative resistance

Transit angle �

� ��

��

2

3. b. Explain the parametric ampli� er with equivalence circuit. (10 Marks)

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Microwave and Radar December 2012 Set 2 - ��

Ans: Parametric ampli� er is a device whose reactance is varied to produce ampli� cation.

N

N

VS

RS CSLS

C (t)

RdR P

L P

C P Ri

Li +

IO

Ci

Pump c

ircuit

VO

VP

IP

Signal Circuit Idler Circuit

The signal frequency fs & pump frequency fp are mixed in the non linear capacitor C. Accordingly a voltage of the fundamental frequencies fs & fp as well as the sum & the difference frequencies mfp ± nfs appear across C.If a resistance load is connected across the terminals of the idler circuit, an output voltage can be generated across the load at the output frequency fo.The output circuit which doesn't require external excitation is called the idler circuit.So, fo = mfp ± nfs (1)where m & n are positive integers from 0 to % If fo > fs the device is called a parametric up-convertorIf fo < fs the device is called a parametric down convertor

Parametric up converter

a) The output frequency is equal to sum of signal frequency & pump frequency i.e fo = fs + fp

b) There is no power� ow in the parametric device at frequencies other than the signal, pump & output frequencies.

( )o

2s

f xGain

f 1 1 x=

+ + (2)

( )

o s p

2s

o

s d

f f f

fx Q

f

1Q 2 f CR

= +

= γ

= π

Rd � Series resistance of a p-n junction diode�Q = Figure of merit for the nonlinear capacitor

( )2x

Gain degradation factor1 1 x

→+ +

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Page 12: Microwaves and Radar - Navodaya Institute of … and Radar Time: 3 hrs. Max. Marks: ... where ZY ZY2 ZY j γ= ⇒γ = ⇒γ ... on smith chart & name that point as 'A'. b) ...

December 2012 Microwave and RadarSet 2 - ��

Parametric down converter:Here fo = fs – fp � fs = fp + fo Down conversion gain is

( )s

2o

f xGain

f 1 1 x=

+ +

4. a. For a two port network, explain the S-parameters and properties of S-parameters. (10 Marks)Ans: S-parameters for 2 port network

Zg Zo

Z S

Zo

OutputInput

ZLa1

b1

Pi

Pi-Pr

PrPo

Device 2 port network ZVg

The incident & re� ected amplitudes of microwaves at any port are used to characterise a microwave circuit.The amplitudes of these are normalised.

2thin n

1Input power at n port P = a

2⇒ (1)

2thr n n

1Reflected power at n port P = b

2⇒ (2)

an � Normalised incident wave amplitude at nth port bn � Normalised re� ected wave amplitude at nth portFor a two port network, the relation between incident & re� ected waves are expressed in terms of S-parameters Sij b1 = S11 a1 + S12 a2 (3)b2 = S21 a1 + S22 a2 (4)S11 = b1/a1|a2

= 0

� Re� ection coef� cient K1 at port 1 when port 2 is terminated with matched load (a2 = 0)

S22 = b2/a2|a1 = 0

� Re� ection coef� cient K2 at port 2 when port 1 is terminated with matched load (a1 = 0)

S12 = b1/a2|a1 = 0

� Attenuation of wave travelling from port 2 to port 1S21 = b2/a1|a2

= 0

� Attenuation of wave travelling from port 1 to port 2

In general [b] = [s] [a]

11 12 1N1 1

21 22 2N2 2

N NN1 N2 NN

S S ............Sb a

S S ............Sb a

b aS S ...........S

⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥

⇒ = ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎣ ⎦

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Microwave and Radar December 2012 Set 2 - ��

Properties of S-parameters:a) Zero diagonal elements for perfect matched network: When any ith port is perfectly matched to the

junction, then there are no re� ections from that port. Thus Sii = 0, if all the ports are perfectly matched, then the diagonal elements are zero.

b) Symmetry property of S-matrix: If a microwave junction satis� es reciprocity condition & if there are no active devices, then the S-parameters are equal to their corresponding transposes.

i.e (Sij = Sji ), (i � j) and [S]t = [S]

c) Unitary property for a lossless junction: It states that for any lossless network, the sum of the products of each term of any one row (or) any one column of the [S] matrix with its complex conjugate is unity.

n*

ki ki ijk 1

i.e S s 1, where 0, if i j=

= δ = ≠∑

d) Zero property:

n*

ki ki ijk 1

S s 0, where 0, if i j=

= δ = ≠∑

e) Phase shift property: For a 2 port network with unprimed reference planes 1 & 2 as shown in � g. S-parameters have de� nite complex values

[ ] 11 12

21 22

S SS

S S

⎡ ⎤= ⎢ ⎥⎢ ⎥⎣ ⎦

DeviceS

1� 1

l2l1

2 2�

When the reference places 1 & 2 are shifted outward to 1� & 2� by electrical phase shifts

�1 = �1 l1 & �2 = �2 l2

[ ] [ ]1 1

2 2

j j

j j

e 0 e 0S S

0 e 0 e

− φ − φ

− φ − φ

⎡ ⎤ ⎡ ⎤∴ = ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦´

4. b. Explain the phase shifter with neat sketches. (10 Marks)Ans: A microwave phase shifter is a two port device which produces a variable shift in phase of the incoming

microwave signal.

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December 2012 Microwave and RadarSet 2 - �

One of phase shifter is precision phase shifter.

TE10

TE11

Ei

TE10

/4 Plate

/2 Plate

/4 Plate

Rotatable section

Dielectric Plates

Fig. (a)

E2Ei El

x

ELE2

E4

E3

E3

E5E4

E6

45&45&x x

�y y

Input /4 plate Out put /4 plateRotary /2 plate

��

Fig. (b)A rotary type phase shifter is as shown in the � gure. It consists of a circular waveguide containing a lossless dielectric plate of length 2l called 'Half-wave-section'. A section of rectangular to circular transition containing a lossless dielectric plate of length 'l', called "quarter wave section", oriented at an angle of 45° to the broader wall of the rectangular waveguide & a circular to rectangular transition, again containing a lossless dielectric plate of same length 'l' oriented at an angle 45°.The incident TE10 mode becomes TE11 mode in circular waveguide section. The halfwave section produces a phase shift equal to twice that produced by the quarter wave section.Let Ei be the maximum electric � eld strength in TE10 mode is resolved into components E1 parallel to plate & E2 perpendicular to E1 as shown in above � gure (b).After propagation through the plate these components are given byE1 = (Ei cos 45°) e-j�1l = E0 e

-j�1l. E2 = (Ei sin 45°) e-j�2l = E0 e

-j�2l.

where, io

EE

2=

PART - B

5. a. With neat sketch, explain the operation of E-planetee & also obtain its S-matrix. (10 Marks)Ans: E-plane is formed when a rectangular slot is cut along the breadth of a long waveguide & a side-arm is

inserted from the top.

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Microwave and Radar December 2012 Set 2 - �

Plane of symmetry

T�10 mode

Side arm (E - arm)

Port (1)

E - � eld linesPort (2)

Collinear armPort (3)

E � eld(T�10 mode)

Port (2)Port (1)

Port (3)

When microwave signal in T�10 mode is allowed to propagate through the E-arm, the E-� eld lines turn at the corner produce a phase shift of 180° between the E-� eld lines coming out of port (1) & (2) as shown in � gure.

S-matrix of E-plane Tee:

1. Because of plane of symmetry, microwave fed into port (3) divides equally between the port (1) & port (2) but with phase shift of 180° between them.

� S32 = – S31

2. Let us assume port 3 is perfectly matched

So, S33 = 0

3. From symmetry of (S) matrix

Sij = Sji

with the above 3 characteristics

11 12 13

12 22 13

13 13

S S S

[S] S S S

S S 0

⎡ ⎤⎢ ⎥

= −⎢ ⎥⎢ ⎥−⎣ ⎦

Since [S21 = S12, S23 = S32 = –S31 = –S13 and S31 = S13]

4. From unitary property

[S] [S]* = [U]

11 12 13

12 22 13

13 13

* * *

11 12 13

* * *12 22 13

* *13 13

S S SS S S 1 0 0

S S S S S S 0 1 0

0 0 1S S 0 S S 0

⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥− − =⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥− ⎣ ⎦⎣ ⎦ −⎣ ⎦

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December 2012 Microwave and RadarSet 2 - �

Considering the product of 1st row & 1st column, we get|S11|

2 + |S12|2 + |S13|

2 = 1 (1)Product of 2nd row & 2nd column yields|S12|

2 + |S22|2 + |S13|

2 = 1 (2)Considering 3rd row with 3rd column elements |S13|

2 + |S13|2 + 0 = 1 (3)

3rd row & 1st column elements* *

13 11 13 12S S S S 0 0− + = (4)

213 13

1From equation (3) 2 | S | 1 S

2⇒ = ⇒ = (5)

Compare eqn (1) & (2)� S22 = S11 (6)From eqn (4) S13 (S11 * – S12 *)=0

* *13 11 12Since S 0, we must have S S≠ =

Remove complex conjugates, S11 = S12 (7)Using eqn (5) & eqn (7) in equation (1)

2 2 2

11 11 11 11 12 22

1 1 1S S 1 2 S S S S

2 2 2+ + = ⇒ = ⇒ = = =

using these values in the matrix

[ ]E

1 1 12 2 2

1 1 1S 2 2 2

1 1 02 2

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥= −⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

5. b. With neat sketch, explain the operation of magic tee & mention its application. (10 Marks)Ans:

E-arm

Collinear

arm

Collinear arm

Port (2)

Port (3)

Port (1)

H - arm

Port (4

)

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Microwave and Radar December 2012 Set 2 - �

A E-H plane tee junction (or) Hybrid tee junction is formed when E & H plane tee junctions are combined as shown in � gure.Rectangular slots are cut both along the width & breadth of a long waveguide & side-arms are attached into it.In the above � g, port (1) & port (2) form the collinear ports, port (3) the H-arm & port (4) the E-arm.Because of the geometry of the junction, the microwave signal fed into port (3) cannot induce the dominant mode in port (4) & vice versa. Therefore port (3) & port (4) behave as isolated ports. Hence the name magic tee.The S-matrix is given by

[ ]

11 12 13 14

21 22 23 24

MT

31 32 33 34

41 42 43 44

1 10 0

2 2S S S S 1 1

0 0S S S S 2 2

S S1 1S S S S

0 02 2S S S S

1 10 0

2 2

⎡ ⎤⎢ ⎥⎢ ⎥

⎡ ⎤ ⎢ ⎥⎢ ⎥ −⎢ ⎥⎢ ⎥ ⎢ ⎥= ⇒ =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥

⎢ ⎥−⎢ ⎥⎣ ⎦

The magic tee has a number of applications in various microwave circuits. They are

i) As an impedance bridge for measurement of impedance at microwave frequencies

ii) As E-H tuner for impedance matching

iii) As balanced mixer in superheterodyne receiver

iv) As power combiner

v) As a duplexer in radar system6. a. With neat schematic diagram, explain the coplanar strip lines. (06 Marks)Ans:

d W S W

Y

Z

X

A coplanar strip line consists of two conducting strips on one substrate surface with one strip grounded as shown in � gure.Two strip conductors each of width 'W' separated by a distance 'S'. The coplanar strip line has advantages over parallel strip line, because its two strips are on the same substrate surface for convenient connections i.e connection of shunt element is easy.

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December 2012 Microwave and RadarSet 2 - �

The coplanar strip lines eliminate the dif� culties involved in connecting shunt elements between the hot & ground strips. As a result reliability is increased & production cost is decreased.The Zo of C.S.L is

avg

O 20

Z PZ

I= (1)

IO � total peak current in one stripPavg � Power flowing in the +ve Z direction

( )avg z

1P Re E H * u .dx.dy

2= ×∫∫

Ex � Electric field intensity in the +ve X directionHy � Magnetic field intensity in the +ve Y direction * � Conjugate

6. b. A lossless parallel strip line has a conducting width w. The substrate dielectric separating the two conducting strips has a relative dielectric constant �r of 6 & a thickness of 4mm. Calculate,

i) The required width w of the conducting strip in order to have a characteristic impedance of 50� ii) The strip-line capacitance iii) The strip-line inductance iv) The phase velocity of the wave in parallel strip line (08 Marks)Ans:

i) O

rc

33

O r

377 dZ for w d

w

377d 377 4 10W W 12.31 10 m

Z 6 50

W 12.31mm

−−

= >>ε

× ×⇒ = ⇒ = = ×∈ ×

=

ii) The stripline capacitance is

12 3d

3

w 8.854 10 6 12.31 10C 163.50pF / m

d 4 10

− −

∈ × × × ×= = =×

iii) The strip line inductance is

7 3c

3

d 4 10 4 10L L 0.41 H / m

w 12.31 10

− −

μ π × × ×= = ⇒ = μ×

iv) The phase velocity is

88

p

rd

C 3 10V 1.22 10 m /s

6

×= = = ×∈

6. c. Write a note on shielded strip lines. (06 Marks)Ans: It consists of a thin conducting strip of width 'w' & thickness 't'. The stripline placed at the centre surrounded

by a low loss dielectric substrate of thickless 'd' between two ground plates.Mode of propagation is pure TEM wave where the E-� eld is perpendicular to strip & concentrated at the centre of the strip. Fringing � elds also exists at the edge of the strip.

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Microwave and Radar December 2012 Set 2 - �

W

t rd

WE � eld

H � eld

Dielectric

Commonly used dielectrics are te� on, polyole� ne, polysterene etc. Operating frequency extends from 100MHz to 30 MHz.ZO for a wide strip (W/d >> 0.35)is

1

fO

rr

C94.15 WZ k

d 8.854

−⎡ ⎤

= +⎢ ⎥∈∈ ⎣ ⎦

1where K

t1 d

=−

t � the strip thickness d � the distance between two ground plates.

( ) ( ) ( )2rf

8.854C 2k ln k 1 k 1 ln k 1

Fringe capacitance in pF / m

∈ ⎡ ⎤= + − − −⎣ ⎦π

7. a. Derive an expression for simple form of the Radar range equation. (05 Marks)Ans: Radar equation relates the range of the radar to characteristics of the transmitter, receiver, antenna, target &

environment.If the transmitter power Pt is radiated by an isotropic antenna.The power density at a distance R from the radar is equal to radiated power divided by the surface area.

t2

PPower densityat range R is W/sq.m

4 R∴

π (1)

But Radar employ directive antenna to concentrate the radiated power Pt in a particular direction.

t2

P GPower density at range R from directive antenna is

4 R∴

π (2)

Wheremax power density radiated by directive antenna

GPower density radiated by lossless isotropicantenna

=

The target intercepts a portion of incident energy and reradiates in various directions.The radar cross section of the target determines the power density returned to the radar.

t2 2

P GReradiated power density back at the radar

4 R 4 R

σ∴ =π π

(3)

The radar antenna captures a portion of the incident eacho signal.The power received by the radar = incident power × effective area

tr e2 2

P GP A

4 R 4 R

σ=π π

(4)

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December 2012 Microwave and RadarSet 2 -

where Ae = �a A. [A: physical area, �a: Antenna aperture ef� ciency]The maximum range of radar Rmax is the distance beyond which target cannot be detected.� Pr is equal to minimum detectable signal Smin

t er min 2 2

P G .AP S

4 R 4 R

σ∴ = =

π π

( )

14

t emax 2

min

P GAR

4 R S

⎡ ⎤σ∴ = ⎢ ⎥

π⎢ ⎥⎣ ⎦ (5)

e2

4 A cBut WKT, G , wavelength fπ

= λ → =λ

( )

142 2

tmax 3

min

P GR

4 r S

⎡ ⎤λ σ∴ = ⎢ ⎥

π⎢ ⎥⎣ ⎦ (6)

OR1

2 4t e

max 2min

P AR

4 S

⎡ ⎤σ= ⎢ ⎥πλ⎣ ⎦

(7)

7. b. With a neat block diagram, explain the conventional pulse radar with a super heterodyne receiver (08 Marks)

Ans:

Duplexer

Low Noise RF ampli� er Mixer

Local oscillator

Threshold detection

IFampli� er

Matched � lter

2nd

detectorVideo

ampli� erOut put

Pulse Modulator

Power ampli� er

RF pulse IF pulse

Wave generator

The Radar block diagram consists of

a. Transmitter

b. Duplexer

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Microwave and Radar December 2012 Set 2 - �

c. Antenna

d. ReceiverTransmitter: It may be power ampli� er such as klystron, travelling wave tube.It may be power oscillator also, like magnetron for a modest capability.The radar signal is produced at low power by a waveform generator. More oftenly modulator turns the transmitter ON & OFF in synchronism with the input pulses.The output of the transmitter is delivered to antenna through duplexer.Duplexer: It is a gas discharge device, while transmitting, duplexer protects receiver from leakage of high powered transmitted signals.While receiving duplexer will not allow echo signal to � ow towards transmitter.It consists of two tubes,TR � Transmit receive tube will work during transmission ATR � Anti transmit receive tube work during reception. Duplexer allows single antenna to be used on time shared basis for both transmitting & receiving.Antenna: Antennas can be mechanically steered parabolic re� ectors, mechanically steered planar arrays (or) electronically steered phased array OR parabola re� ectors, to focus the energy into a narrow beam.Receiver: It is a superheterodyne type. RF stage can be a low noise transistor ampli� er. The mixer & local oscillator convert the RF signal to an intermediate frequency. The signal bandwidth of superheterodyne receiver is decided by bandwidth of IF stage.The IF frequency might be 30 or 60 MHZ. Matched � lter maximizes the detectability of weak echo signals & attenuates unwanted signals.The combination of IF ampli� er, second detector & video ampli� er act as an envelop detector to pass pulse modulation & reject the carrier frequency. At the output of the receiver a decision is made whether target is present (or) not & it is based on the magnitude of receiver output.If the output is large enough to exceed a predetermined thresholds, the decision is that target is present.

7. c. Explain the applications of Radar. (07 Marks)Ans: Radar has been employed to detect targets on the ground, on the sea, in the air, in space & even below

ground.Major areas of Radar applications are

1. Military: Radar is an important part of air-defence systemsIt will be used for surveillance & weapon control. Surveillance includes target detection, target �recognition, target tracking & designation to a weapon system.Weapon control radars track targets, detect the weapon to an intercept �High resolution radars such as SAR are used for detecting � xed & moving targets on the �battle� eld.

2. Remote sensing: All radars are remote sensors.

Ex: a) Weather observation, which is a regular part of TV weather reporting, done by Radar remote

sensingb) Planetary observationc) Short-range below ground probingd) Mapping of sea ice to route shipping

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December 2012 Microwave and RadarSet 2 - �

3. Air traf� c Control: In the vicinity of airports radars have been employed for safety control of air traf� c.

To enroute aircraft from one airport to another ASR (Air survivalnce radars) will be employed �ASR also maps regions of rain so that aircraft can be directed around them �TDWR (Terminal Doppler weather radars) are dedicatedly employed for observing weather in �the vicinity of airports.

4. Law enforcement & highway safety:A radar speed meter is used by police to measure the speed of vehicles �Radars are used to measure the speed of tennis ball �

5. Aircraft safety & Navigation:Some radars outline the regions of dangerous wind shear to allow the pilot to avoid hazardous �conditionsTo measure the height of radar from earth, radar altimeter will be used �

6. Ship Safety: Radar is found on ships & boats for collision avoidance & to enroute the slip when visibility is poor

7. Space: Space vehicles use radars for landing on moon. Large ground based radars are used for detection & tracking of satellites and other space objects.

8. Other applications:In industry for measurement of speed & distance �Radars are employed to study the movement of insects & birds �

8. a. Explain single-delay line canceler and frequency response of the single-delay line canceler and also obtain the expression for blind speeds. (10 Marks)

Ans:

Bipolar video

Bipolar video Unipolar video

A/D Converter

Delay T = 1/fp

SubtractAbsolute

Value

Digital MTI output to automatic

detection & data processing

Analog display

D/Ai/p

The output of the MT1 receiver is the input to the delay line canceler. The delay T is achieved by storing

the radar output from one pulse transmission (or) sweep, in a digital memory for a time equal to the pulse

repetition period so that pp

1T T f= = The output obtained after subtraction of two successive sweeps is

bipolar video.The absolute value of the bipolar video is taken, i.e unipolar video. This will be converted to analog signal by the D/A converter if the signal is to be displayed on PPI.

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Microwave and Radar December 2012 Set 2 - �

Frequency Response of SDLC:

0 fp = 1/TP

Frequency

2fp 3fp

1

2

� �H f

The signal from a target at range Ro at the output of the phase detector isV1 = K sin (2�fdt – �o) (1)fd � Doppler frequency shift

oo

4 RConstant phase

πφ → =

λ � WavelengthK � Amplitude of signalSignal from previous radar transmission isV2 = K sin [2�fd (t – Tp) – �o] (2)

( ) p1 2 d p d o

TV V V 2k sin f T cos 2 f t

2

⎡ ⎤⎛ ⎞= − = π π − − φ⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦ (3)

( ) ( )A B A BUsing sin A sin B 2sin cos

2 2

⎡ − ⎤ ⎡ + ⎤= ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦

Amplitude of above signal is 2sin (�fdTp)� Frequency response function is H (f) = 2 sin (�fdTp) (4)Blind Speeds:The H(f) of SDLC will be zero whenever the magnitude of sin (�fdTp) is zero, which occurs when �fdTp = 0, ± �, ± 2� ..........

d pp

2Vr nf nf n 0,1,2..........

T∴ = = = =

λ (5)

pp n

nf2Vrnf V

2

λ= ⇒ =

λ

8. b. A VHF radar at 220 MHz has a maximum unambigous range of 180 nmi. What is the � rst blind speed? (04 Marks)

Ans: n 1

p p

p 3unun p 8

8

6

1 3

nV ,V

2T 2T

CT 2R 2 180 1852R T 2.221 10

2 C 3 10

3 10C 1.364mf 220 10

1.364V 306.98m / sec

2 2.221 10

λ λ= =

× ×= ⇒ = = = ××

×λ = = =×

= =× ×

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December 2012 Microwave and RadarSet 2 - ��

8. c. With neat block diagram explain the original moving target detector signal processor. (06 Marks)Ans:

3 Pulse canceler

Zero velocity � lter Magnitude

Cluttermap recursive � lter

Memory(clutter map)

Thresholding

8 pulse doppler � lter bank

Weighting & Magnitude

I, Q

The output of receiver IF ampli� er was fed to I & Q phase detectors from there the A/D conversion change analog signals to 10-bit digital words.Coherent processing Interval (CPI): The range was quantized 1

16 nmi intervals, this is equal to range resolution of the pulse. The azimuth angle was quantized into ¾ degree intervals. In each ¾ degree azimath cell there were 10 pulses transmitted at a constant prf, on receive these 10 pulses are called CPI.Filter bank: It is implemented by a FFT. There are 10pulses in the CPI, but only 8 doppler � lters. Since 3 pulse canceler requires all 3 pulses before it cancel the clutter, the � rst 2 pulses are discarded. The frequency response of each � lter of the FFT � lters bank has (sin x)/x shape.Clutter map: The MTI processor eliminates stationary clutter, but it also eliminates aircraft moving on a trajectory (one �

r to radar Los), which causes the aircraft redial velocity to be zero. The clutter map stores magnitude of clutter echoes in a digital memory.3 pulse canceler removed all echoes with zero velocity, the zero velocity � lter had to be restablished in order to produce the cluttermap.

Adaptive thresholds:

Prf - 1

5 6 7 8 1 2 3 4 5

Prf - 2

5 6 7 8 1 2 3 4 5

Radial velocity

RainThe

aircraft velocity

Fig a

aircraft echo

��aircraft echo at � lter 7

��aircraft echo at � lter 8

��rain clutter at � lter 8

aircraft echo

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Microwave and Radar December 2012 Set 2 - ��

Consider the eight doppler � lters as shown in � g a .The adaptive threshold setting for � lter no1 was determined by the value stored in clutter map. The adaptive thresholds for � lter no. 3 through 7 were set by a CFAR. For remaining two � lters 2 & 8, the threshold was selected as the larger of that given by the clutter map & the clutter CFAR.

Unmasking moving targets in moving clutter: For the � g a, at one prf the aircraft echo is at � lter no. 7. At other prf, the aircraft echo is at � lter 8. The rain echo is at � lter 8 & it masks aircraft echo. The masked aircraft echo at � lter 8 is unmasked at � lter no. 7.Post processor Centroiding: The output of MTD processor is a hit report, as many as 20 hit reports generated & produced as one single report.

3-Micro waves and RadarDecember 2012.indd 713-Micro waves and RadarDecember 2012.indd 71 8/13/2013 1:21:34 PM8/13/2013 1:21:34 PM