Lect1

27/01/2014

Objectives
ECE 637 : MIC and Monolithic
Microwave Integrated Circuits
(MMIC)
By Dr. Ghanshyam Kumar Singh
ECE Department, SET
Sharda University

Course Outcomes
After completing this course students will
be able
design Microstrip lines.
analyse different types of dielectric
materials and their effects.
apply optimization technique for
frequency behaviour of MMIC.
solve the problems related to MMIC
Designing.
analyse various types of MMIC devices.

Lecture 1
An Approximate Electrostatic
Solution for Microstrip Line
The Transverse Resonance
Techniques
Wave Velocities and Dispersion

To introduce the design methodology of
Microstrip Lines.
To acquire knowledge of various typs of
dielectric materials and their properties.
To describe the frequency properties of
MMIC.
To explain the fundamentals of various
types of MMIC devices.

Lecture 1
TEM, TE and TM Waves
Coaxial Cable
Grounded Dielectric Slab
Waveguides
Striplines and Microstrip Line
Design Formulas of Microstrip Line

TEM, TE and EM Waves
transmission lines and waveguides
are primarily used to distribute
microwave wave power from one
point to another
each of these structures is
characterized by a propagation
constant and a characteristic
impedance; if the line is lossy,
attenuation is also needed

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27/01/2014

structures that have more than one
conductor may support TEM waves
let us consider the a transmission
line or a waveguide with its cross
section being uniform along the zdirection
b
ρ

a

in a source free region, Maxwell’s
equations can be written as
∇ × E = − jωµH, ∇ × H = jωεE
Therefore,

∂E z
+ jβE y = − jωµH x , ( 3)
∂y
∂E
− jβE x − z = − jωµH y , ( 4 )
∂x
∂E y ∂Ex
−
= − jωµH z , ( 5 )
∂x
∂y

∂H z
+ jβH y = jωεEx , ( 6 )
∂y

− jβH x −
∂H y
∂x

−

the electric and magnetic fields can
be written as
$
$
E( x, y, z) = [e t e t ( x, y ) + ze z ( x, y )]e − jβz − − (1)
$
$
H( x, y, z) = [e t h t ( x, y ) + zh z ( x, y )]e − jβz − − ( 2)

Where e t and ht are the transverse
components and e t and h t are the
longitudinal components

each of the four transverse
components can be written in terms
of and , e.g., consider Eqs. (3) and
(7):
∂E z
+ jβE y = − jωµH x
∂y

∂H z
= jωεE y , ( 7)
∂x

− jβH x −

∂H x
= jωεE z , ( 8)
∂y

each of the four transverse
components can be written in terms
of and , e.g., consider Eqs. (3) and
(7):
∂E z
+ jβE y = − jωµH x
∂y
− jβH x −


∂H z
= jωεE y
∂x

∂H z
= jωεE y
∂x

∂H z
∂E z
) / ( jβ )
= jωε ( − jωµH x −
∂x
∂y
∂E
∂H z
( k 2 − β 2 ) H x = jωε z − jβ
∂y
∂x
− jβH x −

Hx =

j
2
kc

( ωε

∂E z
∂H z
) − − − − − − − ( 9)
−β
∂y
∂x

2
k c = (k 2 − β 2 )

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27/01/2014

Similarly, we have
Hy =
Ex =
Ey =

−j

2
kc

−j
2
kc

j
2
kc

( ωε
(β

∂E z
∂H z
) − − − − ( 10 )
+β
∂x
∂y

∂E z
∂H z
) − − − − ( 11)
+ ωµ
∂x
∂y

( −β

∂E z
∂H z
) − − − ( 12)
+ ωµ
∂y
∂x

k c is called the cutoff wavenumber

now let us consider the Helmholtz’s
equation
2
2
 2


∂
∂
∂
2
( ∇ 2 + k 2 ) E x = 0, 
 2 + 2 + 2 + k  Ex = 0
∂x
∂y
∂z



∂2

2

note that ∂z 2 → −β and therefore, for TEM
wave, we have
 ∂2
∂2 

 Ex = 0
+
 2

 ∂x
∂y 2 

Knowing that ∇ 2 et = 0 and
t
∇ • D = ε∇ t • e t = 0
∇ 2Φ ( x, y) = 0
t

, and we have

2
V12 = Φ 1 − Φ 2 = − ∫1 E • dl − − − − (14 )

while the current flowing on a
conductor is given by I = ∫ H • dl − − − − (15)

Transverse electromagnetic (TEM)
wave implies that both E z and H z
are zero (TM, transverse magnetic,
H z =0, E z ≠ 0 ; TE, transverse electric,
0)
E z =0, H z ≠
the transverse components are also zero
unless k c is also zero, i.e.,
k 2 = ω 2µε = β 2

= ω µε = β

this is also true for E y , therefore,
the transverse components of the
electric field (so as the magnetic
field) satisfy the two-dimensional
Laplace’s equation
∇ 2e t = 0 − − − − (13)
t

this is also true for E y, therefore, the
transverse components of the
electric field (so as the magnetic
field) satisfy the two-dimensional
Laplace’s equation
∇ 2e t = 0 − − − − (13)
t

C

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Knowing that ∇ 2 e t = 0 and ∇ • D = ε∇ t • e t = 0
t
, we have ∇ 2Φ ( x, y) = 0

we can define the wave impedance for the
TEM mode:

t

Z TEM =

the voltage between two conductors is
given by
2
V12 = Φ 1 − Φ 2 = − ∫1 E • dl − − − − (14 )

while the current flowing on a conductor is
I = ∫ H • dl − − − − (15 )
given by

E x ωµ
µ
=
=
= η − − − ( 16 )
Hy
β
ε

i.e., the ratio of the electric field to the
magnetic field, note that the components
must be chosen such that E x H is pointing
to the direction of propagation

C

for TEM field, the E and H are related
by
h ( x, y ) =

1
$
z × e( x , y ) − − − − (17)
Z TEM

why is TEM mode desirable?
a closed conductor cannot support
TEM wave as the static potential is
either a constant or zero leading to
et = 0
if a waveguide has more than 1
dielectric, TEM mode cannot exists as
cannot be zero in all regions

cutoff frequency is zero
no dispersion, signals of different
frequencies travel at the same
speed, no distortion of signals
solution to Laplace’s equation is
relatively easy


sometime we deliberately want to
have a cutoff frequency so that a
microwave filter can be designed

k ci = ( ε ri k 2 − β 2 ) 1/ 2

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TEM Mode in Coaxial Line
a coaxial line is shown here:
b
ρ
V=Vo

a
V=0

the inner conductor is at a potential
of Vo volts and the outer conductor
is at zero volts

use the method of separation of variables, we
let

Φ ( ρ, φ ) = R ( ρ )P( φ ) − − − − ( 21)

substitute Eq. (21) to (18), we have
2


ρ ∂  ρ ∂R  + 1 ∂ P = 0 − − − −(22)
∂φ 2



R ∂ρ  ∂ρ  P



note that the first term on the left only
depends on ρ while the second term only
depends on φ


the electric field can be derived from
the scalar potential Φ; in cylindrical
coordinates, the Laplace’s equation
reads: 1 ∂  ∂Φ  1 ∂ 2Φ
= 0 − − − − (18 )
ρ
+
ρ ∂ρ  ∂ρ  ρ 2 ∂φ 2

the boundary conditions are:
Φ ( a , φ ) = Vo − − − ( 19 ), Φ (b, φ ) = 0 − − − − − ( 20 )

if we change either ρ or φ, the RHS
should remain zero; therefore, each
term should be equal to a constant


ρ ∂  ρ ∂R  = −k 2 − − − (23)
ρ



R ∂ρ  ∂ρ 



2
1 ∂ P = −k 2 − − − −(24), k 2 + k 2 = 0 − − − −(25)
ρ φ
φ
∂φ 2

P


now we can solve Eqs. (23) and (24) in
which only 1 variable is involved, the final
solution to Eq. (18) will be the product of
the solutions to Eqs. (23) and (24)

boundary conditions (19) and (20)
dictates
that
the
potential
is
independent of φ, therefore k φ must be
equal to zero and so as k ρ

the general solution to Eq. (24) is

Eq. (23) is reduced to solving
∂  ∂R 
ρ
 =0
∂ρ  ∂ρ 

P( φ ) = A cos( k φ φ ) + B sin( k φ φ )

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the solution for R(ρ) now reads

the electric field now reads

R ( ρ ) = C ln ρ + D

Vo
 ∂Φ $ 1 ∂Φ 
$
$
e t ( ρ, φ ) = −∇ t Φ ( ρ, φ ) = −  ρ
+φ
 =ρ
 ∂ρ
ρ ∂φ 
ρ ln b / a

Φ ( ρ, φ ) = A ln ρ + B
Φ ( a , φ ) = Vo = A ln a + B

adding the propagation constant back,
we have

Φ ( b, φ ) = 0 = A ln b + B ⇒ B = − A ln b
A = Vo / ln( b / a )

V e − jβz
$
E( ρ, φ ) = e t ( ρ, φ ) e − jβz = ρ o
− − − ( 27)
ρ ln b / a

Vo
(ln ρ − ln b ) − − − ( 26 )
Φ ( ρ, φ ) =
ln(b / a )



the magnetic field for the TEM mode
$
H( ρ , φ ) = ρ

the total current on the inner conductor
2π
is
2π
− jβ z

Voe − jβz
− − − ( 28)
ρη ln b / a

the potential
conductors are

between

I a = ∫ H φ ( ρ, φ )adρ =
0

the

two

b

Vab = ∫ Eρ (ρ, φ)dρ = Voe − jβ z − − − ( 29)

the total current on the outer conductor
is
2π
− 2π
− jβz
0

η ln(b / a )

Vo e

= − Ia

the characteristic impedance can be
calculated as
Zo =

Vo η ln(b / a )
=
− − − ( 31)
Ia
2π

Vo e

− − − ( 30 )

the surface current density on the outer
conductor is
$
J s = − ρ × H( b , φ ) =

a

Ib = ∫ J szbdφ =

η ln( b / a )

$
−z
Vo e − jβz
ηb ln( b / a )

higher-order modes exist in coaxial line but is
usually suppressed
the dimension of the coaxial line is controlled so
that these higher-order modes are cutoff
the dominate higher-order mode is TE11 mode,
the cutoff wavenumber can only be obtained by
solving a transcendental equation, the
approximation k c = 2 / ( a + b )
is often used in
practice

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27/01/2014

Surface Waves on a
a grounded dielectric slab will generate
surface waves when excited

Surface Waves on a
while it does not support a TEM mode, it excites
at least 1 TM mode
x
d
εr

this surface wave can propagate a long
distance along the air-dielectric interface
it decays exponentially in the air region when
move away from the air-dielectric interface

Surface Waves on a
for TM modes, from Helmholtz’s equation
we have  ∂ 2 ∂ 2 ∂ 2
2


 2 + 2 + 2 + k  Ez = 0
∂y
∂z
 ∂x


which reduces to
 ∂2
2
2


 2 + ε r k o − β  E z = 0, 0 ≤ x ≤ d
 ∂x

 ∂2
2
2


 2 + k o − β  E z = 0, d ≤ x ≤ ∞
 ∂x


Surface Waves on a
the general solutions to Eqs. (32) and
(33) are
e z ( x, y ) = A sin k c x + B cos k c x,0 ≤ x ≤ d
e z ( x, y ) = Ce hx + De − hx , d ≤ x ≤ ∞

the boundary conditions are
tangential E are zero at x = 0 and x → ∞
tangential E and H are continuous at x = d

z

assume no variation in the y-direction which
implies that ∂ / ∂y → 0
write equation for the field in each of the two
regions
match tangential fields across the interface

Surface Waves on a
Define

(

)

(

2
2
2
k c = ε rk o − β 2 , h2 = − ko − β 2

)

 ∂2
2


 2 + k c  E z = 0,0 ≤ x ≤ d − − − ( 32)
 ∂x

 ∂2
2


 2 − h  E z = 0, d ≤ x ≤ ∞ − − − ( 33)
 ∂x


Surface Waves on a
tangential E at x=0 implies B =0
tangential E = 0 when x→ ∞
implies
C=0
continuity of tangential E implies
A sin k c d = De − hd − − − ( 34)

tangential H can be obtained from Eq.
(10) with H z = 0

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Surface Waves on a

Surface Waves on a

tangential E at x=0 implies B =0
tangential E = 0 when x → ∞
C=0

continuity of tangential H implies
implies

εr
−h
De − hd − − − ( 35 )
A cos k c d =
kc
− h2

taking the ratio of Eq. (34) to Eq. (35)
we have
k c tan k c d = ε r h − − − ( 36 )

continuity of tangential E implies
A sin k c d = De − hd − − − ( 34)

Surface Waves on a

(

2

2

2

)

2

(

2

note that k c = ε r k o − β , h = − k o − β
lead to
2
2
k c + h 2 = ( ε r − 1) k o − − − ( 37)

Surface Waves on a
2

)

Eqs. (36) and (37) must be satisfied
simultaneously, they can be solved for by
numerical method or by graphical method

Surface Waves on a
Eq. (39) is an equation of a circle with a
radius of ( ε r − 1)k o d , each interception
point between these two curves yields a
solution
hd Eq.(38)
r

Eq. (39)

π/2

π

k cd

to use the graphical method, it is more
convenient to rewrite Eqs. (36) and (37)
into the following forms:
k c d tan k c d = ε r hd − − − ( 38 )
( k c d) 2 + ( hd) 2 = ( ε r − 1)( k o d) 2 = r 2 − − − ( 39 )

Surface Waves on a
note that there is always one intersection
point, i.e., at least one TM mode
the number of modes depends on the radius r
which in turn depends on the d and ε r , k o
h has been chosen a positive real number,
we can also assume that k c is positive
the next TM will not be excited unless
r = ( ε r − 1)k o d = π

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27/01/2014

Surface Waves on a
In general, TM n mode is excited if
r = ( ε r − 1)k o d ≥ nπ

the cutoff frequency is defined as
( ε r − 1)( 2πf c / c)d = nπ →

fc =

nc
, n = 0,1, 2, .. .--- (40)
2d ε r − 1

Surface Waves on a
once k c and h are found, the TM field
components can be written as for
0≤x≤d
E z = A sin k c xe − jβz − − − − ( 41)
− jβ
Ex =
A cos k c xe − jβz − − − − ( 42)
kc
Hy =

Surface Waves on a
For d ≤ x ≤ ∞
E z = A sin k c de − h( x − d) e − jβz − − − − ( 44)
− jβ
Ex =
A sin k c de − h( x − d) e − jβz − − − ( 45)
h
− jωε o
Hy =
A sin k c de − h( x− d) e − jβz − − − − ( 46 )
h

similar equations can be derived for TE
fields

Striplines and Microstrip Lines
the strip line was developed from the
square coaxial

coaxial

rectangular line

square coaxial

flat stripline

− jωε o ε r
A cos k c xe − jβz − − − − ( 43)
kc

various planar transmission line
structures are shown here:
stripline

microstrip
line

slot line

coplanar
line

since the stripline has only 1 dielectric, it
supports TEM wave, however, it is difficult to
integrate with other discrete elements and
excitations
microstrip line is one of the most popular
types of planar transmission line, it can be
fabricated by photolithographic techniques
and is easily integrated with other circuit
elements

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27/01/2014



the following diagrams depicts the
evolution of microstrip transmission line

a microstrip line suspended in air can support
TEM wave
a microstrip line printed on a grounded slab
does not support TEM wave
the exact fields constitute a hybrid TM-TE
wave
when the dielectric slab become very thin
(electrically), most of the electric fields are
trapped under the microstrip line and the
fields are essentially the same as those of the
static case, the fields are quasi-static

+

+

+
two-wire line

single-wire above
ground (with image)

microstrip in air
(with image)

microstrip with
grounded slab


Design Formulas of Microstrip Lines

one can define an effective dielectric constant
so that the phase velocity and the
propagation constant can be defined as

β = k o ε e − − − ( 48 )
the effective dielectric constant is bounded by
1 < ε e < ε r , it also depends on the slab
thickness d and conductor width, W

design formulas have been derived for
microstrip lines
these formulas yield approximate values
which are accurate enough for most
applications
they are obtained from analytical expressions
for similar structures that are solvable exactly
and are modified accordingly



vp =

c
− − − − ( 47)
εe

or they are obtained by curve fitting
numerical data
the effective dielectric constant of a
microstrip line is given by
1
ε + 1 εr − 1
εr = r
+
− − − − ( 49 )
2
2
1 + 12d / W

the characteristic impedance is given by
for W/d≤ 1
Zo =

60  8d W 
ln +  − − − − ( 50 )
ε r  W 4d 

For W/d ≥ 1
Zo =

120 π
− − ( 51)
ε r [ W / d + 1.393 + 0.667 ln( W / d + 1.444 )]

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27/01/2014


for a given characteristic impedance Z o
and dielectric constant ε r , the W/d
ratio can be found as
W/d=

8e A
e 2A − 2

− − − − ( 52)

for W/d<2

for a homogeneous medium with a
complex dielectric constant, the
propagation constant is written as
2
γ = α d + jβ = k c − κ 2
2
γ = k c − ω 2µ o ε o ε r (1 − j tan δ )

note that the loss tangent is usually very
small

2
ε −1
[ B − 1 − ln( 2B − 1) + r
×
2ε r
π
for W/d > 2
0.61
{ln( B − 1) + 0.39 −
}] − − − − ( 53)
εr
W/d=

Where
And

0.11
εr + 1 εr − 1
( 0.23 +
)
+
2
εr + 1
εr

A=

Zo
60

B=

377π
2Z o ε r


Note that

(1 + x ) 1/ 2 = 1 + x / 2

where x is small

therefore, we have
2
γ = kc − k 2 +

jk 2 tan δ
2
2 kc − k2

− − − ( 54 )

2
γ = k c − k 2 + jk 2 tan δ

2
Note that jβ = k c − k 2
for small loss, the phase constant is
unchanged when compared to the
lossless case
the attenuation constant due to
dielectric loss is therefore given by
k 2 tan δ Np/m (TE or TM) (55)
αd =

2β

For TEM wave k = β , therefore
αd =

k tan δ
Np/m (TEM) (56)
2

for a microstrip line that has
inhomogeneous medium, we multiply
Eq. (56) with a filling factor
ε r ( ε e − 1)
ε e ( ε r − 1)

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27/01/2014

k ε tan δ ε r ( ε e − 1) = k o ε r ( ε e − 1) tan δ
αd = o e
2 ε e ( ε r − 1)
ε e ( ε r − 1)
2

(57)

the attenuation due to conductor loss is given by
Rs
αc =
Zo W
(58) Np/m where

R s = ωµ o / ( 2σ )

note that for most microstrip substrate,
the dielectric loss is much more
significant than the conductor loss
at very high frequency, conductor loss
becomes significant

R s is called the surface resistance of the conductor

Solution for Microstrip Lines

d

W

y
εr

-a/2

x
a/2

two side walls are sufficiently far away that
the quasi-static field around the microstrip
would not be disturbed (a >> d)

using the separation of variables and
appropriate boundary conditions, we
write
∞
nπx
nπy
sinh
− − − ( 59 ),0 ≤ y ≤ d
∑ A n cos
a
a
n = 1,odd
∞
nπx − nπy / a
Φ ( x, y ) =
e
− − − ( 60 ), d ≤ y ≤ ∞
∑ B n cos
a
n = 1,odd

we need to solve the Laplace’s equation
with boundary conditions
∇ 2 Φ ( x, y ) = 0,| x| ≤ a / 2,0 ≤ y < ∞
t
Φ ( x, y ) = 0, x = ± a / 2
Φ ( x, y ) = 0, y = 0, ∞
two expressions are needed, one for
each region

the potential must be continuous at y=d
so that
A n sinh

nπd
= B n e − nπd / a
a

Φ ( x, y ) =

note that this expression must be true
for any value of n

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27/01/2014

mπx


nπx

π
π
a/2
cos
dx = 0
due to fact that ∫− a / 2 cos
a
a
if m is not equal to n
Φ ( x, y ) =

∞

∑ A n cos

nπx

sinh

nπy

the normal component of the electric field is
discontinuous due to the presence of surface
charge on the microstrip, E y = − ∂Φ / ∂y
∞
nπ
nπx
nπ y
cos
cosh
,0 ≤ y ≤ d
∑ An
a
a
n = 1,odd a
∞
nπ
nπ x
nπd − nπ ( y − d)/ a
cos
sinh
e
,
Φ ( x, y ) = ∑ A n
a
a
a
n = 1,odd

− − − ( 61),0 ≤ y ≤ d

Ey = −

a
a
n = 1,odd
∞
nπx
nπd − nπ ( y− d)/ a
sinh
e
Φ ( x, y ) =
− ( 62),
∑ A n cos
a
a
n = 1,odd
d≤ y≤ ∞

d≤ y≤ ∞


multiply Eq. (63) by cos mπx/a and
integrate from -a/2 to a/2, we have

ρ s = ε o E y ( x, y = d + ) − ε o ε r E y ( x, y − )

ρs = ε o

∑ A n sinh nπd / a
n = 1,odd

the total charge on the strip is

m πx
2 sin( mπW / 2a )
dx =
a
mπ / a

∞
nπ
nπx
nπd
nπy
cos
(sinh
) dx =
+ ε r cosh
∑ An
a
a
a
n = 1,odd a
nπ
nπd
nπd a / 2
mπx
nπx
(sinh
) ∫ cos
cos
dx =
+ ε r cosh
ε o ∑ An
a
a
a −a/2
a
a

ε o An

nπ
nπd
n πd a
(sinh
) ,m = n
+ ε r cosh
a
a
a 2

An =

4a sin( mπW / 2a )
( nπ ) 2 ε o [sinh( nπd / a ) + ε r cosh( nπd / a )


the voltage of the microstrip wrt the
ground plane is
∞

cos

a/2

assuming that the charge distribution is
given by ρ s = 1 on the conductor and
zero elsewhere

d
V = − ∫0 E y ( x = 0, y )dy =

∫
−W/2

∫− a / 2 ε o

∞
nπ
nπx
nπd
nπd
cos
(sinh
) − − ( 63)
+ ε r cosh
∑ An
a
a
a
n = 1,odd a


W/2

a/2

∫ − a / 2 ρsdx =

the surface charge at y=d is given by

the static capacitance per unit length is
C=

Q
=
V

∞

∑

W
4a sin( mπW / 2a ) sinh( nπd / a )

2
n = 1,odd ( nπ ) ε o [sinh( nπd / a ) + ε r cosh( nπd / a )]

(64)

this is the expression for ε r ≠ 1

W/2
∫− W / 2 dx = W

13

27/01/2014


Techniques

the effective dielectric is defined as
C
εe =
C
Co , where o is obtained from
Eq. (64) with ε r = 1

the transverse resonance technique
employs a transmission line model of
the transverse cross section of the
guide

the characteristic impedance is given by

right at cutoff, the propagation constant
is equal to zero, therefore, wave cannot
propagate in the z direction

Zo =

εe
1
=
v p C cC

Techniques
it forms standing waves in the
transverse plane of the guide
the sum of the input impedance at any
point looking to either side of the
transmission line model in the
transverse plane must be equal to zero
at resonance

Techniques
the characteristic impedance in each of the
air and dielectric regions is given by
k η
k η
k η
Z a = xa o and Z d = xd d = xd o
kd
ε rko
ko

since the transmission line above the
dielectric is of infinite extent, the input
impedance looking upward at x=d is simply
given by Z a

Techniques
consider a grounded slab and its
equivalent transmission line model
to infinity
d

Za, kxa

x
εr

z

Zd,kxd

Techniques
the impedance looking downward is the
impedance of a short circuit at x=0 transfers
Z + jZ o tan βl
to x=d
Z in = Z o L
Z o + jZ L tan βl
Subtituting

Z L = 0, Z o = Z d , β = k xd , l = d

, we have

Z in = jZ d tan βd

Therefore,

k xa ηo
k η
+ j xd o tan k xdd = 0
ko
εrko

14

27/01/2014

Techniques
Note that

k xa = − jh
,

therefore, we have

ε r h = k xd tan k xdd − − − ( 65 )

From phase matching, k yo = k yd
which leads to
2
2
2
2
2
ε r k o − k xd = k o − k xa = k o + h 2 − − − ( 66 )

Eqs. (65) and (66) are identical to that
of Eq. (38) and (39)

if the signal contains a band of
frequencies, each frequency will travel
at a different phase velocity in a nonTEM line, the signal will be distorted
this effect is called the dispersion effect

a plane wave propagates in a medium at the
speed of light 1 / µε
Phase velocity, v p = ω / β , is the speed at
which a constant phase point travels
for a TEM wave, the phase velocity equals to
the speed of light
if the phase velocity and the attenuation of a
transmission line are independent of frequency,
a signal propagates down the line will not be
distorted

if the dispersion is not too severe, a
group velocity describing the speed of
the signal can be defined
let us consider a transmission with a
transfer function of
Z( ω ) = Ae − jβz = | Z( ω )| e − jψ

if we denote the Fourier transform of a timedomain signal f(t) by F(ω), the output signal
at the other end of the line is given by
fo ( t) =

1 ∞
j( ωt − ψ )
dω
∫ F( ω )| Z ( ω )| e
2π − ∞
π

if A is a constant and ψ = aω, the output will
be
fo ( t) =

this expression states that the output
signal is A times the input signal with a
delay of a
now consider an amplitude modulated
carrier wave of frequency ω o
s( t ) = f ( t ) cos ω o t = Re{f ( t ) e jω o t }

∞
1
A ∫ F ( ω )e jω ( t − a ) dω = Af ( t − a )
2π − ∞
π

15

27/01/2014



jω t
the Fourier transform of f ( t ) e o
is given by S(ω ) = F( ω − ω o )

The output signal so ( t ) , is given by

note that the Fourier transform of s(t) is
equal to 1 {F( ω + ω o ) + F( ω − ω o )}
2

if the maximum frequency component of the
signal is much less than the carrier
frequencies, β can be linearized using a
Taylor series expansion
β( ω ) = β( ω o ) +

dβ
|
( ω − ω o ) + ...
dω ω = ω o

note that the higher terms are ignored as the
higher order derivatives goes to zero faster
than the growth of the higher power of

(ω − ω o )

so ( t ) =

∞
1
Re ∫ AF(ω − ω o )e j( ωt − βz) dω
2π − ∞
π

for a dispersive transmission line, the
propagation constant β depends on
frequency, here A is assume to be
constant (weakly depend on ω)

with the approximation of
&&&
β ( ω ) ≈ β ( ω o ) + β'( ω o )( ω − ω o ) = β o + β'o ω
∞
1
&&&z)
so ( t ) =
A Re{ ∫ F(&&&) e j( ωt − β o z− β'o ω dω }
ω
2π
π
−∞
so ( t ) =

∞
1
&&&(
&&&}
A Re{e j( ω o t − β o z) ∫ F(&&&)e jω t − β'o z) dω
ω
2π
π
−∞

s o ( t ) = A Re{ f ( t − β'o z) e j( ω o t − β o z) }

s o ( t ) = Af ( t − β'o z) cos( ω o t − β o z) − − − ( 67)

Eq. (67) states that the output signal is
the time-shift of the input signal
envelope

Thank You!!

the group velocity is therefore defined
as
1
dω
vg =

β' o

=

|ω = ω o
dβ

16

Lect1

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