EECS 117 Homework Assignment 3
Spring 2004
1. Suppose the resonant frequency ω0 is equal to (LC)-0.5. The load impedance ZL is
Z L = R + jω L +
1
1 

= R + j  ωL −

jωC
ωC 

If ω = ω0 +∆ω, ZL is equal to




 ∆ω 
1
1


 −
Z L = R + j (ω 0 + ∆ω ) L +
= R + j ω 0 L1 +

ω
j (ω 0 + ∆ω )C


∆ω 
0 

C 

ω 0 1 +
ω

0  




 ∆ω 
1  ∆ω 
∆ω 
 −
1 −
 = R + j ∆ωL + 2 
≈ R + j ω 0 L1 +
ω 0  ω 0 C  ω 0 
ω 0 C 



The last equality holds because ω0 = (LC)-0.5. Furthermore,

∆ω 
1 
∆ω
∆ω 
(2ω 0 L) = R + j 2∆ωL
R + j ∆ωL + 2  = R + j
ω 0 L +
 = R+ j
ω0 
ω0C 
ω0
ω 0 C 

Using the values of the inductance and capacitance, the length of 2 cm corresponds 1.5π.
β 0l =
ω0
v
l=
1
v LC
In general, β l = ( β 0 + ∆β )l =
l=
2cm
10 8 m / s 2 × 10 −9 × 9.01 × 10 −13
ω 0 + ∆ω
l = β 0 l (1 +
= 1.5π
∆ω
) . Thus Zin is
ω0
v
R + j 2∆ωL + jZ 0 tan(1.5π (1 + ∆ω ω 0 ))
jZ 0 tan(1.5π (1 + ∆ω ω 0 ))
Z in = Z 0
≈ Z0
Z 0 + j (R + j 2∆ωL ) tan(1.5π (1 + ∆ω ω 0 ))
j (R + j 2∆ωL ) tan(1.5π (1 + ∆ω ω 0 ))
2
Z0
=
(R + j 2∆ωL )
The approximation holds because tan(1.5π (1 + ∆ω ω 0 )) >> R, ∆ωL, Z 0 . This expression
has the same form as a parallel RLC circuit, with
2
Z
L
2
1
= Z0 C
Req = 0 , C eq = 2 , Leq =
L
R
Z0
ω 2
0
Z 2
0
Therefore, the input impedance Zin is that of a second order circuit. Also, LeqCeq = LC =
ω0-2, so our assumption is correct, i.e., ω0 = (LC)-0.5. The Q factor is
Q = ω 0 Req C eq =
L R
LC
=
L
R = 94
C
The equivalent circuit is Leq = Z02C = 5 µF, Ceq = L/Z02 = 0.3604 fH, and Req = Z02/R =
1250 in parallel.
2. For a lossy line, the input impedance has a form
Z in = Z 0
Z L + jZ 0 tanh(αl + jβ l )
Z 0 + jZ L tanh(αl + jβ l )
where ZL = 2j∆ωL. Also,
tanh(αl + jβl ) =
sinh(2αl )
sin(2β l )
+ j
cos(2β l ) + cosh(2αl )
cos(2β l ) + cosh(2αl )
sin(2 βl ) = sin(3π (1 + ∆ω / ω 0 )) ≈ sin(3π ) + cos(3π ) * (3π∆ω / ω 0 ) = −3π∆ω / ω 0 . Likewise,
cos(2βl ) ≈ cos(3π ) − sin(3π ) * (3π∆ω / ω 0 ) = −1 . So,
tanh(αl + jβ l ) =
3π∆ω / ω 0
sinh( 2αl )
−j
cosh(2αl ) − 1
cosh(2αl ) − 1
The input impedance is then equal to
 sinh( 2αl )
2 j∆ωL + jZ 0 
−
cosh(2αl ) − 1

Z in = Z 0
 sinh( 2αl )
−
Z 0 + j 2 j∆ωL
 cosh(2αl ) − 1
j
3π∆ω / ω 0 

cosh(2αl ) − 1 
j
3π∆ω / ω 0 

cosh(2αl ) − 1 
3π∆ω / ω 0

sinh( 2αl ) 

+ j  2∆ωL + Z 0
cosh(2αl ) − 1 
cosh(2αl ) − 1 
= Z0
3π∆ω / ω 0
sinh( 2αl )
Z 0 − 2∆ωL
+ j 2∆ωL
cosh(2αl ) − 1
cosh(2αl ) − 1
Z 3π∆ω / ω 0 + j (2∆ωL(cosh(2αl ) − 1) + Z 0 sinh( 2αl ) )
≈ Z0 0
Z 0 (cosh(2αl ) − 1) − 2∆ωL sinh( 2αl )
Z0
The third term in the denominator has ∆ω2 dependence and is thus negligible.
4. This problem is similar to the example 3-16 in the textbook. For matching of the load
with either a short or an open shunt stub, we have the following circuit:
Ytot,in
βl
A
Z0
ZL
ZM
βl’
Yin
Where ZM is either 0 or infinity for a short or an open stub, respectively.
Since we are dealing with shunt stub, admittance would simplify the calculation. The
equivalent admittance at point A (excluding the stub for a moment) is given by
Yin = Y0
YL + jY0 tan( β l )
Y0 + jYL tan( β l )
where Y0 = [Z0]-1= 1/50, and YL = [Z0]-1= 1/100 + j3/100.
Separating the expression into the real and imaginary parts gives
1 + x2
3x 2 − 3x − 3
Yin =
+ j
(10 + 15 x) 2 + 25 x 2
(10 + 15 x) 2 + 25 x 2
where x = tan(βl).
Since a short or an open stub can behave as a purely reactive element, the real part of Yin
above should be equal to Z0 in order for the matching to take place. Thus, we have,
1 + x2
1
= Y0 =
2
2
(10 + 15 x) + 25 x
50
−3− 5
−3+ 5
or
. Both of them are
4
4
negative. This means that βl is larger than π/2. The more negative value represents a
point closer to the load and this value will be used in the following calculation.
The solution of this equation is: tan(βl) =
tan(βl) =
−3− 5
⇔ βl = 2.22315.
4
Substitute this value into the imaginary part of Yin and get
j
3x 2 − 3x − 3
(10 + 15 x ) 2 + 25 x 2
= 0.044721 j
x=
− 3− 5
4
The stub needs to have an impedance opposite to the value above in order to cancel the
reactive part of the Yin for matching. For a short stub, Y = − jY0 cot( β l ' ) . Therefore, we
want
0.044721 = cot( β l ' ) ⇔ βl’ = 0.420534.
Y
0
The input impedance at point A including the stub is
Ytot ,in
1 + tan 2 (2.22315)
=
(10 + 15 tan(2.22315) )2 + 25 tan 2 (2.22315)
 3 tan 2 (2.22315) − 3 tan(2.22315) − 3

Y0

−
+ j 
2
2

tan(
0
.
420534
)
(
)
+
+
10
15
tan(
2
.
22315
)
25
tan
(
2
.
22315
)


The imaginary part is equal to zero. The SWR is given by
1+
Y0 − Yin
Y0 + Yin
1−
Y0 − Yin
Y0 + Yin
S=
In general, for β’ ≠ β, i.e., β’l = βl*n, where n is the ratio of β’ to β, the input impedance
shown above is
1 + tan 2 (2.22315n)
Ytot ,in (n) =
(10 + 15 tan(2.22315n) )2 + 25 tan 2 (2.22315n)
 3 tan 2 (2.22315n) − 3 tan(2.22315n) − 3

Y0

−
+ j 
2
2

n
tan(
0
.
420534
*
)
(
)
+
n
+
n
10
15
tan(
2
.
22315
)
25
tan
(
2
.
22315
)


and the SWR is equal to
1+
Y0 − Yin (n)
Y0 + Yin (n)
1−
Y0 − Yin (n)
Y0 + Yin (n)
S=
SWR can be plotted as a function of n, representing the amount of shift in the signal
frequency / phase constant from the ones used in calculating the numbers above.
S
1.4
1.3
1.2
1.1
0.98
0.99
n
1.01
The values of n where S = 1.2 are 0.989 and 1.01, for a short stub.
Following the same procedure, we have these for an open stub
Y = jY0 tan( β l ' ) ⇔
0.044721
= − tan( βl ' ) ⇔ βl’ = 1.99133.
Y0
1.02
1 + tan 2 (2.22315n)
Ytot ,in (n) =
(10 + 15 tan(2.22315n) )2 + 25 tan 2 (2.22315n)
 3 tan 2 (2.22315n) − 3 tan(2.22315n) − 3


−
+ j 
tan(
1
.
99133
*
Y
n
0
2
2

(
)
+
+
10
15
tan(
2
.
22315
)
25
tan
(
2
.
22315
)
n
n


The plot of S vs. n:
S
1.7
1.6
1.5
1.4
1.3
1.2
1.1
0.98
0.99
1.01
n
1.02
The values of n where S = 1.2 are 0.993 and 1.01.
For impedance matching with a lumped element, we have the following circuit:
βl
Z0
ZM
ZL
To keep the problem simple, let’s make the lumped element ZM a purely reactive
component. With this constraint, the real part of Yin needs to match the characteristic
impedance of the transmission line, just like the cases of short and open stubs above. The
previous calculation gives β l = 2.22315 or 2.95288 . Just as before, we pick the shortest
distance β l = 2.22315 . At this length, Yin is equal to
Yin = 0.02 + 0.0447214 j
The positive imaginary part implies that the lumped element needs to be an inductor in
order to make the impedance matching work. So,
YM = 1 Z M = − j (ωL) = − j /( β vL) = −0.0447214 j
where v is the propagation velocity of the transmission line. The velocity is property of
the transmission line, and does not depend on the signal frequency.
In general, Ytot,in is
1 + tan 2 (2.22315n)
Ytot ,in (n) =
(10 + 15 tan(2.22315n) )2 + 25 tan 2 (2.22315n)
 3 tan 2 (2.22315n) − 3 tan(2.22315n) − 3
0.04427214 

+ j 
−
2
2

n
 (10 + 15 tan(2.22315n) ) + 25 tan (2.22315n)

The plot of S vs. n:
S
1.4
1.3
1.2
1.1
0.98
0.99
n
The values of n where S = 1.2 are 0.989 and 1.01.
1.01
1.02
In summary,
Bandwidth for S ≤ 1.2
Short
0.989 ω0 to 1.01 ω0
Open
0.993 ω0 to 1.01 ω0
Lumped 0.989 ω0 to 1.01 ω0