2/1/2012
Transmission Line Input Impedance present
1/13
Transmission Line
Input Impedance
Consider a lossless line, length  , terminated with a load ZL.
I ( z ) = I +( z ) + I -( z )
+
+
V ( z ) = V +( z ) +V -( z )
VL
ZL
-
-
z  
IL

z =0
 Let’s determine the input impedance of this line!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/1/2012
Transmission Line Input Impedance present
2/13
It’s not ZL and it’s not Z0
Q: Just what do you mean by input impedance?
A: The input impedance is simply the line impedance at the beginning (at
z = - ) of the transmission line, i.e.:
Zin = Z ( z = - ) =
V ( z = - )
I ( z = - )
Note Zin is equal to neither the load impedance ZL, nor the characteristic
impedance Z0 !
Zin ¹ Z L
Zin ¹ Z 0
and
I (z )
IL
+
Zin = Z ( z =-  )
V (z )
-
z  
Jim Stiles

The Univ. of Kansas
+
VL
ZL
-
z =0
Dept. of EECS
2/1/2012
Transmission Line Input Impedance present
3/13
There’s more on the next page…
To determine exactly what Zin is, we first must determine the voltage and
current at the beginning of the transmission line ( z = - ).
V ( z = - ) = V0+ éê e + jβ  + 0 e - jβ  ùú
ë
û
V0+ é + jβ 
-  0 e - jβ  ùú
I ( z = - ) =
e
ê
û
Z0 ë
Therefore:
Zin
æ e + jβ  +  e - jβ  ÷ö
V ( z = - )
ç
0
÷÷
=
= Z 0 çç
+
jβ

jβ

÷ø
I ( z = - )
çè e
- 0 e
We can explicitly write Zin in terms of load ZL using the previously determined
relationship:
L =
Jim Stiles
ZL - Z0
= 0
ZL + Z0
The Univ. of Kansas
Dept. of EECS
2/1/2012
Transmission Line Input Impedance present
4/13
…Zin can be WAY different than ZL
Combining these two expressions, we get:
Zin =
( Z L + Z 0 )e + jβ  + ( Z L - Z 0 )e - jβ 
Z0
( Z L + Z 0 )e + jβ  - ( Z L - Z 0 )e - jβ 
(
(
)
)
(
(
æ Z e + jβ  + e - jβ  + Z e + jβ  - e - jβ 
ç L
0
= Z 0 çç
çç Z e + jβ  + e - jβ  - Z e + jβ  - e - jβ 
0
è L
) ÷÷÷ö
) ÷÷÷ø
Now, recall Euler’s equations:
e + jβ  = cos β  + j sin β 
and
e - jβ  = cos β  - j sin β 
Using Euler’s relationships, we can likewise write the input impedance without the
complex exponentials:
Zin
æ Z L cos β  + j Z 0 sin β  ö÷
æ Z L + j Z 0 tan β  ö÷
ç
= Z0 ç
÷÷ = Z 0 çç
÷
èç Z 0 cos β  + j Z L sin β  ø÷
èç Z 0 + j Z L tan β  ø÷÷
Note that depending on the values of β , Z 0 and  , the input impedance can be
radically different from the load impedance ZL !
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/1/2012
Transmission Line Input Impedance present
5/13
Your brain should be big enough
Now let’s look at the Zin for some important load impedances and line lengths.
 You should commit these results to memory!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/1/2012
Transmission Line Input Impedance present
6/13
1. Line Length is one-half a wavelength
If the length of the transmission line is exactly one-half wavelength (  = λ 2 ),
we find that:
β =
meaning that:
cos β  = cos π = -1
2π λ
=π
λ 2
and
sin β  = sin π = 0
and therefore:
æ Z L cos β  + j Z 0 sin β  ÷ö
æ Z ( - 1) + j Z L (0) ÷ö
÷÷ = Z 0 çç L
÷ = ZL
çè Z cos β  + j Z sin β  ÷ø
çè Z ( - 1) + j Z (0) ÷÷ø
0
L
0
L
Zin = Z 0 çç
In other words, if the transmission line is precisely one-half wavelength long,
the input impedance is equal to the load impedance, regardless of Z0 or 




Z0, 
Zin  Z L
ZL



  
2
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/1/2012
Transmission Line Input Impedance present
7/13
2. Line Length is one-quarter a wavelength
If the length of the transmission line is exactly one-quarter wavelength
(  = λ 4 ), we find that:
2π λ π
β =
=
λ 4 2
meaning that:
cos β  = cos π 2 = 0
and
sin β  = sin π 2 = 1
and therefore:
2
Zin
æ Z cos β  + j Z 0 sin β  ö÷
æ Z (0) + j Z 0 (1) ÷ö ( Z 0 )
= Z 0 çç L
÷÷ = Z 0 çç L
÷=
çè Z cos β  + j Z sin β  ÷ø
çè Z (0) + j Z (1) ÷÷ø
ZL
0
L
0
L
In other words, if the transmission line is precisely one-quarter
wavelength long, the input impedance is inversely proportional to the
load impedance
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/1/2012
Transmission Line Input Impedance present
8/13
A short becomes an open—and vice versa!
 Think about what this means!
Say the load impedance is a short circuit, such that Z L = 0 .
The input impedance at beginning of the λ 4 transmission line is therefore:
2
Zin =
( Z0 )
ZL
2
=
( Z0 )
=¥
0
Zin = ¥ ! This is an open circuit!
The quarter-wave transmission line transforms a short-circuit into an opencircuit—and vice versa!
Zin  
Z0 , 
  
Jim Stiles
ZL=0
4
The Univ. of Kansas
Dept. of EECS
2/1/2012
Transmission Line Input Impedance present
9/13
3. Load is numerically equal to Z0
If the load is numerically equal to the characteristic impedance of the
transmission line (a real value), we find that—regardless of length  (!)—the input
impedance becomes:
æ Z L cos β  + j
èç Z 0 cos β  + j
æ Z cos β  + j
= Z 0 çç 0
èç Z 0 cos β  + j
Zin = Z 0 çç
Z 0 sin β  ÷ö
÷
Z L sin β  ÷÷ø
Z 0 sin β  ÷ö
÷ = Z0
Z 0 sin β  ÷ø
In other words, if the load impedance is equal to the transmission line
characteristic impedance, the input impedance will be likewise be equal to Z0,
regardless of the transmission line length  !!!!
Zin  Z 0
Z0, 
ZL=Z0

Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/1/2012
Transmission Line Input Impedance present
10/13
4. Load is purely reactive (RL=0)
If the load is purely reactive (i.e., the resistive component is zero), the input
impedance is:
æ Z cos β  + j Z 0 sin β  ÷ö
Zin = Z 0 çç L
÷
çè Z cos β  + j Z sin β  ÷÷ø
0
L
æ j X cos β  + j Z sin β  ö÷
L
0
÷÷
= Z 0 ççç
2
èç Z 0 cos β  + j XL sin β  ø÷
æ X cos β  + Z 0 sin β  ö÷
= j Z 0 çç L
÷
çè Z cos β  - X sin β  ø÷÷
0
L
In other words, if the load is purely reactive, then the input impedance will
likewise be purely reactive, regardless of the line length  .
Z in  j X in
Z0 , 
ZL=jXL

Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/1/2012
Transmission Line Input Impedance present
11/13
5. Load is purely real (XL=0)
Q: Hey! If a purely reactive load results in a purely reactive input impedance,
then is seems to reason that a purely resistive load would likewise result in a
purely resistive input impedance.
Is this true? It seems to work for real load Z L = Z 0 !
A: This is definitely not true!!!!
Even if the load is purely resistive (ZL = R), the input impedance will in general be
complex (both resistive and reactive components).
 Do you see why? Why does this make sense? Make sure YOU know!
Z in  Rin  j X in
Z0 , 
ZL = RL

Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/1/2012
Transmission Line Input Impedance present
12/13
6.Line length is much
shorter than a wavelength
If the transmission line is electrically small—its length  is small with respect to
signal wavelength λ --we find that:
β =
and thus:
2π
cos β  = cos 0 = 1
λ
 = 2π
and

λ
»0
sin β  = sin 0 = 0
so that the input impedance is:
Zin
æ Z L cos β  + j Z 0 sin β  ÷ö
æ Z L (1) + j Z L (0) ÷ö
ç
ç
= Z0 ç
÷ = Z0 ç
÷ = ZL
çè Z cos β  + j Z sin β  ÷÷ø
çè Z (1) + j Z (0) ÷÷ø
0
0
L
L
In other words, if the transmission line length is much smaller than a
wavelength, the input impedance Zin will always be equal to the load
impedance Z L .
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/1/2012
Transmission Line Input Impedance present
13/13
Electrically small: A wire is just a wire
This is the assumption we used in all previous circuits courses (e.g., EECS 211,
212, 312, 412)!
In those courses, we assumed that the signal frequency ω is relatively low, such
that the signal wavelength λ is very large ( λ   ).
Note also for this case ( the electrically short transmission line), the voltage and
current at each end of the transmission line are approximately the same!
V (z = -) » V (z = 0)
and
I(z = -) » I (z = 0) if   λ
If   λ , our “wire” behaves exactly as it did in EECS 211 !
Z in  Z L
IL
IL
+
+
VL
Z0, 
VL
ZL=jXL
-
-

Jim Stiles
The Univ. of Kansas
Dept. of EECS