Bird J. Electrical Circuit Theory and Technology
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808 Electrical Circuit Theory and Technology
Z
0T
2C 2C
L
Z
0T
(a)
Z
0p
C
Z
0p
(b)
2L 2L
Figure 42.26
D
!
1
4ω
2
C
2
C
L
C
"
i.e., Z
0T
D
L
C
1
4ω
2
C
2
42.13
Z
0T
will be real when
L
C
>
1
4ω
2
C
2
Thus the filter will pass all frequencies above the point
where
L
C
D
1
4ω
2
C
2
i.e., where ω
2
c
D
1
4LC
42.14
where ω
c
D 2f
c
, and f
c
is the cut-off frequency.
Hence 2f
c
2
D
1
4LC
and the cut-off frequency,
f
c
=
1
4p
p
.LC/
42.15
The same equation for the cut-off frequency is obtained for the high-pass
network shown in Figure 42.26(b) as follows:
From equation (41.3), page 760, the characteristic impedance of a
symmetrical section is given by:
Z
0
D
Z
1
Z
2
2
Z
1
C 2Z
2
From Figure 42.26(b), Z
1
D
1
jωC
and Z
2
D j2ωL
Hence Z
0
D
1
jωC
j2ωL
2
1
jωC
C 2j2ωL
D
j4
ωL
2
C
j
4ωL
1
ωC
D
4L
2
C
4L
1
ω
2
C
i.e., Z
0
D
1
C
L
1
4ω
2
L
2
42.16
Filter networks 809
Z
0
will be real when
C
L
>
1
4ω
2
L
2
and the filter will pass all frequencies
above the point where
C
L
D
1
4ω
2
L
2
, i.e., where ω
2
c
D
1
4LC
as above.
Thus the cut-off frequency for a high-pass network is also given by
f
c
=
1
4p
p
.LC/
(as in equation (42.15)) 42.15
0
(b) Nominal impedance
When the frequency is very high, ω is a very large value and the term
1/4ω
2
C
2
in equations (42.13) and (42.16) are extremely small and may
be neglected.
The characteristic impedance then becomes equal to
p
L/C, this being
the nominal impedance. Thus for a high-pass filter section the nominal
impedance R
0
is given by:
R
0
=
L
C
42.17
the same as for the low-pass filter sections.
Problem 5. Determine for each of the high-pass filter sections
shown in Figure 42.27 (i) the cut-off frequency, and (ii) the
nominal impedance.
(a) Comparing Figure 42.27(a) with Figure 42.26(a) shows that:
2C D 0.2
µF, i.e., capacitance, C D 0.1 µF D 0.1 ð 10
6
F
and inductance, L D 100 mH D 0.1H
(i) From equation (42.15),
cut-off frequency, f
c
D
1
4
p
LC
D
1
4
[0.10.1 ð 10
6
]
i.e., f
c
D
10
3
40.1
D 796 Hz
(ii) From equation (42.17),
nominal impedance, R
0
D
L
C
D
0.1
0.1 ð 10
6
D 1000 Z or1kZ
0.2 µF 0.2 µF
100 mH
(a)
4000 pF
200 µH 200 µH
(b)
Figure 42.27
810 Electrical Circuit Theory and Technology
(b) Comparing Figure 42.27(b) with Figure 42.26(b) shows that:
2L D 200
µH, i.e., inductance, L D 100 µH D 10
4
H
and capacitance C D 4000 pF D 4 ð 10
9
F
(i) From equation (42.15
0
),
cut-off frequency, f
c
D
1
4
p
LC
D
1
4
[10
4
4 ð 10
9
]
D 126 kHz
(ii) From equation (42.17),
nominal impedance, R
0
D
L
C
D
10
4
4 ð 10
9
D
10
5
4
D 158 Z
(c) To determine values of L and C given R
0
and f
c
If the values of the nominal impedance R
0
and the cut-off frequency f
c
are known for a high-pass T or section it is possible to determine the
values of inductance L and capacitance C required to form the section.
From equation (42.17), R
0
D
L
C
D
p
L
p
C
from which,
p
L D R
0
p
C
Substituting in equation (42.15) gives:
f
c
D
1
4
p
L
p
C
D
1
4R
0
p
C
p
C
D
1
4R
0
C
from which,
capacitance C =
1
4pR
0
f
c
42.18
Similarly, from equation (42.17),
p
C D
p
L
R
0
Substituting in equation (42.15) gives: f
c
D
1
4
p
L
p
L
R
0
D
R
0
4L
from which,
inductance, L =
R
0
4pf
c
42.19
Filter networks 811
Problem 6. A filter is required to pass all frequencies above
25 kHz and to have a nominal impedance of 600 . Design (a) a
high-pass T section filter and (b) a high-pass section filter to
meet these requirements.
Cut-off frequency, f
c
D 25 ð 10
3
Hz and nominal impedance,
R
0
D 600
From equation (42.18),
C D
1
4R
0
f
c
D
1
460025 ð 10
3
F D
10
12
460025 ð 10
3
pF
i.e., C D 5305 pF or 5.305 nF
From equation (42.19), inductance,
L D
R
0
4f
c
D
600
425 ð 10
3
H D 1.91 mH
(a) A high-pass T section filter is shown in Figure 42.28(a) where the
series arm capacitances are each 2 C (see Figure 42.26(a)), i.e.,
2 ð 5.305 D 10.61 nF
(b) A high-pass section filter is shown in Figure 42.28(b), where the
shunt arm inductances are each 2 L (see Figure 42.26(b)), i.e.,
2 ð 1.91 D 3.82 mH
10.61 nF
1.91 mH
(a)
5.305 nF
3.82 mH 3.82 mH
(b)
10.61 nF
Figure 42.28
(d) ‘Constant-k’ prototype high-pass filter
It may be shown, in a similar way to that shown in Section 42.5(d), that
for a high-pass filter section:
Z
0T
Z
0p
= Z
1
Z
2
D R
2
0
where Z
1
and Z
2
are the total equivalent series and shunt arm impedances.
The high-pass filter sections shown in Figure 42.26 are thus ‘constant-k’
prototype filter sections.
(e) Practical high-pass filter characteristics
From equation (42.13), the characteristic impedance Z
0T
of a high-pass
T section is given by:
Z
0T
D
L
C
1
4ω
2
C
2
812 Electrical Circuit Theory and Technology
Rearranging gives:
Z
0T
D
!
L
C
1
1
4ω
2
LC
"
D
L
C
1
1
4ω
2
LC
From equation (42.14), ω
2
c
D
1
4LC
Thus
Z
0T
= R
0
1 −
!
c
!
2
42.20
Also, since Z
0T
Z
0
D R
2
0
then Z
0
D
R
2
0
Z
0T
D
R
2
0
R
0
1
ω
c
ω
2
i.e.,
Z
0p
=
R
0
1 −
!
c
!
2
42.21
From equation (42.20),
when ω<ω
c
,Z
0T
is reactive,
when ω D ω
c
,Z
0T
is zero,
and when ω>ω
c
,Z
0T
is real, eventually increasing to R
0
when ω is very large.
Similarly, from equation (42.21),
when ω<ω
c
,Z
0
is reactive,
when ω D ω
c
,Z
0
D1(i.e.,
R
0
0
D1)
and when ω>ω
c
,Z
0
is real, eventually decreasing to R
0
when ω is very large.
Curves of Z
0T
and Z
0
against frequency are shown in Figure 42.29.
Figure 42.4(a), on page 792, showed an ideal high-pass filter section
characteristic of attenuation against frequency. In practise, the character-
istic curve of a high-pass prototype filter section would look more like
that shown in Figure 42.30.
Filter networks 813
Nominal
impedance
Frequency
Pass band
R
0
Z
0T
Z
0Π
Z
0
0
f
c
Attenuation
band
Figure 42.29 Figure 42.30
Problem 7. A low-pass T section filter having a cut-off frequency
of 15 kHz is connected in series with a high-pass T section filter
having a cut-off frequency of 10 kHz. The terminating impedance
of the filter is 600 .
(a) Determine the values of the components comprising the
composite filter.
(b) Sketch the expected attenuation against frequency
characteristic.
(c) State the name given to the type of filter described.
(a) For the low-pass T section filter: f
c
L
D 15000 Hz
From equation (42.6),
capacitance, C D
1
R
0
f
c
D
1
60015000
35.4nF
From equation (42.7),
inductance, L D
R
0
f
c
D
600
15 000
12.73 mH
Thus from Figure 42.17, the series arm inductances are each L/2,
i.e., 12.73/2 D 6.37 mH and the shunt arm capacitance is 35.4 nF.
For a high-pass T section filter: f
C
H
D 10000 Hz
From equation (42.18),
capacitance, C D
1
4R
0
f
c
D
1
460010000
13.3nF
814 Electrical Circuit Theory and Technology
From equation (42.19),
inductance, L D
R
0
4f
c
D
600
410 000
4.77 mH
Thus from Figure 42.26(a), the series arm capacitances are each 2 C,
i.e., 2 ð 13.3 D 26.6 nF, and the shunt arm inductance is 4.77 mH.
The composite filter is shown in Figure 42.31.
6.37 mH 6.37 mH
35.4 nF
26.6 nF 26.6 nF
4.77 mH
600 Ω
Figure 42.31
(b) A typical characteristic expected of attenuation against frequency is
shown in Figure 42.32.
Figure 42.32
(c) The name given to the type of filter described is a band-pass filter.
The ideal characteristic of such a filter is shown in Figure 42.5.
Problem 8. A high-pass T section filter has a cut-off frequency
of 500 Hz and a nominal impedance of 600 . Determine the
frequency at which the characteristic impedance of the section is
(a) zero, (b) 300 , (c) 590 .
Filter networks 815
From equation (42.20), Z
0T
D R
0
1
ω
c
ω
2
(a) When Z
0T
D 0, then ω
c
/ω D 1, i.e., the frequency is 500 Hz, the
cut-off frequency.
(b) When Z
0T
D 300 , R
0
D 600 and f
c
D 500 Hz
300 D 600
1
2500
2f
2
from which
300
600
2
D 1
500
f
2
and
500
f
D
1
300
600
2
D
p
0.75
Thus when Z
0T
D 300 , frequency, f D
500
p
0.75
D 577.4Hz
(c) When Z
0T
D 590 , 590 D 600
1
500
f
2
500
f
D
1
590
600
2
D 0.1818
Thus when Z
0T
D 590 , frequency, f D
500
0.1818
D 2750 Hz
The above three results are seen to be borne out in the characteristic of
Z
0T
against frequency shown in Figure 42.29.
Further problems on high-pass filter sections may be found in
Section 42.10, problems 7 to 12, page 837.
42.7 Propagation
coefficient and time delay
in filter sections
Propagation coefficient
In Figure 42.33, let A, B and C represent identical filter sections, the
current ratios I
1
/I
2
, I
2
/I
3
and I
3
/I
4
being equal.
Although the rate of attenuation is the same in each section (i.e., the
current output of each section is one half of the current input) the amount
of attenuation in each is different (section A attenuates by
1
2
A, B atten-
uates by
1
4
A and C attenuates by
1
8
A). The attenuation is in fact in the
816 Electrical Circuit Theory and Technology
ABC
I
1
= 1 A
I
2
= A
1
2
I
3
= A
1
4
I
4
= A
1
8
Figure 42.33
form of a logarithmic decay and
I
1
I
2
D
I
2
I
3
D
I
3
I
4
D e
42.22
where is called the propagation coefficient or the propagation
constant.
From equation (42.22), propagation coefficient,
D ln
I
1
I
2
nepers 42.23
(See Section 41.3, page 761, on logarithmic units.)
Unless Sections A, B and C in Figure 42.33 are purely resistive there
will be a phase change in each section. Thus the ratio of the current
entering a section to that leaving it will be a phasor quantity having both
modulus and argument. The propagation constant which has no units is a
complex quantity given by:
D ˛ C jˇ
42.24
where ˛ is called the attenuation coefficient, measured in nepers, and ˇ
the phase shift coefficient, measured in radians. ˇ is the angle by which
a current leaving a section lags behind the current entering it.
From equations (42.22) and (42.24),
I
1
I
2
D e
D e
˛Cjˇ
D e
˛
e
jˇ
Since e
x
D 1 C x C
x
2
2!
C
x
3
3!
C
x
4
4!
C
x
5
5!
C ......
then e
jˇ
D 1 C jˇ C
jˇ
2
2!
C
jˇ
3
3!
C
jˇ
4
4!
C
jˇ
5
5!
C ......
D 1 C jˇ
ˇ
2
2!
j
ˇ
3
3!
C
ˇ
4
4!
C j
ˇ
5
5!
C ......
since j
2
D1,j
3
Dj, j
4
DC1, andsoon.
Hence e
jˇ
D
1
ˇ
2
2!
C
ˇ
4
4!
....
C j
ˇ
ˇ
3
3!
C
ˇ
5
5!
.....
D cosˇ C j sin ˇ from the power series for cos ˇ and sin ˇ
Filter networks 817
Thus
I
1
I
2
D e
˛
e
jˇ
D e
˛
cos ˇ C j sin ˇ D e
˛
6
ˇ in abbreviated polar form,
i.e.,
I
1
I
2
= e
a
66
b 42.25
Now e
˛
D
#
#
#
#
I
1
I
2
#
#
#
#
from which
attenuation coefficient, a = ln
#
#
#
#
I
1
I
2
#
#
#
#
nepers or 20 lg
#
#
#
#
I
1
I
2
#
#
#
#
dB
If in Figure 42.33 current I
2
lags current I
1
by, say, 30
°
, i.e., (/6) rad,
then the propagation coefficient of Section A is given by:
D ˛ C jˇ D ln
#
#
#
#
#
#
#
#
1
1
2
#
#
#
#
#
#
#
#
C j
6
i.e., g D .0.693 Y j0.524/
If there are n identical sections connected in cascade and terminated in
their characteristic impedance, then
I
1
I
nC1
D e
n
D e
n
D e
n˛Cjˇ
D e
n˛
6
nˇ, ...... 42.26
where I
nC1
is the output current of the n’th section.
Problem 9. The propagation coefficients of two filter networks are
given by
(a) D 1.25 C j0.52,(b) D 1.794
6
39.4
°
Determine for each (i) the attenuation coefficient, and (ii) the phase
shift coefficient.
(a) If D 1.25 C j0.52
then (i) the attenuation coefficient, ˛, is given by the real part,
i.e., a D 1.25 N
and (ii) the phase shift coefficient, ˇ, is given by the imaginary
part,
i.e., b D 0.52 rad