II. PASSIVE FILTERS
Frequency-selective or filter circuits pass to the output only those input signals that are in a
desired range of frequencies (called pass band). The amplitude of signals outside this range
of frequencies (called stop band) is reduced (ideally reduced to zero). Typically in these
circuits, the input and output currents are kept to a small value and as such, the current
transfer function is not an important parameter. The main parameter is the voltage transfer
function in the frequency domain, Hv (jω) = Vo /Vi . As Hv (jω) is complex number, it has
both a magnitude and a phase, filters in general introduce a phase difference between input
and output signals.
To minimize the number of subscripts, hereafter, we will drop subscript v of Hv . Furthermore,
we concentrate on the the “open-loop” transfer functions, Hvo , and denote this simply by
H(jω). The impact of loading is sperately discussed.
2.1
| H(j ω) |
Low-Pass Filters
An ideal low-pass filter’s transfer function is shown. The
frequency between the pass- and-stop bands is called the
cut-off frequency (ωc ). All of the signals with frequencies below ωc are transmitted and all other signals are
stopped.
In practical filters, pass and stop bands are not clearly
defined, |H(jω)| varies continuously from its maximum
toward zero. The cut-off frequency is, therefore, defined
√
as the frequency at which |H(jω)| is reduced to 1/ 2 =
0.7 of its maximum value. This corresponds to signal
power being reduced by 1/2 as P ∝ V 2 .
A series RL circuit as shown acts as a low-pass filter. For
no load resistance (“open-loop” transfer function), Vo can
be found from the voltage divider formula:
R
Vi
R + jωL
Stop
Band
ωc
ω
| H(j ω) |
Κ
0.7Κ
ωc
ω
L
Low-pass RL filters
Vo =
Pass
Band
→
H(jω) =
+
+
Vi
−
R
Vo
−
R
Vo
1
=
=
Vi
R + jωL
1 + j(ωL/R)
We note
1
|H(jω)| = q
1 + (ωL/R)2
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
23
It is clear that |H(jω)| is maximum when denominator is smallest, i.e., ω → 0 and |H(jω)|
decreases as ω is increased. Therefore, this circuit allows “low-frequency” signals to pass
through while “blocking” high-frequency signals (i.e., reduces the amplitude of the voltage
of the high-frequency signals). The reference to define the “low” and “high”-frequencies is
the cut-off frequency: “low”-frequencies mean frequencies much lower than ωc .
To find the cut-off frequency, we note that the |H(jω)|M ax = 1 occurs at ω = 0 (alternatively find d |H(jω)| /dω and set it equal to zero to find ω = 0 which maximizes |H(jω)|).
Therefore,
|H(jω)|max = 1
1
1
|H(jω)|ω=ωc = √ |H(jω)|max = √
2
2
1
=√
2
1 + (ωc L/R)2
1
q
−→
ωc L
1+
R
2
=2
ωc L
=1
R
→
Therefore,
ωc =
R
L
and
H(jω) =
1
1 + jω/ωc
Input Impedance: Using the definition of the input impedance, we have:
Zi =
Vi
= jωL + R
Ii
The value of the input impedance depends on the frequency ω. For good voltage coupling,
we need to ensure that the input impedance of this filter is much larger than the output
impedance of the previous stage. Since we do not know the frequency of the input signal,
we need to ensure that good voltage coupling criteria is satisfied for all frequencies (or all
possible values of Zi ). As such, the minimum value of Zi is an important number. Zi is
minimum when the impedance of the inductor is zero (ω → 0).
Zi |min = R
Output Impedance: The output impdenace can be
found by “killing” the source and finding the equivalent
impdenace between output terminals:
L
R
Zo
Zo = jωL k R
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
24
where the source resistance is ignored. Again, the value of the output impedance depends on
the frequency ω. For good voltage coupling, we need to ensure that the output impedance
of this filter is much smaller than the input impedance of the next stage for all frequencies,
the maximum value of Zo is an important number. Zo is maximum when the impedance of
the inductor is infinity (ω → ∞).
Zo |max = R
Bode Plots and Decibel
The voltage transfer function of a two-port network (and/or the ratio of output to input
powers) is usually expressed in Bel:
Number of Bels = log10
Po
Pi
or
Vo Number of Bels = 2 log10 Vi
because P ∝ V 2 . Bel is a large unit and decibel (dB) is usually used:
Number of decibels =
Vo 20 log10 Vi
or
Vo Vi
=
dB
There are several reasons why decibel notation is used:
Vo 20 log10 Vi
1) Historically, the analog systems were developed first for audio equipment. Human ear
“hears” the sound in a logarithmic fashion. A sound which appears to be twice as loud
actually has 10 times power, etc. Decibel translates the output signal to what ear hears.
2) If several two-port network are placed in a cascade (output of one is attached to the input
of the next), the overall transfer function, H, is equal to the product of all transfer functions:
|H(jω)| = |H1 (jω)| × |H2 (jω)| × ...
20 log10 |H(jω)| = 20 log10 |H1 (jω)| + 20 log10 |H2 (jω)| + ...
|H(jω)|dB = |H1 (jω)|dB + |H2 (jω)|dB + ...
making it easier to find the overall response of the system.
3) Plot of |H(jω)|dB versus frequency has special properties that make analysis simpler. For
example, the plot asymptotes to straight lines at low and high frequencies as is shown below.
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
25
Also, using dB definition, we see that, there is a 3 dB difference between maximum gain and
gain at the cut-off frequency:
20 log |H(jωc )| − 20 log |H(jω)|max
"
#
|H(jωc )|
1
= 20 log
= 20 log √
|H(jω)|max
2
!
≈ −3 dB
Bode plots are plots of |H(jω)|dB (magnitude) and 6 H(jω) (phase) versus frequency in a
semi-log format (i.e., ω axis is a log axis). Bode plots of first-order low-pass RL filters are
shown below (W denotes ωc ).
|H(jω)|dB
6
H(jω)
At high frequencies, ω/ωc ≫ 1,
1
|H(jω)| ≈
ω/ωc
→
|H(jω)|dB
"
#
1
= 20 log(ωc ) − 20 log(ω)
= 20 log
ω/ωc
which is a straight line with a slope of -20 dB/decade in the Bode plot. It means that if ω
is increased by a factor of 10 (a decade), |H(jω)|dB changes by -20 dB.
At low frequencies, ω/ωc ≪ 1, |H(jω)| ≈ 1 which is also a straight line in the Bode plot.
The intersection of these two “asymptotic” values is at 1 = 1/(ω/ωc) or ω = ωc . Because of
this, the cut-off frequency is also called the “corner” frequency.
The behavior of the phase of H(jω) can be found by examining 6 H(jω) = − tan−1 (ω/ωc). At
low frequencies, ω/ωc ≪ 1, 6 H(jω) ≈ 0 and at high frequencies, ω/ωc ≫ 1, 6 H(jω) ≈ −90◦ .
At cut-off frequency, 6 H(jω) ≈ −45◦ .
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
26
General first-order low-pass filters
As we discussed before, transfer functions characterize a two-port network. As such, it is
useful to group two-port networks into families based on their voltage transfer functions.
To facilitate this grouping, the convention is to simplify the voltage transfer function to a
form such that the “Real” part of the denominator of H(jω) is unity (i.e., the denominator
should be 1 + j · · · or 1 − j · · · ). As we will see later in this section, this grouping will also
help reduce the math that we do in analyzing various circuits.
The low-pass RL filter discussed before is part of the family of first-order low-pass filters
(first order means that ω appears in the denominator with an exponent of 1 or −1. In
general, the voltage transfer function of a first-order low-pass filter is in the form:
H(jω) =
K
1 + jω/ωc
The maximum value of |H(jω)| = |K| is called the filter gain. Note that the exponent of ω
in the denominator is +1 so that |H(jω)| decreases with frequency (thus,a low-pass filter):
|K|
|H(jω)| = q
1 + (ω/ωc)2
6
ω
|K|
tan−1
H(jω) = −
K
ωc
For RL filter, K = 1, and ωc = R/L. Note that K can be negative, and in that case, the
“minus” sign adds 180◦ phase shift to the transfer function as is denoted by |K|/K factor
above.
Low-pass RC filters
A series RC circuit as shown also acts as a low-pass filter.
For no load resistance (“open-loop” transfer function), Vo
can be found from the voltage divider formula:
1/(jωC)
1
Vi =
Vi
R + 1/(jωC)
1 + j(ωRC)
1
H(jω) =
1 + jωRC
Vo =
R
+
Vi
−
+
C
Vo
−
We see that the voltage transfer function of this circuit is similar to transfer function of a
general first-order low-pass filter. So, this is a low-pass filter with K = 1 and ωc = 1/RC.
(Note: we identified the circuit and found the cut-off frequency without doing any math!).
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
27
We could, of course, do the math following the procedure in analyzing the low-pass RL filter
to get the same answer. (Exercise: Show this.).
Following the same procedure as for RL filters, we find input and output Impedances
1
jωC
1
Zo = R k
jωC
Zi = R +
and
Zi |min = R
and
Zo |max = R
R
Terminated RL and RC low-pass filters
+
Now let us examin the effect of a load on the performance of our RL and RC filters. For this example,
a resistive load is considered but the analysis can be
easily extended to an impedance load. For example,
consider the terminated RC filter shown:
+
Vi
C
−
Vo
RL
−
From the circuit,
H(jω) =
1/(jωC) k RL
R′ /R
Vo
=
=
Vi
R + [1/(jωC) k RL ]
1 + j(ωR′ C)
with
R′ = R k RL
This is similar to the transfer function for unterminated RC filter but with resistance R
being replaced by R′ . Therefore,
ωc =
1
1
=
′
RC
(R k RL )C
and
H(jω) =
R′ /R
1 + jω/ωc
We see that the impact of the load is to reduce the filter gain (K = R′ /R < 1) and to shift
the cut-off frequency to a higher frequency as R′ = R k RL < R.
Input Impedance: Zi = R +
1
k RL
jωC
Output Impedance: Zo = R k
Zi |min = R
1
jωC
Zo |max = R
We could have arrived at the same results using the the relationship between open-loop,
Ho (jω), and terminated, H(jω), transfer functions of a two-port network:
H(jω) =
ZL
Ho (jω) =
ZL + Zo
RL
RL + R k
1
jωC
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
×
1
1 + jωRC
28
(Exercise: show this.) Also, note that the output impdenace of the terminated circuit is
exactly the same as the open-loop version.
Furthermore, it can be seen that as long as RL ≫ Zo or RL ≫ Zo |max = R (our condition
for good voltage coupling), R′ ≈ R and the terminated RC filter will look exactly like an
unterminated filter – The filter gain is one, the shift in cut-off frequency disappears, and
input and output resistances become the same as before.
Terminated RL low-pass filters
The parameters of the terminated RL filters can be found similarly:
Voltage Transfer Function: H(jω) =
1
Vo
=
,
Vi
1 + jω/ωc
ωc = (R k RL )/L.
Input Impedance: Zi = jωL + R k RL ,
Zi |min = R k RL
Output Impedance: Zo = (jωL) k R,
Zo |max = R
Here, the impact of load is to shift the cut-off frequency to a lower value. Filter gain is not
affected. Again for RL ≫ Zo or RL ≫ Zo |max = R (our condition for good voltage coupling),
the shift in cut-off frequency disappears and the filter will look exactly like an unterminated
filter.
Exercise: Derive above equations for the transfer function and input and output impdenacess.
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
29
2.2
First-order high pass filters
In general, the voltage transfer function of a first-order high-pass filter is in the form:
K
1 − jωc /ω
H(jω) =
It is a first-order filter because ω appears in the denominator with an exponent of −1. It
is a high-pass filter because |H| = 0 for ω = 0 and |H| is constant for high-freqnecies.
√
Paramter ωc is the cut-off freqnecy of the filter (Exercise: prove that |H(jωc )| is 1/ 2 = 0.7
of |H(jω)|M ax.)
The maximum value of |H(jω)| = |K| is called the filter gain.
|K|
|H(jω)| = q
1 + (ωc /ω)2
6
ωc
|K|
tan−1
H(jω) = +
K
ω
Bode Plots of first-order high-pass filters (K = 1) are shown below. The asymptotic behavior
of this class of filters is:
At low frequencies, ω/ωc ≪ 1, |H(jω)| ∝ ω (a +20dB/decade line) and 6 H(jω) = 90◦
At high frequencies, ω/ωc ≫ 1, |H(jω)| ∝ 1 (a line with a slope of 0) and 6 H(jω) = 0◦
|H(jω)|
6
H(jω)
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
30
High-pass RC filters
C
A series RC circuit as shown acts as a high-pass filter.
The open-loop voltage transfer function of this filter is:
+
+
Vi
R
−
Vo
R
1
H(jω) =
=
=
Vi
R + 1/(jωC)
1 − j(1/ωRC)
Vo
−
Therefore, this is a first-order high-pass filter with K = 1 and ωC = 1/RC. Input and output
impdenaces of this filter can be found similar to the procedure used for low-pass filters:
Zi = R +
1
jωC
and
Zi |min = R
Output Impedance: Zo = R k
1
jωC
and
Zo |max = R
Input Impedance:
High-pass RL filters
R
A series RL circuit as shown also acts as a high-pass filter.
Again, we find the open-loop tranfunction to be:
ωc =
R
L
Input Impedance:
H(jω) =
Vi
−
1
1 − jωc /ω
Zi = R + jωL
and
Zi|min = R
Output Impedance: Zo = R k jωL
and
Zo |max = R
+
+
L
Vo
−
Exercise: Compute the voltage transfer function and input and output impdenaces of
terminated RC and RL filters.
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
31
2.3
Band-pass filters
A band pass filter allows signals with a range of frequencies (pass band) to pass through and
attenuates signals with frequencies outside this range.
ωl :
Lower cut-off frequency;
ωu :
ω0 ≡
| H(j ω) |
Upper cut-off frequency;
√
ωl ωu :
B ≡ ωu − ωl :
ω0
:
Q≡
B
Pass
Band
Center frequency;
Band width;
ωl
Quality factor.
ωu
ω
As with practical low- and high-pass filters, upper and lower cut-off frequencies of practical
band pass filter are defined as the frequencies at which the magnitude of the voltage transfer
√
function is reduced by 1/ 2 (or -3 dB) from its maximum value.
Second-order band-pass filters:
Second-order band pass filters include two storage elements (two capacitors, two inductors,
or one of each). The transfer function for a second-order band-pass filter can be written as
K
ω0
ω
−
1 + jQ
ω0
ω
|K|
|H(jω)| = s
ω
ω0 2
2
1+Q
−
ω0
ω
H(jω) =
6
H(jω) = −
|K|
ω
ω0
tan−1 Q
−
K
ω0
ω
The maximum value of |H(jω)| = |K| is called the filter gain. The lower and upper cut-off
√
frequencies can be calculated by noting that |H(jω)|max = K, setting |H(jωc )| = K/ 2 and
solving for ωc . This procedure will give two roots: ωl and ωu .
1
K
K
|H(jωc)| = √ |H(jω)|max = √ = s
2
2
ωc ω0 2
1 + Q2
−
ω0 ωc
ωc ω0 2
Q
−
=1
ω0 ωc
ωc ω0
=0
ωc2 − ω02 ±
Q
2
→
ωc ω0
−
Q
ω0 ωc
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
= ±1
32
The above equation is really two quadratic equations (one with + sign in front of fraction
and one with a − sign). Solving these equation we will get 4 roots (two roots per equation).
Two of these four roots will be negative which are not physical as ωc > 0. The other two
roots are the lower and upper cut-off frequencies (ωl and ωu , respectively):
ωl = ω0
s
ω0
1
−
1+
2
4Q
2Q
s
ωu = ω0 1 +
ω0
1
+
2
4Q
2Q
Bode plots of a second-order filter is shown below. Note that as Q increases, the bandwidth
of the filter become smaller and the |H(jω)| becomes more picked around ω0 .
|H(jω)|db
6
H(jω)
Asymptotic behavior:
At low frequencies, ω/ω0 ≪ 1,
|H(jω)| ∝ ω (a +20dB/decade line), and 6 H(jω) → 90◦
At high frequencies, ω/ω0 ≫ 1, |H(jω)| ∝ 1/ω (a -20dB/decade line), and 6 H(jω) → −90◦
At ω = ω0 , H(jω) = K (purely real) |H(jω)| = K (maximum filter gain), and 6 H(jω) = 0◦ .
There are two ways to solve second-order filter circuits. 1) One can try to write H(jω) in
the general form of a second-order filters and find Q and ω0 . Then, use the formulas above
to find the lower and upper cut-off frequencies. 2) Alternatively, one can directly find the
√
upper and lower cut-off frequencies and use ω0 ≡ ωl ωu to find the center frequency and
B ≡ ωu − ωl to find the bandwidth, and Q ≡= ω0 /B to find the quality factor. The two
examples below show the two methods. Note that one can always find ω0 and k rapidaly as
H(jω0 ) is purely real and |H(jω0 )| = k
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
33
Series RLC Band-pass filters
L
Using voltage divider formula, we have
H(jω) =
H(jω) =
Vo
R
=
Vi
R + jωL + 1/(jωC)
C
+
+
Vi
−
R
Vo
−
R
1
R + j ωL −
ωC
There are two approaches to find filter parameters, K, ω0, ωu , and ωl .
Method 1: We transform the transfer function in a form similar to general form of the
transfer function for second order bandpass filters:
H(jω) =
K
ω
ω0
1 + jQ
−
ω0
ω
Note that the denominator of the general form is in the form 1 + j . . . Therefore, we divide
top and bottom of transfer function of series RLC bandpass filters by R:
H(jω) =
1
ωL
1
1+j
−
R
ωRC
Comparing the above with the general form of the transfer function, we find K = 1. To find
Q and ω0 , we note that the imaginary part of the denominator has two terms, one positive
and one negative (or one that scales as ω and the other that scales as 1/ω) similar to the
general form of transfer function of 2nd-order band-pass filters (which includes Qω/ω0 and
−Qω0 /ω). Equating these similar terms we get:
ωL
Q
L
Qω
=
→
=
ω0
R
ω0
R
Qω0
1
1
=
→ Qω0 =
ω
ωRC
RC
We can solve these two equations to find:
1
ω0 = √
LC
ω0
Q=
=
R/L
s
L
R2 C
The lower and upper cut-off frequencies can now be found from the formulas on page 33.
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
34
Method 2: In this method, we directly calculate the filter parameters similar to the procedure followed for general form of transfer function in page 32. Some simplifications can be
made by noting: 1) At ω = ω0 , H(jω) is purely real and 2) K = H(jω = jω0 ).
Starting with the transfer function for the series RLC filter:
H(jω) =
R
1
R + j ωL −
ωC
We note that the transfer function is real if coefficient of j in the denominator is exactly
zero (note that this happens for ω = ω0 ), i.e.,
ω0 L −
1
=0
ω0 C
ω0 = √
−→
1
LC
Also
K = H(jω = jω0 ) =
R
=1
R
The cut-off frequencies can then be found by setting:
K
1
|H(jωc)| = √ = √
2
2
1
ωc L
−
1+
R
ωc RC
2
=2
which can be solved to find ωu and ωl .
Input and Output Impedance of band-pass RLC filters
1
1
Zi = jωL +
+R
+ R = j ωL −
jωC
ωC
Zi |min = R
occurs at ω = ω0
1
Zo = jωL +
jωC
!
kR
→
Zo |max = R
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
35
Wide-Band Band-Pass Filters
Band-pass filters can be constructed by putting a high-pass and a low-pass filter back to
back as shown below. The high-pass filter sets the lower cut-off frequency and the low-pass
filter sets the upper cut-off frequency of such a band-pass filter.
| H (j ω) |
| H (j ω) | X | H (j ω) |
| H (j ω) |
2
1
1
ω c2 ω
ω
ω c1
An example of such a band-pass filter is
two RC low-pass and high-pass filters put
back to back. These filters are widely
used (when appropriate, see below) instead
of an RLC filter as inductors are usually
bulky and take too much space on a circuit board.
2
ω l = ωc2
R2
C1
+
+
+
C2
Vi
ω
ω u = ωc1
R1
V1
−
−
Low−Pass
Vo
−
High−Pass
In order to have good voltage coupling in the above circuit, the input impedance of the
high-pass filter (actually Zi |min = R1 ) should be much larger than the output impedance of
the low-pass filter (actually Zo |max = R2 ), or we should have R1 ≫ R2 . In that case we can
use un-terminated transfer functions:
H(jω) = H1 (jω) × H2 (jω) =
ωc1 = 1/(R1 C1 )
H(jω) =
1
1
×
1 + jω/ωc2 1 − jωc1 /ω
ωc2 = 1/(R2 C2 )
1
1
=
(1 + jω/ωc2)(1 − jωc1/ω)
(1 + ωc1/ωc2 ) + j(ω/ωc2 − ωc1 /ω)
Again, we can find the filter parameters by either of two methods above. Transforming the
transfer function to a form similar to the general form (left for students) gives:
K=
1
1 + ωc1 /ωc2
Q=
q
ωc1 /ωc2
1 + ωc1 /ωc2
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
ω0 =
√
ωc1 ωc2
36
One should note that the Bode plots of previous page are “asymptotic” plots. The real
H(jω) differs from these asymptotic plots, for example, |H(jω)| is 3 dB lower at the cutoff frequency. A comparison of “asymptotic” Bode plots for first-order high-pass filters are
given in page 30. It can be seen that |H1 (jω)| achieves its maximum value (1 in this case)
only when ω/ωc1 < 1/3. Similarly for the low pass filter, |H2 (jω)| achieves its maximum
value (1 in this case) only when ω/ωc2 > 3. In the band-pass filter above, if ωc2 ≫ ωc1 (i.e.,
ωc2 ≥ 10ωc1), the center frequency of the filter will be at least a factor of three away from
both cut-off frequencies and |H(jω)| = |H1 | × |H2 | achieves its maximum value of 1. If ωc2
is not ≫ ωc1 (i.e., ωc2 < 10ωc1), H1 and H2 will not reach their maximum of 1 and the filter
|H(jω)|max = |H1 | × |H2 | will be less than one. This can be seen by examining the equation
of K above which is always less than 1 and approaches 1 when ωc2 ≫ ωc1.
More importantly, we can never make a “narrow” band filter by putting two first-order highpass and low-pass filters back to back. When ωc2 is not ≫ ωc1 , |H(jω)|max becomes smaller
than 1. Since the cut-off frequencies are located 3 dB below the maximum values, the cut-off
frequencies will not be ωc1 and ωc2 (those frequencies are 3 dB lower than |H(jω)|max = 1).
The lower cut-off frequency moves to a value lower than ωc1 and the upper cut-off frequency
moves to a value higher than ωc2 . This can be seen by examining the quality factor of this
filter at the limit of ωc2 = ωc1
Q=
q
ωc1 /ωc2
1 + ωc1 /ωc2
=
1
= 0.5
1+1
while our asymptotic description of previous page indicated that when ωc2 = ωc1 , band-width
becomes vanishingly small and Q should become very large.
Because these filters work only when ωc2 ≫ ωc1, they are called “wide-band” filters. For
these wide-band filters (ωc1 ≪ ωc2 ), we find from above:
K=1
Q=
H(jω) =
q
ωc1 /ωc2
ω0 =
√
ωc2 ωc1
1
1 + j(ω/ωc2 − ωc1 /ω)
We then substitute for Q and ω0 in the expressions for cut-off frequencies (page 33) to get:
ωu = ω0
s
s
1
ω0
ω0 q
2
1+
1 + 4Q + 1
+
=
4Q2 2Q
2Q
ωl = ω0 1 +
1
ω0
ω0 q
2 −1
−
=
1
+
4Q
4Q2 2Q
2Q
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
37
Ignoring 4Q2 term compared to 1 (because Q is small),we get:
√
ωc2ωc1
ω0
q
= ωc2
ωu =
=
Q
ωc1 /ωc2
For ωl , if we ignore 4Q2 term compared to 1, we will find ωl = 0. We should, therefore,
expand the square root by Taylor series expansion to get the first order term:
ωl ≈
ω0
1
ω0
1 + 4Q2 − 1 =
× 2Q2 = ω0 Q = ωc1
2Q
2
2Q
What are Wide-Band and Narrow-Band Filters? Typically, a wide-band filter is
defined as a filter with ωc2 ≫ ωc1 (or ωc2 ≥ 10ωc1). In this case, Q ≤ 0.35 (prove this!). A
narrow-band filter is usually defined as a filter with B ≪ ω0 (or B ≤ 0.1ω0 ). In this case,
Q ≥ 10.
Example: Design a band-pass filter with cut-off frequencies of 160 Hz and 8 kHz. The load
for this circuit is 1 MΩ.
As this is wide-band, band-pass filter (ωu /ωl = fu /fl = 50 ≫ 1), we use two low- and
high-pass RC filter stages similar to circuit above. The prototype of the circuit is shown
below:
R2
The high-pass filter sets the lower cut-off
frequency, and the 1 MΩ load sets the output impedance of this stage. Thus:
Zo |max = R1 ≪ 1 MΩ
ωc (High-pass) = ωl =
→ R1 ≤ 100 kΩ
1
= 2π × 160
R1 C1
C1
+
Vi
+
+
C2
−
V1
R1
−
Low−Pass
Vo
−
High−Pass
→ R1 C1 = 1 × 10−3 kΩ
One should choose R1 as close as possible to 100 kΩ (to make the C1 small) and R1 C1 =
1×10−3 using commercial values of resistors and capacitors. A good set here are R1 = 100 kΩ
and C1 = 10 nF.
The low-pass filter sets the upper cut-off frequency. The load for this component is the input
resistance of the high-pass filter, Zi |min = R1 = 100 kΩ. Thus:
Zo |max = R2 ≪ 100kΩ
ωc (Low-pass) = ωu =
→ R2 ≤ 10 kΩ
1
= 2π × 8 × 103
R2 C2
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
→ R2 C2 = 2 × 10−5
38
As before, one should choose R2 as close as possible to 10 kΩ and R2 C2 = 2 × 10−5 using
commercial values of resistors and capacitors. A good set here are R2 = 10 kΩ and C2 = 2 nF.
In principle, we can switch the position of low-pass and high-pass filter stages in a wideband, band-pass filter. However, the low-pass filter is usually placed before the high-pass
filter because the value of capacitors in such an arrangement will be smaller. (Try redesigning
the above circuit with low-pass and high-pass filter stages switched to see that one capacitor
become much smaller and one much larger.)
Exercise: Design an RLC filter with the specifications in the previous example. (Hint: Do
not set R = 100 kΩ as this would make the value of the inductor very large.)
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
39
2.4
Exercise Problems
In circuit design, use 5% commercial resistor and capacitor values (1, 1.1, 1.2, 1.3, 1.5, 1.6,
1.8, 2, 2.2, 2.4, 2.7, 3., 3.3, 3.6, 3.9, 4.3, 4.7, 5.1, 5.6, 6.2, 6.8, 7.5, 8.2, 9.1 × 10n where n is
an integer). Youcan also use 10 mH inductors
Problem 1. Design a RLC bandpass filter with a lower cut-off frequency of 1 kHz and a
bandwidth of 3 kHz. What is the center frequency and Q of this filter?
Problem 2. We have an amplifier that amplifies a 1 kHz signal from a detector. The load
for this amplifier can be modeled as a 50 kΩ resistor. The amplifier output has a large
amount of 60 Hz noise. We need to reduce the amplitude of noise by a factor of 10. Design
a first-order passive filter which can be placed between the amplifier and the load and does
the job. Would this filter affect the 1 kHz signal that we are interested in? If so, by how
much?
Problem 3. The tuner for an FM radio requires a band-pass filter with a central frequency
of 100 MHz (frequency of a FM station) and a bandwidth of 2 MHz. a) Design such a filter
(Use a 1 µH inductor). b) What are its cut-off frequencies?
Problem 4. A telephone line carries both voice band (0-4 kHz) and data band (25 kHz to
1 MHz). Design a filter that lets the voice band through and rejects the data band. The
filter must meet the following specifications: a) For the voice band, the change in transfer
function should be at most 1 dB; and b) The transfer function should be as small as possible
at 25 kHz, the low end of the data band.
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
40
2.5
Solution to Exercise Problems
Problem 1. Design a RLC bandpass filter with a lower cut-off frequency of 1 kHz and a
bandwidth of 3 kHz. What is the center frequency and Q of this filter?
L
The circuit prototype is:
For a 2nd order band-pass filter:
B(Hz) = fu − fl
C
+
+
Vi
fu = 1 + 3 = 4 kHz
−
R
Vo
−
B(rad/s) = 2πB(Hz) = 1.88 × 104
ωu = 2πfu = 2.51 × 104
√
ω0 = ωu ωl = 1.26 × 104
B(rad/s) =
ω0
Q
→
Q=
ωl = 2πfl = 6.28 × 103
1.26 × 104
= 0.67
1.88 × 104
For the series RLC circuit:
1
1
→ LC = 2
ω0
LC
1
1
C=
=
= 0.63 µF
Lω02
10 × 10−3 × (1.26 × 104 )2
ω0 = √
Q=
ω0
R/L
→
R
ω0
=
= B(rad/s)
L
Q
R = LB = 10 × 10−3 × 1.88 × 104 = 188 Ω
Therefore, using commercial values, the design values are L = 10 mH, R = 180 Ω, and
C = 0.68 µF.
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
41
Problem 2. We have an amplifier that amplifies a 1 kHz signal from a detector. The load
for this amplifier can be modeled as a 50 kΩ resistor. The amplifier output has a large
amount of 60 Hz noise. We need to reduce the amplitude of noise by a factor of 10. Design
a first-order passive filter which can be placed between the amplifier and the load and does
the job. Would this filter affect the 1 kHz signal that we are interested in? If so, by how
much?
We want to have 1 kHz signals to go through but reduce 60 Hz signals, so we need a high-pass
filter. The prototype of the circuit is shown below. For this circuit:
C
H(jω) =
+
+
1
Vo
=
Vi
1 − jωc /ω
Inverting
Amp.
1
RC
Zi |min = R
ωc =
Vi
R
Vo
RL
−
−
Zo |max = R
As the output impedance of the inverting amplifier circuit is “zero”, we do not need to worry
about the input impedance of our filter. The output impedance of the filter is restricted by
Zo |max = R ≪ 50 kΩ
→
R ≤ 5 kΩ
This filter should reduce the amplitude of 60 Hz (ω60 = 2π × 60 = 120π rad/s) signal by a
factor of 10, i.e.,
|H(jω = jω60 )| =
Vo Vi
1 + (ωc /ω60 )2 = 100
60
1
= 0.1
=q
Hz
1 + (ωc /ω60 )2
→
1
= ωc ≈ 10ω60 = 3751 rad/s
RC
→
RC = 2.67 × 10−4
Reasonable choices are R = 3.9 kΩ (to keep it below 5 kΩ) and C = 68 nF (fc ≈ 600 Hz).
The impact on 1 kHz signal (ω1000 = 2000π rad/s) can be found from:
1
1
=q
= 0.86
|H(jω = jω1000 )| = q
1 + (ωc /ω1000 )2
1 + (3751/6283)2
So the amplitude of 1 kHz signal is reduced by 14% (or by -1.3 dB).
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
42
Problem 3. The tuner for an FM radio requires a band-pass filter with a central frequency
of 100 MHz (frequency of a FM station) and a bandwidth of 2 MHz. a) Design such a filter
(Use a 1 µH inductor). b) What are its cut-off frequencies?
L
Because this is not a wide-band filter, the
simplest filter will be an RLC filter as is
shown. For this filter:
s
+
+
Vi
−
1
= 2π100 × 106
ω0 = √
LC
ω0
Q=
=
B
C
R
Vo
−
L
2π100 × 106
=
= 50
R2 C
2π × 2 × 106
Using a L = 1 µH inductor:
1
= 4π 2 1016
LC
1
= 4π 2 1016 × 10− 6
C
→
→
C = 2.5 × 10−12 F
Choose: C = 2.2 pF
L
= 2, 500
R2 C
→
R2 =
L
10−6
=
= 182 →
2, 500C
2, 500 × 2.2 × 10−12
R = 13.5 Ω
Choose: R = 13 Ω (L = 1 µH and C = 2.2 pF).
To find the cut-off freqnecies, we not:
B = fu − fl = 2 MHz
f0 =
q
fu fL = 100 MHz
Solution of the above two equations in two unknowns will give fl ≈ 99 MHz and fu ≈
101 MHz.
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
43
Problem 4. A telephone line carries both voice band (0-4 kHz) and data band (25 kHz to
1 MHz). Design a filter that lets the voice band through and rejects the data band. The
filter must meet the following specifications: a) For the voice band, the change in transfer
function should be at most 1 dB; and b) The transfer function should be as small as possible
at 25 kHz, the low end of the data band.
We need a low-pass filter as it should allow low-frequency signals (voice band) to go through
while eliminating high-frequency signals (data band). The prototype of an RC low-pass filter
is shown and its transfer function is:
R
1
1
H(jω) =
=
1 + jω/ωc
1 + jf /fc
+
Vi
+
C
Vo
−
The cut-off frequency of the filter is not given and
it should be found from the specifications. First,
we need the change in transfer function to be at
most 1 dB for the frequency range of 0-4 kHz.
The transfer function of filters that satisfy this
constraint is the curve labeled “1” in the figure
and any transfer function located to the right of
this curve (such as transfer function labeled “2”).
−
| H(j ω) |
0
2
−1dB
1
f (kHz)
4
25
Second, the transfer function should be as small as possible at 25 kHz. This requires that
we choose the cut-off frequency as small as possible. Therefore, the transfer function of our
filter should be curve labeled “1” as it has the smallest possible value at 25 kHz:
20 log (|H(jf = 4 kHz)|) = −1 dB
→
|H(jf = 4 kHz)| = 0.891
Using the expression for H(jω), we have:
1
|H(jf = 4 kHz)| = q
= 0.891
1 + (f /fc )2
f /fc = 0.509
fc =
→
1
= 7.85 × 103
2πRC
fc =
f
= 7.85 kHz
0.509
→
RC = 2.0 × 10−5
Choosing C = 1 nF, we have R = 2.0 × 104 . The commercial values then are C = 1 nF and
R = 20 kΩ.
ECE65 Lecture Notes (F. Najmabadi), Winter 2009
44