ECET 303
Lecture 9 “Band-pass and Band-stop Filters”
9.1. The Band-Pass Filters
9.1.1 Series Resonant Band-Pass Filter
9.1.2 Parallel Resonant Band-Pass Filter
9.2 The Band-Stop Filters
9.2.1 Series Resonant Band-Stop Filter
9.2.2 Parallel Resonant Band-Stop Filter
9.1 The Band-Pass Filters
A band-pass filter (BPF) is a circuit that transmits [from input to output] group of frequencies
from the certain spectrum and reject frequencies outside of that group. One of the important
characteristics of BPF is its bandwidth BW, i. e., the range of frequencies that the filter transmits
and which have the output voltage equal or greater than 0.7071 of the input voltage.
BW = ωC2 - ωC1
(9.1)
where ωC2 and ωC1 are upper and lower cutoff frequencies, respectively, in rad/s.
BW is also expressed in Hz:
BW = fC2 - fC1
(9.2)
It is easy to see that BPF can be built using two-stage circuit: the low-pass (LPF) and the highpass (HPF). While building such a filter one should consider loading effect of HPF on LPF. If
the cutoff frequency of the low-pass filter, fC1, is bigger than the cutoff frequency of the highpass filter, fC2, the frequency responses of the filters overlap and create the bell-shape region
(Figure 9-1) with two critical frequencies, fC1 and fC2. Such a filter transmits all frequencies
between fC1 and fC2 and suppresses frequencies beyond that range.
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Figure 9-1
Frequency response of a BPF built as a combination of LPF and HPF
Possible schematic diagram of the band-pass filter built as a combination of LPF-HPF is shown
in Figure 9-2.
Figure 9-2.
BPF That Consists of Two Stages HPF and LPF
In Figure 9-2 the first stage (C1R1) and the second stage (R2C2) are easily recognizable HPF and
LPF, respectively. The transfer function 1 is
H(ω)1 =
The transfer function 2:
H(ω)2 =
𝑽𝑹𝟏
𝑽𝒊𝒏
𝑽𝒐𝒖𝒕
𝑽𝑹𝟏
=
𝑗𝜔𝑅1𝐶1
1+ 𝑗𝜔𝑅1𝐶1
=
1
1+ 𝑗𝜔𝑅2𝐶2
(9.3)
(9.4)
From (9.3) and (9.4) follows TF for the whole circuit:
H(ω) = (H(ω)1)( H(ω)2)
(9.5)
(9.5) can be presented differently. Using time constants τ1 and τ2 we can write:
2
H(ω) =
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛
=
𝑗𝜔𝜏1
(1+ 𝑗𝜔𝜏1)(1+ 𝑗𝜔𝜏2)
(9.6)
τ1 = R1C1 = 2 ms, τ2 = R2C2 = 50 µs, ωC1 = 500 rad/s, and ωC2 = 20 krad/s.
The Bode plot can be built using three components of (9.6).
Figure 9-3
The Bode Plot of the Band-Pass Filter
The plot for the phase angle of the BPF can be built using the same approach, combining three
components: the constant of 90o (due to the term jωτ1) and graphs of the phase angles for HPF
and LPF. While building the Bode plot keep in mind the following:
-
For LPF: at frequency ω ≤ 0.1ωC the phase angle is approximately zero; at frequency
ω ≥ 10ωC the phase angle is - 90o, and at ω = ωC the phase angle is – 45o.
-
For HPF: at frequency ωC ≤ 0.1ωC the phase angle is approximately +90o; at frequency
ω ≥ 10ωC the phase angle is 0o, and at ω = ωC the phase angle is + 45o.
-
BW of the filter: 20 krad/s – 0.5 krad/s = 19.5 krad/s or approximately
3,184.7 Hz – 79.6 Hz = 3,105 Hz ≈ 3.1 kHz.
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9.1.1 Series Resonant Band-Pass Filter
Another practical approach to build BPF is using frequency characteristics of the series resonant
circuits, Figure 9-4.
Figure 9-4
Series Resonant Band-Pass Filter
At the resonant frequency, fr (which is also called the center frequency fo), circuit in Figure 9-4
will have minimum impedance since XL and XC have equal magnitudes but the phase shift of 180o
with respect to each other. At the resonant frequency magnitudes of the current and the output
voltage, VR, across resistor R will be at their maximum. The true power dissipated by the circuit
at resonance will be also at its maximum. Bandwidth, BW, of this filter is defined by the quality
factor Q = XL/RT, where RT is total resistance of the circuit.
BW = fr/Q
(9.7)
Example. For the circuit in Figure 9-4 assume L = 0.5 mH, coil resistance, RW = 14 Ω, and input
signal v(t) = 30 sin ωt. Determine the following:
(a) The values of R and C for the circuit to have a resonant frequency of 1000 kHz and a
bandwidth of 10 kHz;
(b) Output voltage vout(t) at resonance;
(c) The power dissipated by the circuit at resonance.
Solution. From (9.7) Q = fr/BW = 1000 kHz/10 kHz = 100;
XL = 2πfL = 2π(1000 kHz)(0.5 mH) = 3,140 Ω;
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XL = XC; C = 1/(2πfXL) = 1/[2π(1000 kHz)( 3,140 Ω)] = 4.8 pF;
RT = R + RW = XL/Q = 314 Ω;
R = RT - RW = 314 Ω - 14 Ω = 300 Ω;
Amplitude of the current Ip in the circuit at resonance: Ip = (Vp)/(RT) ≈ 95.5 mA;
The output voltage can be found using the voltage divider rule:
Vout = [R/(R + RW)]×Vin = (300 Ω/314 Ω)(30 sin ωt) = 28.66 sin ωt;
The power, Ptrue, dissipated at resonance:
Ptrue = (Voutrms)2/RT = (30 V/1.4142)2/ RT = (21.21 Vrms)2/314 Ω ≈ 1.43 W.
9.1.2 Parallel Resonant Band-Pass Filter
Another realization of band-pass filter that uses a tank circuit is shown in Figure 9-5.
Figure 9-5
Parallel Resonant Band-Pass Filter
At resonance impedance of the tank circuit is much greater than the resistance of the circuit and
output voltage will be at its maximum. Above or beyond the resonance frequency impedance of
the tank circuit decreases creating a resonance response of a band-pass filter.
Example. Determine the center frequency and BW for the band-pass filter in Figure 9-5, if
R = 82 Ω, RW = 15 Ω, L = 50 mH, and C = 0.01 µF.
Solution. Using equation for the resonant frequency of a non-ideal tank circuit:
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𝐶
fr =
2 ( )]
√1−[𝑅𝑊
𝐿
2𝜋√𝐿𝐶
=
√1−(15 Ω)2 (0.01 µ𝐹)/(50 𝑚𝐻) ]
2𝜋√(50 𝑚𝐻)(0.01 µ𝐹)
= 7.12 kHz;
The quality factor Q of the coil at resonance is
Q=
𝑋𝐿
𝑅𝑊
=
2𝜋(7.12 kHz)(50 mH)
The bandwidth, BW, of the filter BW =
15 Ω
𝑓𝑟
𝑄
=
7.12 𝑘𝐻𝑧
149
= 149;
= 47.7 Hz.
9.2 The Band-Stop Filters
A band-stop filter rejects range of frequencies specified by its technical characteristics and
transmits all frequencies below and above that range. Frequency response is shown in
Figure 9-6. A band-stop filter can be also constructed as a combination of a low-pass and a
Figure 9-6
a high-pass filters (Figure 9-8), at that, contrary to the design of a band-pass filter, the cutoff
frequency fCl of a low-pass filter should be lower than the cutoff frequency fCh of a highpass filter.
Figure 9-7
Frequency response of the band-stop filter shown in Figure 9-7
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Figure 9-8
Block Diagram of a Band-stop Filter as a Combination of a Low-pass
and a High-pass Filters
9.2.1 Series Resonant Band-Stop Filter
RLC circuits are also used to build band-stop filters. A series resonant band-stop filter is
shown in Figure 9-9.
Figure 9-9
Series Resonant Band-Stop Filter
At the resonant frequency, the impedance is at its minimum, thus output voltage is at its
minimum. As a frequency goes below fCl or above fCh the impedance of LC circuit
increases, which causes increase in output voltage. Figure 9-10 presents Bode plots for the
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frequency response and the phase angle of the band-stop filter obtained through the
computer simulation.
Figure 9-10
Bode plots for the frequency response and the phase angle
of the band-stop filter shown in Figure 9-9
Example. For the circuit shown in Figure 9-11, determine the following:
(a) Type of the filter;
(b) Q;
(c) Bandwidth, BW, and the cutoff frequencies fcl and fch;
(d) What is the output voltage at resonance?
(e) What will be the output voltage if RL = 2 kΩ is connected?
Figure 9-11
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Solution. (a) This is a series resonant band-stop filter.
(b) Q = (XL)/(R + Ri) = 5 kΩ/410 Ω = 12.195;
(c) BW = fr/Q = (5 kHz)/12.195 = 410 Hz;
cutoff frequencies:
fcl = 5 kHz – (410 Hz)/2 = 4,795 Hz;
fch = 5 kHz + (164 Hz)/2 = 5,205 Hz.
(d) At resonance Vout = [(10 Ω/400 Ω)]×VIN = 0.024 VIN;
(e) At resonance with RL = 2 kΩ
Vout = [(10 Ω)||(2 kΩ)]/[(10 Ω)||(2 kΩ) + (400 Ω)]×VIN = 0.024 VIN.
9.2.2 Parallel Resonant Band-Stop Filter
A parallel resonant band-stop filter is shown in Figure 9-12 and its frequency response
obtained through the computer simulation in Figure 9-13.
Figure 9-12
Parallel Resonant Band-Stop Filter
Example 18-12. For the circuit in Figure 9-12 find the central frequency. Draw the output
response curve showing the min and max voltages.
𝐶
Solution. fr =
2 ( )]
√1−[𝑅𝑊
𝐿
2𝜋√𝐿𝐶
=
√1−(8 Ω)2 (150 pF)/(5 µH) ]
2𝜋√(5 µH)(150 𝑝F)
= 5.81 MHz
At the resonant frequency XL = 2πfrL = 2π(5.81 MHz)(5 µH) = 183 Ω
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Q = XL/RW = 183 Ω/8 Ω = 22.8
Zr = RW(Q2 + 1) = (8 Ω)[(22.8)2 + 1] = 4.16 kΩ (pure resistive)
Using the voltage divider formula to find Vout(min):
Vout(min) =
𝑅𝐿
𝑅𝐿 + 𝑍𝑟
×(VIN) =
560 Ω
4.73 kΩ
×(10 V) = 1.18 V
At zero frequency, the impedance of the tank circuit is RW because XC = ∞ and
XL = 0 Ω. Therefore, the maximum output voltage below resonance is
Vout(max) =
𝑅𝐿
𝑅𝐿 + 𝑅𝑊
For f >> fr approaches XC
×(VIN) =
560 Ω
×(10 V) = 9.86 V
568 Ω
0 Ω, and VOUT
VIN (10 V).
Figure 9-13
Magnitude Bode Plot for Parallel Resonant Band-Stop Filter
(Multisim Simulation for the example above, 2 MHz ≤ 𝑓 ≤ 10 MHz)
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