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Electric Circuit Theory

Nam Ki Min

010-9419-2320 [email protected]

Active Filter Circuits

Chapter 15

Nam Ki Min

010-9419-2320 [email protected]

CIEN346 Electric Circuits Nam Ki Min 010-9419-2320 [email protected]

Chapter 15 Active Filter Circuits 3 Contents and Objectives

Chapter Contents

15.1 First-Order Low-Pass and High-Pass Filters

15.2 Scaling

15.3 Op Amp Bandpass and Bandreject Filters

15.4 Higher Order Op Amp Filters

15.5 Narrowband Bandpass and Bandreject Filters

Chapter Objectives

1. Know the op amp circuits that behave as first-order low-pass and high-pass filters and be able to calculate component values for these circuits to meet specifications of cutoff frequency and passband gain.

2. Be able to design filter circuits starting with a prototype circuits and use scaling to achieve desired frequency response characteristics and component values.

3. Understand how to use cascaded first- and second-order Butterworth filters to implement low-pass, high-pass, bandpass, and bandreject filters of any order.

4. Be able to use the design equations to calculate component values for prototype narrowband, bandpass, and bandreject filters to meet desired filter specifications.


Chapter 15 Active Filter Circuits 4

Three Major Limits to The Passive Filters

• First, they cannot generate gain greater than 1; passive elements cannot add energy to the network.

• Second, they may require bulky and expensive inductors.

• Third, they perform poorly at frequencies below the audio frequency range (300 Hz < f <3000 Hz). Nevertheless, passive filters are useful at high frequencies.

Advantages and Disadvantages of Active Filters over Passive RLC Filters

Active filters consist of combinations of resistors, capacitors, and op amps. They offer some advantages over passive RLC filters.

• First, they are often smaller and less expensive, because they do not require inductors. This makes feasible the integrated circuit realizations of filters.

• Second, they can provide amplifier gain in addition to providing the same frequency response as RLC filters.

• Third, active filters can be combined with buffer amplifiers (voltage followers) to isolate each stage of the filter from source and load impedance effects. This isolation allows designing the stages independently and then cascading them to realize the desired transfer function.

However, active filters are less reliable and less stable. The practical limit of most active filters is about 100 kHz—most active filters operate well below that frequency.,

Some Preliminaries



Types of Active Filters

Filters are often classified according to their order (or number of poles) or their specific design type.

- First-order lowpass filter

- First-order highpass filter

- Bandpass filter

- Bandreject (or Notch) filter

Some Preliminaries



Bode Plots

Bode Plots

Some Properties of Logarithms

• A more systematic way of obtaining the frequency response is to use Bode plots.

• Before we begin to construct Bode plots, we should take care of two important issues: the use of logarithms and decibels in expressing gain.

log 𝑃1𝑃2 = log 𝑃1 + log 𝑃2

log 𝑃1/𝑃2 = log 𝑃1 − log 𝑃2

log 𝑃𝑛 = 𝑛 log 𝑃

log 1 = 0

• In communications systems, gain is measured in bels. Historically, the bel is used to measure the ratio of two levels of power or power gain G; that is,

𝐺 = 𝑁𝑢𝑚𝑣𝑒𝑟 𝑜𝑓 𝑏𝑒𝑙𝑠 = log10

𝑃2

𝑃1

• The decibel (dB) provides us with a unit of less magnitude. It is 1/10th of a bel and is given by

𝐺𝑑𝐵 = 10log10

𝑃2

𝑃1 (1)

Decibel Scale

Some Preliminaries



Decibel Scale

• When P1 = P2, there is no change in power and the gain is 0 dB.


𝑃2

𝑃1

If P2 = 2P1, the gain is


2𝑃1

𝑃1= 10log102 = 3 dB

If P2 = 0.5P1, the gain is


0.5𝑃1

𝑃1= 10log100.5 = −3 dB

(1)

(3)

(2)

Equations (1) and (2) show another reason why logarithms are greatly used: The logarithm of the reciprocal of a quantity is simply negative the logarithm of that quantity.

Bode Plots

Some Preliminaries




𝑃2

𝑃1 (1)

• Alternatively, the gain G can be expressed in terms of voltage or current ratio.

𝐺𝑑𝐵 = 10 log10

𝑃2

𝑃1 = 10 log10

𝑉22/𝑅2

2

𝑉12/𝑅1

2

= 10 log10

𝑉2

𝑉1

2

+ 10 log10

𝑅1

𝑅2


𝑉2

𝑉1 − 10 log10

𝑅2

𝑅1

(4)

(5)

• For the case when R2 = R1, a condition that is often assumed when comparing voltage levels, Eq.(5) becomes


𝑉2

𝑉1 (6)

Bode Plots

Decibel Scale

Some Preliminaries



• If 𝑃1 = 𝐼12𝑅1 and 𝑃2 = 𝐼2

2𝑅2, for R2 = R1, we obtain


𝐼2

𝐼1 (7)

• The dB value is a logarithmic measurement of the ratio of one variable to another of the same type. Therefore, it applies in expressing the transfer function H in Eqs.(8) and (9), which are dimensionless quantities,

𝐻 𝜔 = 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑔𝑎𝑖𝑛 =𝑉𝑜(𝜔)

𝑉𝑖(𝜔)

𝐻 𝜔 = 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑔𝑎𝑖𝑛 =𝐼𝑜(𝜔)

𝐼𝑖(𝜔)

(8)

(9)

but not in expressing H in Eqs.(10) and (11).

𝐻 𝜔 = 𝑇𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝐼𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 =𝑉𝑜(𝜔)

𝐼𝑖(𝜔)

𝐻 𝜔 = 𝑇𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑎𝑑𝑚𝑖𝑡𝑡𝑎𝑛𝑐𝑒 =𝐼𝑜(𝜔)

𝑉𝑖(𝜔)

(10)

(11)

Bode Plots

Decibel Scale

Some Preliminaries



Bode Plots

Bode Plots

• The frequency range required in frequency response is often so wide that it is inconvenient to use a linear scale for the frequency axis(Fig.a) .

• Also, there is a more systematic way of locating the important features of the magnitude and phase plots of the transfer function.

• For these reasons, it has become standard practice to use a logarithmic scale for the frequency axis and a linear scale in each of the separate plots of magnitude and phase.

• Such semilogarithmic plots of the transfer function—known as Bode plots—have become the industry standard(Fig.b).

Some Preliminaries

Fig.a: A linear scale for the frequency axis



Bode Plots

Bode Plots

Bode plots differ from the frequency response plots in Chapter 14 in two important ways.

• Instead of using a linear axis for the frequency values, a Bode plot uses a logarithmic axis. This permits us to plot a wider range of frequencies of interest. Normally we plot three or four decades of frequencies, say from 102 kHz to 106 kHz , or 1 kHz to 1 MHz, choosing the frequency range where the transfer function characteristics are changing. If we plot both the magnitude and phase angle plots, they again share the frequency axis.

Some Preliminaries



Bode Plots

Bode Plots

• The Bode magnitude is plotted in decibels (dB) versus the log of the frequency.

Briefly, if the magnitude of the transfer function is 𝐻(𝑗𝜔) , its value in dB is given by

𝐴𝑑𝐵 = 20 log10 𝐻(𝑗𝜔)

- When 𝐴𝑑𝐵 = 0,

𝐻(𝑗𝜔) = 1 20 log10 𝐻(𝑗𝜔) = 0 →

- When 𝐴𝑑𝐵 < 0,

0 < 𝐻(𝑗𝜔) < 1 20 log10 𝐻(𝑗𝜔) < 0 →

- When 𝐴𝑑𝐵 > 0,

1 < 𝐻(𝑗𝜔) 20 log10 𝐻(𝑗𝜔) > 0 →

- When 𝐻(𝑗𝜔) = 1/ 2 ,

20 log10

1

2= −3 dB

Some Preliminaries



Bode Plots

Bode Plots

• At cutoff frequency 𝜔𝑐 = 2𝜋𝑓𝑐,

20 log10

1

2= 20 log10 2

−1 = 20 log102−

12

= −20 ×1

2 log102

= −20 ×1

2× 0.3

= −3 dB

Some Preliminaries

𝐻(𝑗𝜔) =1

2

The lowpass filter reduces the overall voltage gain of an amplifier by 3 dB when the frequency is reduced to the cutoff frequency 𝜔𝑐 = 2𝜋𝑓𝑐.

Bode plot

Actual response of a RC filter



Bode Plots

Bode Plots

Some Preliminaries

• As the frequency continues to increase beyond 𝜔𝑐 = 2𝜋𝑓𝑐, the overall voltage gain also continues to decrease.

The rate of decrease in voltage gain with frequency is called roll-off.

For each ten times increase in frequency beyond 𝑓𝑐 , there is a 20 dB reduction in voltage gain.

𝑉𝑜(𝑠) =𝑍𝑐

𝑅 + 𝑍𝑐𝑉𝑖(𝑠) 𝐻 𝑗𝜔 =

1

1 + 𝑗𝜔𝑅𝐶

20 log10 0.1 = −20 dB

A ten-times change in frequency is called a decade.

The attenuation is reduced by 20 dB for each decade that the frequency increases beyond the cutoff frequency. This causes the overall voltage gain to drop 20 dB per decade.



Bode Plots

Bode Plots

Some Preliminaries

• The -20dB/decade roll-off rate for the gain of a basic RC filter means that at a frequency of 10𝑓𝑐, the output will be -20dB(10%) of the input.

• This roll-off rate is not a particularly good filter characteristic because too much of the unwanted frequencies (beyond the passband) are allowed through the filter.

• In order to produce a filter that has a steeper transition region (and hence form a more effective filter), it is necessary to add additional circuitry to the basic filter.

• Responses that are steeper than -20dB/decade in the transition region cannot be obtained by simply cascading identical RC stages (due to loading effects).

• However, by combining an op-amp with frequency -selective feedback circuits, filters can be designed with roll-off rates of -40, -60, or more dB/decade.

• Filters that include one or more op-amps in the design are called active filters.

• These filters can optimize the roll-off rate or other attribute (such as phase response) with a particular filter design. In general, the more poles the filter uses, the steeper its transition region will be. The exact response depends on the type of filter and the number of poles.



Summary

• Bode plots describe the behavior of a filter by relating the magnitude of the filter's response (gain) to its frequency.

• The key feature of this graph is that both axes have logarithmic scales.

• An example of this type of plot is shown in Figure

Pass band

stop band

Pass band

stop band

Gain

Pass-band ripple

Frequency (Hz)

Attenuation rate

Cutoff frequency (𝜔𝑐 = 2𝜋𝜔𝑓𝑐 )

A filter's Bode plot can show key features of a filter,

such as cutoff frequency, attenuation rate, and pass-

band ripple.

Some Preliminaries

Bode Plots


Chapter 15 Active Filter Circuits 17 First-Order Low-Pass and High-Pass Filters

First-Order Low-Pass Filter

Active Low-Pass Filter Circuits




Qualitative Analysis

• At very low frequencies (𝜔𝑜 → 0), the capacitor acts like an open circuit, and the op amp circuit acts like an amplifier with a gain of −𝑅2/𝑅1 .

• At very high frequencies(𝜔𝑜 → ∞), the capacitor acts like a short circuit, thereby connecting the output of the op amp circuit to ground.

• The op amp circuit in Fig. 15.1 thus functions as a low-pass filter with a passband gain of − 𝑅2/𝑅1 .



Quantitative Analysis

• Output voltage

• Transfer function:

𝐻 𝑠 =𝑉𝑜

𝑉𝑖= −

𝑍𝑓

𝑍𝑖= −

𝑅21 + 𝑠𝑅2𝐶

𝑅1= −

𝑅2

𝑅1

1

1 + 𝑠𝑅2𝐶

𝑉𝑜 = −𝑍𝑓

𝑍𝑖𝑉𝑖 𝑍𝑓 =

𝑅21

𝑠𝐶

𝑅2 +1

𝑠𝐶

=𝑅2

1 + 𝑠𝑅2𝐶

= −𝐾𝜔𝑐

𝑠 + 𝜔𝑐

𝐾 =𝑅2

𝑅1

𝜔𝑐 =1

𝑅2𝐶

= −𝑅2

𝑅1

1𝑅2𝐶

𝑠 +1

𝑅2𝐶

𝐻 𝑗𝜔 = −𝑅2

𝑅1

1

1 + 𝑗𝜔𝑅2𝐶

1 + 𝑗𝜔𝑅2𝐶 = 1 + 𝑗

𝐴𝑡 𝜔 = 𝜔𝑐

1 + 𝑗 = 2

(15.1)

(15.3)

(15.2)




First-Order High-Pass Filter

Active High-Pass Filter Circuits

A first-order(single pole) high-pass filter




Qualitative Analysis

• At very low frequencies (𝜔𝑜 → 0), the capacitor acts like an open circuit, thereby connecting the input of the op amp circuit to ground(Fig.b).

• At very high frequencies(𝜔𝑜 → ∞), the capacitor acts like a short circuit, and the op amp circuit acts like an amplifier with a gain of −𝑅2/𝑅1.(Fig.c)

• The op amp circuit in Fig. 15.4 thus functions as a high-pass filter with a passband gain of − 𝑅2/𝑅1 .

𝑣𝑜 → 0



Quantitative Analysis

• Output voltage

• Transfer function:

𝐻 𝑠 =𝑉𝑜

𝑉𝑖= −

𝑍𝑓

𝑍𝑖= −

𝑅2

𝑅1 +1

𝑠𝐶

=

𝑉𝑜 = −𝑍𝑓

𝑍𝑖𝑉𝑖 𝑍𝑖 = 𝑅1 +

1

𝑠𝐶

𝐾 =𝑅2

𝑅1

𝜔𝑐 =1

𝑅1𝐶

−𝑅2

𝑅1

𝑠

𝑠 +1

𝑅1𝐶

𝐻 𝑗𝜔 = −𝑅2

𝑅1

1

1 +1

𝑗𝜔𝑅1𝐶

1 + 1

𝑗𝜔𝑅1𝐶= 1 − 𝑗

𝐴𝑡 𝜔 = 𝜔𝑐

1 − 𝑗 = 2

(15.4)

(15.6)

(15.5)


= −𝐾𝑠

𝑠 + 𝜔𝑐


Chapter 15 Active Filter Circuits 23 Scaling

Scaling

• In the design and analysis of both passive and active filter circuits, working with element values such as 1Ω, 1 H, and 1 F is convenient.

• After making computations using convenient values of R, L, and C, the designer can transform the convenient values into realistic values using the process known as scaling.

• There are two types of scaling: magnitude and frequency.

𝑅′ = 𝑘𝑚𝑅, 𝐿′ = 𝑘𝑚𝑅, 𝐶′ =𝐶

𝑘𝑚

𝑘𝑚 = 𝑠𝑐𝑎𝑙𝑒 𝑓𝑎𝑐𝑡𝑜𝑟: 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑡ℎ𝑎𝑡 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑖𝑡ℎ𝑒𝑟 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑜𝑟 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 1

- Unprimed variables represent the initial values of the parameters. - Primed variables represent the scaled values of the variables

• Magnitude scaling

We scale a circuit in magnitude by multiplying the impedance at a given frequency by the scale factor 𝑘𝑚.

(15.7)



Scaling

𝑘𝑓 = 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑠𝑐𝑎𝑙𝑒 𝑓𝑎𝑐𝑡𝑜𝑟: 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑡ℎ𝑎𝑡 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑖𝑡ℎ𝑒𝑟 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑜𝑟 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 1

• Frequency scaling

We change the circuit parameters so that at the new frequency, the impedance of each element is the same as it was at the original frequency.

Because resistance values are assumed to be independent of frequency, resistors are unaffected by frequency scaling.

Both inductors and capacitors are multiplied by the frequency scaling factor,𝑘𝑓.

𝑅′ = 𝑅, 𝐿′ = 𝐿/𝑘𝑓, 𝑎𝑛𝑑 𝐶′ = 𝐶/𝑘𝑓

• Circuit scaling

A circuit can be scaled simultaneously in both magnitude and frequency.

The scaled values (primed) in terms of the original values (unprimed) are

𝐿′ =𝑘𝑚

𝑘𝑓𝐿

𝑅′ = 𝑘𝑚𝑅

𝐶′=1

𝑘𝑚𝑘𝑓𝐶

(15.8)

(15.9)



The Use of Scaling in the Design of Op Amp Filters

To use the concept of scaling in the design of op amp filters,

(a) First select the cutoff frequency, 𝜔𝑐 , to be 1 rad/s (if you are designing low- or highpass filters), or select the center frequency, 𝜔𝑜 , to be 1 rad/s (if you are designing bandpass or bandreject filters).

(b) Then select a 1 F capacitor and calculate the values of the resistors needed to give the desired passband gain and the 1 rad/s cutoff or center frequency.

(c) Finally, use scaling to compute more realistic component values that give the desired cutoff or center frequency.


Chapter 15 Active Filter Circuits 26 Op Amp Bandpass and Bandreject Filters

Op Amp Bandpass Filters

The Easiest Way of Creating a Bandpass Filter; Initial Approach

• While there is a wide variety of such op amp circuits, our initial approach is motivated by the Bode plot construction shown in Figure.

• The bandpass filter consists of three separate components:

A unity-gain low-pass filter whose cutoff frequency is 𝜔𝑐2, the larger of the two cutoff frequencies.

A unity-gain high-pass filter whose cutoff frequency is 𝜔𝑐1, the smaller of the two cutoff frequencies.

A gain component to provide the desired level of gain in the passband.

• These three components are cascaded in series.

𝜔𝑐2 𝜔𝑐1




The Easiest Way of Creating a Bandpass Filter; Initial Approach

• These three components are cascaded in series.

𝜔𝑐2 𝜔𝑐1 𝜔𝑐1 𝜔𝑐2

Op Amp Bandpass and Bandreject Filters

𝜔𝑐2 𝜔𝑐1 𝐾




Analysis of the Cascaded Bandpass Filter

• The transfer function of the cascaded bandpass filter is the product of the transfer functions of the three cascaded components:

𝐻 𝑠 =𝑉𝑜

𝑉𝑖 = −

𝜔𝑐2

𝑠 + 𝜔𝑐2−

𝑠

𝑠 + 𝜔𝑐1−

𝑅𝑓

𝑅𝑖 =

−𝐾𝜔𝑐2𝑠

𝑠 + 𝜔𝑐1 𝑠 + 𝜔𝑐2

=−𝐾𝜔𝑐2𝑠

𝑠2 + 𝜔𝑐1 + 𝜔𝑐2 𝑠 + 𝜔𝑐1𝜔𝑐2 (15.10)






• Eq.(15.10) is not in the standard form for the transfer function of a bandpass filter discussed in Chapter 14.

𝐻 𝑠 = −𝐾𝜔𝑐2𝑠

𝑠2 + 𝜔𝑐1 + 𝜔𝑐2 𝑠 + 𝜔𝑐1𝜔𝑐2 (15.10) 𝐻BP =

𝛽𝑠

𝑠2 + 𝑠𝛽 + 𝜔𝑜2 ↔

• In order to convert Eq.(15.10) into the form of the standard transfer function for a bandpass filter, we require that

𝜔𝑐2 ≫ 𝜔𝑐1 (15.11)

• When Eq.(15.11) holds,

𝜔𝑐1 + 𝜔𝑐2 ≈ 𝜔𝑐2

and the transfer function for the cascaded bandpass filter in Eq.(15.10) becomes

𝐻 𝑠 =−𝐾𝜔𝑐2𝑠

𝑠2 + 𝜔𝑐2𝑠 + 𝜔𝑐1𝜔𝑐2 (1)






Once we confirm that Eq.(15.11) (𝜔𝑐2 ≫ 𝜔𝑐1) holds for the cutoff frequencies specified for

the desired bandpass filter, we can design each stage of the cascaded circuit independently and meet the filter specifications.

𝜔𝑐2 ≫ 𝜔𝑐1

- Low-pass filter: 𝜔𝑐2 =1

𝑅𝐿𝐶𝐿

𝜔𝑐1 =1

𝑅𝐻𝐶𝐻

• Now we compute the values of 𝑅𝐿 and 𝐶𝐿 in the low-pass filter and the values 𝑅𝐻 and 𝐶𝐻 in the high-pass filter to provide the desired cutoff frequencies.

- High-pass filter:

(15.12)

(15.13)

𝜔𝑐2 𝜔𝑐1






To do this, we consider the magnitude of the bandpass filter’s transfer function(Eq.1), evaluated at the center frequency, 𝜔𝑜:

(15.14)

• We compute the values of 𝑅𝑖 and 𝑅𝑓 in the inverting amplifier to provide the desired passband gain.

𝐻 𝑗𝜔0 =−𝐾𝜔𝑐2 𝑗𝜔0

𝑗𝜔02 + 𝜔𝑐2 𝑗𝜔0 + 𝜔𝑐1𝜔𝑐2

=𝐾𝜔𝑐2

𝜔𝑐2

= 𝐾

• Therefore,

𝐾 =𝑅𝑓

𝑅𝑖: 𝑡ℎ𝑒 𝑔𝑎𝑖𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑣𝑒𝑟𝑡𝑖𝑛𝑔 𝑎𝑚𝑝

𝐻 𝑗𝜔0 =𝑅𝑓

𝑅𝑖 (15.15)




Op Amp Bandreject Filters

A component approach to the design of op amp bandreject

• Like the bandpass filter, the bandreject filter consists of three separate components. There are important differences, however:

A unity-gain low-pass filter has a cutoff frequency of 𝜔𝑐1, which is the smaller of the two cutoff frequencies.

A unity-gain high-pass filter has a cutoff frequency of 𝜔𝑐2, which is the larger of the two cutoff frequencies.

The gain component provides the desired level of gain in the passbands.

• The most important difference is that these three components cannot be cascaded in series, because they do not combine additively on the Bode plot.

• Instead, we use a parallel connection and a summing amplifier, as a circuit in Fig. 15.13(a).






𝐻 𝑠 = −𝑅𝑓

𝑅𝑖

−𝜔𝑐1

𝑠 + 𝜔𝑐1+

−𝑠

𝑠 + 𝜔𝑐2

=𝑅𝑓

𝑅𝑖

𝜔𝑐1 𝑠 + 𝜔𝑐2 + 𝑠 𝑠 + 𝜔𝑐1


=𝑅𝑓

𝑅𝑖

𝑠2 + 2𝜔𝑐1𝑠 + 𝜔𝑐1𝜔𝑐2

𝑠 + 𝜔𝑐1 𝑠 + 𝜔𝑐2 (15.16)


• Again, it is assumed that the two cutoff frequencies are widely separated, so that the resulting design is a broadband bandreject filter, and 𝜔𝑐2 ≫ 𝜔𝑐1.

• Then each component of the parallel design can be created independently, and the cutoff frequency specifications will be satisfied.

• The transfer function of the resulting circuit is the sum of the low-pass and high-pass filter transfer functions.





• The cutoff frequencies are given by the equations

(15.17) 𝜔𝑐1 =1

𝑅𝐿𝐶𝐿

𝜔𝑐2 =1

𝑅𝐻𝐶𝐻 (15.18)

• In the two passbands (as 𝑠 → 0 and 𝑠 → ∞), the gain of the transfer function is

𝐾 =𝑅𝑓

𝑅𝑖 (15.19)






• The magnitude of the transfer function in Eq.(15.16) at the center frequency, 𝜔𝑜 = 𝜔𝑐1𝜔𝑐2;

𝐻 𝑗𝜔0 =𝑅𝑓

𝑅𝑖

𝑗𝜔02 + 2𝜔𝑐1 𝑗𝜔0 + 𝜔𝑐1𝜔𝑐2

𝑗𝜔02 + 𝜔𝑐1 + 𝜔𝑐2 𝑗𝜔0 + 𝜔𝑐1𝜔𝑐2

=𝑅𝑓

𝑅𝑖

2𝜔𝑐1

𝜔𝑐1 + 𝜔𝑐2

≈𝑅𝑓

𝑅𝑖

2𝜔𝑐1

𝜔𝑐2

(15.19)

• If 𝜔𝑐2 ≫ 𝜔𝑐1, then

So, the magnitude at the center frequency is much smaller than the passband magnitude. Thus the bandreject filter successfully rejects frequencies near the center frequency, again confirming our assumption that the parallel implementation is meant for broadband bandreject designs.

𝐻 𝑗𝜔0 ≪2𝑅𝑓

𝑅𝑖 (15.20)

𝐻(𝑠) =𝑅𝑓

𝑅𝑖

𝑠2 + 2𝜔𝑐1𝑠 + 𝜔𝑐1𝜔𝑐2


(15.16)

≈𝑅𝑓

𝑅𝑖

2𝜔𝑐1

𝜔𝑐2

𝐾 =𝑅𝑓

𝑅𝑖




Higher Order Op Amp Filters

How can we realize a filter with a sharper transition at the cutoff frequency?

Cascading Identical Filters

• All of the filter circuits we have examined so far, both passive and active, are nonideal.

• An ideal filter has a discontinuity at the point of cutoff, which sharply divides the passband and the stopband.

• Although we cannot hope to construct a circuit with a discontinuous frequency response, we can construct circuits with a sharper, yet still continuous, transition at the cutoff frequency.

Idea filter

• In order to produce a filter that has a steeper transition region (and hence form a more effective filter), it is necessary to add additional circuitry to the basic filter.

• Responses that are steeper than -20dB/decade in the transition region cannot be obtained by simply cascading identical RC stages (due to loading effects).

• However, by combining an op-amp with frequency-selective feedback circuits, filters can be designed with roll-off rates of -40, -60, or more dB/decade.




A Cascade of Identical Low-pass Filters

• The Bode magnitude plots of a cascade of identical low-pass filters includes plots of just one filter, two in cascade, three in cascade, and four in cascade.

• It is obvious that as more filters are added to the cascade, the transition from the passband to the stopband becomes sharper.

- One filter : 20 dB/dec

- Two filters : 20+20=40 dB/dec

- Three filters : 60 dB/dec

- Four filters : 80 dB/dec

⋯

- n-element cascade of identical low-pass filters: n20dB/dec





A Cascade of Identical Low-pass Filters

• The transfer function for a cascade of n prototype low-pass filters.

𝐻 𝑠 =−1

𝑠 + 1

−1

𝑠 + 1⋯

−1

𝑠 + 1 =

−1 𝑛

𝑠 + 1 𝑛 (15.21)

• The order of a filter is determined by the number of poles in its transfer function. From Eq.(15.21), we see that a cascade of first-order low-pass filters yields a higher order.

• In fact, a cascade of 𝑛 first-order filters produces an 𝑛𝑡ℎ -order filter, having n poles in its transfer function and a final slope of 20n dB/dec in the transition band.





Cutoff frequency of 𝒏𝒕𝒉-order low-pass filter

• As the order of the low-pass filter is increased by adding prototype low-pass filters to the cascade, the cutoff frequency also changes.

• If we start with a cascade of n low-pass filters, we can compute the cutoff frequency for the resulting nth-order low-pass filter.

𝐻 𝑠 =−1 𝑛

𝑠 + 1 𝑛

𝐻 𝑗𝜔𝑐𝑛 =1

𝑗𝜔𝑐𝑛 + 1 𝑛=

1

2

1

𝜔𝑐𝑛2 + 1

𝑛 =1

2

1

𝜔𝑐𝑛2 + 1

=1

2

2/𝑛

(15.21)

1

𝜔𝑐𝑛2 + 1

=1

2

1/𝑛

𝜔𝑐𝑛2 + 1 = 2

𝑛

𝜔𝑐𝑛 = 2𝑛

− 1

1

𝜔𝑐𝑛2 + 1

=1

22/𝑛

=1

2 1/𝑛=

1

2𝑛

(15.22)


𝑗𝜔𝑐𝑛 + 1 = 𝜔𝑐𝑛2 + 1




Cutoff frequency of 𝒏𝒕𝒉-order low-pass filter

• To demonstrate the use of Eq.(15.22), let’s compute the cutoff frequency of a fourth-order unity-gain low-pass filter constructed from a cascade of four prototype low-pass filters:

𝜔𝑐𝑛 = 2𝑛

− 1

𝜔𝑐4 = 24

− 1 = 0.435 rad/s

(15.22)

• Thus, we can design a fourth-order low-pass filter with any arbitrary cutoff frequency by starting with a fourth-order cascade consisting of low-pass filters and then scaling the components by 𝑘𝑓 = 𝜔𝑐/0.435 to place the cutoff frequency at any value of desired.

(15.23)





Nonideal passband behavior of low-pass nth-order cascade

• The transfer function for a unity-gain low-pass nth-order cascade.

𝐻 𝑠 =𝜔𝑐𝑛

𝑛

𝑠 + 𝜔𝑐𝑛𝑛

H jω =ωcn

n

ω2 + ωcn2

n

=1

𝜔/𝜔𝑐𝑛2 + 1

𝑛

- When 𝜔 ≪ 𝜔𝑐𝑛, 𝐻 𝑗𝜔 ≈ 1

- When 𝜔 → 𝜔𝑐𝑛, 𝐻 𝑗𝜔 < 1

• Because the cascade of low-pass filters results in this nonideal behavior in the passband, other approaches are taken in the design of higher order filters. One such approach is examined next.

(15.24)





Filter Response Characteristics.

Butterworth Filters

• The filter frequency responses are typically characterized by the shape of the response curve. A general comparison of the three response characteristics for a low-pass filter response curve is shown in Figure.

Comparative plots of three types of filter response characteristics.

• The Butterworth(버터워스) characteristic provides a very flat amplitude response in the passband and a roll-off rate of -20dB/decade/pole. Filters with the Butterworth response are normally used when all frequencies in the passband must have the same gain. The Butterworth response is often referred to as a maximally flat response.

• Filters with the Chebyshev(체비셰프) response characteristic are useful when a rapid roll-off is required because it provides a roll-off rate greater than -20dB/decade/pole. This type of filter response is characterized by overshoot or ripples in the passband (depending on the number of poles) and an even less linear phase response than the Butterworth.

• The Bessel response exhibits a linear phase characteristic, meaning that the phase shift increases linearly with frequency. The result is almost no overshoot on the output with a pulse input. For this reason, filters with the Bessel response are used for filtering pulse waveforms without distorting the shape of the waveform.




Butterworth low-pass filter

Butterworth Filters

• Butterworth filters are characterized by a maximally flat pass-band frequency response characteristic.

• A unity-gain Butterworth low-pass filter has a transfer function whose magnitude is given by

Butterworth low-pass filter frequency response

𝐻 𝑗𝜔 =1

1 + 𝜔/𝜔𝑐2𝑛

n: the order of filter

- The cutoff frequency is 𝜔𝑐 for all values of 𝑛.

- If 𝑛 is large enough, the denominator is always close to unity when 𝜔 < 𝜔𝑐.

- In the expression for 𝐻 𝑗𝜔 the exponent of 𝜔/𝜔𝑐is always even.





Butterworth Filters

• Given an equation for the magnitude of the transfer function, how do we find 𝐻 𝑗𝜔 ?

𝐻 𝑗𝜔 2 = 𝐻(𝑗𝜔) 𝐻(−𝑗𝜔) = 𝐻 𝑠 𝐻(−𝑠)

=1

1 + 𝜔2𝑛

𝑠 = 𝑗𝜔

=1

1 + (𝜔2)𝑛 =

1

1 + (−𝑠2)𝑛 𝑠2 = −𝜔2 =

1

1 + (−1)𝑛𝑠2𝑛

or

𝐻 𝑠 𝐻(−𝑠) =1

1 + (−1)𝑛𝑠2𝑛

1 + (−1)𝑛𝑠2𝑛 : Butterworth polynomials, given in Table 15.3 in factored form.





Butterworth Filters

• They are the product of first- and second-order factors; therefore, we can construct a circuit whose transfer function has a Butterworth polynomial in its denominator by cascading op amp circuits, each of which provides one of the needed factors.

• A block diagram of such a cascade is shown in Fig. 15.20, using a fifth-order Butterworth polynomial as an example.





Butterworth Filters

• All odd-order Butterworth polynomials include the factor (𝑠 + 1), so all odd-order Butterworth filter circuits must have a subcircuit that provides the transfer function.

𝐻 𝑠 =1

𝑠2 + 𝑏1 + 1

• So what remains is to find a circuit that provides a transfer function of the form

𝐻 𝑠 = −𝜔𝑐

𝑠 + 𝜔𝑐= −

1

𝑠 + 1

• This is the transfer function of the prototype low-pass op amp filter from Fig. 15.1.

• Such a circuit is shown in Fig. 15.21.

𝜔𝑐 =1

𝑅1𝐶




Analysis of a Butterworth Filter Circuit

Butterworth Filters

• Output voltage

𝑉𝑎 − 𝑉𝑖

𝑅+ 𝑉𝑎 − 𝑉𝑜 𝑠𝐶1 +

𝑉𝑎 − 𝑉𝑜

𝑅= 0


𝑅= 𝑉𝑜𝑠𝐶2

2 + 𝑅𝐶1𝑠 𝑉𝑎 − 1 + 𝑅𝐶1𝑠 𝑉𝑜 = 𝑉𝑖

−𝑉𝑎 + 1 + 𝑅𝐶2𝑠 𝑉𝑜 = 0

𝑉𝑜 =

2 + 𝑅𝐶1𝑠 𝑉𝑖

−1 02 + 𝑅𝐶1𝑠 − 1 + 𝑅𝐶1𝑠

−1 1 + 𝑅𝐶2𝑠

=𝑉𝑖

𝑅2𝐶1𝐶2𝑠2 + 2𝑅𝐶2𝑠 + 1

(15.31)

(15.32)

(15.33)

Using Cramer’s rule with Eqs. 15.31 and 15.32, we




Butterworth Filters

• Transfer function

𝑉𝑜 =

2 + 𝑅𝐶1𝑠 𝑉𝑖

−1 02 + 𝑅𝐶1𝑠 − 1 + 𝑅𝐶1𝑠

−1 1 + 𝑅𝐶2𝑠

=𝑉𝑖

𝑅2𝐶1𝐶2𝑠2 + 2𝑅𝐶2𝑠 + 1

(15.33) 𝐻 𝑠 =

𝑉𝑜

𝑉𝑖=

1𝑅2𝐶1𝐶2

𝑠2 +2

𝑅𝐶1𝑠 +

1𝑅2𝐶1𝐶2

(15.34)

𝐻 𝑠 =

1𝐶1𝐶2

𝑠2 +2

𝐶1𝑠 +1

𝐶1𝐶2

𝐻 𝑠 =1

𝑠2 + 𝑏1 + 1

𝑏1 =2

𝐶1 1 =

1

𝐶1𝐶2

If we set 𝑅 = 1Ω

Analysis of a Butterworth Filter Circuit

In order to get a transfer function of the form

we choose capacitor values so that

(15.35)

(15.36)



Butterworth Filters

𝐻 𝑠 =1

𝑠2 + 𝑏1 + 1

• The second-order Butterworth low-pass filter has a transfer function of the form

Butterworth High-Pass Filter

• An nth-order Butterworth high-pass filter has a transfer function with the nth-order Butterworth polynomial in the denominator, just like the nth-order Butterworth low-pass filter. But in the high-pass filter, the numerator of the transfer function is 𝑠𝑛 whereas in the low-pass filter, the numerator is 1.

• To produce the second-order factors in the Butterworth polynomial, we need a circuit with a transfer function of the form

𝐻 𝑠 =𝑠2

𝑠2 + 𝑏1𝑠 + 1





Butterworth Filters

Butterworth High-Pass Filter


𝑉𝑎 − 𝑉𝑖

1𝑠𝐶

+𝑉𝑎 − 𝑉𝑜

𝑅1+


1𝑠𝐶

= 0


1𝑠𝐶

=𝑉𝑜 − 0

𝑅2

1 + 2𝑅1𝐶𝑠 𝑉𝑎 − 1 + 𝑅1𝐶𝑠 𝑉𝑜 = 𝑠𝐶𝑉𝑖

−𝑅2𝐶𝑠 𝑉𝑎 − 1 + 𝑅2𝐶𝑠 𝑉𝑜 = 0

𝐻 𝑠 =𝑉𝑜

𝑉𝑖=

𝑠2

𝑠2 +2

𝑅2𝐶 𝑠 +1

𝑅1𝑅2𝐶2

𝑉𝑜 =𝑠2

𝑠2 +2

𝑅2𝐶 𝑠 +1

𝑅1𝑅2𝐶2

𝑉𝑖

(15.47)

Setting 𝐶 = 1 F yields

𝐻 𝑠 =𝑠2

𝑠2 +2

𝑅2𝑠 +

1𝑅1𝑅2

(15.48)




Butterworth Filters

𝑏1 =2

𝑅2 1 =

1

𝑅1𝑅2

Butterworth High-Pass Filters

𝐻 𝑠 =𝑠2

𝑠2 +2

𝑅2𝑠 +

1𝑅1𝑅2

(15.48) 𝐻 𝑠 =𝑠2

𝑠2 + 𝑏1𝑠 + 1

(15.49)

• First, the high-pass circuit in Fig.15.25 was obtained from the low-pass circuit in Fig. 15.21 by interchanging resistors and capacitors.

• Second, the prototype transfer function of a highpass filter can be obtained from that of a low-pass filter by replacing s in the low-pass expression with 1/𝑠 (see Problem 15.46).



Narrowband Bandpass and Bandreject Filters

Narrowband Bandpass Filters

• The narrowband filter requires a high-Q.

Active High-Q Bandpass Filter

• The transfer functions for cascaded bandpass and parallel bandreject filters have discrete real poles. With discrete real poles, the highest quality bandpass filter (or bandreject filter) we can achieve has

𝑄 =𝜔𝑜

𝛽=

1

2

• To build active filters with high quality factor values, we need an op amp circuit that can produce a transfer function with complex conjugate poles.

• Figure 15.26 depicts one such circuit for us to analyze







𝑉𝑎 =−𝑉𝑜

𝑠𝑅3𝐶

𝑉𝑖 − 𝑉𝑎

𝑅1=


1/𝑠𝐶+

𝑉𝑎 − 0

1/𝑠𝐶+

𝑉𝑎 − 0

𝑅2

𝐻 𝑠 =

−𝑠𝑅1𝐶

𝑠2 +2

𝑅3𝐶 𝑠 +1

𝑅𝑒𝑞𝑅3𝐶2

𝑅𝑒𝑞 =𝑅1𝑅2

𝑅1 + 𝑅2

𝑉𝑖 = 1 + 2𝑠𝑅1𝐶 + 𝑅1/𝑅2 𝑉𝑎 − 𝑠𝑅1𝐶𝑉𝑜

Node b 𝑉𝑎 − 0

1/𝑠𝐶=

0 − 𝑉𝑎

𝑅3

Node a

(15.54)

(15.55)

Substituting Eq.(15.54) into Eq.(15.55) and then rearranging, we get an expression for the transfer function

(15.56)




• The standard form of the transfer function for a bandpass filter is



𝐻 𝑠 =−𝐾𝛽𝑠

𝑠2 + 𝛽𝑠 + 𝜔𝑜2

Equating terms and solving for the values of the resistors give

𝐻 𝑠 =

−𝑠𝑅1𝐶

𝑠2 +2

𝑅3𝐶 𝑠 +1


(15.56)

𝛽 =2

𝑅3𝐶

𝐾𝛽 =1

𝑅1𝐶

𝜔𝑜2 =

1


𝑅𝑒𝑞 =𝑅1𝑅2

𝑅1 + 𝑅2

(15.57)

(15.58)

(15.59)

𝜔𝑜: a specified center frequency 𝑄: quality factor 𝐾: passband gain




• The prototype version of the circuit in Fig. 15.25



𝐾𝛽 =1

𝑅1𝐶 →

𝜔𝑜2 =

1

𝑅𝑒𝑞𝑅3𝐶2 →

𝜔𝑜: a specified center frequency 𝑄: quality factor 𝐾: passband gain

𝜔𝑜 =1

𝐿𝐶

𝜔′𝑜 =1

𝐿𝐶𝜔𝑜

𝑄 =𝜔𝑜

𝛽=

1

𝛽

𝜔𝑜 = 1 rad/s 𝐶 = 1 F

𝑅1 =1

𝐾𝛽=

𝑄

𝐾

𝛽 =2

𝑅3𝐶 → 𝑅3 =

2

𝛽= 2𝑄

𝑅𝑒𝑞 =1

𝑅3=

1

2𝑄 =

𝑅1𝑅2

𝑅1 + 𝑅2 → 𝑅2 = 𝑄/(2𝑄2 − 𝐾)

From Eq.(15.58),

From Eq.(15.57),

From Eq.(15.59),



15.5 Narrowband Bandpass and Bandreject Filters

Narrowband Bandreject Filters

Active High-Q Bandreject Filter

• The parallel implementation of a bandreject filter that combines low-pass and high-pass filter components with a summing amplifier has the same low-Q restriction as the cascaded bandpass filter.

• An active high-Q bandreject filter known as the twin-T notch filter because of the two T-shaped parts of the circuit at the nodes labeled a and b.






• Analysis of twin-T notch filter

𝑉𝑎 − 𝑉𝑖 𝑠𝐶 + 𝑉𝑎 − 𝑉𝑜 𝑠𝐶 +2(𝑉𝑎 − 𝜎𝑉𝑜)

𝑅= 0

𝑉𝑎 2𝑠𝐶𝑅 + 2 − 𝑉𝑜 𝑠𝐶𝑅 + 2𝜎 = 𝑠𝐶𝑅𝑉𝑖

Node a :

Node b :

𝑉𝑏 − 𝑉𝑖

𝑅+

𝑉𝑏 − 𝑉𝑜

𝑅+ 𝑉𝑏 − 𝜎𝑉𝑜 2𝑠𝐶 = 0

𝑉𝑏 2 + 2𝑅𝐶𝑠 − 𝑉𝑜 1 + 2𝜎𝑅𝐶𝑠 = 𝑉𝑖

(15.60)

(15.61)

Noninverting input terminal of the top op amp :

𝑉𝑎 − 𝑉𝑜 𝑠𝐶 +𝑉𝑏 − 𝑉0

𝑅= 0

−𝑠𝑅𝐶𝑉𝑎 − 𝑉𝑏 + 𝑠𝑅𝐶 + 1 𝑉𝑜 = 0 (15.62)







𝑉𝑎 2𝑠𝐶𝑅 + 2 − 𝑉𝑜 𝑠𝐶𝑅 + 2𝜎 = 𝑠𝐶𝑅𝑉𝑖

𝑉𝑏 2 + 2𝑅𝐶𝑠 − 𝑉𝑜 1 + 2𝜎𝑅𝐶𝑠 = 𝑉𝑖

(15.60)

(15.61)

−𝑠𝑅𝐶𝑉𝑎 − 𝑉𝑏 + 𝑠𝑅𝐶 + 1 𝑉𝑜 = 0 (15.61)

Using Cramer’s rule to solve for 𝑉𝑜 gives

𝑉𝑜 =

2 𝑅𝐶𝑠 + 1 0 𝑠𝐶𝑅𝑉𝑖

0 2(𝑅𝐶𝑠 + 1) 𝑉𝑖

−𝑅𝐶𝑠 −1 02(𝑅𝐶𝑠 + 1) 0 −(𝑅𝐶𝑠 + 2𝜎)

0 2(𝑅𝐶𝑠 + 1) −(2𝜎𝑅𝐶𝑠 + 1)−𝑅𝐶𝑠 −1 𝑅𝐶𝑠 + 1

=𝑅2𝐶2𝑠2 + 1 𝑉𝑖

𝑅2𝐶2𝑠2 + 4𝑅𝐶 1 − 𝜎 𝑠 + 1 (15.63)







Rearranging Eq.(15.63), we can solve for the transfer function:

(15.64) 𝐻 𝑠 =𝑉𝑜

𝑉𝑖=

𝑠2 +1

𝑅2𝐶2

𝑠2 +4 1 − 𝜎

𝑅𝐶𝑠 +

1𝑅2𝐶2

The standard form for the transfer function of a bandreject filter:

𝐻 𝑠 =𝑠2 + 𝜔0

2

𝑠2 + 𝛽𝑠 + 𝜔02 (15.65)

Equating Eqs.(15.64) and (15.65) gives

𝜔𝑜2 =

1

𝑅2𝐶2

𝛽 =4 1 − 𝜎

𝑅𝐶

(15.66)

(15.67)







In this circuit, we have three parameters (R, C, and 𝜎) and two design constraints (𝜔𝑜 and 𝛽).

Thus one parameter is chosen arbitrarily; it is usually the capacitor value because this value typically provides the fewest commercially available options.

Once C is chosen,

𝑅 =1

𝜔𝑜𝐶

𝜎 = 1 −𝛽

4𝜔𝑜= 1 −

1

4𝑄

And

(15.68)

(15.69)

Electric Circuit Theoryelearning.kocw.net/KOCW/document/2015/korea_sejong/... · Advantages and...

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