Definitions and Symbols

1

Definitions and Symbols

VI = VIe

jI = VI

I is the vector quantity corresponding to an input sinusoidal waveform

)2cos()( III tfVtVI ; VI is the amplitude of the input sinusoidal signal; I is the phase

of the input sinusoidal signal.

VO = VOejO = VO

O is the vector quantity corresponding to the output sinusoidal waveform

)2cos()( OOO tfVtVO ; VO is the amplitude of the output sinusoidal signal; O is the

phase of the output sinusoidal signal.

G(jw) is the transfer function of the filter, which depends on the input sinusoidal frequency

fI=wI/2. w is the angular frequency of the input sinusoidal signal. G(jw) amplitude is |G(jw)|

while its phase is G(jw): G(jw)= |G(jw)| G(jw)

=RC is the time constant of the first order RC filter

fC is the cutoff frequency of the filter, which defines the pass band B of the filter.

B is the pass band of the filter and it is the range of frequencies for which the input sinusoidal

signal is not significantly attenuated.

Bw is the bandwidth of the filter and it is the width of its pass band B.

A Decade is the distance between two major ticks in a frequency logarithmic scale axis.

Decibel (dB) is a measurement unit defined as 20*Log of the value of the quantity to be

measured. Log is the logarithmic with base 10.

The Bode plot is the transfer function amplitude diagram in full logarithmic scale: both |G|

(transfer function amplitude) axis and f (frequency) axis are represented in logarithmic scale.

OPAMP stands for OPerational AMPlifier, and it is an integrated circuit (IC) used to realize

active filters.

2

1

Lesson 1: RC low pass filter schematic and transfer function (2hrs) 2

3

Objectives 4

To know the schematic of an RC low-pass filter. 5

To calculate the transfer function of an RC low-pass filter. 6

To calculate the time constant of an RC low-pass filter. 7

To draw the schematic of a circuit with CAD tools in English. 8

9

10

Today we're going to study a new circuit: the RC low-pass filter in the frequency domain. We‟ll start 11

with the schematic of the filter, shown in figure 1.1. 12

13 Fig. 1.1: schematic of a first-order passive RC low pass filter. 14

15

We are going to study the frequency response of this simple circuit. Since we are interested in the 16

frequency response the input voltage VI is a sinusoidal waveform with amplitude VI, frequency fI and 17

phase I: )2cos( III tfVVI . The output voltage VO is sinusoidal too with frequency fI because 18

there are only linear components in the circuit (a resistor R and a capacitor C) that cannot cause 19

distortion. So VO is a sinusoidal waveform with amplitude VO, phase O and the same frequency fI as 20

VI: )2cos( OIO tfVVO . We can represent both VI(t) and VO(t) with the corresponding vector 21

quantities: VI=VIejI and VI=VOe

jO. Assuming the input signal VI is known (the circuit stimulus) the 22

only unknown variables of the output signal VO are its amplitude VO and its phase O. How can we 23

find the mathematical relation between VO and VI and O and I? 24

25

The RC filter can be seen as an impedance divider, as shown in figure 1.2: the input signal VI is 26

divided between the resistor impedance (Z1=R) and the capacitor impedance (Z2=1/jwC). 27

VO(jw) VI(jw)

R

C

stimulus or input sinusoidal voltage

output sinusoidal voltage

3

1 Fig. 1.2: the schematic of a first-order passive RC low pass filter can be seen as an impedance divider. 2

3

Thus, applying the simple impedance divider formula we get: 4

Eq. 1.1) VIZR

ZVO

2

2 , whereCfjCwj

Z

2

112 is the capacitor impedance. As it can 5

be seen, Z2 depends on w=2f, the frequency of the input signal. 6

7

The transfer function G(jw) of the RC filter is defined as: 8

Eq. 1.2) VI

VOjwG )( . 9

Note that by definition )()()( jwVIjwGjwVO . We can therefore determine the amplitude and phase 10

of VO(jw) using the properties of complex numbers: 11

Eq. 1.3.a) Amplitude of VO(jw): IO VjwGV )( (the amplitude of the output is the product of the 12

amplitude of the input and the amplitude of the transfer function at the stimulus given frequency). 13

Eq. 1.3.b) Phase of VO(jw): IO jwG )](arg[ , where arg[G(jw)]=G(jw), is the phase of the 14

transfer function (the phase of the output is the sum of the phase of the input and the phase of the 15

transfer function at the stimulus given frequency). 16

Using equations 1.3 a) and b) we can calculate VO and O once VI, I and G(jw) are known. Note that 17

VI, I are the given inputs, so the only unknown variable of our problem is G(jw). However, using 18

equations 1.1 and 1.2 G(jw) is given by: 19

Eq. 1.4) 2

2)(ZR

Z

VI

VOjwG

20

By substituting Z2=1/jwC into equation 1.4 we get: 21

1

1

1

11

1

1

)(

jwRC

jwCR

jwC

jwCR

jwCjwG ; =RC is defined as the time constant of the RC 22

filter; hence we can write the transfer function of the low-pass RC filter as: 23

Eq. 1.5)

fjjw

jwG21

1

1

1)( 24

VO(jw)

VI(jw)

Z1=R

Z2=1/jwC

4

Equation 1.5 shows that G(jw) is a complex number whose value depends only on the filter time 1

constant and the frequency f of the input signal. From equation 1.5, using the properties of complex 2

numbers, we can easily calculate the amplitude )( jwG and the phase G(jw)= arg[G(jw)] of the filter 3

transfer function: 4

Eq. 1.6.a) 2)(1

1)(

wjwG , 5

Eq. 1.6.b) )()( wartgjwG 6

By substituting equation 1.6.a into 1.3.a and equation 1.6.b into 1.3.b we reach our final resolution 7

formulas: 8

Eq. 1.7.a) 22 )(1)(1

1

w

VV

wV I

IO 9

Eq. 1.7.b) )()( wartgwartg IIO 10

VI, I and w=2f are the amplitude, phase and angular frequency of the input signal and are all known 11

quantities. =RC depends on the RC filter components and can be calculated from the filter schematic. 12

Hence, eq. 1.7.a and 1.7.b are the resolution formulas that allow us to calculate amplitude and phase of 13

the output for any given input and any given RC filter. 14

For instance, let‟s consider an RC low pass filter with R=1K and C=159.24nF. We can easily 15

determine its constant time sss 3693 1015924.01024.159101024.1591 , i.e. about 16

0,16ms. Now let‟s consider two different cases: 17

Case A) input signal VI with amplitude 1V, phase 0 degrees (00) and frequency 1KHz; 18

Case B) input signal VI with amplitude 1V, phase 0 degrees (00) and frequency 15KHz. 19

In case A the input frequency is 1KHz, hence the value of the amplitude of the transfer function is (see 20

eq. 1.6.a): 21

707,02

1

)005,1(1

1

)16,02(1

1

)16,012(1

1

)(1

1)(

2222

msKHzwjwG22

Therefore, the amplitude of the output voltage is VVVV IO 06.15.1707.0707.0 . 23

The phase of the transfer function at 1KHz is instead (see eq. 1.6.b): 24 0

)( 45)005.1()16.012()( artgmsKHzartgwartgjwG . 25

Therefore, the phase of the output voltage is 000 45450 O . 26

Considering now case B the input frequency is 15KHz, so the values of the amplitude and of the phase 27

of the transfer function are: 28

066.0)08.15(1

1

)16.0152(1

1)(

22

msKHzjwG

29

0

)( 2.86)08,15()( artgwartgjwG . 30

Consequently the amplitude of the output voltage is VVVV IO 15.1066.0066.0 , while the 31

phase of the output voltage is 000 2.862.860 O . 32

Of course different results are obtained if we change input frequency (w value changes) and/or the 33

resistor or the capacitor of the RC filter ( value changes). 34

35

English is an important language for electronics since modern electronics was born in the United States 36

with the invention of the transistor in 1958. English is the common language of the electronic engineers 37

community and all electronics components are described by datasheets written in English. Commonly 38

5

used CAD (Computer Aided Design) tools, useful to design electronic circuits, are in English language, 1

too. Hence, the knowledge of technical English is a necessary prerequisite to design any innovative 2

electronic circuit. Just to supply an example we will now use a CAD tool called LTSpiceIV that can be 3

downloaded freely from the Internet (http://www.linear.com/designtools/software) to build the 4

schematic of the RC low-pass filter we have just analyzed. 5

6

7

6

1

Lesson 2: Frequency response graphical representation. (2 hrs) 2

3

Objectives 4

To be able to use the frequency response graphs of a filter to calculate the output voltage. 5

To know the cutoff frequency (fC) and the low-pass filter pass-band (B) and the relationship between 6

fC and . 7

To be able to recognize and draw the graphs of the frequency response of a low-pass filter in 8

logarithmic scale. 9

To be able to design a first-order RC low-pass filter with a given pass-band B. 10

To draw logarithmic and semilogarithmic graphs of the frequency response of an RC low-pass filter 11

with MS-excel. 12

13

14

In the previous lesson we found the algebraic expressions of the amplitude and phase of the transfer 15

function G(jw) of a first order RC low-pass filter: 16

Eq. 2.1) 2)(1

1)(

wjwG , )()( wartgjwG 17

Both functions depend on the frequency

2

wf of the input signal VI. We also showed how to 18

calculate VO using Eq. 2.1). 19

In this lesson we will show a more straightforward approach for the calculation of VO, i.e. the 20

graphical approach. By definition of transfer function it is: 21

Eq. 2.2) IO VjwGV )( , )( jwGIO 22

The graphical approach uses Eq. 2.2) but does not calculate |G(jw)|and G(jw) analytically. On the other 23

hand, the amplitude and phase of the transfer function at the given input frequency are directly read 24

from the amplitude and phase diagrams of the filter. The amplitude diagram shows |G(jw)| vs. f while 25

the phase diagram shows G(jw) vs. f. An example of amplitude and phase plots for an RC low pass 26

filter with =1ms is given in figure 2.1. Hence, once we know the amplitude and phase plots of a given 27

filter it is very easy to determine the value of |G(jw)| and G(jw) graphically from the plot at the given 28

input frequency without any analytical calculation (without calculating Eq. 2.1). 29

For instance, let us assume the input is a sinusoidal waveform with the following characteristics: VI 30

=1.5V, I=00 and f=100Hz. 31

From the amplitude plot at f=100Hz we read |G|=0.8 while from the phase plot at 100Hz we read G=-32

350. 33

Hence, using Eq. 2.2 we derive that VO=0.8*1.5V=1.2V while O=00-35

0=-35

0. 34

Let‟s assume now that the same input sinusoidal waveform has a higher frequency, 1KHz instead of 35

100Hz. At 1KHz we read |G|=0.18 and G=-810 from the amplitude and phase plots, respectively. 36

Hence at 1KHz the output of the filter is given by VO=0.18*1.5V=0.27V while O=00-81

0=-81

0. 37

The examples above show two clear advantages of the graphical method: 38

1) it is valid for any filter topology: we need only to know the amplitude and phase plots 39

associated to the filter to determine its output voltage. 40

2) output voltage is very easily calculated for any input frequency since we need only to solve Eq. 41

2.2 where |G| and G are directly read from the filter plots. 42

43

44

7

1

2

3

4

5

6

7

8

9

10

11

12

Fig. 2.1: examples of amplitude and phase plots for a first order RC low-pass filter with =1ms. 13

14

Let‟s now analyze more in detail the filter amplitude diagram (|G| vs. f), shown in figure 2.2: 15

16

17

18

19

20

21

22

23

24

25

26

27

28

29 Fig. 2.2: example of amplitude diagram of a passive first order RC filter. Note the cutoff frequency fC 30

can be easily read and that the frequency scale of the diagram is logarithmic. Note also the pass band 31

(B) of the filter. 32

33

The cutoff frequency fC of the filter is defined as the frequency value at which |G|=0.7. So when input 34

signal frequency f is less than the cutoff frequency (f<fC) the output signal amplitude is equal or only 35

slightly lower than input amplitude (no attenuation or only small attenuation of the input). On the 36

contrary, when f>fC the output signal amplitude is significantly lower than input amplitude (significant 37

attenuation). To sum up we can roughly approximate an ideal low-pass filter behavior as follows: 38

- input with f<fC no attenuation (the input signal passes) VO=VI 39

- input with f>fC great attenuation (the input signal is cut) VO0 40

In fact, a low pass filter is typically used to eliminate (attenuate) all high frequency components of the 41

input signals. Its fC represents the frequency threshold at which the filter starts to be active. 42

The pass band B of the filter is the range of frequencies for which |G| 0.7, i.e. the input signal is not 43

significantly attenuated (the input signal “passes”). Of course, it is B= [0,fC] for the low pass RC filter. 44

The bandwidth Bw of the filter is the width of the pass band B, hence: Bw=fC-0=fC. 45

The following relationship between fC and can be easily demonstrated (do it yourself to exercise!): 46

47

Eq. 2.3) 48

Amplitude diagram

0

0,1

0,2

0,3

0,4

0,5

0,6

0,7

0,8

0,9

1

0,1

1 10

10

0

10

00

10

00

0

10

00

00

10

00

00

0frequency

|G(j

w)|

Phase diagram

-90

-75

-60

-45

-30

-15

0

0,1

1 10

10

0

10

00

10

00

0

10

00

00

10

00

00

0frequency

G(j

w)

0.8

0.18

-350

-810

Amplitude diagram

0

0,1

0,2

0,3

0,4

0,5

0,6

0,7

0,8

0,9

1

0,1

1 10

10

0

10

00

10

00

0

10

00

00

10

00

00

0frequency

|G(j

w)|

fC=160Hz

|G|=0,7

Note the

frequency

scale is

logarithmic

1Decade

B=[0,fC]

CRfC

2

1

2

1

8

1

Finally, note the frequency scale of the plot is logarithmic and not linear. In fact, the distance between 2

0.1 and 1Hz is the same between 100 and 1000Hz. The frequency logarithmic scale allows to read the 3

values of the whole amplitude waveform for a wide range of frequencies. The distance between f and 4

10*f in the logarithimic scale is called a Decade. So, when we move from 1Hz to 10Hz we increase the 5

frequency of a Decade. When we move from 1Hz to 100Hz we increase the frequency of 2 Decades. 6

Figure 2.3 shows the comparison between a linear and a logarithmic scale: the logarithmic scale is 7

much wider than the linear one. 8

9

10 Fig. 2.3: comparison between logarithmic and linear frequency axis scale. In the linear scale the 11

distance between two ticks is constant =1Hz. In the logarithmic scale the distance between 10*f and f is 12

constant = 1 Decade. Hence, the logarithmic scale features a much wider range of frequency values. 13

14

Figure 2.4 shows the comparison between the same two amplitude diagrams, one with logarithmic 15

frequency scale and the other with linear frequency scale. The frequency axis of the two plots are the 16

same (from 0.1Hz to 1000000Hz=1MHz). Only the plot with logarithmic scale can be used to read the 17

transfer function amplitude values at different frequencies. 18

19 Fig. 2.4: the same two amplitude diagrams with and without logarithmic scale are compared. The 20

values of fC and of the transfer function amplitude can be read only from the logarithmic plot. 21

22

Figure 2.5 shows the filter phase diagram (G vs. f). We can divide the plot into three different 23

frequency ranges: 24

- f<<fC G=00, the filter does not introduce any phase shift, i.e. the output waveform is not 25

delayed with respect to the input. 26

- f=fC G=-450, at the cutoff frequency the filter phase shift is -45

0, corresponding to a 27

delay=0.125/fC of the output waveform with respect to the input. 28

- f>>fC G=-900, the filter introduces the maximum phase shift. The output waveform exhibits 29

the maximum delay with respect to the input =0.25/f. 30

Finally, we can roughly approximate an ideal low-pass filter behavior as follows: 31

- input with f<fC no delay (the input signal passes without any delay) O=I 32

Amplitude diagram

0

0,2

0,4

0,6

0,8

1

0,1

1 10

100

1000

10000

100000

1000000frequency

|G(j

w)|

fC=160Hz

Amplitude diagram

0

0,2

0,4

0,6

0,8

1

0,1

100000

200000

300000

400000

500000

600000

700000

800000

900000

1000000

frequency

|G(j

w)|

linear scale

log. scale

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1Hz

1 2 3 4 5 6 7 8

9

10

1 decade

100

16 17

200

18 19 20 21 22 23 24 25 26

20

linear scale

logarithmic scale

9

- input with f>fC 900 delay (the input signal is delayed) O=I-90° 1

2

3 Fig. 2.5: example of phase diagram of a passive first order RC filter. The cutoff frequency fC is the 4

frequency value at which G=-450. The frequency scale is logarithmic. 5

6

Let‟s now rewrite Eq. 2.2 for the calculation of the output voltage of a filter: 7

Eq. 2.4) IO VjwGV )( . 8

The equation becomes dimensionless if we divide both its members by 1V: 9

Eq. 2.5) V

VjwG

V

V IO

1)(

1 . 10

We can rewrite equation 2.5 in decibel (dB) by applying the function 20*Log() to both its members: 11

Eq. 2.6) )1

)((20)1

(20V

VjwGLog

V

VLog IO . 12

Remembering the properties of the logarithm Log(a*b)=Log(a)+Log(b) we can rewrite equation 2.6 as 13

follows: 14

Eq. 2.7) )1

(20))((20)1

(20V

VLogjwGLog

V

VLog IO . 15

Since by definition of dB it is: OO VV

VLog )

1(20 [dBV], |)(|))((20 jwGjwGLog [dB] and 16

II V

V

VLog )

1(20 [dBV], equation 2.7 gives the resolution formula to calculate the filter output in dB: 17

Eq. 2.8) ][][)(][ dBVVdBjwGdBVV IO . 18

The filter output voltage amplitude (VO) in dBV is equal to the filter input voltage amplitude (VI) in 19

dBV plus the value in dB of the filter transfer function amplitude (|G|) in dB calculated at the given 20

input frequency f=w/2. 21

Eq. 2.8 is almost the same of Eq. 2.2: the only difference is that the product in Eq. 2.2 becomes a sum 22

in Eq. 2.8. Since the addition is a mathematical operation much simpler than the multiplication, 23

engineers prefer to use Eq. 2.8 rather than Eq. 2.2. 24

This means that engineers prefer to plot |G(jw)| in dB vs. frequency. When |G| is expressed in dB the 25

transfer function amplitude plot of figure 2.2 becomes as shown in figure 2.6: 26

Phase diagram

-90

-75

-60

-45

-30

-15

0

0,1

1 10

10

0

10

00

10

00

0

1E

+0

5

1E

+0

6frequency

G(j

w)

frequency scale is logarithmic

fC=160Hz

10

1 Fig. 2.6: this is the same amplitude diagram of figure 2.2 with the only difference that the amplitude |G| 2

is expressed in dB. This is a full logarithmic plot and it is called Bode plot of the filter. 3

4

The transfer function amplitude (|G|) plot in dB is called the Bode plot of the filter (any system 5

described by a transfer function has its own Bode plot). The Bode plot is a full logarithmic scale plot, 6

since the scale in dB=20*Log() is logarithmic too. 7

The cutoff frequency fC in a Bode plot is the frequency at which |G|[dB]=-3dB. In fact, when f=fC 8

|G|=0.7 and hence |G|[dB]=20*Log(0.7)=-3 decibel. 9

When f>fC the Bode plot can be approximated with a straight line of slope -20dB/decade. In fact, 10

|G(jw)|[dB] decreases of 20dB for each increase of a factor of 10 of the frequency, i.e. for each decade 11

of increase of frequency. Thanks to this property the Bode plot graphical drawing is very simple since 12

it can be approximated with only two straight lines: 13

1) f<fC |G(jw)|=0 (horizontal line along x-axis) 14

2) f<fC |G(jw)|=-20*Log(f/fC) (straight line with slope -20dB/dec). 15

Once the filter is given, fC can be calculated using Eq. 2.3. When fC is known the drawing of the Bode 16

plot of the filter is straightforward, as shown in figure 2.7: 17

18 Fig. 2.7: example of simplified Bode plot drawing of a first order RC low pass filter. The only 19

significant approximation of this plot is that |G(fC)|=0dB while it is -3dB in reality. 20

21

Amplitude diagram in dB (Bode Plot)

-80

-70

-60

-50

-40

-30

-20

-10

0

0,1

1 10

10

0

10

00

10

00

0

10

00

00

10

00

00

0frequency

|G(j

w)|

[d

B]

fC=160Hz

1dec

20dB

-3dB

|G(jw)| [dB]

f (log)

fC 10fC 100fC 1000fC 0dB

-20dB

-40dB

-60dB

11

In a few words, the Bode plot of the filter presents the following advantages with respect to the 1

traditional plot with a linear scale along |G| axis: 2

1) very simple to draw using the -20dB/dec rule. 3

2) VO can be calculated as a sum of VI and the |G| value read from the plot. 4

For instance, let‟s consider an input signal of amplitude =0dBV (corresponding to 1V amplitude) at the 5

frequency f=10000Hz=10KHz. From the Bode plot of Fig. 2.5 we read that |G|=-35dB when f=10KHz. 6

Hence, we can immediately calculate VO=0dBV-35dB=-35dBV using equation 2.8). Engineers of some 7

years ago were familiar with voltage values in dB. For us it is a little more difficult to associate a 8

voltage value to -35dBV. In this case it is sufficient to remember the inverse formula: 9

Eq. 2.9) 20

][

10][

dBVV

O

O

VV , 10

from which VO=10-1.75

V=0.018V (remember these general rules: 0dBV corresponds to 1V. when dBV 11

values are negative it means voltage amplitude is between 1 and 0. The positive dB values are for all 12

voltages >1V). 13

14

Eq. 2.3) is used to calculate fC when the filter is known. However, it is often required to design a filter 15

for a given fC or pass band B. First of all we can use the inverse formula of Eq. 2.3) to calculate the 16

time constant of the filter: 17

18 Eq. 2.10) . 19

20

Then we can design the filter following the four steps algorithm proposed below: 21

22

1) calculate using equation 2.10) 23

24

2) choose a capacitor among those available in the laboratory, for instance C=1F 25

26

3) calculate the corresponding value of R according to the formula: 27

28

4) if R is within the range [100,100K] OK, otherwise go back to point 2) choosing another 29

C value 30

31

It is suggested to choose the capacitor value among those available because almost all resistors values 32

are commonly available. It is recommended not to have R<100 because parasitic resistors may affect 33

filter parameters. R>100K is not recommended either because the filter impedance may become 34

comparable to that of multimeters, thus affecting filter behavior by multimeter insertion. 35

For instance, the desing of a filter with fC =1KHz can be done as follows: 36

37

1) ; 2) C=1F; 3) ; 4) R[100,100K]OK! 38

39

And now we‟ll conclude the lesson by showing how to use an electronic sheet as MS-excel to draw 40

amplitude and phase plots of first order low pass RC filters. The starting point is equation 2.1) with the 41

analytical expression of |G| and G. MS-excel is used to calculate those analytical expressions at the 42

various frequency values of interest and to plot the corresponding graphs. 43

44

BfC

2

1

2

1

CR

sfC

41059.12

1

15910

1059.16

4

F

sR

12

1 2

13

1

Lesson 3: Analysis of the CR high pass filter. (2 hrs) 2

3

Objectives 4

5

To be able to draw the schematic of a high-pass CR filter and to know its transfer function. 6

To be able to draw amplitude and phase diagrams of the high-pass filter. 7

To know the relationship among the time constant (), the cutoff frequency (fC) and the filter pass 8

band (B). 9

To be able to calculate the output voltage with the graphical method. 10

To be able to design the filter with a given cutoff frequency fC. 11

To be able to simulate the filters with CAD tools in English. 12

13

14

In the previous lesson we saw how to represent the transfer function of a low pass RC filter graphically. 15

Figure 3.1 shows the schematic of a first-order passive CR high pass filter: 16

17 Fig. 3.1: schematic of a first order passive CR high pass filter. 18

19

The schematic is very similar to that of a low pass filter (compare figure 3.1 with figure 1.1 of lesson 20

1). The only difference is that R and C are interchanged. So we can follow the same approach used for 21

the RC low pass filter to determine the high pass filter transfer function: 22

Eq. 3.1) 21

2

)(

)(|)(|

ZZ

Z

jwV

jwVjwG

I

O

, 23

where the only difference with respect to the low pass filter is that: Z1=1/jwC while Z2=R (Z1 and Z2 24

are interchanged). Substituting Z1 and Z2 expression into equation 3.1 after a few mathematical 25

passages we get the analytical expression of the high pass filter transfer function G(jw): 26

27

Eq. 3.2) 28

29

is the time constant of the filter and it is defined as in the low pass filter: =RC. Using the properties 30

of complex numbers, we can calculate the amplitude and phase of G(jw) from equation 3.2: 31

32

Eq. 3.3) , . 33

C

1µ

R

159

V1

SINE(0 1 1K)

OI

VO(jw) VI(jw)

VI=VIeji

VO=VOejo

wj

wjjwG

1)(

2)(1)(

w

wjwG )(900

)( wartgjwG

14

Equation 3.3 can be used to calculate the output of the filter once the input is known. However, 1

equation 3.3 can also be used to represent the frequency response diagrams of the filter graphically by 2

means of MS-excel, as shown in the previous lesson. Figure 3.2 shows the amplitude and phase 3

diagrams of a CR high pass filter with =0.16ms obtained using equation 3.3 and MS-excel: 4

5 Fig. 3.2: amplitude and phase diagrams of a CR high pass filter with =0.16ms. 6

7

The filter cutoff frequency fC is defined as the frequency at which |G|=-3dB (corresponding to a 0.7 8

factor of amplitude attenuation), as for the low pass filter. However note that for the high pass filter 9

when f= fC it is G=450 and not -45

0 as for the low pass filter. This means that the high pass filter 10

output is phase led with respect to the input, and not delayed as occurs in the low pass filter case. 11

The high pass filter pass band B is defined as the frequency range for which |G|>-3dB (or 0.7), i.e. the 12

frequency range where filter amplitude attenuation can be considered negligible. This is exactly the 13

same definition used in the low pass filter case. However, B=[fC,] for the high pass filter, as it can be 14

seen from figure 3.2. Consequently, the bandwidth Bw (i.e. the width of B) of the filter =. 15

Another important property of the Bode plot of the high pass filter is that |G| increases of 20dB for each 16

decade of increase of frequency (i.e. for a factor 10 of frequency increase). 17

With regard to the phase plot, the maximum phase shift of +900 occurs when f<<fC and corresponds to 18

an output voltage with 0.25/f lead with respect to input voltage. On the contrary when f>>fC there is no 19

phase delay. 20

In conclusion, input „high‟ frequencies (where „high‟ means higher than filter fC) are not significantly 21

attenuated nor phase shifted – the output and the input of the filter are practically the same. On the 22

contrary, input „low‟ frequencies (lower than fC) are attenuated and phase led, so that the output 23

amplitude is negligible and it is phase led with respect to the input. In a high pass filter, input high 24

frequencies “pass”, input low frequencies are “cut”. 25

The relationship between and fC is the same as the low pass filter: 26

Eq. 3.4) , CR

fC

2

1

2

1 27

28 Hence, the design of the high pass filter can be done exactly the same way as that of a high pass filter 29

(the four step algorithm): 30

31

1) calculate corresponding to the required fC using equation 3.4) 32

33

2) choose a capacitor among those available in the laboratory, for instance C=1F 34

35

3) calculate the corresponding value of R according to the formula: 36

37

Phase diagram

0

15

30

45

60

75

90

0,1

1 10

100

1000

10000

100000

1000000frequency

G(j

w)

Amplitude diagram in dB (Bode Plot)

-80

-70

-60

-50

-40

-30

-20

-10

0

0,1

1 10

10

0

10

00

10

00

0

10

00

00

10

00

00

0frequency

|G(j

w)|

[d

B]

fC=1KHz 1dec

20dB B=[fC,]

fC=1KHz

G(fc)=+450

|G(fC)| = -3dB

Cf

2

1

CR

15

4) if R[100,100K] OK, otherwise go back to point 2) choosing another C value 1

2

The Bode plot of the filter is easy to draw, too. In fact, we can approximate it as two straight lines: 3

1) one horizontal line when f>fC: |G(f)|=0dB (=1) when f>fC 4

2) one straight line with slope 20dB/dec when f< fC: |G(f)|=20*Log(f/ fC) 5

Figure 3.3 shows the Bode plot of a generic high pass filter obtained following the two rules above: 6

7 Fig. 3.3: Bode plot drawing of a first order high pass filter. The only significant approximation is 8

|G|=0dB when f=fC (|G(fC)|=-3dB in reality). 9

10

Let‟s now show some examples of output calculation for the high pass filter of figure 3.1. We‟ll 11

calculate the ouput using the graphical method, based on the knowledge of the two graphs of figure 3.2 12

and of the output resolution formulas found in lesson 2 (valid independently of filter topology and 13

order): 14

Eq. 3.5) ][][)(][ dBVVdBjwGdBVV IO , )( jwGIO . 15

The input voltage amplitude is VI=1V (corresponding to 0dBV), the input phase voltage is I=00 and 16

we will consider three different cases of input frequency: 17

Case 1): f=100Hz. 18

In this case we read from the graphs of figure 3.2 that |G(100Hz)|=-20dB while 19

G(100Hz)=830. Hence, using equation 3.5 we derive: 20

VO=-20dB+0dBV=-20dBV corresponding to 0.1V (using equation 2.9); 21

O=830+0

0=83

0 (i.e.output with 83/(360*100Hz)=2.3ms lead). 22

Case 2): f=1KHz. 23

In this case we read from the graphs of figure 3.2 that |G(1KHz)|=-3dB while 24

G(1KHz)=450. Hence, using equation 3.5 we derive: 25

VO=-3dB+0dBV=-3dBV corresponding to 0.7V (using equation 2.9); 26

O=450+0

0=45

0 (i.e.output with 45/(360*1KHz)=125s lead) 27

Case 3): f=10KHz. 28

In this case we read from the graphs of figure 3.2 that |G(10KHz)|=-0.09dB while 29

G(10KHz)=70. Hence, using equation 3.5 we derive: 30

VO=-0.09dB+0dBV=-0.09dBV corresponding to 0.99V (using equation 2.9); 31

O=70+0

0=7

0 (i.e.output with 7/(360*1KHz)=1.9s lead) 32

As expected, increasing the frequency the output amplitude increases while the phase shift decreases 33

with respect to the input. 34

35

Up to now we have shown you some useful instruments to calculate the output of an RC filter subject 36

to a sinusoidal stimulus. In particular, we have shown the full analytical method characterized by many 37

|G(jw)| [dB]

f (log)

fC 0,1fC 0,01fC

0,001fC

0dB

-20dB

-40dB

-60dB

16

hand calculations and the graphical method, with fewer calculations but requiring the knowledge of 1

amplitude and phase diagrams. What is the actual method used in electronics industry? 2

Electronic industries use a third method, really straightforward because it doesn‟t need any hand 3

calculation at all: the simulation of electronic circuit. Circuit simulation requires only the knowledge of 4

the circuit schematic and of the models of the components used in the circuit. Then a CAD (Computer 5

Aided Design) tool performs all the calculations and provides the circuit output waveforms directly in a 6

GUI (Graphical User Interface). Since the models are usually provided within the CAD tool, it may 7

seem it doesn‟t need an electronic technician to perform these simple tasks. It may seem you only need 8

to know the schematic and how to use the CAD tool. And in many simple cases it is really so! 9

However, in real application a very skilled designer is required for many reasons: simulations should be 10

run at different operating conditions (temperature, supply); parasitic effects layout dependent should be 11

accounted for in the simulation schematic; weaknesses in the component models should be masked by 12

robust design techniques. 13

Simulations are widely used in electronic industry because they allow to dramatically decrease the time 14

to market of any electronic product: circuit behavior can be predicted without any need to physically 15

implement and measure it; modifications to circuit behavior can also be predicted without experiments; 16

design alternatives can be evaluated without physically implementing any of them. This permits to 17

improve the design and sell electronic products minimizing time and the number of bugs. 18

Just as an example of simulation, we‟ll conclude this lesson by showing you how to use the CAD 19

LTspiceIV (downloadable free from the Internet: http://www.linear.com/designtools/software/) to 20

simulate an RC filter. Note that LtspiceIV is in English, like many other CAD tools. 21

The Figure below shows a typical simulation output obtained with LTspiceIV. 22

23

17

As expected, the simulation results are in perfect agreement with the calculated ones. The RC filter is a 1

simple circuit that can be easily solved by means of hand-calculations. However, there are electronics 2

circuits that cannot be solved by means of hand-calculations and simulation is the only instrument to 3

predict the behaviour of the circuit before its actual physical implementation and characterization. 4

5

6

18

1

Lesson 4: An introduction to active filters and simulation examples. (2 hrs) 2

3

Objectives 4

To know the main differences between active and passive filters. 5

To be able to draw the schematic and the frequency response of active RC low-pass and high-pass 6

filters. 7

To be able to draw the schematic of an RC band-pass filter and to explain how it works. 8

To be able to simulate active filters with CAD tools. 9

To know some practical filter applications. 10

To know the main difference between first and second order filters. 11

12

13

Figure 4.1 shows the schematic of a low pass filter connected to an RL load equal to 1K. 14

15 Fig. 4.1: low-pass filter schematic with an RL load on the output. 16

17

The RL load affects the transfer function of the filter since it is connected in parallel to the capacitor C. 18

Hence: 19

20

Eq. 4.1) 21 22

23

The maximum amplitude of the output voltage is less than 1 (in fact, it is 1/(1+R/RL)) and the cutoff 24

frequency of the filter is dependent on RL too: fC=(1+R/RL)/2. 25

In general, the frequency response of a passive filter is dependent on the load. Furthermore, the 26

maximum amplitude (or gain) of passive filters is 1 even when the load is a high impedance (i.e. 27

RL=∞). 28

Active filters provide: 29

1) a frequency response independent of any load. 30

2) a maximum amplitude that can be higher or lower than 1. 31

Active filters are obtained by simply adding an active component (typically an OP.AMP.=OPerational 32

AMPlifier) to a passive filter. 33

Figure 4.2 shows the schematic of an RC low-pass active filter: 34

C

1µ

R

159V1

SINE(0 1 1K)

RL

1K

OI

VO(jw)

VO=VOejo

VI(jw)

VI=VIeji

wjRL

RjwG

1

1)(

19

1 Fig. 4.2: first-order active RC low-pass filter schematic. OP05 is an OPAMP with dual supply 2

VCC=+15V and –VSS=-15V. R1 is only used to fix the maximum amplitude value. 3

4

The OPAMP of figure 4.2 is connected in inverting input configuration. Hence G(jw)=VO/VI=-Z2/Z1, 5

where Z2=R2//C=R2/(1+jw2) and Z1=R1. 2=R2*C is the time constant of the filter.By substituting 6

Z2 and Z1 into G(jw) expression we get: 7

8

Eq. 4.2) 9 10

G(jw) is independent of any RL load connected to VO thanks to the opamp low output resistance. Only 11

when RL becomes comparable to opamp output resistance (RL<100) G(jw) is affected by RL. 12

Equation 4.2 shows that maximum amplitude of the filter is R2/R1. So when R2>R1 the gain is higher 13

than 1, while if R2<R1 maximum gain will be lower than 1. Note also the minus sign above the transfer 14

function: this means that the output signal VO is inverted with respect to the input VI, or, equivalently, 15

that they are phase shifted of 1800. 16

Figure 4.3 shows the amplitude and phase diagrams of the low pass active filter designed using MS-17

excel, with the corresponding analytical expressions: 18

19 Fig. 4.3: frequency response of the active low pass filter of figure 4.2. Note that the low frequency 20

amplitude value of the filter is 20dB (because R2=10*R1) and fC=1KHz. 21

22

Figure 4.3 shows that the amplitude diagram of the active filter is the same as the equivalent passive 23

filter shifted up of 20dB (=20Log(R2/R1)). Also, the phase diagram is the same as the equivalent 24

V1

-15

V2

15

R1

159

R2 1.59K

C 100n

V3

SINE(0 1 1K)

U1

OP05

VCC -VSS

-VS

SV

CC

VOVI

VI

.tran 4m

.lib ./lib/sub/LTC.lib

Amplitude diagram in dB (Bode Plot)

-40

-30

-20

-10

0

10

201 1

0

10

0

10

00

10

00

0

1E

+0

5

1E

+0

6frequency

|G(j

w)|

[d

B]

Phase diagram

90

105

120

135

150

165

180

1 10

10

0

10

00

10

00

0

1E

+0

5

1E

+0

6frequency

G(j

w)

))(1

1(20)

1

2(20][)(

2

2

wLog

R

RLogdBjwG )(1800

)( wartgjwG

fC=1KHz

|G(fC)|=17dB G(fC)=1350

21

1

1

2)(

wjR

RjwG

20

passive filter shifted up of 1800 (to take into account the inversion of the output with respect to the 1

input). 2

The cutoff frequency fC of the filter is defined as the frequency value at which the amplitude is 3dB 3

lower than its low frequency value, i.e. |G(fC)|=|G(0)|-3dB. Also it is: G(fC)=G(0)-450. fC can be 4

calculated analytically by using the formula: 5

6

Eq. 4.3) , where 2=R2*C is the time constant of the filter. 7

8

The first-order high-pass active RC filter is shown in figure 4.4: 9

10 Fig. 4.4: first-order active RC high-pass filter schematic. R2 is only used to fix the maximum 11

amplitude value. 12

13

Its frequency response is shown in figure 4.5 with the analytical expressions of amplitude and phase: 14

15 Fig. 4.5: frequency response of the active high pass filter of figure 4.4. Note that the high frequency 16

amplitude value of the filter is 20dB because R2=10*R1 and fC=1KHz. 17

18

fC can be determined from the plots or analytically using the expression: 19

20

Eq. 4.4) , where 1=R1*C is the time constant of the filter. 21

22

In certain applications we‟re interested in selecting only the input frequency components included 23

within a given frequency band B=[f1,f2]. This problem can be easily solved using the cascade of a low-24

pass and a high-pass filters with cutoff frequencies f2 (low pass) > f1 (high pass), as shown in figure 25

4.6: 26

V1

-15

V2

15

R1

159

R2 1.59K

C 1µ

V3

SINE(0 1 1K)

U1

OP05

VCC -VSS-V

SS

VC

C

VOVI

VI

.tran 4m

.lib ./lib/sub/LTC.libPhase diagram

180

195

210

225

240

255

270

0,1

1 10

100

1000

10000

100000

1000000frequency

G(j

w)

Amplitude diagram in dB (Bode Plot)

-60

-50

-40

-30

-20

-10

0

10

20

0,1

1 10

100

1000

10000

100000

1000000frequency

|G(j

w)|

[d

B]

fC=1KHz

|G(fC)|=17dB

G(fC)=2250

))(1

(20)1

2(20][)(

2

1

1

w

wLog

R

RLogdBjwG )(270 1

0

)( wartgjwG

22

1

Cf

12

1

Cf

21

1 Fig. 4.6: the simplest band-pass filter is the cascade of a low-pass and a high-pass active filter. Whether 2

the low-pass comes first or second in the cascade does not matter since both filters are active. Only the 3

fact that fC2<fC1 is important. The filter pass-band is B=[fC2,fC1]. 4

5

An active band-pass filter can be designed as the cascade of the schematic 4.2 and 4.4. However, if we 6

want to spare an OPAMP we may use the active RC band-pass schematic of figure 4.7: 7

8 Fig. 4.7: second-order active RC band-pass filter schematic. 2=R2*C2 should be < 1=R1*C1 to 9

maximize amplitude in the pass-band. 10

11

The frequency response of this filter is shown in figure 4.8: 12

13 Fig. 4.8: frequency response of the active band-pass filter of figure 4.7. The amplitude diagram is 14

exactly the sum of the Bode plots of a low-pass and a high-pass filter with the same time constants. 15

Looking at the amplitude analytical expression in figure 4.8, we understand it is the sum of an active 16

high pass-filter with gain R2/R1 and time constant 1=R1*C1 and a passive low-pass filter with time 17

constant 2=R2*C2. The phase plot is the sum of the phase plots of the two single filters. Hence, the 18

schematic of figure 4.7 represents the most compact RC band-pass active filter we may use. 19

V1

-15

V2

15

R1

1K

R2 1.59K

C130n

V3

SINE(0 1 1K)

U1

OP05

C2 6n

VCC -VSS

-VS

SV

CC

VOVI

VI

.tran 4m

.lib ./lib/sub/LTC.lib

fC1 INPUT SIGNAL

fC2 OUTPUT SIGNAL

frequencies above fC1 are attenuated

frequencies above fC1 and below fC2 are attenuated

low-pass with fC=fC1

high-pass with fC=fC2

Phase diagram

90105120135150165180195210225240255270

1 10

10

0

10

00

10

00

0

10

00

00

10

00

00

0frequency

G(j

w)

Amplitude diagram in dB (Bode Plot)

-50

-40

-30

-20

-10

0

10

10

10

0

10

00

10

00

0

10

00

00

10

00

00

0

1E

+0

7

frequency

|G(j

w)|

[d

B]

fC1=5,3K

G(fC1)=2050

))(1

1(20)

)(1(20)

1

2(20][)(

2

2

2

1

1

wLog

w

wLog

R

RLogdBjwG

)()(270 21

0

)( wartgwartgjwG

fC2=16,7K G(fC2)=1550

22

A band-pass filter provides an interesting example of circuit to be studied with the aid of simulation. A 1

particularly interesting case is that of a square wave input with frequency in the middle of the pass-2

band B of the filter. What is the filter output to the square wave stimulus? 3

The analytical approach is too difficult. The graphical approach is practicable but still difficult. In fact, 4

we need to find all harmonic components of the input square wave. Then we need to find the output 5

corresponding to each harmonic component; finally we need to sum all the output to find the final 6

output waveform. Simulation, instead, provides us with the straightforward solution for each kind of 7

filter. 8

9

10 11

We have simulated two different band-pass filters with the same input square waveform. The output of 12

the two filters is different because their frequency response is different. The active filter has a wider 13

pass-band: so the fundamental harmonic “passes”, but also the third harmonic and higher order 14

harmonics “pass”, even if attenuated. The final output result is the blue waveform that is still the 15

combination of many harmonic components. The passive series-resonant filter has a very narrow band: 16

only the fundamental harmonic “passes”, while the third harmonic and all higher order harmonics are 17

“cut”. Hence, the final output result is the red waveform that is sinusoidal with frequency equal to the 18

input square wave: in practice a highly selective band-pass filter is able to select the fundamental 19

harmonic of the input square waveform. 20

21

What are the practical applications of filters? They are actually used in each electronic circuit; however 22

we will name only two applications we just find in our houses: the ADSL and the Hi-Fi (High-Fidelity) 23

system. 24

If you have an ADSL Internet connection, you realize that you can browse the Internet while 25

simultaneously some other members of your family are speaking on the phone. That‟s not so trivial 26

since the two signals (the ADSL and the phone signal) share the same cable. How is it possible they do 27

not disturb each other? 28

The reason is that the phone signal is in the frequency range [0,8KHz] while the ADSL signal is above 29

26KHz. In particular, a low pass filter placed before the phone receiver is used to make sure all 30

23

frequency components higher than 8KHz are “cut” and do not disturb the phone communication as 1

high frequency noises. On the other hand, a high-pass filter is embedded in the ADSL modem to make 2

sure all frequency components lower than 26KHz do not disturb the ADSL modulation/demodulation 3

and, hence, the Internet communication. Figure 4.9 shows the schematic representation of the filters 4

used in ADSL: 5

6 Fig. 4.9: A) schematic representation of the low-pass and high-pass filters used to avoid disturb when 7

browsing the Internet while speaking on the phone. B) example of low-pass filter actually placed before 8

the phone receiver. 9

10

Filters are also used in any Hi-Fi system. In fact, the minimum configuration for any Hi-Fi system is 11

two speakers, one woofer to reproduce the bass sounds and a tweeter to reproduce the high sounds. 12

Audio signal high frequency components are “cut” by a low-pass cross-over filter before they reach the 13

woofer. On the contrary, audio signal low frequency components are “cut” by a high-pass cross-over 14

filter before they reach the tweeter. So both speakers can work fine at the maximum power for which 15

they have been desinged. Figure 4.10 shows the simplified schematic of this simple audio system. 16

17 Fig. 4.10: simplified schematic of a simple audio system made of two speakers: a woofer and a tweeter, 18

each with the corresponding cross-over filter. 19

20

audio signal

[0,20KHz]

cross-over

low-pass

[0,3KHz]

cross-over

high-pass

[3K,20K]

woofer (bass)

tweeter (high)

A)

B) example of sixth order low-pass filter

placed before the phone receiver to avoid

high frequency noise present in the phone

line.

24

As in the case of the ADSL modem, also Hi-Fi cross-over filters are not first-order. This is because 1

first-order filters are not sufficiently “selective”: frequency components close to the pass-band are not 2

sufficiently attenuated. Figure 4.11 shows the simplified Bode plots of 1st, 2

nd and 3

rd order low-pass 3

filters. When f=10*fC the first order filter offers only an attenuaton of a factor 0.1 (-20dB). You need a 4

second order filter to have an attenuation factor of 0.01 (-40dB) and a third order filter to reach a 0.001 5

attenuation factor. Finally a higher order filter is more able to select only the frequencies within its 6

pass-band only high-order filters are usually used in practical applications. 7

8 Fig. 4.11: simplified Bode plots of low-pass filters of different orders but with the same pass-band. 9

10

|G(jw)| [dB]

f (log)

fC 10fC 100fC 1000fC 0dB

-20dB

-40dB

-60dB

1st order: -20dB/dec

2nd

order: -40dB/dec

3rd

order: -60dB/dec

Note: the order of a

filter is equal to the

number of reactive

components in its

schematic (i.e. the

number of poles of its

transfer function).

25

BIBLIOGRAPHY

1. The Analysis and Design of Linear Circuits, 6th Edition, R. E. Thomas, A. J. Rosa, G. J.

Toussaint, ISBN: 978-0-470-38330-8, 2009.

(basic textbook on linear circuits analysis with a complete chapter on AC analysis and

Bode plots of analog filters).

2. http://www.animations.physics.unsw.edu.au/jw/RCfilters.html

(basic analysis of RC filters with animations).

3. http://www.play-hookey.com/ac_theory/filter_basics.html

(basic theory of active and passive filters).

4. http://www.educypedia.be/electronics/analogfil.htm

(link to several resources on analog filters, both passive and active).

5. http://www.usna.edu/Users/ee/kintzley/ee303sp09/lectures/EE303Sp09_L05_Filters.pdf

(lecture on basic filter topologies and solutions).

6. http://dev.emcelettronica.com/performing-ac-analysis-ltspice

(method to perform an AC analysis of an RC filter with LTspice)

7. http://www.swarthmore.edu/NatSci/echeeve1/Ref/DataSheet/IntroToFilters.pdf

(basic introduction to analog filters from National Semiconductor Corp., with examples of

real analog filters IC's and criteria to select among different filter topologies).

8. Active Filter Cookbook, D. Lancaster, 2nd ed. Thatcher, Synergetics Press, 1995

(a comprehensive textbook on active filter design and analysis).

9. http://www.wisc-online.com/objects/index_tj.asp?objID=ACE2803

(analysis of the Bode plots of an RC low pass filter).

10. http://www.che.ttu.edu/pcoc/software/ppt/Chap09.ppt

(chapter on Bode plot analysis of linear systems with stability criteria).

11. http://eprints.iisc.ernet.in/13500/1/lec_5_web.pdf

(lecture on Bode plots and second order linear systems).