Filters2

filters of vechicals

1. Active Filters

2. Filters

o A filteris a system that processes a signal
in some desired fashion.
A continuous-time signal or continuous signal of

X(t) is a function of the continuous variable t. A
continuous-time signal is often called an analog
signal.

A discrete-time signal or discrete signal x(kT) is
defined only at discrete instances t=kT, where k
is an integer and Tis the uniform spacing or
period between samples

3. Types of Filters

e There are two broad categories of filters:
An analog filter processes continuous-time signals
A digital filter processes discrete-time signals.

e The analog or digital filters can be subdivided
into four categories:
Lowpass Filters
Highpass Filters
Bandstop Filters
Bandpass Filters

4. Analog Filter Responses

ana y ADAM Practical filter

5. Ideal Filters
Owpass 1lter ignpass 1lter

Stopband Stopband | Passband

6. + There are a number of ways to build
filters and of these passive and active
filters are the most commonly used in

voice and data communications.

7. Passive filters

o

o

Passive filters use resistors, capacitors, and
inductors (RLC networks).

To minimize distortion in the filter

characteristic, it is desirable to use inductors
with high quality factors (remember the model
of a practical inductor includes a series
resistance), however these are difficult to
implement at frequencies below 1 kHz.

They are particularly non-ideal (lossy)

They are bulky and expensive

8. + Active filters overcome these drawbacks
and are realized using resistors,
capacitors, and active devices (usually

op-amps) which can all be integrated:

Active filters replace inductors using op-amp
based equivalent circuits.

9. Op Amp Advantages

e Advantages of active RC filters include:
reduced size and weight, and therefore parasitics
increased reliability and improved performance
simpler design than for passive filters and can realize
a wider range of functions as well as providing voltage
gain
in large quantities, the cost of an IC is less than its
passive counterpart

10. Op Amp Disadvantages

e Active RC filters also have some disadvantages:
limited bandwidth of active devices limits the highest
attainable pole frequency and therefore applications above
100 kHz (passive RLCfilters can be used up to 500 MHz)
the achievable quality factor is also limited
require power supplies (unlike passive filters)
increased sensitivity to variations in circuit parameters
e by environmental changes compared to passive
ilters
+ For many applications, particularly in voice and data
communications, the economic and performance
advantages of active RC filters far outweigh their
disadvantages.

11. Bode Plots

e Bode plots are important when
considering the frequency response
characteristics of amplifiers. They plot

the magnitude or phase of a transfer
function in dB versus frequency.

12. The decibel (dB)

Two levels of power can be compared using a
unit of measure called the bel.

P
B=log,,—
#10 p

The decibel is defined as:

1 bel = 10 decibels (dB)

13. dB =10log,, —

A common dB term is the half power point
which is the dB value when the P, is one-
half P..

10log 5 =-3.01dB = -3dB

14. Logarithms

e A logarithm is a linear transformation used
to simplify mathematical and graphical
operations.

+ A logarithm is a one-to-one correspondence.

15. Any number (N) can be represented as a
base number (b) raised to a power (x).

N =(b)*

The value power (x) can be determined by
taking the logarithm of the number (N) to
base (b).

x=log, N

16. e Although there is no limitation on the
numerical value of the base, calculators are
designed to handle either base 10 (the
common logarithm) or base e (the natural
logarithm).

e Any base can be found in terms of the
common logarithm by:

1
log, w= —— log y w
logio 4

17. Properties of Logarithms

e The common or natural + The log of the quotient
logarithm of the number of two numbers is the
lis 0. log of the numerator
The log of any number minus the denominator.
less than 1 is a negative The log a number
number. taken to a power is
The log of the product of equal to the product of
two numbers is the sum the power and the log
of the logs of the of the number.
numbers.

18. Poles & Zeros of the transfer
function

+ pole value of s where the denominator
goes to zero.

+ zero value of s where the numerator

goes to zero.

19. Single-Pole Passive Filter

Vout Zo 1/sC

v, R+Z. R+i/sC

R
Vin | + C | Vout 1 1/RC

~ sCR+1 s+1/RC

e First order low pass filter
e Cut-off frequency = 1/RC rad/s

e Problem : Any load (or source)
impedance will change frequency
response.

20. Single-Pole Active Filter

R
m| C | | ,
out

+ Same frequency response as passive
filter.

+ Buffer amplifier does not load RC
network.

e Output impedance is now zero.

21. Low-Pass and High-Pass

Designs

High Pass Low Pass

1
“TT T+sRC
— +] y
sCR sCR E

1/RC
sRC 5 Vi, S+I/RC
TRC(S+I/RC) (s+1/RC)

22. To understand Bode plots, you need to
use Laplace transforms!

The transfer function
of the circuit is: V,,(8)

V,(s) 1/sC 1
V.(s) R+l/sC sRC+l

in

A =

23. Break Frequencies

Replace s with jo in the transfer function:

1 1 1
j@RC +1 1+ j2aRG
JORC+ j2aRCf ft

Af)

where f, is called the break frequency, or corner
frequency, and is given by:

1

~ 2nRC

Fe

24. Corner Frequency

The significance of the break frequency is that
it represents the frequency where

Af) = 0.7072-45°.
This is where the output of the transfer function
has an amplitude 3-dB below the input
amplitude, and the output phase is shifted by
-45° relative to the input.
Therefore, f, is also known as the 3-dB
frequency or the corner frequency.

25. 4 PERFORMANCE CRITERIA

AMPLITUDE RESPONSE

jay Pass band 20 logy, JA] = Gain in dB

ripple

348 point 1¿= Cut-off Frequency

JAI
Gain at 348 point (atf,)= —=

Va

stop band ripple

ph e——

pass band stop band

@ RIPPLE IN PASS BAND CAUSES NON-LINEARITY
® POSSIBLE TO DESIGN WITH NO RIPPLE
e RIPPLE IN STOP BAND IS LESS IMPORTANT

e FALL OFF dB / Decade (Gain in dB / Decade of f}

A

STOP BAND ATTENUATES (SAY - 40dB}

26. Bode plots use a logarithmic scale for
frequency.
One decade

+ tt

10 20 30 40 50 60 70 80 90 100 200

where a decade is defined as a range of
frequencies where the highest and lowest
frequencies differ by a factor of 10.

27. e Consider the magnitude of the transfer
function:

1
AN ==
ku Vi+(f/ 5)

Expressed in dB, the expression is

|A,(f)|,, = 20log1—20log J1+(f/ £, Y
= 20108 1+(£/f,Y =-1010ef +(f/f,Ÿ |

=-20l0g(f / f,)

28. e Look how the previous expression
changes with frequency:
at low frequencies f<< f,, |Av/yg = 0 dB

low frequency asymptote
at high frequencies f>>f,,
|AV(/ag = -20log f/f,
high frequency asymptote

29. | RE weonitude |
O EEE ee man m =

LL TENTE TITN
0.1 1 10 100

20-log(| PCo ) |)
-40

Ta ye

se

30. e The technique for approximating a filter
function based on Bode plots is useful
for low order, simple filter designs

e More complex filter characteristics are
more easily approximated by using some
well-described rational functions, the
roots of which have already been
tabulated and are well-known.

31. Real Filters

e The approximations to the ideal filter are
the:
Butterworth filter

Chebyshev filter
Cauer (Elliptic) filter
Bessel filter

32. Standard Transfer Functions
e Butterworth
Flat Pass-band.
20n GB per decade roll-off.
Chebyshev

Pass-band ripple.
Sharper cut-off than Butterworth.
Elliptic
Pass-band and stop-band ripple.
Even sharper cut-off.
Bessel

Linear phase response __ i.e. no signal distortion in
pass-band.

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34. Butterworth Filter

The Butterworth filter magnitude is defined by:

M(o)=|H(jo) =

1
(+0)

where n is the order of the filter.

35. From the previous slide:

M(0)=1

1
M (1) = Y for all values of n

For large 00:

1
MOS

36. And

20log,, M(@) = 20log,, 1— 20 log, a”

=-20nlog,,®

implying the M(o) falls off at 20n db/decade for large values
of a.

37. 10

40 ib/des

ide

10

38. To obtain the transfer function H(s) from the magnitude
response, note that

M°(o)=|H(jo) = H(jo)H(-jo) =

1+(0?)

39. Because s =j@ for the frequency response, we have s? = — 0”.

1 1

H(sSH(=s)=

1+(-s2) REN

The poles of this function are given by the roots of

141) 5% =-1=6 10407, &=1,2,...

40. The 2n pole are:

| ejl(2k-1/2n1% y even, k = 1,2,...,2n
ge

estr n odd, k=0,1,2,...,2n-1

Note that for any n, the poles of the normalized Butterworth
filter lie on the unit circle in the s-plane. The left half-plane
poles are identified with H(s). The poles associated with
H(-s) are mirror images.

42. =
3
=
oath

43. Recall from complex numbers that the rectangular form
of a complex can be represented as:

Z=x+ jy

Recalling that the previous equation is a phasor, we can
represent the previous equation in polar form:

z=r(cos6+ ¡sinQ)

where

x=rcos@ and y=rsin®

44. Definition: If z = x + jy, we define e* = e**” to be the
complex number

e* =e* (cos y+ ¡sin y)
Note: When z = 0 + jy, we have

e” =(cos y+ ¡sin y)

which we can represent by symbol:

jo
e?

45. el =(cos@ + jsin 6)

Note that

cos(-#)=cos(-0) even function
sin(-@)=—sin@) odd function

46. This implies that

e =(cos0— jsin@)

This leads to two axioms:

nn jo
ef +e”} e” —e

cos @ = and Sing = |
2 2j

47. + Observe that ef represents a unit vector
which makes an angle 6 with the
positivie x axis.

48. Find the transfer function that corresponds to a third-order
(n = 3) Butterworth filter.

Solution:

From the previous discussion:

5, = el, k=0,1,2,3,4,5

49. Therefore,
So

50. The roots are:

P, -1 Pe --1

p, --5+0.8668) ps --.5- 0.866

p, -.5- 0.866) Py --.5+ 0.866)

52. Using the left half-plane poles for H(s), we get

H(s) !

~ (s+ D(s+1/2— jv3/ 2541/24 jN312)

which can be expanded to:

1
~ (s+1(s? +541)

H(s)

53. + The factored form of the normalized
Butterworth polynomials for various
order n are tabulated in filter design

tables.

54. Denominator of H(s) for Butterworth Filter

s+l

s?+ 1.41458 +1

(s?+s+1)(s+1)

(s2 + 0.765 + 1)(s? + 1.848s + 1)

(s + 1) (s? + 0.618s + 1)(s? + 1.6185 + 1)

(s? + 0.517s + 1)(s? + 1.4145 + 1 )(s? + 1.932s + 1)

(s + 1)(s? + 0.4455 + 1)(s? + 1.2475 + 1 )(s? + 1.802s + 1)

n
1
2
3
4
5
6
2
8

(s? + 0.390s + 1)(s? + 1.111s + 1 )(s? + 1.6635 + 1 )(s? + 1.962s + 1)

55. Frequency Transformations

56. So far we have looked at the Butterworth filter
with a normalized cutoff frequency

©, = 1 rad / sec

By means of a frequency transformation, we
can obtain a lowpass, bandpass, bandstop, or
highpass filter with specific cutoff frequencies.

57. Lowpass with Cutoff Frequency
(a)

u

e Transformation:

s,=s/®,

58. Highpass with Cutoff Frequency
0;

e Transformation:

s, =@,1S

59. Bandpass with Cutoff
Frequencies o and o,

e Transformation:

2 2
LS +05 _% o, |

O o 0

Do = yO, O;

B=0,-0,

u

60. Bandstop with Cutoff
Frequencies o and o,

e Transformation:

Bs B
S = =
Ms +05 ot
meee)
o Ss