CT Filters - Passive

8/17/2019 CT Filters - Passive

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1/18

EECE 301Signals & SystemsProf. Mark Fowler

Note Set #35

• C-T Systems: CT Filters - Passive


2/18

2/18

Introductory Comments

Recall that we already talked about ideal CT filters:• |H()| is Constant in Pass band • |H()| is zero in Stop band (Transition Band has zero width)• H() is linear in Pass band

We also saw that such ideal filters can not really exist because they would needto be non-causal!!

Here we’ll take a brief look at some of the kinds of CT filters that can be made…• Note… all CT filter behavior exploits the fact that capacitors and

inductors have an impedance that varies with frequency!

And we’ll illustrate how to describe such filters using:• Transfer Function• Frequency Response

• Pole-Zero Diagrams

Also… keep in mind that although DT filters only need to be examined over – to rad/sample (their Freq Resp repeats outside of that)… CT filters needto be examined for how they behave over – to rad/second . Thus, we will

mostly plot them on a log frequency axis… with dB for the magnitude.

“CT Filters” are also

called “Analog Filters”


3/18

Practical Filter Specification

LPF Spec – Version 1

To make filter “more ideal”: p 0, s 0, s p

3/18Pass-band Stop-bandTransition

Specs for HPF, BPF, & BSF are similar…

Unlike for DT… for CT we

need look all the way up to


4/18

| H ( )|

0 dB

c

–3 dB

“Decade” = 10x Change

dB

LPF Spec – Version 2

4/18Pass-band

Specs for HPF, BPF, & BSF are similar…

Passband cutoff frequency c is

defined at the “–3 dB point”.

If passband is at other than 0 dB

the cutoff is at “3 dB down” from

the passband level.

Log scale

Filter Specs• Cutoff Freq @ “–3 dB point”

• dB per Decade (Rolloff Rate)


5/18

5/18

CT Filter Types

Recall that DT filters were categorized as recursive (IIR) vs. non-recursive (FIR).

CT filters don’t have a corresponding categorization… they all have infiniteduration impulse responses!!!

Instead the main way to categorize CT filters is: Passive vs. Active

Passive: These filters use only “passive components” (resistors, capacitors, andinductors) and do not contain any op amps or transistors.

• One main advantage of such filters is that they can be used in places

where access to a power supply is not available (e.g., inside a stereospeaker to separate the audio into bass and treble before sending it to thewoofer & tweeter).

Active: These filters use op amps (and/or transistors) together with resistors,capacitors, and inductors.• Allows filters to be designed without inductors• Op amp characteristics enable design by cascading several “stages”

Heavy, Bulky,

Expensive

•

Large Input Impedance• Small Output Impedance


6/18

6/18

“First-Order” Lowpass Filter: RC Circuit

We already analyzed this

filter using phasor ideas…

but we’ll take another look

here.

R

1/Cs

( )Y s

( ) X s

To analyze this filter in the s-domain:• Replace input and output by their LT symbols• Replace components by their s-domain impedances• Solve for output Y (s) in terms of input X (s)… the thing that multiplies X (s)

is the TF H (s)

By voltage divider (the best approach here) we get this

1( ) ( )

1

CsY s X s

Cs R

1( )1

Cs H sCs R

1 1( )

1 1

RC H s

RCs s RC

1 Pole @s = -1/ RC j

1 RC

RC RC

1st Order

1( ) 1

RC H s s RC


7/18

7/18

1 1( )

1 1

RC H s

RCs s RC

j

1 c RC

RC

RC

1( )

1 H

jRC

2 2

1( )

1 H

RC

When =1 then

Magn is 1 2 ( 3dB)

RC

So… c = 1/ RC ( ) c

c

H ss

j

100

j

1000

101

102

103

104

105-60

-50

-40

-30

-20

-10

0

10

CT Frequency (rad/sec)

| H ( ) | ( d B

)

100 rad/secc

1000 rad/secc

>> w=logspace(1,5,1000);>> wc=100;H=freqs(wc,[1 wc],w);>> semilogx(w,20*log10(abs(H)))


8/18

8/18

“First-Order” Highpass Filter: RC Circuit

1/Cs

R

( )Y s( ) X s

By voltage divider (the best approach here) we get this

( ) ( )1

RY s X s

Cs R

( )1

R H s

Cs R

( )1 1

RCs s H s

RCs s RC

1 Pole @

s = -1/ RC

1 Zero @

s = 0 j

1 RC

RC RC

1st Order

( )

1

s H s

s RC


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9/18

( )1 1

RCs s H s

RCs s RC

( )

1

jRC H

jRC

2 2

( )1

RC H

RC

When =1 then

Magn is 1 2 ( 3dB)

RC

So… c = 1/ RC ( )c

s H ss

j

1 c RC

RC

RC

j

10,000

j

1000>> w=logspace(1,5,1000);>> wc=1000;H=freqs([1 0],[1 wc],w);>> semilogx(w,20*log10(abs(H)))

101

102

103

104

105-60

-50

-40

-30

-20

-10

0

10


| H ( ) | ( d B

)

10,000 rad/secc

1000 rad/secc


10/18

10/18

Lowpass Filter Highpass Filter

1( )

1 H

jRC

( ) c

c

H ss

( )

c

s H s

s

( )

1

jRC H

jRC

At high freqs… C is like a short…

Stops high frequencies!!

At low freqs… C is like an open…

Stops low frequencies!!

j

1 c RC

RC RC

j

1 c RC

RC RC

101

102

103

104

105-60

-50

-40

-30

-20

-100

10


| H ( ) | ( d B )

101

102

103

104

105-60

-50

-40

-30

-20

-100

10


| H ( ) | ( d B )


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11/18

A “Second-Order” Lowpass Filter: RLC Circuit

By voltage divider (the best approach here) we get this:

1( )

1

Cs H s

Cs Ls R

21

( )1

LC H s

s R L s LC

2 Poles

(3 Possible Ways)2nd Order

1

( ) ( )1

Cs

Y s X sCs Ls R

j

Distinct “Real” Poles

j

2 poles

Repeated “Real” Poles

j

Complex-Conjugate Poles

1n

LC

Natural Freq.

2

2 2( )

2n

n n

H ss s

2

R

L

Damping Ratio


12/18

12/18

2

1,2 2 2 1 p R L R L LC The poles are the roots of s2 + ( R / L) s + 1/ LC :

2R 1

2L LC

Complex Roots

1,2 02

R

p j L

2

0 2

1

L

R

LC

j

2

2

1

L

R

LC

2

2

1

L

R

LC

L R 2/

20

L R

LC

When R = 0, poles

are on j

axis

0 1

2R 1

2L LC

Repeated Real Roots

1,22

R p

L

2 L R

LC

j

(2 poles)1

2R 12L LC

Distinct Real Roots

2 L R LC

j

1


13/18

13/18

j

R = 0

R = 0

LC

L R

2

Pole Positions as R is Varied

Repeated

Roots

LC

L R

20

LC

L R

20

LC

L R

2

LC L R 2

0 1

0

0

0 1

1

1

1


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14/18

-3 -2 -1 0 1

x 104

-5

-4

-3

-2

-1

0

1

2

3

4

5

x 104

101

102

103

104

105

-60

-40

-20

0

20


| H ( )

| ( d B )

–500 + j49,749

–500 – j49,749

Freq Resp Magnitude for Three Cases

Complex Poles (1)

•

Two Breaks @ “Poles”• -20 dB/dec then -40 dB/dec

–5000

–5000

–858

–29,142

2nd Order has faster rolloff vs 1st Order

(-40 dB/dec vs. -20 dB/dec)


15/18

15/18

A “Second-Order” Highpass Filter: RLC Circuit


( )1

Ls H s

Cs Ls R

2

2( )

1

s H s

s R L s LC

2 Poles

(3 Possible Ways)2nd Order

( ) ( )1

Ls

Y s X sCs Ls R

1n

LC

Natural Freq.

2

2 2( )

2n n

s H s

s s

2

R

L

Damping Ratio

2 Zeros @ Origin

Same

Denominator!!

j


2

j

2 poles


2

j


2


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16/18

101

102

103

104

105

-100

-80

-60

-40

-20

0

20


| H (

) | ( d B )

11000 rad/sec

n LC

–172–5828

0.1

1 3

–1000–1000

–100+ j995

–100- j995

2nd Order has faster rolloff vs 1st Order

(-40 dB/dec vs. -20 dB/dec)

What we are seeing is that we get 20 dB of slope for each order!!!

(For LPF and HPF… But see next for BPF…)

So a 3rd Order LPF would (eventually) rolloff at -60 dB/decade!!!

So… the main advantage of higher order filters is that your stop

band is better due to the faster rolloff!!


17/18

17/18

A “Second-Order” Bandpass Filter: RLC Circuit


( )1

R H s

Cs Ls R

2

( )1

R L s H s

s R L s LC

2 Poles(3 Possible Ways)

2nd Order

( ) ( )1

R

Y s X sCs Ls R

1n

LC

Natural Freq.

2 2

2( )

2n

n n

s H s

s s

2

R

L

Damping Ratio

1 Zero @ Origin

Same

Denominator!!

j


j

2 poles


j


v(t ) RC y(t ) Output SignalInput Signal

L


18/18

18/18

101

102

103

104

105

-60

-50

-40

-30

-20

-10

0

10


| H ( ) | ( d B )

11000 rad/sec

n LC

–172

–5828

0.1

1

3

–1000

–1000

–100+ j995–100- j995

Note: All slopes are 20 dB/decade!

So… unlike for LPF & HPF… 2nd Order

BPF does NOT have the faster rolloff…

But, 1st

Order can’t even GIVE a BPF!!!

What is happening is that the second order gives you two 20 dB/dec

slopes “available”…

But for a BPF you need one going up and one going down… so

each only gets one of the two 20 dB slopes!

CT Filters - Passive

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