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EECE 301

Signals & Systems

Prof. Mark Fowler 

Note Set #15

• C-T Systems: CT Filters & Frequency Response

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Ideal Filters

Often we have a scenario where part of the input signal’s spectrum comprises“what we want” and part comprises something we “do not want”. We can use a

filter to remove (or filter out) the “bad part”.

 H ( )

( ) x t  ( ) y t 

Called “a Filter” in this case

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Case #1: )(  X 

   

Spectrum of the

Input Signal

In this case, we want

a filter like this:

)(  H 

 

 

otherwise H  ,0

,1

)(

  Mathematically:

“Passband”

“Stopband”

A filter that

“passes”

low

frequencies

is called a“low-pass

filter”

Undesired

Part

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Case #2:

 

otherwise H  ,1

,0

)(

 

 “Passband”

“Stopband”

)(  X 

   

Input SignalSpectrum

Undesired

Part

We then want:)(  H 

 

1

A filter that

“passes”

high

frequencies

is called a

“high-pass

filter”

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( ) X    

 

)(  H 

 

1

Case # 3:

A filter that “stops” middle

frequencies is called a

“band-stop filter”

Undesired Part

Case #4:

A filter that “passes” middle

frequencies is called a“band-pass filter”

( ) X    

 

)( 

 H 

 1

Desired Part

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What about the phase of an IDEAL filter’s  H ( )?

Well…we could tolerate a small delay in the output so…

From the time-shift property of the FT then we need:

d t  jg   e X Y 

        )()(

 H ( )

)(t  xg )()( d g   t t  xt  y   Put in the signal

we want “passed”

Want to get out the

signal we want“passed”… but we can

accept a “small” delay

d t  j

t  j

t e H 

e H 

d 

d 

  

 

 

 

)(

1)( For in the“pass band”

of the filter 

Thus we should treat the exponential term here as  H ( ), so we have:

Line of slope – t d 

“Linear Phase”5/14

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So… for an ideal low-pass filter (LPF) we have:

 

otherwise

e H 

d t  j

,0

,1)( 

 

 

)(  H 

 

Slope = d t 

)(  H 

 

Phase is undefined in stop band:

?0

00

    je

i.e. phase is undefined

for frequencies outside

the ideal passband

Summary of Ideal Filters

1. Magnitude Response:

a. Constant in Passband

 b. Zero in Stopband

2. Phase Response

a. Linear in Passband (negative slope = delay)

 b. Undefined in Stopband

 p2( )

2( ) ( )  d  j t  H p e   

  

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)32cos()22cos(3)2cos(59)(   t t t t  x        

 H ( )

0 Hz 1 Hz 2 Hz 3 Hz

Filter has

 Non-Linear Phase

Filter has FLAT

Passband

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)10

32cos()22cos(3)4

2cos(59)(   

    

      t t t t  y

Point of this Example

A filter with an ideal magnitude response but non-

ideal phase response can still degrade a signal!!!

Example of the effect of a nonlinear phase but an ideal magnitude

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Are Ideal Filters Realizable? (i.e., can we actually MAKE one?)Sadly… No!!

So… a big part of CT filter design focuses on how to get close to the ideal.

Can’t Get an Ideal Filter… Because they are Non-Causal!!!

For the ideal LPF we had2( ) ( )

  d  j t  H p e   

  

2 sinc 2 / 2t      From the FT Table: 22 ( ) p  

 Now consider applying a delta function as its input:  x(t ) = (t )   X ( ) = 1

Then the output has FT 2( ) ( ) ( ) ( )  d  j t Y X H p e   

  

Linear Phase

Imparts Delay

Ideal

LPFt 

 x(t ) =(t )   y(t )

t 

Starts before input starts…

Thus, system is non-causal!8/14

So the response to a delta (applied at t = 0) is:   ( ) ( / ) sinc ( / )( )d  y t t t   

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a logarithmic unit of measure for a ratio between two powersDecibel:

   

decibel

 bel

2

110log10

 

  

 P

P

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Plotting Frequency Response of Practical Filters

Although we’ve previously shown the plots of Freq. Resp. using the actualnumerical values of | H ( )| it is VERY common to plot its decibel values.

P1/P2(non-dB)

P1/P2(dB)

1000 = 103 30 dB

100 = 102 20 dB

10 = 101 10 dB

1 = 100 0 dB

0.1 = 10-1  –10 dB

0.01 = 10-2  –20 dB

0.001 = 10-3  –30 dB

P1/P2 = 2   ~ 3 dB

Another “Rule” to Know!!

P1/P2 = 1/2   ~ -3 dB

0 dB is “P ratio of 1”

10 dB is “P ratio of 10”

20 dB is “P ratio of 100”

30 dB is “P ratio of 1000”

-10 dB is “P ratio of 0.1”

-20 dB is “P ratio of 0.01”

-30 dB is “P ratio of 0.001”

Decibel Power Rules

Know

These!

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But… | H ( )| relates Voltages (or current)… not POWER!!!

h(t)

 H( )

)cos( 0        t  A ))(cos()( 000          H t  H  A  

Output voltage amplitude =  A| H ( 

0)|

AmplitudeVoltageInput

AmplitudeVoltageOutput|)(| 0     H 

Input voltage amplitude =  A

Convert to PowersInput Power =  A2/2

Output Power =  A2|H(0)|2/2

22

10 10 2

2

10

10

( ) / 210 log 10 log

/ 2

10 log ( )

20 log ( )

out 

in

 A H P

P A

 H 

 H 

 

 

 

   

20 log10

(| H ( )|)Decibel value for | H (

 0)|

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In addition to using decibels for the | H ( )| it is also common to use a

logarithmic scale for the frequency axis

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24

68 20

4060

80

10 100

200

We may be just as interested in 0 – 1 kHz as we are in 1 – 10 kHz ─  But the linear axis plot has the 0 – 1 kHz region all “scrunched up”

 ─  However… the log axis allows us to expand out the lower

frequencies to see them better!

Linear Axis:

Log Axis:

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 100000

f (Hz)

100

101

102

103

104

0

f (Hz)

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Simplest Real-World Lowpass Filter: RC Circuit

v(t )  R C  y(t ) Output SignalInput Signal

1. Convert capacitor into impedance:

C  j

 Z c 

 1

)(   Small impedance at high  

Large impedance at low  

 x Ae j    2. Imagine input as phasor:

3. Now analyze the circuit as if it were a DC circuit with a complex voltage

in (the phasor) and complex resistors (the impedances):

 R

1/ j C 

 y

 x Ae  j 

 

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 Now find the output phasor

as a function of the input

 phasor… the thing that

multiplies the input phasor

is ALWAYS the Freq Resp !

( ) 1

( ) 1

c

c

 Z  y x x

 R Z j RC 

 

 

Here…

use

Voltage

Divider.

1( )

1 H 

 j RC  

 

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 Now… we can plot this

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 1 0000

0

0.5

1

f (Hz)

      |      H      (      f      )      | 

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000-1.5

-1

-0.5

0

f (Hz)

     <     H      (      f      )

RC=1.5915e-4;

f=0:10:10000;

H=1./(1 + j*2*pi*f*RC);

subplot(2,1,1)

 plot(f,abs(H))grid

xlabel('f (Hz)')

ylabel('|H(f)|')

subplot(2,1,2)

 plot(f,angle(H))

gridxlabel('f (Hz)')

ylabel('

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Instead… we can plot this using dB and log axes…

RC=1.5915e-4;

f=1:10:10000;

H=1./(1 + j*2*pi*f*RC);subplot(2,1,1)

semilogx(f,20*log10(abs(H)))

grid

xlabel('f (Hz)')

ylabel('|H(f)|')

subplot(2,1,2)semilogx(f,angle(H))

grid

xlabel('f (Hz)')

ylabel('