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Introduction to Filters

Section 14.1-14.2

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Application of Filter

Application: CellphoneCenter frequency: 900 MHzBandwidth: 200 KHz

Adjacentinterference

Use a filter to removeinterference

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Filters

Classification Low-Pass

High-Pass

Band-Pass Band-Reject

Implementation

Passive Implementation (R,L, C) Active Implementation (Op-Amp, R, L, C)

Continuous time and discrete time

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Filter Characteristics

Must not alterthe desired signal!

Sharp Transition

in order to attenuatethe interference

Not desirable.Alter Frequency content.

Affect selectivity

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Low-Pass Example

How much attenuation is provided by the filter?

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Answer

How much attenuation is provided by the filter?40 dB

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High-Pass Filter

What filter stopband attenuation is necessary in orderto ensure the signal level is 20 dB above the interference?

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High-Pass Filter (Solution)

What filter stopband attenuation is necessary in orderto ensure the signal level is 20 dB above the interference? 60 dB @60 Hz

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Bandpass

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Replace a resistor with a

capacitor!

How do you replace a resistor with a switch and a capacitor?

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Resistance of a Switched

Capacitor Circuit

(315A, Murmann, Stanford)

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What is the equivalent

continuous time filter?

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Filter Transfer Function

(Increase filter order in order to increase filter selectivity!)

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Low Pass Filter Example

1 1

1( )

1aH s

R C s

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Adding a Zero

1 1

1( )

1aH s

R C s

1

11 2

1

( )1 1b

C sH s

RC s C s

1 2

1 1 2

1

( ) 1

R C s

R C C s

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Complex Poles and Zero at the

Origin

1 1

1 1

1

( )( )

1( )

c

L s RH s

L s RC s

1

2

1 1 1 1 1

C s

R L C s L s R

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RC Low Pass (Review)

A pole: a root of the denomintor1+sRC=0S=-RC

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Laplace Transform/Fourier

Transform

p=1/(RC)

(Fourier Transform)

(Laplace Transform)

-p

Location of the zero in the left complexplane

Complex s plane

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Rules of thumb: (applicable to a pole)Magnitude:1. 20 dB drop after the cut-off frequency2. 3dB drop at the cut-off frequencyPhase:1. -45 deg at the cut-off frequency

2. 0 degree at one decade prior to the cut-frequency3. 90 degrees one decade after the cut-off frequency

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RC High Pass Filter (Review)

A zero at DC.A pole from the denominator.1+sRC=0S=-RC

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Laplace Transform/Fourier

Transform

p=1/(RC)Zero at DC.

(Fourier Transform)

(Laplace Transform)

-p

Location of the zero in the left complexplane

Complex s plane

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Zero at the origin.Thus phase(f=0)=90 degrees.The high pass filter has a cut-off frequency of 100.

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RC High Pass Filter (Review)

R12=(R1R2)/(R1+R2)A pole and a zero in the left complex plane.

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Laplace Transform/Fourier

Transform (Low Frequency)

z=1/(RC)p=1/(R12C)

(Fourier Transform)

(Laplace Transform)

-p

Location of the zero in the left complexplane

Complex s plane

-z

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Laplace Transform/Fourier

Transform (High Frequency)

z=1/(RC)p=1/(R12C)

(Fourier Transform)

(Laplace Transform)

-p

Location of the zero in the left complexplane

Complex s plane

-z

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Stability Question

Why the poles must lie in the left half plane?

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Answer

Recall that the impulse response of a system contains terms such as .

If , these terms grow indefinitely with time while oscillating ata frequency of

exp( ) exp( ) exp( )k k kp t t j t