Introduction to Filters

Section 14.1-14.2

Application of Filter

Application: CellphoneCenter frequency: 900 MHzBandwidth: 200 KHz

Adjacent interference

Use a filter to removeinterference

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

Filter Characteristics

Must not alter the desired signal!

Sharp Transitionin order to attenuatethe interference

Not desirable.Alter Frequency content.

Affect selectivity

Low-Pass Example

How much attenuation is provided by the filter?

Answer

How much attenuation is provided by the filter? 40 dB

High-Pass Filter

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

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

Bandpass

Replace a resistor with a capacitor!

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

Resistance of a Switched Capacitor Circuit

(315A, Murmann, Stanford)

What is the equivalent continuous time filter?

Filter Transfer Function

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

Low Pass Filter Example

1 1

1( )

1aH sRC s

Adding a Zero

1 1

1( )

1aH sRC s

1

11 2

1

( )1 1b

C sH s

RC s C s

1 2

1 1 2

1

( ) 1

RC s

R C C s

Complex Poles and Zero at the Origin

1 1

1 11

( )( )

1( )

c

L s RH s

L s RC s

12

1 1 1 1 1

C s

R LC s L s R

RC Low Pass (Review)

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

Laplace Transform/Fourier Transform

p=1/(RC)

(Fourier Transform)

(Laplace Transform)

-p

Location of the zero in the left complexplane

Complex s plane

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 frequency2.0 degree at one decade prior to the cut-frequency3.90 degrees one decade after the cut-off frequency

RC High Pass Filter (Review)

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

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

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

RC High Pass Filter (Review)

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

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

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

Stability Question

Why the poles must lie in the left half plane?

Answer

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

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

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

k