Transcript Document

ECE 8443 – Pattern Recognition

LECTURE 18:

FOURIER ANALYSIS OF CT SYSTEMS

Objectives:

Response to a Sinusoidal Input Frequency Analysis of an RC Circuit Response to Periodic Inputs Response to Nonperiodic Inputs Analysis of Ideal Filters

Resources:

Wiki: The RC Circuit CN: Response of an RC Circuit CNX: Ideal Filters URL: Audio:

Response of an LTI System to a Sinusoid

• • • • •

Consider an LTI CT system with impulse response h(t):

y

(

t

) 

h

(

t

) *

x

(

t

)     

h

(  )

x

(

t

  )

d

We will assume that the Fourier transform of h(t) exists:

H

(  )     

h

(

t

)

e

j

t dt

The output can be computed using our Fourier transform properties:

Y

(  ) 

H

(  )

X

(  ) and

Y

(  ) 

H

(  )

X

(  ) 

Y

(  )  

H

(  )  

X

(  )

Suppose the input is a sinusoid:

x

(

t

) 

A

cos(  0

t

  )

Using properties of the Fourier transform, we can compute the output:

X

(  ) 

A

 

e

j

      0  

e j

    0  

Y

(  ) 

y

(

t

)     

F

H

(  )

X A

H A

 

H

(  ( ) (    

e

)  0 )

j

e

A

A

-

1

H H

Y

(  (  (  ) 0 0  ) )   

e e

 

A

j j j

  

H

  

H

 

H

   (  0 (  0 (  0 )   

e

0 )   )      

j

 0    

e

cos  

j

     0

t

H

    0  (     0  0     0 )

e e

j

e j

 

H j

   

H

 (     

H

(  0 (  0 0 )   )   )  

e

  0

j

          0     0  

ECE 3163: Lecture 18, Slide 1

Example: RC Circuit

dy

(

t

) 

dt

1

RC y

(

t

)  1

RC x

(

t

) •

Using our FT properties:

j

Y

 1

RC Y

 1

RC X

(  )

Y

(  ) 

H

(  ) 

j

 1 /

RC

 1 /

RC Y X

(    ) 

X j

 1 /

RC

 1 /

RC H

(  )  1 /

RC

H

(  )   2  tan   1 ( 1 / 

RC

) 2 

RC

 •

Compute the frequency response: RC = 0.001; W=0:50:5000; H=(1/RC)./(j*w+1/RC); magH=abs(H); angH=180*angle(H)/pi; ECE 3163: Lecture 18, Slide 2

Example: RC Circuit (Cont.)

• • • We can compute the output for RC =0.001

y

(

t

) 

A

( 0 .

707 ) cos  1000

t

   45   We can compute the output for RC =0.001

y

(

t

) 

A

( 0 .

316 ) cos  3000

t

   71 .

6  

and

 0 =1000 rad/sec

: and

 0 =3000 rad/sec

: Hence the circuit acts as a lowpass filter. Note the phase is not linear.

If the input was the sum of two sinewaves:

x

(

t

)  cos  100

t

  cos  3000

t

describe the output.

ECE 3163: Lecture 18, Slide 3

Response To Periodic Inputs

• • •

We can extend our example to all periodic signals using the Fourier series:

x

(

t

) 

a

0 

k

  1

A k

cos 

k

 0  

k

 (a variant of the trigonome tric Fourier series)

The output of an LTI system is:

y

(

t

) 

a

0

H

  

k

  1

A k H

k

 0  cos 

k

 0

t

 

k

 

H

k

 0  

We can write the Fourier series for the output as:

y

(

t

) 

a

0

y

k

   1

A k y

cos 

k

 0

t

 

k y

 where, a y 0 

a

0

x H

( 0 )

A k y

A k x H

(

k

 0 ) 

k y

 

k x

 

H

(

k

 0 ) also, c y k  1 2

A k x H

(

k

 0 ) and  c k y  

k x

 

H

(

k

 0 ) •

It is important to observe that since the spectrum of a periodic signal is a line spectrum, the output spectrum is simply a weighted version of the input, where the weights are found by sampling of the frequency response of the LTI system at multiples of the fundamental frequency,

 0

.

ECE 3163: Lecture 18, Slide 4

Example: Rectangular Pulse Train and an RC Circuit

Recall the Fourier series for a periodic rectangular pulse:

x

(

t

) 

a

0 

k

   1

a k

cos   • where,

a

0  0 .

5

a k

 sin (

k

k

  / / 2 ) 2 

Also recall the system response was:

H

(  ) 

j

 1 /

RC

 1 /

RC

The output can be easily written as:

y

(

t

) 

a

0

y

k

   1

A k y

cos   0

t

 

k y

 where,

a

0 y

A k y

a

0

x H

( 0 )  0 .

5 

A k x H

(

k

 0 )  sin (

k

k

  / / 2 ) 2  1 /

RC

(

k

 ) 2  ( 1 /

RC

) 2    2

k

 1 /

RC

(

k

 ) 2 0  ( 1 /

RC

) 2

k

odd

k

even

ECE 3163: Lecture 18, Slide 5

Example: Rectangular Pulse Train (Cont.)

We can write a similar expression for the output:

y

(

t

) 

a

0

y

k k

   1

odd

2

k

 1 /

RC

(

k

 ) 2  ( 1 /

RC

) 2 cos 

k

t

 tan  1

k

RC

 1/

RC

= 1 •

We can observe the implications of lowpass filtering this signal.

What aspects of the input signal give rise to high frequency components?

What are the implications of increasing

1/ RC in the circuit?

Why are the pulses increasingly rounded for lower values of

1/ RC?

What causes the oscillations in the signal as

1/ RC is increased? 1/

RC

= 10 1/

RC

= 100

ECE 3163: Lecture 18, Slide 6

Response to Nonperiodic Inputs

• •

We can recover the output in the time domain using the inverse transform:

y

(

t

)  1 2     

H

(

e j

 )

X

(

e j

 )

e j

t d

These integrals are often hard to compute, so we try to circumvent them using transform tables and combinations of transform properties.

Consider the response of our RC circuit to a single pulse:

X

(

e j

 )  sin( (   / / 2 ) 2 )

H

(

e j

 ) 

Y

(

e j

 )  

j

 1 /

RC

 1 /

RC X

(

e j

 )

H

(

e j

 ) sin( (   / / 2 ) 2 )

j

 1 /

RC

 1 /

RC

MATLAB code for the frequency response: RC=1; w=-40:.3:40; X=2*sin(w/2)./w; H=(1/RC)./(j*w+1/RC); Y=X.*H; magY=abs(Y); ECE 3163: Lecture 18, Slide 7

Response to Nonperiodic Inputs (Cont.)

We can recover the output using the inverse Fourier transform: syms X H Y y w X = 2*sin(w/2)./w; H=(1/RC)./(j*w+1/RC); Y=X.*H; Y=ifourier(Y); ezplot(y,[-1 5]); axis([-1 5 0 1.5])

1/

RC

= 1 1/

RC

= 1 1/

RC

= 10 1/

RC

= 10

ECE 3163: Lecture 18, Slide 8

Ideal Filters

The process of rejecting particular frequencies or a range of frequencies is called filtering. A system that has this characteristic is called a filter .

An ideal filter is a filter whose frequency response goes exactly to zero for some frequencies and whose magnitude response is exactly one for other ranges of frequencies.

• •

To avoid phase distortion in the filtering process, an ideal filter should have a linear phase characteristic. Why?

H

(

e j

 )   

t d

for all ω in the filter passband

We will see this “ideal” response has some important implications for the impulse response of the filter.

Lowpass

Highpass

Bandstop

Bandpass ECE 3163: Lecture 18, Slide 9

Ideal Linear Phase Lowpass Filter

Consider the ideal lowpass filter with frequency response:

H

(

e j

 )  

e

  

j

t d

0 , 

B

   

B

,   

B B

Using the Fourier transform pair for a rectangular pulse, and applying the time-shift property:

h

(

t

) 

B

 sin

c B

 (

t

t d

Is this filter causal?

The frequency response of an ideal bandpass filter can be similarly defined:

H

(

e j

 )  

e

  

j

t d

0 , ,

B

1   

B

2 elsewhere •

Will this filter be physically realizable?

Why?

ECE 3163: Lecture 18, Slide 10

Phase Response

Impulse Response

Summary

Showed that the response of a linear LTI system to a sinusoid is a sinusoid at the same frequency with a different amplitude and phase.

Demonstrated how to compute the change in amplitude and phase using the system’s Fourier transform.

Demonstrated this for a simple RC circuit.

Generalized this to periodic and nonperiodic signals.

Worked examples involving a periodic pulse train and a single pulse.

Introduced the concept of an ideal filter and discussed several types of ideal filters.

Noted that the ideal filter is a noncausal system and is not physically realizable. However, there are many ways to approximate ideal filters, and that is a topic known as filter design.

ECE 3163: Lecture 18, Slide 11