1 Fourier Series – LTI

ResponseofLTISystemstoFourierSeries

Review

Consider an LTI system with the unit impulse response h t or h n .

When the input signal is orst ne z , the output signal is a complex exponential same as the

input, multiplied by a constant factor that depends on s or z:

and are referred to as the of the system.H s H z system function

Contents

Response of LTI Systems to Fourier Series

Review

Response to Fourier Series

Continuous-time LTI Systems

RC Low-pass Filter

RC High-pass Filter

Differentiator

Discrete-time LTI Systems

Two-point Average

Properties – Periodicity in Frequency

First-Order Recursive Discrete-Time Filters

Ideal Filters

Continuous-time Systems

Discrete-time Systems

ste stH s e h t

stH s h t e dt

nz nH z z h n

n

n

H z h n z

2 Fourier Series – LTI

ResponsetoFourierSeries

For a continuous-time system with a periodic input signal with period T,

0 02

Select and .s jkT

For a discrete-time system with a periodic input signal with period N,

00

2Select and .jkz e

N

and are referred to as the of the system jH jw H e frequency response .

Magnitude : H

Phase : H

is periodic in with period 2 . jH e

0jk tk

k

x t a e

0

0jk t

kk

y t a H jk e

H j

j tH j h t e dt

0jk nk

k N

x n a e

0 0jk jk n

kk N

y n a H e e

jH e

j j n

n

H e h n e

3

Conti

RC Lo

Review

For the

For the

Derive

Input:

Output

Since v

Find th

Suppos

Since

H j

e

Eq.e1 m

inuous‐

ow-pass F

w:

e capacitor,

1( )

e resistor,

( )

c

r

v tC

dvi t C

v t R i

the differ

( )

: ( )

s

c

v t

v t

( )

( )

s r

r

c

v t v

v t R i

dv tRC

dt

he frequen

: the Frequ

se is th

is an ei

j t

j t

e

e

must satisfy

dRC H

dt

‐timeLT

Filter

( ) ,

( ).

.

t

c

i dr

v t

dt

i t

ential equa

.c

c

c s

t v t

dvt RC

d

v t v

ncy respons

uency Resp

he input sign

igenfunction

the input-o

j tj e

TISyste

ation decsr

( ),

.

c

s

t

dt

t

se of the sy

ponse of the

nal.

n of the syst

output relatio

jH j e

ms

ribing the

ystem usin

system.

tem, the out

on.

t j te

system

g eigenfun

tput of the s

F

nctions

system must

Fourier Seri

1e

t be H j

es – LTI

.j te

4 Fourier Series – LTI

.

1.

1

j t j t j tRC H j j e H j e e

H jRCj

5 Fourier Series – LTI

Find the frequency response of the system using the unit impulse response

1

1 1 1

We will find the unit impulse reponse first.

From eq.e1, substituting ,

1 1. 2

Multiply to both sides of eq. e2,

1 1

s

tRC

t t tRC RC RC

v t t

d h th t t e

dt RC RC

e

d h te h t e e t

dt RC RC

1

We have

13

tRCd

h t e t edt RC

1

1 1

1

From 3

1 + , some constant

1 + .

1 + .

We assume the system is causal and demand 0 for 0.

0,

1.

ttRC

t tRC RC

tRC

e

h t e dRC

u tRC

h t e u t eRC

h t t

h t e u tRC

6 Fourier Series – LTI

Next we will find the frequency response of the system.

1

0

1

0

1

0

1

1

1 11

1

1

j t

t j tRC

j tRC

j tRC

H j h t e dt

e e dtRC

e dtRC

j e dtRC

RC jRC

RCj

Magnitude and Phase of the Frequency Response

2

1

1.

1

1 .

1

tan .

H jRCj

H jRC

H j RCw

7 Fourier Series – LTI

For RC=1,

For RC=10,

The RC filter, in the case of taking the capacitor voltage as the output, acts as a lowpass filter.

As the value of RC increases, the attenuation gets higher.

0

0.2

0.4

0.6

0.8

1

1.2

‐4 ‐2 0 2 4 6 8 10 12

magnitude of frequency response

‐2

‐1.5

‐1

‐0.5

0

0.5

1

1.5

‐4 ‐2 0 2 4 6 8 10 12

Phase of Frequency Response

0

0.2

0.4

0.6

0.8

1

1.2

‐4 ‐2 0 2 4 6 8 10 12

magnitude of frequency response

‐2

‐1.5

‐1

‐0.5

0

0.5

1

1.5

2

‐4 ‐2 0 2 4 6 8 10 12

Phase of Frequency Response

8

RC Hi

Input:

Output

Derive

Since v

Find th

Suppos

Since

H j

e

Eq.e1 m

When t

igh-pass F

( )

: ( )

s

r

v t

v t

the differ

1( )

1

s r

tc

r

r

v t v

v tC

v tRC

dRC v t

dt

he frequen

: the Frequ

se is th

is an ei

j t

j t

e

e

must satisfy

the input

1

sv

dRC H

dt

RC H j

H j

Filter

ential equa

( )

c

tr

r

t v t

i dr

v dC

v t R

ncy respons

uency Resp

he input sign

igenfunction

the input-o

is ,

.1

j t

j t

i t

t e

j e

j e

j RC

j RC

ation decsr

1 tr

s

s

vRC

dr v t

dRC v t

dt

se of the sy

ponse of the

nal.

n of the syst

output relatio

the output

jH j e

H j

ribing the

,dr

ystem usin

system.

tem, the out

on.

t mustr

t

i t

v t

dRC e

dt

e j R

system

g eigenfun

tput of the s

t be

.

.

j t

i t

H j

e

C e

F

nctions

system must

.j te

Fourier Seri

( 1)e

t be H j

es – LTI

.j te

9 Fourier Series – LTI

Magnitude and Phase of the Frequency Response

2

22 1

2

1.

1 1 1 1

1 and tan . for 0.

21

j RC j RC j RC RCH j RC j

j RC j RC j RC RC

RCH j H j H j

RCRC

For 0.5RC ,

For 5RC ,

The RC filter, in the case of taking the resistor voltage as the output, acts as a highpass filter.

As the value of RC increases, the attenuation gets higher.

0

0.2

0.4

0.6

0.8

1

1.2

‐4 ‐2 0 2 4 6 8 10 12

magnitude of frequency response

‐2

‐1.5

‐1

‐0.5

0

0.5

1

1.5

2

‐4 ‐2 0 2 4 6 8 10 12

Phase of Frequency Response

0

0.2

0.4

0.6

0.8

1

1.2

‐4 ‐2 0 2 4 6 8 10 12

magnitude of frequency response

‐1

‐0.8

‐0.6

‐0.4

‐0.2

0

0.2

0.4

0.6

0.8

1

‐4 ‐2 0 2 4 6 8 10 12

Phase of Frequency Response

10 Fourier Series – LTI

Also note that

1

11

1

The first term, 1, corresonds to the identity system.

1The second term, , corresponds to a low-pass filter.

1

As the result, the Identity minus a LPF must be

j RCH j

j RC

j RC

j RC

a High-pass filter.

1

The unit impulse response must be

1.

tRCh t t e u t

RC

11 Fourier Series – LTI

Differentiator

Consider a continuous-time system which differentiates its input.

Find the frequency response of the system using the unit impulse response.

The unit impulse response of the system is the unit doublet:

1d

h t u t tdt

.

The frequency response is

.

j t

j t

H j h t e dt

d te dt

dt

1

Consider

. 1

Since , eq.e1 can be written as

. ( 2)

j tj t j t

j t

d t e d tdt e dt j t e dt e

dt dt

t e t

u t dt H j j e

1

From the lemma below,

0

and thus

.

u t dt

H j j

D y t x t

12 Fourier Series – LTI

Lemma

1

1

The unit doublet has the following properties.

0

u t

u t dt

1

1

1

1

1

.

is the unit impulse response of the differentiator.

Therefore, for any signal , .

Since LTI systems are commutative, .

Thus .

Let 1.

Then 0.

proof

u t

x t x t u t x t

u t x t x t

u x t d x t

x t

u d

13

Find th

The ou

On the

is an

The ou

D

Therefo

Magnit

:

H j

Magnit

Phase

he frequen

tput of wD

other hand,

n LTI system

tput must b

ore it must b

jwH j e

H j j

tude and P

.

:

j

tude H j

H j

ncy respons

when the inp

m and thus

be j

e

H j e

be that

wt j tj e

j

Phase of th

2 2

1

.

tan0

j

se of the sy

put is ij te

is an ei

when th

j t

j t

e

he Frequen

,2

,2

H j

ystem usin

is .j tj e

igenfunctio

he input is e

ncy Respon

0

0

g an eigen

on.

.j t

nse

F

function.

Fourier Series – LTI

14 Fourier Series – LTI

Find the output signal using the frequency response

Suppose the input signal is

1 2cos 2 2cos 4 .x t t t

0

2

0

1 1

2 2

We notice 1 , and thus 2 .

Let denote the Fourier coefficients of , that is,

.

Then 1,

1,

1.

k

j ktk

k

T

a x t

x t a e

a

a a

a a

2

0 0

1 1 1 1

2 2 2 2

Let denote the Fourier coefficients of the output, that is,

.

Then remembering ,

0 0,

2 2 , 2 2

4 4 , 4 4 .

k

j ktk

k

b

y t b e

H j j

b a H

b a H j j b a H j j

b a H j j b a H j j

2 2 4 4

2 2 4 4

2 2 4 4

4 82 2

4 sin 2 8 sin 4

must equal .

j t j t j t j t

j t j t j t j t

y t j e j e j e j e

e e e e

j j

t t

dx t

dt

15 Fourier Series – LTI

Discrete‐timeLTISystems

Two-point Average

Consider the following discrete-time system

11

2

x n x ny n t

Find the frequency response of the system using the unit impulse response.

The impulse response of the system is

1

2

n nh n

.

The frequency response of the system is

12

12

1

1

j j n

n

j n

n

j

H e h n e

n n e

e

Find the frequency response of the system using an eigenfunction.

112

From 1 , when the input is , the output of the system is .j nj n j nt e e e

112

112

12

On the other hand, since is an eigenfunction, the output must be .

Therefore it must be that .

1

jwn j jwn

j nj jwn j n

j nj j n j n

j

e H e e

H e e e e

H e e e e

e

16

Magnit

Note H

Magnit

Phase

For the

j

Note.

H e

Phase

tude and P

12

j

j

H e

H e

e

:

: j

tude H e

H e

2

2

e interval

cos 02

co

:

j

j

j

e

e

H e

Phase of th

1 2

2

2

1

cos2

j

j j

j

e

e e

e

cos

, 2

je

3 ,

.

os2

1 cos

+ .2

he Frequen

2 2

is periodic

je

2

2

ncy Respon

in with p

.

nse

period 2 .

FFourier Series – LTI

17 Fourier Series – LTI

Properties – Periodicity in Frequency

For a discrete-time system, the frequency response of the system is periodic in frequency with period of 2 . We need to consider frequencies only within any one interval of length 2 .

Low frequencies near

0, 2 , .

High frequencies near

, 3 , .

18 Fourier Series – LTI

First-Order Recursive Discrete-Time Filters

Consider the following system.

1 , 1y n x n a y n a

Find the frequency response of the system using eigenfunctions.

1

is an eigenfunction of the system.

When , must be

Since the system is time invariant, 1 must be the output when the input is 1 .

1

j n

j n j j n

j nj

e

x n e y n H e e

y n x n

y n H e e

1

Combining the above arguments,

1 .

1.

1

j nj j n j n j

j j j

jj

H e e e aH e e

H e aH e e

H ea e

Find the frequency response of the system using the unit impulse response

The unit impulse response of the system is

.nh n a u n

The frequency response of the system is

0

0

1 for 1

1

j j n

n

n j n

n

nj

n

j

H e h n e

a e

ae

aae

D

x n y n

a

19 Fourier Series – LTI

0

0.5

1

1.5

2

2.5

3

‐4 ‐2 0 2 4 6 8 10 12

magnitude of frequency response

0

2

4

6

8

10

12

‐4 ‐2 0 2 4 6 8 10 12

magnitude of frequency response

Magnitude and Phase of the Frequency Response

2 2 2

1

1 for 1.

1

Let 1 .

1Then and 1 .

1 cos sin .

1 cos sin .

sintan .

1 cos

jj

j

j j

H e aae

C ae

H e H e CC

C a ja

C a a

aC

a

0.6a 0.9a

20 Fourier Series – LTI

0.6a 0.9a

For 0, the averager acts as a low pass filter.

For 0, the averager acts as a high pass filter.

a

a

For a larger value of , the attenuation gets higher. a

0

0.5

1

1.5

2

2.5

3

‐4 ‐2 0 2 4 6 8 10 12

magnitude of frequency response

0

2

4

6

8

10

12

‐4 ‐2 0 2 4 6 8 10 12

magnitude of frequency response

21 Fourier Series – LTI

Speed of Response in the time domain

Drive the system with the unit step signal as the input. The speed of the response in the time domain is measured by how fast the output signal approaches the long-term value.

In this filter, .nh n a u n

The unit step response is

1

0

*

1

1

n

k

k

nnk

k

s n u n h n h n u n

a u n u n

a u k u n k

aa

a

0.8a 0.1a

For smaller values of a , the system responds faster.

0

0.2

0.4

0.6

0.8

1

1.2

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

y[n]

0.84

0.86

0.88

0.9

0.92

0.94

0.96

0.98

1

1.02

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

y[n]

22

Ideal

Contin

Ideal L

Ideal H

Ideal B

Discre

Ideal L

Ideal H

Ideal B

lFilters

nuous-tim

Low Pass Fi

High Pass F

Band Pass F

ete-time S

Low Pass Fi

High Pass F

Band Pass F

s

me System

ilter

Filter

Filter

Systems

ilter

Filter

Filter

ms

FFourier Series – LTI