Frequency Response of FIR Filtersjoelsd/Fundamentals/coursework/BME... · 2017. 7. 24. · BME 310 Biomedical Computing - J.Schesser 231 Frequency Response of a FIR 1. When the input

BME 310 Biomedical Computing -J.Schesser

226

Frequency Response of FIR Filters

Lecture #9Chapter 6


227

What Is this Course All About ?

• To Gain an Appreciation of the Various Types of Signals and Systems

• To Analyze The Various Types of Systems

• To Learn the Skills and Tools needed to Perform These Analyses.

• To Understand How Computers Process Signals and Systems


228

Frequency Response• Study FIR systems, where the input is a

discrete-time sinusoidal signal, in particular, a discrete-time complex exponential signal with normalized radian frequency

• Let’s substitute this signal into the general expression of a FIR.

ˆ

( )ˆ[ ] ( ) ;

s

j j t

j nTjs s

j j n

x t Ae ex n x nT Ae e T

Ae e n


229

Frequency Response Function ˆ

0

ˆ ( )

0

[ ]

[ ] [ ]

j j n

M

kk

Mj j n k

kk

x n Ae e

y n b x n k

b Ae e

• We define this function as the Frequency Response Function or the Frequency Response

ˆ ˆ

0ˆ

ˆ ˆ

0

[ ] ( )

ˆ( )

ˆ( ) ( )

Mj k j j n

kk

j j n

Mj j k

kk

y n b e Ae e

H Ae e

H H e b e

)( jeH


230

Frequency Response FunctionAnother Approach using Convolution

ˆ ˆ ( )

0

ˆ ˆ ˆ( )

ˆ ˆ ˆ ˆ

0

[ ] [ ]

[ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ]

( [ ] ) ( [ ] [ ] )

Since [ ] 0 for and

j j n j j n l

M

k

j j n l j j n j l

l l l

Mj l j j n j l j j n

l l k

x n Ae e x n l Ae e

h n h k n k

y n h l x n l h l Ae e h l Ae e e

h l e Ae e h k l k e Ae e

l k l k

ˆ ˆ

0

[ ] 1 for , then the outer summation can be removed.

( [ ] )M

j k j j n

k

l k l k

h k e Ae e


231

Frequency Response of a FIR

1. When the input is a discrete-time complex exponential signal, the output of an FIR filter is also a discrete-time complex exponential signal with a different amplitude but same frequency

2. The Frequency Response Function only applies to inputs which are discrete-time complex exponential signals

3. The Frequency Response is a complex number which can be represented in Cartesian or polar form

4. Therefore, the amplitude of the output is product of the amplitude of the input signal times the amplitude of the Frequency Response (polar form)

5. The phase of the output is the sum of the phase of the input and the phase of the Frequency Response

M

k

kjM

k

kjk

j ekhebeH0

ˆ

0

ˆˆ ][)(


232

Polar Form of the Frequency Responseˆ

ˆ

ˆ

ˆ ˆ ( )

ˆ ˆ

ˆ ˆ( )

ˆ ˆ( ( ) )

( ) ( )

[ ] ( )

( )

( ( ) )

j

j

j

j j j H e

j j j n

j j H e j j n

j j H e j n

H e H e e

y n H e Ae e

H e e Ae e

A H e e e

ˆ

ˆ

Note that this: [ ] ( ) [ ]

is only meaningful for [ ]

j

j j n

y n H e x n

x n Ae e


233

An Example

ˆ)(

ˆcos22)(

)ˆcos22()2(

121

)(

}1 ,2 ,1{}{Let

ˆ

ˆ

ˆ

ˆˆˆ

2ˆˆ

2

0

ˆˆ

j

j

j

jjj

jjk

kjk

j

k

eH

eH

eeee

ee

ebeH

b


234

Example continued

njj

njj

njjjnjjj

j

j

jj

njj

ee

ee

eeeeeeH

nxeHny

e

eeH

eenx

312

3)43(

343343

3

3

33

34

6

6

232)(

][)(][

3

)3cos22()(

,3ˆ since

2][let system For this


235

Using Superposition• We can use superposition to find the outputs of more

complex signals.1 1 1 1

1 1 1 1

1 1 1 1 1 1

1 1

0 1 1 1

ˆ ˆ0 10

ˆ ˆˆ 0 10

ˆ ˆ ˆ ˆ0 0 1 10

ˆ ˆ0 0 10

ˆ[ ] cos( )

( )2

[ ] ( ) { ( )}2

( ) ( ) ( )2 2

( ) ( )2

j j n j j nj n

j j n j j nj j n

j j j n j j j nj j n

j j jj j n

x n A A nAA e e e e e

Ay n H e A e e e e e

A AH e A e H e e e H e e e

AH e A e H e e e

1 1 1 1

ˆ ˆ1 11 1 1 1 1 1

ˆ ˆ1 11 1 1 1 1

1

ˆ ˆ1

ˆ ˆ ˆ ˆ0 0 ( ) ( )1 10

ˆ ˆ ˆ0 ( ) ( )10

ˆ0 10

*( )2

( ) ( ) ( )2 2

( ) ( ) { }2

( ) ( )2

j j

j j

n j j j n

j j j n j j j nj j n j H e j H e

j j j n j j nj j H e j H e

jj

AH e e e

A AH e A e H e e e e H e e e e

AH e A H e e e e e e e

AH e A H e

ˆ ˆ1 1

1 1 1 1

1 1

ˆ ˆ( ( )) ( ( ))

ˆ ˆ00 1 1 1

{ }

ˆ( ) ( ) cos( ( ))

j jj n H e j n H e

j jj

e e

H e A A H e n H e


236

An Example

4 4 4 4

For this system let [ ] 2cos( )4 2

ˆ since ,4

( ) (2 2cos ) (2 2) 3.414

[ ] 2 3.41cos(4

j j j j

x n n

H e e e e

y n

)2 4

6.82cos( ( 1) )4 2

This system is a time shifter

n

n


237

Superposition

• Using superposition, find the output using the frequency response function for each component of the input

• Add up the component outputs to yield the complete output solution.


238

Another Example

))1(87cos(31522.0)2)1(3cos(3344][

1522.0)87cos22()(

3)3cos22()(

4)0cos22()(8

7 ,3 ,0ˆfor

)87cos(3)23cos(34][let system For this

87

87

87

333

00

nnny

eeeH

eeeH

eeH

nnnx

jjj

jjj

jj


239

Cascaded System• Two systems where the output of the first is the

input to the second.

• This states that a cascaded system can be replaced a single system with frequency response,

)( ˆ2

jeHLTI 1 LTI 2x[n]

)( ˆ1

jeH

][)(

][ˆ

1

1

nxeH

nyj

ˆj j nAe e

][)()(

][)(][ˆ

2ˆ

1

1ˆ

2

nxeHeH

nyeHnyjj

j

)()()( ˆ2

ˆ1

ˆ jjj eHeHeH


240

Summary• The Frequency Response is associated a FIR system

with complex exponential signals as inputs• The Frequency Response can be used to calculate the

output signal of a system• Superposition can be used to find the output signal of

each component and the complete response can be obtained by adding the outputs of the components

• This type of analysis is called frequency domain analysis and we can work exclusively in the frequency– Solving for the output in the time domain (the difference

equation or impulse response) is not necessary.


241

The Unit-Step Signal

• Before we continue, let’s define another important sequence called the unit step signal:

• This signal can be used represent real world signals since signals start as some point in time.

• A signal like x[n]=cos(.2πn) is defined for all -∞ < n < ∞ , while x[n]=cos(.2πn) u[n] is more realistic since it is defined only for n ≥ 0

0for 1 ][

0for 0

nnu

n


242

Unit Step Example

cos[.2n]

-15 -10 -5 0 5 10 15 20

cos[.2n]*u[n]

-15 -10 -5 0 5 10 15 20


243

Steady-State and Transient Response

• Let’s examine how systems responds to signals which begin at a particular time.

• We will see that the output show two phases:– A Transient Response or a period of time where the signal

“builds up” from its starting value (usually zero) to its final “state”

– A Steady State Response or the period where the transient response has “completed”

• Examples:– Turning on a light or a motor– Starting a car


244

FIR System Transient and Steady-State Response

• The input to the system is:

• Our FIR system:

• The output:

0for 0 ][][

0for ˆ

ˆ

nnuXenx

nXenj

nj

M

k

kjk

j ebeH0

ˆˆ )(

M

k

knjk

M

kk knuXebknxbny

0

)(ˆ

0][][][


245


• Let’s study the output:

0}0

000

00{

]}[

)]1([][)]1([

]1[]0[{]0[

]}[

)]1([][)]1([

]1[]0[{

][][

)(ˆ

))1((ˆ1

)(ˆ))1((ˆ1

)1(ˆ1

)0(ˆ0

)(ˆ

))1((ˆ1

)(ˆ))1((ˆ1

)1(ˆ1

)0(ˆ0

)(ˆ

))1((ˆ1

)(ˆ))1((ˆ1

)1(ˆ1

)0(ˆ0

0

)(ˆ

MnjM

lnjl

lnjl

lnjl

njnj

MnjM

lnjl

lnjl

lnjl

njnj

MnjM

lnjl

lnjl

lnjl

njnj

M

k

knjk

eb

ebebeb

ebebX

Mnueb

lnueblnueblnueb

nuebnuebXny

Mnueb

lnueblnueblnueb

nuebnuebX

knuXebny


246


0at input thestarting toduemainly isIt !!response! transient theis This

}0

011

11{

]}[

)]1([

][

)]1([

]1[][{]0[

0

ˆˆ)(ˆ

))1((ˆ1

)(ˆ))1((ˆ1

)1(ˆ1

)0(ˆ0

)(ˆ

))1((ˆ1

)(ˆ

))1((ˆ1

)1(ˆ1

)0(ˆ0

n

eXebeb

ebebeb

ebebX

Mnueb

lnueb

lnueb

lnueb

nuebnuebXMlny

n

k

njkjk

MnjM

lnjl

lnjl

lnjl

njnj

MnjM

lnjl

lnjl

lnjl

njnj


247


ˆ ˆ( 0) ( 1)0 1

ˆ ˆ ˆ( ( 1)) ( ) ( ( 1))1 1

ˆ ( )

ˆ0

[ ] { [ 0] [ 1]

[ ( 1)] [ ] [ ( 1)]

[ ]}

{

j n j n

j n l j n l j n ll l l

j n MM

j

y n M X b e u n b e u n

b e u n l b e u n l b e u n l

b e u n M

X b e

ˆ( 0) ( 1)

1ˆ ˆ ˆ( ( 1)) ( ) ( ( 1))

1 1

ˆ ˆ ˆ( )

0

1 1

1 1 1

1}

This is the steady-state response!!!(Looks like the respo

n j n

j n l j n l j n ll l l

Mj n M j k j n

M kk

b e

b e b e b e

b e b Xe e

nse due to the input when the input is non-zero for all .)n


248

An Example of the Transient and Steady State Response

x[n]

-1.5

-1

-0.5

0

0.5

1

1.5

-10 -5 0 5 10 15 20 25

y[n]

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

-10 -5 0 5 10 15 20 25


249

Homework

• Exercises:– 6.1

• Problems:– 6.1, 6.2, – 6.4, 6.6 Use Matlab to plot the Frequency Response; show

your code– 6.7, – 6.12 Use Matlab to plot the Frequency Response; show

your code– 6.13

Frequency Response of FIR Filtersjoelsd/Fundamentals/coursework/BME... · 2017. 7. 24. · BME 310 Biomedical Computing - J.Schesser 231 Frequency Response of a FIR 1. When the input

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