5/3/2018 © 2003, JH McClellan & RW Schafer 1

Signal Processing First

Chapter 6Frequency Response

of FIR Filters

5/3/2018 © 2003, JH McClellan & RW Schafer 2

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5/3/2018 © 2003, JH McClellan & RW Schafer 3

Lecture Objectives

Sinusoidal input signal Determine the FIR filter output

Frequency response of FIR Plotting vs. Frequency Magnitude vs. Freq. Phase vs. Freq. )(ˆˆ ˆ

)()( jeHjjj eeHeH

MAG

PHASE

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6-1 Sinusoidal Response ofFIR Systems

Input: x[n] = Sinusoid Output: y[n] will also be a Sinusoid Different Amplitude and Phase

SAME Frequency

Amplitude & Phase change vs. frequency Called the FREQUENCY RESPONSE

5/3/2018 © 2003, JH McClellan & RW Schafer 5

Complex Exponential Input

M

k

M

kk knxkhknxbny

00][][][][

FIR DIFFERENCE EQUATION

neAenx njj ˆ][x[n] is the input signal—a complex exponential

5/3/2018 © 2003, JH McClellan & RW Schafer 6

Complex Exponential Output

Use the FIR “Difference Equation”

njj eAeH ˆ)ˆ(

M

k

knjjk

M

kk eAebknxbny

0

)(ˆ

0][][

njjM

k

kjk eAeeb ˆ

0

)(ˆ

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Frequency Response

Complex-valued formula Magnitude vs. frequency Phase vs. frequency

Notation:

FREQUENCYRESPONSE

At each frequency, we can DEFINE

)ˆ(of place in )( ˆ HeH j

kjM

kkebH ˆ

0)ˆ(

kjM

kk

j ebeH ˆ

0

ˆ )(

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Example 6.1

Exploitsymmetry

}1,2,1{}{ kb

)ˆcos22()2(

21)(

ˆ

ˆˆˆ

ˆ2ˆˆ

j

jjj

jjj

eeee

eeeH

ˆ)( is Phaseand

)ˆcos22()( is Magnitude0)ˆcos22( Since

ˆ

ˆ

j

j

eH

eH

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PLOT of FREQ RESPONSE

ˆˆ )ˆcos22()( jj eeH RESPONSE at /3

}1,2,1{}{ kb

(radians)

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Example 6.2

y[n]x[n])( jeH

ˆˆ )ˆcos22()( jj eeH

njj

j

eenxeHny

)3/(4/

ˆ

2][ andknown is )( when][ Find

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njj eenxny )3/(4/2][ when][ Find

Example 6.2 (answer)

3/ˆat )( evaluate-Step One ˆ jeH ˆˆ )ˆcos22()( jj eeH

3/ˆ@3)( 3/ˆ jj eeH

njjj eeeny )3/(4/3/ 23][ njj ee )3/(12/6

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6-2 Superposition and theFrequency Response

)cos(2][ andknown is )( when][ Find

43

ˆ

nnxeHny j

ˆˆ )ˆcos22()( jj eeH

y[n]x[n] )( jeH

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Example 6-3: Cosine Input

)cos(2][ when][ Find 43 nnxny

][][][)cos(2

21

)4/3/()4/3/(43

nxnxnxeen njnj

][][][)(][

)(][

21

)4/3/(3/2

)4/3/(3/1

nynynyeeHny

eeHnynjj

njj

UseLinearity

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Ex: Cosine Input (ans-2)

)cos(2][ when][ Find 43 nnxny

)4/3/()3/()4/3/(3/2

)4/3/()3/()4/3/(3/1

3)(][3)(][

njjnjj

njjnjj

eeeeHnyeeeeHny

)cos(6][33][

123

)12/3/()12/3/(

nnyeeny njnj

ˆˆ )ˆcos22()( jj eeH

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Sinusoid Output of FIR

If Multiply the Magnitudes Add the Phases

))(ˆcos()(][

)ˆcos(][11 ˆ

1ˆ

1

jj eHneHAny

nAnx

)()( ˆˆ* jj eHeH

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LTI Systems

LTI: Linear & Time-Invariant

Completely characterized by: FREQUENCY RESPONSE, or IMPULSE RESPONSE h[n]

Sinusoid Input -----> Sinusoid Output At the SAME Frequency

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6-4 Properties of the FrequencyResponse

Get Frequency Response from h[n] Here is the FIR case:

kjM

k

kjM

kk

j ekhebeH ˆ

0

ˆ

0

ˆ ][)(

IMPULSE RESPONSE

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Block Diagrams

Equivalent Representations

y[n]x[n]

y[n]x[n])( jeH

][nh

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Time & Frequency Domain

FIR DIFFERENCE EQUATION is the TIME-DOMAIN

M

k

M

kk knxkhknxbny

00][][][][

M

k

kjj ekheH0

ˆˆ ][)(

ˆ3ˆ2ˆˆ ]3[]2[]1[]0[)( jjjj ehehehheH

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Unit-Delay System

y[n]x[n] ]1[ n

y[n]x[n] je)( jeH

]1[][for )( and ][ Find ˆ nxnyeHnh j

}1,0{}{ kb

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First Difference System

y[n]x[n] 1 je

y[n]x[n] ]1[][ nn

)( jeH

]1[][][ :Equatione Differencfor the )( and ][ Find ˆ

nxnxnyeHnh j

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Example: Delay by 2 System

y[n]x[n] )( jeH

y[n]x[n] ][nh

]2[][for )( and ][ Find ˆ nxnyeHnh j

]2[][ nnh

}1,0,0{kb

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Delay by 2 System

k = 2 ONLY

y[n]x[n] ]2[ n

]2[][for )( and ][ Find ˆ nxnyeHnh j

M

k

kjj ekeH0

ˆˆ ]2[)(

)( jeHy[n]x[n] 2je

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General Delay Property

Only one non-ZERO termfor k at k = nd

dnjM

k

kjd

j eenkeH ˆ

0

ˆˆ ][)(

][][for )( and ][ Find ˆd

j nnxnyeHnh

][][ dnnnh

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Frequency domain --> Time domain

Start with kj bnheH or find and ][)(

)ˆcos(7)( ˆ2ˆ jj eeH

y[n]x[n] ][nh ?][ nh

y[n]x[n])( jeH

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Euler’s Formula)ˆcos(7)( ˆ2ˆ jj eeH )5.05.0(7 ˆˆˆ2 jjj eee

)5.35.3( ˆ3ˆ jj ee

]3[5.3]1[5.3][ nnnh

}5.3,0,5.3,0{kb

Frequency domain --> Time domain

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6-6 Cascaded LTI Systems

Does the order of S1 & S2 matter? No, LTI systems can be rearranged ! What are the filter coefficients? {bk} What is the overall frequency response?

S1 S2][n ][1 nh ][][ 21 nhnh ][1 nh ][2 nh

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Cascade Equivalent

Multiply the Frequency Responses

y[n]x[n] )( jeH

x[n] )( ˆ1

jeH y[n])( ˆ2

jeH

)()()( ˆ2

ˆ1

ˆ jjj eHeHeH EQUIVALENTSYSTEM

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6-7 Running-Average Filtering

Spectrum of x(t) (Sum of sinusoids) Spectrum of x[n] Is ALIASING a PROBLEM ?

Spectrum y[n] (FIR Gain or Nulls) Output y(t) = Sum of sinusoids

D-to-AA-to-Dx(t) y(t)y[n]x[n] )( jeH

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Frequency Scaling

Time sampling: If NO aliasing: Frequency scaling

D-to-AA-to-Dx(t) y(t)y[n]x[n] )( jeH

sfsT ˆsnTt

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11-point Average Example

D-to-AA-to-Dx(t) y(t)y[n]x[n] )( jeH

?

250 Hz

25 Hz ?))250(2cos())25(2cos()( 2

1 tttx

ˆ5

21

211

ˆ

)ˆsin(11)ˆsin(

)( jj eeH

10

0111 ][][

kknxny

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D-A Frequency Scaling

Reconstruct up to 0.5fs Frequency scaling

D-to-AA-to-Dx(t) y(t)y[n]x[n] )( jeH

Time sampling:

sf ˆ

ss ftnnTt

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Track the Frequencies

D-to-AA-to-Dx(t) y(t)y[n]x[n] )( jeH

Fs = 1000 Hz

0.5

.05

0.5

.05

250 Hz

25 Hz

NO new freqs

250 Hz

25 Hz )(

)(

05.0

5.0

j

j

eH

eH

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11-point Average

NULLS or ZEROS

5.0ˆ 05.0ˆ

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Evaluate Frequency Response

ˆ5

21

211

ˆ

)ˆsin(11)ˆsin(

)( jj eeH

)5.0(5

21

211

ˆ

))5.0(sin(11))5.0(sin(

)(

jj eeH

5.0ˆAt

5.2

)25.0sin(11)75.2sin( je

5.00909.0 je

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Evaluate Frequency Response

fs = 1000MAG SCALE

PHASE CHANGE

)( 1000/)25(2jeH

)( 1000/)250(2jeH

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EFFECTIVE RESPONSE

DIGITAL FILTER

LOW-PASS FILTER

)( jeH

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Filter Types

Low-pass Filter (LPF) Blurring Attenuates high frequencies

High-pass Filter (HPF) Sharpening for images Boots high frequencies Removes DC

Band-pass Filter (BPF)

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Black & White Image

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B&W Image with CosineFILTERED: 11-pt AVG

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Filtered B&W Image

LPF:Blur

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Row of B&W Image

BLACK = 255

WHITE = 0

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Row of Filtered Image

Adjusted Delay by 5 samples