Chap6_Frequency Response of FIR Filters

8/2/2019 Chap6_Frequency Response of FIR Filters

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ECE 2610 Signal and Systems 61

Frequency Responseof FIR FiltersThis chapter continues the study of FIR filters from Chapter 5,

but the emphasis isfrequency response, which relates to how the

filter responds to an input of the form

.

A fundamental result we shall soon see, is that the frequency

response and impulse response are related through an operation

known as the Fourier transform. The Fourier transform itself is

not however formally studied in this chapter.

Sinusoidal Response of FIR Systems

Consider an FIR filter when the input is a complex sinusoid

of the form

, (6.1)

where it could be that was obtained by sampling the

complex sinusoid and

From the difference equation for an tap FIR filter,

(6.2)

x n ej0n n =

x n Aejej0n n =

x n x t Aeje

j0t= 0 0Ts=

M 1+

y n bk x n k k 0=

M

bkAej

ej0 n k

k 0=

M

= =

Chapter

6


2/32


ECE 2610 Signals and Systems 62

(6.3)

where we have defined for arbitrary

(6.4)

to be thefrequency response of the FIR filter

Note: The notation is used rather than say , to

be consistent with the z-transform which will defined in

Chapter 7, and to emphasize the fact that the frequency

response is periodic, with period (more on this later)

Note: , where sine and cosine are

both functions

Returning to (6.2), the implication is that when the input is a

complex exponential at frequency , the output is also a

complex exponential at frequency

The complex amplitude (magnitude and phase) of the input is

changed as a result of passing through the system

The frequency response at multiplies the input ampli-tude to produce the output

It is not true in general that , but

only for the special case of a complex sinusoid

applied starting at

y n Aej

ej0n

bkej 0k

k 0=

M

=

Aejej0n

H ej

0 n =

H ej bke

j k

k 0=

M

=

H ej H

2

ej j sin+cos=mod 2

00

0

n H ej0 x n =x n
http://-/?-http://-/?-


3/32



The frequency response is a complex function of that is

generally viewed in either polar or rectangular form

(6.5)

The polar or magnitude and phase form is perhaps the

most common

The polar form offers the following interpretation of in

terms of , when the input is a complex sinusoid

(6.6)

Here we see that the input amplitude is multiplied by the

frequency response amplitude, and the input phase has

added to it the frequency response phase The output amplitude expression means that is

also termed the gain of an LTI system

Example:

The frequency response of this FIR filter is

H e

j

H ej

e j H e

j

=Re H e

j jIm H ej +=

n x n

y n H ej e j H ej

Aej

ejn

=

H ej

Ae j H e

j + e

jn=

H ej

bk 1 1 3 1 1 =

H ej

bke j

k

k 0=

4

=

1 ej

3ej 2

ej 3

ej 4

+ + + +=


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We have used the inverse Euler formula for cosine twice

For this particular filter we have that

Why?

Use MATLAB to plot the magnitude and phase response

>> w = 0:2*pi/200:2*pi;

>> H = exp(-j*2*w).*(3 + 2*cos(w) + 2*cos(2*w));

>> subplot(211)

>> plot(w,abs(H))

>> axis([0 2*pi 0 8])

>> grid

>> ylabel('Magnitude')>> subplot(212)

>> plot(w,angle(H))

>> axis([0 2*pi -pi pi])

>> grid

>> ylabel('Phase (rad)')

H ej e j2

e

j2e

j3 e

je

j2+ + + + =

ej2

2 2 2 3+cos+cos =

H ej 3 2 cos 2 2 cos+ +=

H ej 2=


5/32



>> xlabel('hat(\omega) (rad)')

Example: Find for Input

The input frequency is rad, the amplitude is 5, and

the phase is

Assuming is input to the 4-tap FIR filter in the previous

example, the filter output is

The amplitude response or gain at is =3.248; why?

0 1 2 3 4 5 60

2

4

6

8

Magnitude

0 1 2 3 4 5 6

2

0

2

Phase(rad)

hat() (rad)

2

2

3.248

-2

y n x n 5ej 1 n =

0 1=

0=x n

y n 3 2 1 cos 2 2 1 cos+ + ej 1 2

5ej 1 n

=

16.2415ej2

ejn

=

0 1= H ej1


6/32

Superposition and the Frequency Response



We can use the linearity of the FIR filter to compute the out-

put to a sum of sinusoids input signal

As a special case we first consider a single real sinusoid

(6.7)

Using Eulers formula we expand (6.7)

(6.8)

The filter output due to each complex sinusoid is known from

(6.3), so now using superposition we can write

(6.9)

We can simplify this result to a nice compact form, if we

make the assumption that the FIR filter has real coefficients Special Result: It will be shown in a later section of this

chapter that an FIR filter with real coefficients has conjugate

symmetry

(6.10)

What does this mean?

x n A 0n + cos=

x n A2---e

j 0n + A

2---e

j 0n + +=

n A2---H e

j0 ej 0n + A2---H e

j 0 e j 0n + +=

H ej H* e

j =

H ej Re H ej jIm H ej =

H ej H ej =


7/32



We now use (6.10) to simplify (6.9)

(6.11)

We see that when a real sinusoid passes through an LTI sys-

tem, such as an FIR filter (having real coefficients), the out-

put is also a real sinusoid which has picked up the magnitude

and phase of the system at

The generalization (sum of sinusoids) of this result is when

, (6.12)

then the corresponding LTI system output is

(6.13)

y n A

2---H e

j0 ej 0n + A

2---H* e

j0 e j 0n +

+=

A H ej0 =

e

j 0n H ej + +

ej 0n H e

j + + +

2-------------------------------------------------------------------------------------------------

y n A H ej0 0

n H ej0 + + cos=

0=

x n X0 Xk kn Xk+ cosk 1=

N

+=

y n X0H ej0 =

Xk H ejk

kn Xk H ejk

+ + cosk 1=

N

+


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Example: Three Inputs with

The input is

The frequency response is

The input frequencies are

Thus

>> w = 0:pi/200:pi;

>> H = 1 - exp(-j*w) + exp(-j*2*w);>> subplot(211)

>> plot(w,abs(H))

>> grid

>> hold

>> plot([pi/4 pi/4],[0 3],'r')

bk 1 1 1 =

x n 10 4 4---n

8---+ cos 3 3

---n 4--- cos+ +=

H ej 1 e j e

j2+=

ej

1 2 cos+ =

k 0 4 3 =

H ej0 e j0

1 2 0 cos+ 1= =

H ej 4 e j 4 1 2 4 cos+ 0.4142e

j 4= =

H ej 3 e j 3

1 2 3 cos+ 0= =

n 10 1 4 0.4142 4---n

8---

4---+

cos+=

10= 1.65694---n

8---

cos+


9/32



>> plot([pi/3 pi/3],[0 3],'r')

>> ylabel('Magnitude')

>> subplot(212)

>> plot(w,angle(H))

>> grid>> hold

>> plot([pi/4 pi/4],[-2 3],'r')

>> ylabel('Phase (rad)')

>> xlabel('Digital Frequency \omega')

We have used frequency domain analysis to complete this

example

We could also perform a time domain analysis using say

MATLABs filter function (more on this in the next section)

0 0.5 1 1.5 2 2.5 3 3.50

0.5

1

1.5

2

2.5

3

Magnitude

0 0.5 1 1.5 2 2.5 3 3.52

1

0

1

2

3

Digital Frequency

Phase(rad)

H ej

4---

3---

4---

0.4142 0


10/32

Steady-State and Transient Response



The frequency response definition relies on the fact that the

support interval for the complex sinusoid input is

In a computer simulation or in a real-time signal processing

application, it is not practical to consider input signals which

begin at

Consider an input that begins at using the unit step

function to turn on the input

(6.14)

where and

(6.15)

For an LTI FIR system, the convolution sum formula yields

(6.16)

n 0=

x n Xejnu n =

Aej

=

u n 1, n 00, otherwise

n bkXej n k

u n k k 0=

M

=

h k

k0 M

kn 0

. . .

u n k


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The sum of(6.16) is composed of three cases

(6.17)

When the input is not present, so the output is zero

When the input has partially engaged the filter and

the output produced is the transient response

When , the output is in steady-state, since the input has

fully engaged the filter

Note that the steady-state response is also equivalent to

saying

(6.18)

Note also that the steady-state output assumes that the

input is always for

Example: Revisit Three Inputs with

The frequency domain analysis used in the previous example

obtained just the steady-state portion of the output

We can evaluate the convolution sum or use MATLABs filter

function to obtain a solution of the form described in (6.17)

n

0, n 0

bkejk

k 0=

n

Xejn, 0 n M

bkejk

k 0=

M

Xejn

, n M

=

n 0

0 n M

n M

n H ej Xej n M=

Xejn

n 0

bk 1 1 1 =


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>> n = -5:20;

>> x = (10 + 4*cos(pi/4*n+pi/8) + ...

3*cos(pi/3*n-pi/4)).*ustep(n,0);

>> y = filter([1 -1 1],1,x);

>> subplot(211)>> stem(n,x,'filled')

>> grid

>> ylabel('x[n]')

>> subplot(212)

>> stem(n,y,'filled')

>> hold

Current plot released

>> stem(n,y,'filled')>> grid

>> ylabel('y[n]')

>> xlabel('Sample Index n')

5 0 5 10 15 200

5

10

15

20

x[n]

5 0 5 10 15 205

0

5

10

15

20

y[n]

Sample Index n

TransientInterval

Steady-state


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The function ustep() is defined as:

function x = ustep(n,n0)

% function x = ustep(n,n0)

%

% Generate a time shifted unit step sequence

x = zeros(size(n));

i_n0 = find(n >= n0);

x(i_n0) = ones(size(i_n0));

If we consider just the steady-state portion of the output we

should be able to discern the same output as obtained from

the frequency domain analysis of the previous example

The measurements above agree with the earlier example

5 10 15 208

8.5

9

9.5

10

10.5

11

11.5

12

y[n]

Sample Index n

2 1.6568 3.3138=

x[n] cos() peak

y[n] peaklag 1/2sample =/8

DC = 10


14/32

Properties of the Frequency Response



Relation to Impulse Response and Difference Equation

In the definition of the frequency response, , for anFIR filter, the coefficients were utilized

Similarly given the frequency response, the impulse response

can be found

In Chapter 5 we saw that the impulse response and difference

equation are directly related, so in summary given one the

other two are easy to obtain

Example:

Given say

we immediately see that

The difference equation is


H ej

bk

Time Domain Frequency Domain

h n h k n k k 0=

M

= H ej

h k ejk

k 0=

M

=

h n H ej

h n 2 n 3 n 1 2 n 2 +=

bk 2 3 2 =

y n 2x n 3x n 1 2x n 2 +=

H ej 2 3e j 2e j 2+=


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Example: Difference Equation from

Suppose that

We immediately can write that

and

It could have been that

started in this form

Example:

We need to convert the sine form back to complex exponen-

tials

The impulse response is thus?

H ej

H ej 1 4e j 2 4e j 4+ +=

h n n 4 n 2 4 n 4 + +=

n x n 4x n 2 4x n 4 + +=

H ej

1 8ej2 e

j2e

j2+

2-----------------------------

+=

1 8ej2

2 cos+=

H ej 2j 2 e j 2sin=

H ej 2je j 2 e

j 2e

j 2

2j----------------------------------

=

1 ej

=


16/32



The difference equation is?

Periodicity of

For any discrete-time LTI system, the frequency response is

periodic, that is

(6.19)

The proof for FIR filters follows

(6.20)

This result is consistent with the fact that sinusoidal signals

are insensitive to frequency shifts, i.e.,

In summary, the frequency response is unique on at most a

interval, say in particular the interval

Conjugate Symmetry

The frequency response is in general a complex

quantity, in most cases obeys certain symmetry properties

H ej

H ej 2+ H e

j

H ej 2+

bkej 2+

k 0=

M

=

bkej

ej2

k 0=

M

H ej

==

1

2

x n Xej 2+ n

Xejn

ej2n

Xejn

= = =

2

H ej


17/32



In particular if the coefficients of a digital filter are real, that

is or , then

(6.21)

proof

(6.22)

A consequence of conjugate symmetry is that

(6.23)

We say that the magnitude response is an even function of

and the phase is an odd function of

It also follows that

(6.24)

We say that the real part response is an even function ofand the imaginary part is an odd function of

When plotting the frequency response we can use the above

symmetry to just plot over the interval

bk bk*= h k h* k =

Conjugate Symmetry: H ej H* ej =

H* ej bke

jk

k 0=

M

*

bk*

ejk

k 0=

M

= =

bkej k

k 0=

M

H e

j ==

H ej

H ej

=

H ej H ej

=

Re H ej

Re H ej

=

Im H ej Im H ej =

0


18/32

Graphical Representation of the Frequency Response


Graphical Representation of the Frequency

Response

A useful MATLAB function for plotting the frequencyresponse of any discrete-time (digital) filter is freqz()

The interface to freqz() is similar to filter() in that a

and b vectors are again required

Recall that the b vector holds the FIR coefficients

and for FIR filters we set a = 1.>> help freqz

FREQZ Digital filter frequency response.

[H,W] = FREQZ(B,A,N) returns the N-point complex frequency response

vector H and the N-point frequency vector W in radians/sample of

the filter:

jw -jw -jmw

jw B(e) b(1) + b(2)e + .... + b(m+1)e

H(e) = ---- = ------------------------------------

jw -jw -jnw

A(e) a(1) + a(2)e + .... + a(n+1)e

given numerator and denominator coefficients in vectors B and A. The

frequency response is evaluated at N points equally spaced around the

upper half of the unit circle. If N isn't specified, it defaults to512.

[H,W] = FREQZ(B,A,N,'whole') uses N points around the whole unit circle.

H = FREQZ(B,A,W) returns the frequency response at frequencies

designated in vector W, in radians/sample (normally between 0 and pi).

[H,F] = FREQZ(B,A,N,Fs) and [H,F] = FREQZ(B,A,N,'whole',Fs) return

frequency vector F (in Hz), where Fs is the sampling frequency (in Hz).

H = FREQZ(B,A,F,Fs) returns the complex frequency response at the

frequencies designated in vector F (in Hz), where Fs is the samplingfrequency (in Hz).

FREQZ(B,A,...) with no output arguments plots the magnitude and

unwrapped phase of the filter in the current figure window.

See also filter, fft, invfreqz, fvtool, and freqs.

bk


19/32



Example: Delay System

This filter contains one coefficient, , so

>> [H,w] = freqz([0 0 0 0 1],1);

>> subplot(211)

>> plot(w,abs(H))

>> axis([0 pi 0 1.2]); grid

>> ylabel('Magniude Response')

>> subplot(212)

>> plot(w,angle(H))>> axis([0 pi -pi pi]); grid

>> ylabel('Phase Response (rad)')

>> xlabel('hat(\omega)')

n x n n0 =

bn01=

H ej e j

n0 1 n0= =

0 0.5 1 1.5 2 2.5 30

0.5

1

Magnitude

Response

0 0.5 1 1.5 2 2.5 3

2

0

2

PhaseResponse(rad)

hat()

n0 4=

4

Constant magnitude (gain) response

Linear Phase

MATLABwraps thephase mod2


20/32



Example: First-Difference System


>> w = -pi:pi/100:pi;

>> H = freqz([1 -1],1,w); %use a custom w axis

>> subplot(211)

>> plot(w,abs(H))

>> axis([-pi pi 0 2]); grid; ylabel('Magniude Response')

>> subplot(212)

>> plot(w,angle(H))>> axis([-pi pi -2 2]); grid;



y n x n x n 1 =

H ej 1 e j 1 cos j sin+= =

3 2 1 0 1 2 30

0.5

1

1.5

2

MagnitudeResponse

3 2 1 0 1 2 32

1

0

1

2

PhaseRe

sponse(rad)

hat()

2 2 sin

sin1 cos----------------------

atan


21/32



Example: A Simple Lowpass Filter


>> w = -pi:pi/100:pi;

>> H = freqz([1 2 1],1,w);

>> subplot(211)

>> plot(w,abs(H))

>> axis([-pi pi 0 4]); grid;

>> ylabel('Magnitude Response')>> subplot(212)

>> plot(w,angle(H))

>> axis([-pi pi -pi pi]); grid;



n x n 2x n 1 x n 2 + +=

H ej

1 2ej

ej2

+ + ej

2 2 cos+ = =

3 2 1 0 1 2 30

1

2

3

4

MagnitudeRes

ponse

3 2 1 0 1 2 3

2

0

2

PhaseResponse(rad)

hat()

2 2 cos+


22/32

Cascaded LTI Systems



Cascaded LTI system were discussed in Chapter 5 from the

time-domain standpoint

In the frequency-domain there is an alternative view of the

input/output relationships

To begin with consider again

(6.25)

With LTI #1 followed by LTI #2, the output at frequency is

(6.26)

x n ejn n =

y1 n H2 ej H1 e

j ejn =

H2 ej

H1 ej

ejn

=


23/32



With LTI #2 followed by LTI #1, the output at frequency is

(6.27)

Since , it follows that

, and we have established that the frequency

response for a cascaded system is simply the product of the

frequency responses

(6.28)

We have also shown that a convolution in the time-domain is

equivalent to a multiplication in the frequency-domain

(6.29)

Example: Two System Cascade

Consider

The cascade frequency response is

2 n H1 ej H2 e

j ejn =

H1 ej H2 ej ejn=

H1 ej H2 e

j H2 ej H1 e

j =y1 n y2 n =

Hcascade ej H ej H1 ej H2 ej = =

Time Domain Frequency Domain

h n h1 n *h2 n = H1 ej

H2 ej

H ej

=

H1 ej 1 e j +=

H2 ej

1 2ej

ej2

+ +=

H ej 1 e j

+ 1 2e

je

j2+ + =

1 3ej

3ej2

ej3

+ + +=


24/32

Moving Average Filtering


Given the cascade frequency response we can easily convert

back to the impulse response of the cascade difference equa-

tion

and

Check


The moving average filter occurs frequency enough that we

should consider finding a general expression for the fre-quency response

The difference equation for anL-point averager is

(6.30)


(6.31)

h n n 3 n 1 3 n 2 n 3 + + +=

n x n 3x n 1 3x n 2 x n 3 + + +=

h1 n *h2 n

n 1

L--- x n k

k 0=

L 1

=

H ej

1

L--- e

jk

k 0=

L 1

=


25/32



This sum is known as afinite geometric series

It can be shown (discussed in class) that

(6.32)

Apply (6.32) to (6.31) by setting

(6.33)

Notice that the phase response is composed of a linear term

and due to the sign changes of /

In MATLAB

can be used for analyzing moving average filters

k

k 0=

L 1

1 L1 ---------------, 1

L, 1 (why?)=

=

ej

=

H e

j

1

L---

1 ejL

1 ej

----------------------

=

1

L---

ejL 2

ejL 2

ej L 2

ej 2

ej 2

ej 2

----------------------------------------------------------------

=

L 2 sin

L

2 sin

--------------------------- e j L 1

2=

ej L 1 2 L 2 sin

L 2 sin

DL ej

diric L L 2 sin

L 2 sin---------------------------= =


26/32



Plotting the Frequency Response

The frequency response can be plotted most easily using

MATLABs freqz() function

Consider a 10-point moving average

>> w = -pi:pi/500:pi;

>> H = freqz(ones(1,10)/10,1,w);

>> subplot(211)

>> plot(w,abs(H))

>> grid; axis([-pi pi 0 1])

>> ylabel('Magnitude Response')

>> subplot(212)

>> plot(w,angle(H))

>> grid; axis([-pi pi -pi pi])



3 2 1 0 1 2 30

0.5

1

Magnitude

Response

3 2 1 0 1 2 3

2

0

2

PhaseRes

ponse(rad)

hat()

L = 10

ej4.5

on this segment

2L------=

5--- 2

5------

35

------45

------


27/32

Filtering Sampled Continuous-Time Signals



Consider the system model as shown below

Suppose that

(6.34)

We know that after sampling we have

(6.35)

The key relationship to connect the continuous-time and the

discrete-time quantities is

(6.36)

As long as the input analog frequency satisfies the sampling

theorem, i.e., or , the output of the dis-

crete-time system will be

x t Xejt

= t

x n x nTs XejnTs

Xejn

= = =

Ts 2f

fs---= =

Ts----- fs= =

f

2Ts------------

2------

fs= =

Ts f fs 2


28/32



(6.37)

We can write in terms of the analog frequency variable

via

(6.38)

At the output of the D-to-C converter we have the recon-

structed output

(6.39)

wherefis the input analog sinusoid frequency (perhaps a bet-

ter notation would be )

Example: Lowpass Averager

Consider a 5-point moving average filter wrapped up

between a C-to-D and D-to-C system

We assume a sampling rate of 1000 Hz and an input com-

posed of two sinusoids

Find the system frequency response in terms of the analog

frequency variablef, and find the steady-state output

We will use freqz() to obtain the frequency response>> w = -pi:pi/100:pi;

>> H = freqz(ones(1,5)/5,1,w);

>> subplot(211)

y n H ej Xejn=

y n

Ts=

y n H ejTs Xe

jTsn=

t H ejTs Xe

jt=

H ej2f fs Xe

j 2f fs t=

0 0 0

x t 2 100 t cos 3 2 300 t cos+=

t


29/32



>> plot(w,abs(H))

>> axis([-pi pi 0 1]); grid

>> ylabel('Magnitude Response')

>> subplot(212)

>> plot(w,angle(H))>> axis([-pi pi -pi pi]); grid



>> subplot(211)

>> plot(w*1000/(2*pi),abs(H))

>> grid

>> ylabel('Magnitude Response')>> subplot(212)

>> plot(w*1000/(2*pi),angle(H))

>> grid


>> xlabel('f (Hz)')

3 2 1 0 1 2 30

0.5

1

Magnitude

Response

3 2 1 0 1 2 3

2

0

2

PhaseRes

ponse(rad)

hat()

Magnitude and Phase Plots ofH ej


30/32



The output will be of the same form as the input ,

except the sinusoids at 100 and 300 Hz need to have the filterfrequency response applied

Note 100 Hz and 300 Hz < 1000/2 = 500 Hz (no aliasing)

To properly apply the filter frequency response we need to

convert the analog frequencies to the corresponding discrete-

time frequencies

(6.40)

500 400 300 200 100 0 100 200 300 400 5000

0.5

1

MagnitudeR

esponse

500 400 300 200 100 0 100 200 300 400 5004

2

0

2

4

PhaseResponse(rad)

f (Hz)

Magnitude and Phase Plots ofH ej2f fs

t x t

100Hz 2100

1000------------ 2 0.1 0.2= =

300Hz 2300

1000------------ 2 0.3 0.6= =


31/32



The frequency response for in the general moving

average filter is

(6.41)

The frequency response at these two frequencies is

(6.42)

>> diric(0.2*pi,5)

ans = 0.6472

>> diric(0.6*pi,5)

ans = -0.2472 % also = 0.2472 at angle +/- pi

The filter output is

(6.43)

and the D-to-C output is

(6.44)

L 5=

H e

j

2.5 sin5 2 sin--------------------------- e

j2

=

H ej0.2

2.5 0.2 sin5 0.2 2 sin---------------------------------------e

j2 0.2 0.6472e

j0.4= =

H ej0.6

2.5 0.6 sin

5 0.6 2 sin---------------------------------------e

j2 0.6 0.2472e

j1.2= =

0.2472e j0.2

y n

n 0.6472 0.2n 0.4 cos=

0.7416+ 0.2n 1.2 + cos

y t 0.6472 2 100 t 0.4 cos=

0.7416+ 2 300 t 0.2 cos


32/32


Interpretation of Delay

In the study of the moving average filter, it was noted that

where is a purely real function

The pure delay system

has impulse response

and has frequency response

We also know that the impulse response of two systems in

cascade is

The L-point moving average filter thus incorporates a delay

of samples

Viewed as a cascade of subsystems

(6.45)

IfL is odd then the delay is an integer

IfL is even then the delay is an odd half integer

H e

j

DL ej

ej L 1 2

=

DL ej

n x n n0 =

h n n n0 =

H ej e

jn0=

h n h1 n *h2 n =

H ej H1 e

j H2 ej =

L 1 2

H ej DL e

j e j L 1 2=

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Chap6_Frequency Response of FIR Filters