Ch10-1(FIR Filter Design) U

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FIR Filt D iFIR Filt D iFIR Filter DesignFIR Filter Design

1 Finite Impulse Response Filter Format1.Finite Impulse Response Filter Format

2.Fourier Transform Design

3.Window Method

FIR Filter Format (1)FIR Filter Format (1)

Input-output relationshipK

b FIR filt ffi i t

( ) 1K

y n b x n i b x n b x n b x n K

• bi: FIR filter coefficients• K+1: FIR filter length

Transfer Function– 1

KY z b X z b z X z b z X z

Y zH z b b z b z

Ch7. Finite Impulse Response Filter Design 2

FIR Filter Format (2)FIR Filter Format (2)

Ex 7.1 Given the following FIR filter:Ex 7.1 Given the following FIR filter:

Y(n)=0.1x(n)+0.25x(n-1)+0.2x(n-2)Determine the transfer function filter length nonzero Determine the transfer function, filter length, nonzero coefficients, and impulse response.

Properties from FIR filter format Properties from FIR filter format– All the poles are at the origin STABLE– Operations includep

• Multiplying the filter inputs by the corresponding filter coefficients and accumulating them

Fourier Transform DesignIdeal Low Pass Filter (1)Ideal Low Pass Filter (1)

Frequency Response

– 1,0,

Fourier Transform DesignIdeal Low Pass Filter (1)

Periodicity of the ideal lowpass frequency

Ideal Low Pass Filter (1)

response

Fourier Transform DesignImpulse Response (1)Impulse Response (1)

Discrete-time Fourier Transform

– 12

j j nx n X e e d

– j j n

X e x n e

Impulse response of the ideal lowpass filter

– 1 1 for 2 2

cj j n j nh n H e e d e d n

, 0c n

– sin

Fourier Transform DesignImpulse Response (2)Impulse Response (2)

– Plot

– Theoretically h(n) exists for -∞<n<∞ and is symmetrical about n=0. (h(n) = h(-n))

– As n increases, |h(n)| decreases.As n increases, |h(n)| decreases.

Fourier Transform DesignCausal FIR Filter (1)Causal FIR Filter (1)

Infinite length of filter coefficients– Truncated for FIR filter–N l

11 0 1M MH z h M z h z h h z h M z Noncausal

Causal FIR FilterDelay the truncated impulse response h(n) by M – Delay the truncated impulse response h(n) by M samples

– 1 20 1 2

MMH z b b z b z

for 0,1, , 2nb h n M n M

Fourier Transform DesignCausal FIR Filter (2)Causal FIR Filter (2)

Fourier Transform DesignExample 7 2 (1)Example 7.2 (1)

a. Calculate the filter coefficients for a 3-tap FIR lowpass filter with a cutoff frequency of 800 Hz and a sampling rate of 8,000 Hz using the Fourier transform methodFourier transform method.

b. Determine the transfer function and difference equation of the designed FIR systemequation of the designed FIR system

c. Compute and plot the magnitude frequency response for Ω = 0, π/4, π/2, 3π/4, and πresponse for Ω 0, π/4, π/2, 3π/4, and πradians.

Fourier Transform DesignObservationsObservations

Gibbs effect– The oscillations exhibited in the passband (main lobe)

and stop band (side lobes) of the magnitude frequency responseresponse

– Originates from the abrupt truncation of the infinite impulse response

– Window functions will be used to remedy the problem

A large number of the filter coefficientsSh ll ff h t i ti f th t iti b d– Sharp roll-off characteristics of the transition band

– Increased time delay and increased computational complexity for implementing the designed FIR filter.p y p g g

The phase response is linear in the passband– Symmetrical coefficients (odd number)

Window MethodIntroductionIntroduction

Fourier transform design with window functions– Remedy the undesirable Gibbs oscillations in the

passband and stopband of the designed FIR filterGradually weight the designed FIR coefficients down to – Gradually weight the designed FIR coefficients down to zeros at both ends for the range of -M≤n≤M

FIR filter coefficients– hw(n) = h(n)w(n)

• w(n): window function• h(n): ideal impulse response

Window MethodWindow function (1)Window function (1)

Rectangular window( ) 1 M≤ ≤M (7 15)– wrec(n) = 1, -M≤n≤M (7.15)

Triangular (Bartlett) window

– (7.16)

Hanning window

1 , tri

nw n M n M

Hanning window

– (7.17)

Hamming window

0.5 0.5cos , han

nw n M n M

Hamming window

– (7.18) 0.54 0.46 cos , ham

nw n M n M

Blackman window

– (7.19) 20.42 0.5 cos 0.08 cos , black

n nw n M n M

Window MethodWindow function (2)Window function (2)

Window MethodExample 7 4 (1)Example 7.4 (1)

Given the calculated filter coefficientsh(0)=0.25, h(-1)=h(1)=0.22508, h(-2)=h(2)=0.15915,

h(-3)=h(3)=0.007503( ) ( )

a. Apply the Hamming window function to obtain windowed coefficients hw(n).

b Plot the impulse response h(n) and windowed impulse b. Plot the impulse response h(n) and windowed impulse response hw(n).

Window MethodDesign procedureDesign procedure

1. Obtain the FIR filter coefficients h(n) via the Fourier transform method (Table 7.1)

2. Multiply the generated FIR filter coefficients by h l d i d the selected window sequence

where w(n) is chosen to be one of the window functions listed in , , ,0,1, , ,wh n h n w n n M M where w(n) is chosen to be one of the window functions listed in eqs. (7.15) to (7.19) in page 22.

3. Delay the windowed impulse sequence hw(n) by M l t t th i d d FIR filt M samples to get the windowed FIR filter coefficients:

, 0, 1, , 2n wb h n M n M

a. Design a 3-tap FIR lowpass filter with a cutoff frequency of 800 Hz and a sampling rate of 8,000 Hz using the Hamming window function.

b. Determine the transfer function and difference equation of the designed FIR systemequation of the designed FIR system.

C t d l t th it d f c. Compute and plot the magnitude frequency response for Ω=0, π/4, π/2, 3π/4, and π radians.

Window MethodE l 7 6Example 7.6

a. Determine a 5-tap FIR band reject filter with a lower cutoff frequency of 2,000 Hz, an upper frequency of 2,400 Hz, and a sampling rate of 8 000 Hz using the Hamming window function8,000 Hz using the Hamming window function.

b Determine the transfer functionb. Determine the transfer function.

Window MethodComparison of magnitude frequency responsesComparison of magnitude frequency responses

Window MethodHow to choose a window? (1)How to choose a window? (1)

Specifications: – Fig. 7.14– Table 7.7 FIR filter length estimation

; normalized transition widthstop pass

Filter length for Hamming window: N=3.3/Δf Passband ripple

dB 20l 1 –

Stopband attenuation 10 dB = 20log 1p p

dB 20log –

Cut-off frequency– f = (f + f )/2

10 dB = 20logs s

– fc = (fpass + fstop)/2

Window MethodH t h i d ? (2)How to choose a window? (2)

Window MethodHow to choose a window? (3)How to choose a window? (3)

Window MethodExample (1)Example (1)

Ex. A bandpass filter must be designed according to the following specifications:passband 150 – 250 Hztransition width 50 Hzpassband ripple 0.1 dBstopband attenuation 60 dBsampling frequency 1 kHz

Window MethodExample (2)

a) Specify the desired frequency response of filter, (H ( ))

Example (2)

(HD(ω)).b) Obtain hD[n].c) Find all the window functions which satisfy the given c) Find all the window functions which satisfy the given

specifications from the following table.d) Find the required value for N to use each window

f ti f th lt f ( )function from the results of (c).e) Mention the window function having the smallest value

of N, and find h(0) for the window function., ( )

Ch10-1(FIR Filter Design) U

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