This document discusses the design of finite impulse response (FIR) filters. It begins by describing the basic FIR filter model and properties such as filter order and length. It then covers topics such as linear phase response, different filter types (low-pass, high-pass, etc.), deriving the ideal impulse response, and filter specification in terms of passband/stopband edges and ripple levels. The document concludes by outlining the common FIR design method of windowing the ideal impulse response, describing popular window functions, and providing a step-by-step example of designing a low-pass FIR filter using the Hamming window.

1. EC533: Digital Signal Processing
5 l l
Lecture 8
FIR Filter Design

2. 8.1 – FIR Filter
The basic FIR filter is characterized by the following:
N −1
y (n) = ∑ h(k ) x(n − k )
k =0
N −1
H ( z) = ∑ h(k ) z −k
k =0
where, h(k); k=0,1,…, N-1 are the impulse response coefficients of the filter.
• Filter order= N-1.
• Filter Length N
Length=N
• FIR filters can have an exactly linear phase response.

3. 8.2 – Linear Phase Response & Its Implications
• A filter is said to have linear phase response if its phase response satisfies one of
the following relationships;
θ (ω ) = −αω 1
θ (ω) = β −αω 2
•where α, β are constants.
,
If condition 1 is satisfied, the filter will have a constant group & constant
phase delay responses & the impulse response of the filter must have positive
(even) symmetry. i
( ) i.e., ⎧ N −1 ⎫
⎪n = 0,1........ if N is odd ⎪
2
⎪ ⎪
⎪ ( i bl f all types of fil ) ⎪
(suitable for ll f filters)
h(n) = h( N − n − 1); ⎨ ⎬
⎪n = 0,1........ N - 1 if N is even⎪
⎪ 2 ⎪
⎪ ⎪
α = ( N − 1) 2 ⎩ (not suitable for HPF) ⎭

4. 8.2 – Linear Phase Response & Its Implications
– cont.
When condition 2 is satisfied, the filter will have a constant group delay only
& the impulse response of the filter will have a negative (odd) symmetry. i.e.,
⎧ N −1 ⎫
⎪ n = 0,1........ if N is odd ⎪
2
⎪ ⎪
⎪ N ⎪
h(n) = −h( N − n − 1); ⎨ n = 0,1........ - 1 if N is even ⎬
⎪ 2 ⎪
⎪(suitable for Hilbert Transformers and Diffrentiators) ⎪
⎪ ⎪
⎩ ⎭
α = ( N − 1) 2 ;
β =π /2

5. 8.3 - Filter Types
• The desired (Ideal) frequency response of different filters:
LPF HPF
ω (normalized)
BPF BSF
ω1 ω2 ω1 ω2
The amplitude response is periodic in frequency with a period ωs & symmetric around the
Nyquist frequency ωN = ωs /2.

6. 8.4 – Deriving Filter Impulse Response
As HD(ω) is the desired amplitude response & it is periodic in frequency with
period ωs ,We can obtain the filter impulse response hD(n) – filter coefficients ak’s-
s
by evaluating the IDTFT of HD(ω) as follows;
∞
− jω n
H D (ω ) = ∑ h D ( n ).e DTFT
m = −∞
1 π jω n 1 ωc jω n
hD ( n ) =
2π ∫− π H D (ω ).e d ω =
2π ∫− ω c1.e d ω
ωc
1 ⎡e jω n
⎤ fc ⎡ 2 sin( n ω c ) ⎤
= ⎢ ⎥ = ⎢ ⎥
2 π ⎣ jn ⎦ − ω
j 2π ⎣ nf c
f ⎦
c
2 f c sin( n ω c )
=
nω c
Where fc is the normalized cutoff frequency

7. 8.5 – Ideal Impulse Response for Different Filter Types
where fc,f1 and f2 are the normalized passband or stopband edge frequencies.

8. 8.6 - FIR Filter Specification
jw
H(e )
• ωp ‐ passband edge frequency
• ωs ‐ stopband edge frequency
• δp ‐ peak ripple value in the passband
• δs ‐ peak ripple value in the stopband
Ap = 20 log10 (1 + δp) dB Peak passband ripple (dB)
p pp ( )
As = -20 log10 (δs) dB Stopband attenuation (dB)
∆(ω)= ωs - ωp≡ Transition width Transition Width

9. 8.7 – FIR coefficient Calculation Methods
• The filter coefficients - h(n)- can be calculated by using
several methods as the window, optimal, & frequency
sampling methods.
li h d
• All of these methods can lead to linear phase FIR filter.

10. 8.8 – Window Method
H(ejw )
1
-∞ ∞
ω
W(ejw ) w(nT) 2
noncausal
ω ω
− (N - 1 ) 2 (N - 1 ) 2
Gibb’s jωT jωT h w (n) = h(nT).w(nT)
Oscillations H (e ) ⊗ W (e )
Noncausal change
to causal by shifting
by (N‐1)/2

11. 8.8.1 – Effect of the Window Size
As we increase the width of the window (n), its spectrum will be sharper & the
result of convolution will be near to ideal.
As th
A the resulting ideal impulse response does not lead to FIR filter, the ideal
lti id l i l d tl dt filt th id l
impulse response can be truncated by setting hD(n)=0; for n>M (say). But this
introduces the so called Gibb’s oscillations.
Filter designed by windowing method has equal passband and stopband ripples
M=5 M=20 M=50
This is done by multiplying the ideal impulse response by rectangular window of
the form:
⎧1 ; n = 0,1,..., M ; M = ( N − 1) 2
w(n) = ⎨
⎩0 ; else where

12. 8.9 – Some Common Window Functions
• A practical approach to design an FIR filter is to multiply the ideal impulse
response by a window function, w(n), whose duration is finite other than the
rectangular window. This l d to the resulting impulse response decaying
l d Th leads h l l d
smoothly towards zero, so the Gibb’s oscillations will be reduced. But the
transition width will be wider than the case when the rectangular window is used.

13. 8.9.1 – Rectangular Window
⎧1 ; n = 0,1,..., M ; M = ( N − 1) 2
w(n) = ⎨
⎩0 ; else where
time frequency

14. 8.9.2 – Hanning Window
⎧ 0 .5 + 0 .5 cos( 2π n ) ⎧− (N − 2) 2 ≤ n ≤ (N − 2) 2 (N odd)
⎪ ⎨
w(n) = ⎨ N −N 2≤n≤ N 2
⎪ 0 ; else where ⎩ (N even)
⎩
time frequency

15. 8.9.3 – Hamming Window
⎧0.54 + 0.46 cos(2πn ) ⎨ ⎧− (N − 2) 2 ≤ n ≤ (N − 2) 2 (N odd)
⎪ (
w(n) = ⎨ N −N 2≤n≤ N 2
; else where ⎩
(N even)
⎪
⎩ 0
time frequency

16. 8.9.4 – Blackman Window
⎧ 2πn 4πn
⎪0.42 + 0.5 cos( ) + 0.08cos( )
w(n) = ⎨ N −1 N −1
⎪
⎩ 0 ; else where
⎧−(N−2) 2≤n≤(N−2) 2 (N odd)
⎨
⎩ −N 2≤n≤ N 2 (N even)
time frequency

17. 8.10 – Summary of Important Features of Common
Window Functions

18. 8.11 – FIR Design Steps by Windowing
Step 1: Specify the ‘ideal’ or desired frequency response of filter, HD(ω).
Step 2: Obtain the impulse response, hD(n), of the desired filter by evaluating
p p p ( ) f f y g
the inverse DTFT. For the standard frequency selective filters the expressions for
hD(n) are summarized in table stated in slide 7
Step 3: Select a window function that satisfies the passband or attenuation
specifications & then determine the number of filter coefficients using the
appropriate relationship between the filter length & the transition width, ∆f
(expressed as a fraction of the sampling frequency).
Step 4: Obtain values of w(n) for the chosen window function & the values of
the actual FIR coefficients, h(n), by multiplying hD(n) by w(n) h(n)= hD(n) w(n)
Step 5: shift the filter coefficients by (N-1)/2 to achieve a causal filter
It is clear that the window method is straightforward & involves a minimal amount of
computational effort. On the other hand, it should be said that the resulting filter is
not optimal, that is in many cases a filter with a smaller number of coefficients can
be designed using other methods.

19. Example
Obtain the coefficients of an FIR LPF to meet the specifications
given below;
• Passband edge frequency 1.5 kHz
• Transition width 0.5 kHz
• Stopband attenuation >50 dB
• Sampling frequency 8 kHz
Solution sin(nωc )
Select hD(n) for LPF which is given by hD (n) = 2 f c ;n ≠ 0
nω c
hD (n) = 2 f c ; n = 0
The table in slide 17 indicates that the Hamming, Blackman, or Kaiser window
will satisfy the stopband attenuation requirements. We will use the Hamming
ill i f h b d i i W ill h H i
window for simplicity.
Now ∆f =0.5/8=0.0625.
From N=3.3/ ∆f =3.3/0.0625=52.8, l t N=53
F N 3 3/ 3 3/0 0625 52 8 let N 53 odd

20. Example – cont.
The filte coefficients are obtained from h(n)= hD(n) w(n);
he filter ae f om h(n) -26 ≤ n ≤ 26
6 6
where sin( nωc )
hD (n) = 2 f c ;n ≠ 0
nω c
hD (n) = 2 f c ; n = 0
w(n) = 0.54 + 0.46 cos(2πn );−26 ≤ n ≤ 26
53
Because of the smearing effect of the window on the filter response, the cutoff
frequency of the resulting filter will be different from that given in the specifications.
To account for this, we will use fc that is centered on the transition band:
′
fc = fc + ∆f = (1.5 + 0.25)kHz = 1.75kHz 1.75/8 = 0.21875
2
Nothing that h(n) is symmetrical, we need only to compute values for h(0),h(1), …,h(26)
g ( ) y , y p f ( ), ( ), , ( )
& then use the symmetry property to obtain the other coefficients.
n=0: hD(0) = 2 fc = 2 * 0.21875 = 0.4375
w(0) = 0.54 + 0.46 cos(0) = 1
h(0) = hD(0) w(0) = 0.4375

21. Example – cont.
2 × 0.21875 sin(360 × 0.21875)
n 1:
n=1: hD (1) = sin( 2π × 0.21875) = = 0.31219
2π × 0.21875 π
w(1) = 0.54 + 0.46 cos(2π/53) = 0.54 + 0.46 cos(360/53) = 0.99677
h(1) = h(-1) = hD(1) w(1) = 0 31118
h( 1) 0.31118
2 × 0.21875 sin(157.5)
n=2: hD (2) = sin( 2 × 2π × 0.21875) = = 0.06013
2 × 2π × 0.21875 2π
w(2) = 0.54 + 0.46 cos(2π*2/53) = 0.54 + 0.46 cos(720/53) = 0.98713
h(2) = h(-2) = hD(2) w(2) = 0.06012
2 × 0.21875
n=26: hD ( 26) = sin( 26 × 2π × 0.21875) = −0.01131
26 × 2π × 0.21875
w(26) = 0.54 + 0.46 cos(2π*26/53) = 0.54 + 0.46 cos(9360/53) = 0.08081
h(26) = h(-26) = hD(26) w(26) = - 0 000914
h( 26) 0.000914

22. Example – cont.
• We note that the indices of the filter coefficients run from -26 to 26. To make the filter
causal (necessary for implementation) we add 26 to each index so that the indices start at
zero. The filter coefficients, with indices adjusted, are listed in the following table. The
spectrum of the filter (is plotted) would indicate that the specifications were satisfied.