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FIR FILTER DESIGN: THE WINDOW DESIGN METHOD Page by: Nguyen Huu Phuong

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FIR FILTER DESIGN: THE WINDOW DESIGN METHOD Page by: Nguyen Huu Phuong

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▸ Summary

The design of a FIR filter starts with its specifications in either discrete-time domain or DTFT

frequency domain, or both. In the time domain, the design objective is the impulse response. In the

frequency domain, the requirement is on various parameters of the magnitude response, shown in

Fig.5.8 for a lowpass filter. Important parameters are band edge frequencies and , passband

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ωp ωs

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, transition band , stopband , cutoff frequency ,

passband ripple (or deviation) and stopband ripple . The two ripples are usually assumed

equal . Notice that for ideal (desired) filter, the passband frequency magnitude is normalized to 1 and

the stopband to 0 , and the frequency response of the designed filter oscillates between the high

amplitude of 1 or the low amplitude of 0.

Figure illustrates the specifications of bandpass filter. Here there are two sets of edge frequencies,

lower and upper. The rising bandwith and the falling bandwidth are usually assumed equal.

Even the initial requirement is imposed on magnitude response but we must design the filter having

linear phase or generalized linear phase.

Overall filter design

From the filter specifications, the first step is to choose between FIR and IIR filters based on their

advantages and disadvantages. This chapter only concerns FIR filters. The next step is to select the

proper linear phase FIR filters (section 5.2.3). A rather complete procedure for the design of FIR filters

are as follows.

• Specifications of the filter

• Choosing an appropriate linear phase filter type (section 5.2.3)

• Choosing the mothod of design such as window, optimal, frequency sampling…

• Calculation the filter coefficients (impulse response)

• Finding suitable structure

• Analysis of the finite wordlength effects (chapter 7)

• Implementation of the filter in hardware and/or software (chapter 7).Fig.5.9: Bandpass filter

specifications

Figure 1. Typical normalized magnitude response specfications of a lowpass filter

[0, ]ωp ∆ω = −ωs ωp =( + )/2ωc ωp ωc

δp δs

∆ωl ∆ωu

Figure 2. Bandpass filter specifications



Fixed windowsThe impulse response of ideal filters is infinite duration (IIR). We cannot evaluate the corresponding

frequency response and, especially, implement the filter by hardware and/or software. Thus we must

truncate the impulse response at both ends with respect to the central. Even we truncate the impulse

response when it is small enough, but such a sudden cutoff will cause some undesired effects. The

window method will reduce them.

In the time domain, windowing means that the we multiply the desired (usually ideal) infinite

duration impulse response by a finite duration window (or window function) to

get a soft truncation. The resulted impulse response of the designed filter is the product

Here we assume that all desired impulse responses and windows are causal, from to

, i.e. window length is samples (time indices). Many authors start the design with

and bi-sided (noncausal), i.e. defined in the interval then shift to the right

indices to make them causal.

Multiplying in time domain corresponds to convolution in frequency domain. Thus the frequency

response of the designed filter (corresponding to the windowed impulse response ) is

where is the Fourier transform (DTFT) of the window .

In the window design method , we first evaluate the desired filter impulse response from the

given desired frequency response , and then apply an appropriate window . Thus the method

should be called the Fourier – window method.

Retangular window

Actually, the sudden truncation mentioned earlier is the simplest window called rectangular window,

defined as (Figure)

(n)hd w(n)h(n)

h(n) = (n)w(n), 0 ≤ n ≤Mhd

n = 0n =M M + 1(n)hd w(n) M/2

h(n)

H(ω)= (ω)∗W(ω)= (ω')W(ω− ω')dω'Hd

1

2π∫

π

−π

Hd

W(ω) w(n)

(n)hd(ω)Hd

Figure 3. Rectangular window of order M or length M + 1



The Fourier transform of this window is

This is the same as Equation (3.54) when replacing by . Also, instead of the above

expression we can use the sum of cosines (Equations (3.46), (3.53)). has includes a phase

factor showing the time shift of a two-sided symmetric window to a causal window. The magnitude

and phase responses are, respectively,

where is 0 or as discussed in previous section. Thus the window is linear phase. Fig.3.27

illustrates the variation of the amplitude of (remember to replace by ). It is

maximum and equals to at . The zero-crossing points are multiples of

w(n)= , 0 ≤ n ≤M1

, otherwise0

W(ω) = w(n) = =∑n=−∞

∞

e−jωn ∑n=0

M

e−jωn1 − e−jω(M+1)

1 − e−jω

, ω ≠ 0= e−j ωM

2sin ω(M+1)/2

sin ω/2

M + 1 , ω = 0

M 2NW(ω)

Figure 4. Normalized frequency response of rectangular window for M = 30 and 50

|W(ω)| = , ω ≠ 0∣ ∣ sin ω(M+1)/2

sin ω/2∣ ∣

ω = 0M + 1 ,

Φ(ω) = − ω+ βM

2

β π∣W(ω)∣ 2N M

M + 1 ω = 0



. The response consists of a main lobe and many sidelobes. As gets larger the

main-lobe gets smaller, and there are more sidelobes, also narrower but the amplitude of the first

sidelobe remains the same (Fig.5.11). As , tends to be an unit sample .

The oscillatory when convolved with the ideal lowpass response will result in a

response having a non-zero transition width and ripples is both the passband and stopband.

Fig.5.12 shows the designed lowpass filter when using rectangular window having to

abruptly truncate the impulse response of an ideal lowpass filter with cutoff frequency .

Fig.5.13 shows the magnitude in linear scale for two case of filter order, M = 10 and M = 18.

In order that approaches the rectangular window must be of infinite duration, i.e. we

must take into account the whole impulse response without truncation. In frequency domain this

means is an unit sample as said. We can reason in the reverse direction: Since

then must be a unit sample and the windows must be of

infinite duration. When we simulate on a computer using Maltlab sofware, or else, with in

the hundreds we will still see the non-zero transition width and ripples due to the Gibbs phenomenon

(section 3.1.4). An infinitely long windows is not practical, so the idea is to look for finite duration

windows which perform better than the rectangular.

2π/(M + 1) MM →∞ W(ω) δ(ω)

Figure 5. Magnitude response of designed lowpass filter when using rectangular window with M = 44

∣W(ω)∣ (ω)Hd

H(ω)M = 44

= π/2ωc

H(ω) (ω)Hd

W(ω) δ(ω)H(ω)= (ω)∗δ(ω)= (ω)Hd Hd δ(ω)

H(ω) M



Actually in examples 5.2.1 and 5.2.2 the truncation of the impulse responses of the ideal filters meant

the rectangular window had been applied.

Other windows

The Gibbs phenomenon can be reduced considerably by the use of a softer (less abrupt) truncation,

i.e. by tapering the rectangular smootly to zero at both ends. Unfortunately, the reduction in amplitude

of sidelobes of is accompanied by a widening of its mainlobe. For comparision, the window

responses are usually plotted in scale.

Figure 6. Magnitude response of designed filter for M = 10 and M = 18

H(ω)W(ω) dB



Several smooth window functions have been proposed and used. It might come to our mind that the

first applicant would be the triangular window, also called Bartlett window , depicted in Fig.5.14.

From the figure we can write the window function ( table 5.1). Table 5.1 lists common fixed windows

and Fig.5.15 plots their functions. Fig.5.16 plots the dB magnitude for and Fig.5.17 for M = 50.The first

sidelobe is at -25dB compared to the -13dB of the corresponding (same length) rectangular window.

However the mainlobe width is about twice as large.

The sidelobe of the triangular window is still high because the tapering is still rather coarse. For

smoother tapering, cosinusoid is incorporated into the window function. This observation has led to

the three well known windows: Hanning (or von Hann), Hamming, and Blackman, all defined in the

interval , otherwise zero (table 5.1). We can check that the functions are normalized

(peak value of 1 at ) and zero at both ends except the rectangular and the Hamming

(0.08 instead of 0). Besides, there are many other less-used fixed windows.

Notice that all windows mentioned, from rectangular to Blackman, are simple functions easily

evaluated, and their frequency responses concentrate around (the mainlobe) as expected.

Also notice that all the windows mentioned are symmetric about the mid-point , this

when combined with the symmetry or antisymmetry of the filter’s impulse response will make the

corresponding designed filter linear phase or generalized linear phase.

Figure 7. Triangular window (Bartlett window)

Figure 8. Common window plotted as function of continuous time n (whereas actual windows are function

of discrete index n)

0 ≤ n ≤Mn =M/2

ω = 0n =M/2



Figure 9. dB magnitude frequency response of fixed windows for M = 20 (a) Bartlett, (b) Hanning, (c)

Hamming, (d) Blackman



Table 5.2 lists various features of common windows for comparison. The passband ripple and the

stopband ripple and the stopband ripple , are taken equal (the smaller of the two):

Fig.5.16, Fig.5.17 and Table 5.2 show that no window is the best in all aspects but there is a tradeoff

of features, and the choice of an appropriate window depending on our requirement. For smallest

mainlobe width it’s the rectangular window, for best sidelobe attenuation it’s the Blackman. In

between, the Hamming is a good choice. The Bartlett is the transition from the rectangular to the other

windows. See section 5.3.3 for design examples.

Figure 10. dB magnitude frequency response of various windows for M = 50 (a) Bartlett, (b) Hanning, (c)

Hamming, (d) Blackman

δs δs

A = −20 [min( , )]log10 δp δs



The Kaiser windowThe windows discussed so far are fixed windows, only the length is adjustable. The

Kaiser window has an additional ripple parameter , enabling the designer to tradeoff the transition

and ripple. It is defined in the interval , otherwise zero:

where is the modified zeroth-order Bessel function, computed by a power series

expansion:

Following is the values of for small :

The upper limit of the summation does not need to be too large, say about 20 at the most.

(M + 1)β

0 ≤ n ≤M

w(n) = , 0 ≤ n ≤M

{ }I0 1 − ( )n−M/2

M/2

− − − − − − − − − − − √β 2

(β)I0

(x)I0

(x)= 1 + = 1 + + ⋅ ⋅ + ⋅ ⋅ + ⋅I0 ∑k=1

∞ [ ]x(2)k

k!

2

x2

4

1

22x2

4

x2

4

1

z2x2

4

x4

16

(x)I0 x

X (n) X (n)I0 I0

0.0 1.000 1.2 1.3938

0.1 1.002 2.0 2.2796

0.2 1.010 3.0 4.3306

0.4 1.040 4.0 11.302

0.6 1.092 5.0 27.240

0.8 1.166 6.0 67.235

1.0 1.266 7.0 168.59



For both the numerator and denominator are 1 and the Kaiser window becomes the

rectangular window, and for it is near to the Hamming window. The parameter is so

chosen that the magnitude response of the filter lies in the allowed region (see Fig.5.1). Fig.5.18

shows the Kaiser for various values of and . Notice that the Kaiser is also symmetric. It has

minimum stopband attenuation from 50dB (for ) to 90dB (for ), so we can

choose a Kaiser window to have maximum stopband atternuation equal to the Blackman window (74,

Table 5.2) or higher.

Figure 11. (a) Kaiser window for and 6 and M = 20, (b) Magnitude response corresponding to

the windows in (a), (c) magnitude response for and M = 10, 20,

β = 0, 3β = 6

β = 0β = 5.44 β

β Mβ = 4.54 β = 8.96



For ideal filters (Fig.5.1 and Fig.5.2) the Kaiser window has equal ripple in both the passband and

stopband. For real filters the ripples are slightly different . In design we take the common ripple as

of the desired values . As said earlier, actually is usually specified in term

of dBs: , for example then , then

. Since is a fraction of 1 then A is always ponsitive.

Kaiser has developed empirical formulae and guide for the application of the window. Following is

design steps for lowpass filters (similarly for other frequency selective filters).

1/ Specifications of edge frequencies ; ripples

2/Evaluate the minimum ripple , the cutoff frequency

, the transition width

3/ Determine the filter filter order M

4/ Determine the parameter , depending on A:

5/ Computer the window coefficients from the formula (5.40)

6/ Compute the impulse response of the desired filter which is linear phase

7/ Compute the designed impulse response .

8/ Find the frequency response to check with specifications. For bandpass

filter the two transition bandwidths are usually equal, if not we take the smaller one.

Design examples

Example

Design a linear phase lowpass filter using the Hamming window to meet the specifications

• Cutoff frequency 2.5 kHz

• Transition width 1.65 kHz

• Stopband attenuation >50dB

• Sampling frequency 10kHz

δδ = min( , )δp δs ,δp δs δ

A = −20 δlog10

δ = 0.01 A = 40dB δ = 0.001A = 60dB δ

,ωp ωs ,δp δs

A = −20 [min( , )]log10

δp δs= ( + )/2ωc ωp ωs △ω = −ωs ωp

M = A−7.952.285∆ω

, A > 21

M = , A ≤ 215.79∆ω

β

β = 0, A ≤ 21

β = 0.5842 + 0.07886(A− 21), 21 ≤ A ≤ 50(A− 21)0.4

β = 0.1102(A− 8.7), A ≥ 50

w(n)

(n)hd

h(n)= (n)w(n)hd

H(ω) = IDTFT [h(n)]



Solution

The relation between the analog linear frequency samples/sec (or Hz) and the digital angular

frequency radians/sample is (Equation 1.40) is

where is the sampling frequency (samples/sec, or Hz). Thus

• Cutoff frequency rad/sample

• Transition width rad/sample

Table 5.2 shows that the Hamming satisfy (so can the Blackman) the maximum stopband attenuation

( requirement. Table 5.2 includes the relation between transition width and

window order :

Now we compute the causal impulse response of ideal lowpass filter (Equation (5.32)) for

. Notice that for even and symmetric we have linear phase FIR-1

(Fig.5.4a). The result is plotted in Fig.5.19a.

Next we evaluate the Hamming window

The result is plotted in Fig.5.19b.

The impulse response of the designed filter is the product

We can evaluale w(n) and separately as above then take the product, or evaluate the

product directly . The result is plotted in Fig.5.19c. We then take the inverse transform of to get

the frequency response :

Fig.5.19d shows , and Fig.5.19c shows .

Example

Fω

ω = 2πF

fs

fs= 2π = 0.5πωc

2.510

∆ω = 2π = 0.33π1.6510

−20 )log10δs ∆ωM

∆ω = M = = 206.6π

M⇒

6.6π

0.33π

(n)hLP

0 ≤ n ≤ 20 M h(n)

w(n) = 0.54 − 0.46 cos , 0 ≤ n ≤M = 202πn

M

h(n) = (n)w(n) , 0 ≤ n ≤M = 20hLP

(n)hLPh(n)

H(ω)

H(ω)= h(n) = h(n)∑∞

n=−∞ e−jωn ∑M

n=0 e−jωn , M = 20

|H(ω)| |H(ω)|dB



We need ro extract a low frequency signal having bandwidth 200Hz and center frequency 300Hz

from a blackground noise. Design a bandpass filter having transition width 100Hz, passband

ripple 40dB, stopband ripple 60dB. The sampling frequency is 1200Hz.

Solution

The passband of the signal is from 300 – 200/2 to 300 + 200/2, i.e. from 200Hz to 400Hz. We are to

design a bandpass FIR filter to meet the specifications

• Passband: 200Hz

• Transition width: 100Hz

Figure 12. Example(linear phase lowpass filter with ) using the Hamming

window

= 0.5π, ∆ω = 0.33πωc



• Passband ripple:

• Stopband ripple:

• Sampling frequency: 1200Hz

From Table 5.2 see that only the Blackman window can satisfy the stopband ripple (so can the Kaiser

window, see next subsection). Since the passband and stopband attenuations are different, we take

the smaller one in the design, i.e. = 60dB. The transition width and the filter order are, respectively,

Figure 13. Example(linear phase bandpass filter using Blackman window

(−20 )= 40dBlog10δp ( = 0.01)δp

(−20 ) = 60dB⇒ = 0.001log10δp δs

A

∆ω = 2π =100

1200

π

6



Thus the Blackman window is

Ideal bandpass filter impulse response (Equation 5.10b)

where

Since M is odd we have linear phase type 2 (FIR-2) . The designed impulse response is

We then take the inverse transform of h(n) to get or the frequency response of the designed filter

(Fig.5.20).

Example

Design a highpass FIR filter using Kaiser window to meet the specifications :

• Cutoff frequency 2,5 kHz

• Transition width 0.5 kHz

• Passband ripple 0.001

• Stopband attenuation 40dB

• Sampling frequency 10kHz

Solution

First let’s derive the expression of the impule response of ideal highpass filter (Fig.5.1b) with

generalized linear phase (Fig.5.4) (Fig.5.1b) with generalized linear phase (Fig.5.4). The frequency

response of the causal highpass filter in the interval is

M = = 66.6 ≈ 6711.1π

π/6

W(n) = 0.42 − 0.5 cos + 0.08 cos , 0 ≤ n ≤M = 672πn

M

4πn

M

(n) = −hBPsin nωu

πn

sin nωl

πn

= 2π =ωu

400

1200

2π

3

= 2π =ωl200

1200

π

3

h(n)= (n)w(n)hBP

ω = [0,π]

(ω) = 0, 0 ≤ ω ≤HHP ωc

, < ω ≤ πe−j ωM

2 ωc



or related to the frequency response of the lowpass filter (Equation (5.32)) as

Taking the inverse DTFT of either above expression we will get the impulse response of the ideal

highpass filter

Notice that we cannot just shift the result of the two-sided symmetric impulse response Equation

(5.10a) to get the above causal result (but must replace by before shifting).

The given specifications are interpreted as

• Cutoff frequency : rad/sample

• Transition width : rad/sample

• Passband ripple :

• Stopband ripple : dB

Thus the attenuation is

Knowing and we can determine the window order

Since dB, the ripple parameter is given by

With and , the Kaiser window (Equation (5.40)) is

The impulse response of the FIR-1 highpass filter is computed for , and the

designed impulse response is

As usual, the frequency response is the inverse DTFT of above impulse response.. Using Matlab we

(ω)HLP

(ω)= − (ω)HHP e−j ωM

2 HLP

h = − , n ≠ 0(n)

HP

sin π(n−M/2)

π(n−M/2)

sin (n−M/2)ωc

π(n−M/2)

1 − , n = 0ωcπ

δ(n) sinπn/πn

= 2π = 0.5πωc

2.510

∆ω = 2π = 0.1π0.510

= 0.001δp−20 δ = 40log10 ⇒ = 0.01δs

A = −20 [min( , )] = −20 = 60 dBlog10 δp δs log10δp

∆ω A

M = = = 72.5 ≈ 72A− 7.95

2.285∆ω

60 − 7.95

2.285(0.1π)

A > 50 β

β = 0.1102(A− 8.7)= 5.65

M/2 = 33 β = 5.65

W(n) ={ }I0 1 − ( )n−36

36

2− − − − − − − − − − √5.65(5.65)I0

(n)hHP 0 ≤ n ≤ 66

h(n)= (n)w(n)hHP



can simulate all the responses. ■

Error in the Fourier design methodLet’s briefly consider the error in the Fourier design. Just like the Fourier series expansion of a signal

(section 3.1), the Fourier design method give the least mean square (LMS) error as presented here.

This also applies to the case of abupt truncation of the desired impulse , that is using the rectangular

window. Let be the error between the desired ideal frequency response and the

designed frequency response :

This error varies with frequency and is a complex quantity , the same as the frequency response. We

are interested in the error amplitude, so we take the square

Since still depends on frequency, we then take its average value:

The problem is to find so that the mean square error is smallest (i.e. least mean square -

LMS). Because the frequency response can be expressed in term of the designed filter impulse

response h(n) , we thus look for the filter impulse response for LMS error. To this end we take the

derivative and equate it to zero to find the root of h(n) for the minimum

. After some mathematics (not presented here) , we will get the impulse response

This result means that when we start the design by evaluating the impulse response from the given

(desired) filter frequency response then we will get the LMS error (i.e. no other design method can

give a smaller mean square error) .

In LMS error we emphasized the large errors ( at local frequencies). This criterion is not always

adopted , other criterion may be more privileged, such as the transition width must be as narrow as

possible, or the error must be distributed evenly in both the passband and the stopband…

ε(ω) (ω)Hd

H(ω)

ε(ω) = (ω) −H(ω)Hd

ε(ω) = ε(ω) + ε(ω) = ε(ω) (ω)∣ ∣ ∣ ∣ 2

Re2 Im2 ε∗

|ε(ω)|2

= |ε(ω) dω = ε(ω) (ω)dωε̄ 12π

∫ π

−π

|2 12π

∫ π

−π

ε∗

= [ (ω) −H(ω)] dω12π

∫π

−π

Hd [ (ω) −H(ω)]Hd2

H(ω) H(ω)H(ω)

d /dh(n)ε̄

ε̄

h(n) = (ω) dω = (n)1

2π∫ π

−π

Hd ejωn hd



Digital differentiatorIn digital signal processing , sometimes we need to take the differentiation or integration of a signal .

For example the differentiation (derivation) of displacement is speed, and the differentiation of speed

is acceleration. It is interesting that the former can just be realized by a FIR filter or by an IIR filter,

whereas the latter by an IIR filter. In this section we will discuss digital differentiators (Fig.5.2)

designed by linear phase FIR-3 and FIR-4 associated with windowing. Other FIR design method ,

such as the optimal method discussed in the subsequent section , can also be used.

In an analog differentiator the output signal y(t) is related to the input signal x(t) by

For a LTI (LSI) digital system having frequency response (i.e.DTFT) , an input

will give the output (section 3.7.3):

On the other hand the derivative of the input is

By comparing the above two expressions we obtain

This is the frequency response of ideal digital differentiator. The response is purely imaginary, the

magnitude response is proportional to the frequency and the phase response is ,

independent of frequency (Fig.5.2) . Thus differentiators are generalized linear phase (Equation (5.17)

with ). Fig.5.21 shows the variation of .

y(t) =dx(t)

dt

H(ω) x(n) = ejnω

H(ω)ejωn

y(n) = H(ω)ejωn

x(n) = ejωn

y(n) = = jωdx(n)

dnejωn

H (ω) = jω , − π ≤ ω ≤ π

|H(ω)| Φ(ω) π/2

α = 0 , β = π/2 |H(ω)|



Even looked so simple but the such a differentiator cannot be realized exactly in practice . Thus

approximate ciruits are used, the simplest of which is the first order differentiator which is just the

difference

Its frequency response can be found to be

whose magnitude is

This is plotted in Fig.5.21 . The approximation is surprisingly good especially at low frequency. Notice

that for small ,

Actually a differentiator is usually designed to use at the low frequency because the high response at

high frequency will enhance noise. This again means that the simple first order differentiator is very

good. The linear phase FIR filter implementing the differentiator is of FIR-4 type (

)

In order to reduce the high response at high frequency, the central differentiator is the candidate :

Figure 14. Frequency response of various types digital differentiators

y(n) = x(n) − x(n− 1)

H(ω) = 1 − = (1 − cosω) + j sinωe−jω

H(ω) = = 2 sin∣

∣ ∣

∣

∣ ∣ [ + ω](1 − cosω)2 sin2

1/2 ω

2

H(ω) ≈ (1 − 1) + jω = jω (small ω)

H(ω) = 0at ω = 0 and ≠ 0 at ω = π

y(n) = 0.5x(n) − 0.5x(n− 2)



having frequency response

The magnitude response is

The response is also plotted in the same Fig.5.21. Notice also that at low . The linear

phase FIR filter implementing the differentiator is of FIR-3 type ( at both and

).

Example

For the ideal differentiator.

2. Find the expression of its two-sided impulse response . Evaluate and plot the response for

with M = 20

4. Find the expression of its causal impulse response.

6. Design a differentiator using the Hamming window of length 21.

Solution

Taking the inverse DTFT of the known frequency response we will get the two-

sided impulse response.

H(ω) = 0.5 − 0.5 = 0.5 [(1 − cos 2ω) + j sin 2ω]e−j2ω

|H(ω)| = sinω

H(ω) ≈ 0 ω

H(ω) = 0 ω = 0ω = π

−M/2 ≤ n ≤M/2

H(ω) = −jω

(n) = (ω) dωhDIF12π

∫ π

−π

Hd ejωn

H(ω) cosωndω + j (ω) dω= 12π

∫π

−π

1

2π∫

π

−π

Hd sinωn



Because is an odd function while cosine is even and sine is odd, we only need to take the last

term of the above expression:

or

The last case is obtained by taking the original integral. Fig 5.22a depicts the impulse response which

is of FIR-3.

2. We cannot just shift the noncausal result Equation (5.49a) M/2 time indices , i.e. by replacing n by n

– M/2 , together the causal impulse response. Let’s start from the frequency response of a causal

linear phase differentiator ( the two-sided response is delayed by M/2 samples):

Taking the inverse transform to get the causal impulse response

Figure 15. Example impulse response having M = 20 of ideal differentiator ((a) two-sided (noncausal), (b)

causal)

(ω)Hd

(n) = j (ω) sinωndω = − ω sinωndωhDIF12π

∫π

−π

Hd1π ∫

π

0

= , n ≠ 0cos nπn

0 , n = 0

(n) = − , n = ±1,±3,±5...hDIF1n

, n = ±2,±4,±6...1n

0 , n = 0

jω

H(ω) = jω , − π ≤ ω ≤ πe−j ωM

2



When putting M/2 = 0 in above Equation we will get back Equation (5.49a) as expected.

In both impusle response (5.49a) and (5.50) we can notice that , thus

the differentiator is linear phase type 3 when M is even (Fig.5.4c) or type 4 when M is odd (Fig.5.4d).

3. Window length of 21 means M + 1 = 21 and the filter order is M = 20. We can use the two-sided

impulse response in (a) and shift M/2 = 10 time indices to the future to get the causal response , or

use the causal response in (b) . Fig.5.23a is the frequency response of the differentiator with

truncation (equivalent to using rectangular window), Fig.5.23b is the frequency response of the

differentiator when using the Hamming window. The optional method (later section) is a better method

to design digital differentiator.

Downloads + History + Attribution +

(n) = − , n ≠M/2hD/Fcos(n−M/2)π

(n−M/2)

sin(n−M/2)π

π(n−M/2)2

0 , n =M/2

(n) = − (M − n)hD/F hD/F

Figure 16. Example (magnitude response of differentiator for M = 20 when (a) using truncation

(rectangular window) and (b) using the Hamming window)



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