continuous-time analog filtersmbingabr/dsp/ch2... · continuous-time analog filters chapter 2. dr....

Continuous-Time Analog FiltersChapter 2

Dr. Mohamed BingabrUniversity of Central Oklahoma

ENGR 4333/5333: Digital Signal Processing

Outline

• Frequency Response of an LTIC System

• Signal Transmission through LTIC Systems

• Ideal and Realizable Filters

• Data Truncation by Windows

• Specification of Practical Filters

• Analog Filter Transmissions

• Practical Filter Families

2.1 Frequency Response of an LTIC System

Frequency Response of an LTIC System

h(t)h(t)δ (t)

x(t) y(t) = x(t) ∗ h(t)

Time Domain

Y(s) = X(s) H(s)H(s)X(s)

Laplace Domain

Y(ω) = X(ω) H(ω)H(ω)X(ω)

Frequency Domain

Time Domain Verses Frequency Domain

Frequency Response of Common SystemsAn Ideal Time Delay of T seconds 𝑦𝑦 𝑡𝑡 = 𝑥𝑥(𝑡𝑡 − 𝑇𝑇)

An Ideal Differentiator 𝑦𝑦 𝑡𝑡 =𝑑𝑑𝑥𝑥𝑑𝑑𝑡𝑡

An Ideal Integrator 𝑦𝑦 𝑡𝑡 = �𝑥𝑥 𝜏𝜏 𝑑𝑑𝜏𝜏

Pole-Zero PlotsLTIC system is described by the constant-coefficient linear differential Equation.

ExampleConsider a system whose transfer function is

Determine and plot the poles and zeros of H(s), and use the pole-zero information to predictoverall system behavior. Confirm this predicted behavior by computing and graphing the system’s frequency response (magnitude and phase). Lastly, find and plot the system response y(t) for the inputs (a) xa(t) = cos(2t) (b) xb(t) = 2sin(4t − π/3)

𝐻𝐻 𝑠𝑠 =2𝑠𝑠

𝑠𝑠2 + 2𝑠𝑠 + 5

Continue Example

2.2 Signal Transmission through LTIC Systems

Distortionless Transmission

Group Delay tg

If group delay is constant then all frequency components are delayed by the same time interval.

For distortionless transmission, we require that the system possess linear phase characteristics. For real systems, the phase is not only a linear function of ω but also passes through the origin (or ± π if a is negative) at ω = 0 .

ExampleHilbert transformers are LTIC systems that, among other things, are useful in the study of communication systems. Ideally, a Hilbert transformer is described by

h(t) =1/πt ⇐⇒ H(ω) = −jsgn(ω).Compute and plot the magnitude response |H(ω)|, the phase response ∠H(ω), and the group delay tg(ω) of a Hilbert transformer. Compute and plot the output y(t) of this system in response to the pulse input x(t) = Π(t), and comment on the results.Solution

Nature of Distortion in Audio and Video SignalsHuman ear readily perceives amplitude distortion but is relatively insensitive to phase distortion. For phase distortion to become noticeable delay should be comparable to signal length. In audio signals, the average duration of a spoken syllable is on the order of 0.01 to0.1 seconds the maximum variation in ∠H(ω) is only a small fraction of a millisecond.

A human eye is sensitive to phase distortion but is relatively insensitive to amplitude distortion.

Real Bandpass Systems and Group DelayFor distortionless transmission of bandpass signals, the system need satisfy

Generalized Linear Phase (GLP)Example: What is the output of an input xbp(t) = x(t) cos(ωct) to the above bandpass filter.

The envelope and carrier are both delayed by tg, although the carrier acquires an additional phase shift of φ0. The delay tg is termed the group (or envelope) delay.

2.3 Ideal and Realizable Filters

Ideal and Realizable Filters

Ideal and Realizable Filters

Example: Determine a suitable output delay td for a practical audio filter designed to pass frequencies audible to humans.

This delay is quite small compared with phoneme lengths, which are typically in the tens to hundreds of milliseconds.

2.4 Data Truncation by Windows

Data Truncation by Windows

x(t): is the signalw(t): is the windowxw(t): is the truncated x(t).

The two side effects of truncation is spectral spreading and spectral leakage.

Example: What is the effect of truncating a dc signal?

Lowpass Filter Design Using Windows

Remedies for Truncation Impairments1. The spectral spread (main lobe width) of a

truncated signal is equal to the bandwidth of the window function w(t).

2. The smoother the signal, the faster is the decay of its spectrum (smaller leakage).

Common Window Functions

2.5 Specification of Practical Filters

A stopband is a frequency band over which the gain is below some small number δs (the stopband ripple parameter),

A passband is a frequency band over which the gain is between 1 − δp and 1.

ωP = Passband edgeωS = Stopband edge

Specification of Practical Filters (Gain)

Frequently, it is more convenient to work with filter attenuation rather than filter gain. Filterattenuation, expressed in dB, is the negative of filter gain, also expressed in dB.

The half-power gain of 1/ 2 , for example, is 20log10 1/ 2 ≈ −3.01 dB, which corresponds to 3.01 dB of attenuation.

αp : The maximum passband attenuationαs : The minimum stopband attenuation

Specification of Practical Filters (Attenuation)

2.6 Analog Filter Transformations

Filter design start with a prototype filter (usually normalized lowpass) and then transform it to the final desired filter type. A normalized lowpass filter has unity critical frequencies such as ωp and ωs.

Frequency Transformation starts with a normalized lowpass response to produce, respectively, a lowpass, highpass, bandpass, or bandstop response.

Analog Filter Transformation

ωo is the half-power prototype frequencyω1 is the half-power frequency of the desired filter

Lowpass-to-Lowpass Transformation

Example: A lowpass prototype filter is given as Hp(s)= 2/s+2. Determine the lowpass-to-lowpass transformation rule to produce a lowpass filter Hlp(s) with a half-power (3 dB) frequency of ω1 = 3 rad/s. Plot the magnitude response of the resulting lowpass filter.

ωo

ω1

ω

H(ω)

Lowpass-to-Highpass Transformation

Example: Apply the lowpass-to-highpass transformation ω → 2/ −ω to the lowpass prototype Hp(ω) = 1/ jω+1. Plot the magnitude response, and determine the 3-dB frequency of the resulting filter.

In the lowpass-to-highpass transformation the frequencies undergo reciprocal transformation.

Lowpass-to-Bandpass TransformationThe lowpass-to-bandpass transformation utilizes a one-to-two mapping. In this way, the lowpass filter’s single cutoff frequency becomes both the upper and lower cutoff frequencies required of the bandpass response.

Or

Example

A lowpass prototype filter is given as Hp(s)= 2/s+2. Determine the lowpass-to-bandpasstransformation rule to produce a bandpass filter Hbp(s) with half-power (3 dB) cutofffrequencies ω1 = 1 rad/s and ω2 =3 rad/s. Plot the magnitude response of the resulting bandpassfilter.

Lowpass-to-Bandstop TransformationMuch as the lowpass-to-highpass transformation reciprocally relates to the lowpass-to-lowpass transformation, the lowpass-to-bandstop transformation reciprocally relates to the lowpass-to-bandpass transformation.

Or

Example

Apply a lowpass-to-bandstop transformation to the lowpass prototype Hp(s)= 1/ s2+√2s+1. The resulting bandstop filter should have 3-dB cutoff frequencies of 1 and 3 rad/s. Plot the magnitude response of the resulting filter.

Hbs = @(s) (s.^2+3).^2./(s.^4+2*sqrt(2)*s.^3+10*s.^2+6*sqrt(2)*s+9); omega = 0:.01:6; plot(omega,abs(Hbs(1j*omega))); xlabel('\omega'); ylabel('|H_{\rm bs}(\omega)|');

Solution

2.7 Practical Filter Families

Practical Filter Families

Family of practical filters that are realizable (causal) and that possess rational transfer functions are:

• Butterworth Filters• Chebyshev Filters• Inverse Chebyshev Filters• Elliptic Filters• Bessel-Thomson Filters

Butterworth FiltersThe K poles of a Kth order Butterworth filter are spaced equally by π/K radians around a half circle in the left hand side of the complex plane of radius ωcand centered at the origin.

K = 4 K = 5

Butterworth Filtersωc = 1

Example Find the transfer function H(s) of a fourth-order lowpass Butterworth filter with a 3-dB cutoff frequency ωc= 10 by (a) frequency scaling the appropriate Table 2.2 entry and (b) direct calculation. Plot the magnitude response to verify filter characteristics..

Solution(a)

(b) Find the poles using the equation in previous slide and then find the polynomial equationK = 4; k=1:K; omegac = 10;p = 1j*omegac*exp(1j*pi/(2*K)*(2*k-1))A = poly(p)omega = linspace(0,35,1001); H = 10^4./polyval(A,1j*omega);plot(omega,abs(H)); xlabel('\omega'); ylabel('|H(j\omega)|');

Example

(b) Find the poles using the equation

K = 4; k=1:K; omegac = 10;p = 1j*omegac*exp(1j*pi/(2*K)*(2*k-1))A = poly(p)omega = linspace(0,35,1001); H = 10^4./polyval(A,1j*omega);plot(omega,abs(H)); xlabel('\omega'); ylabel('|H(j\omega)|');

and then find the polynomial equation.

p = -3.8268 + 9.2388i -9.2388 + 3.8268i -9.2388 - 3.8268i -3.8268 - 9.2388i

A = 1.0000 26.1313 341.4214 2613.1259 10000.0000

Determination of Butterworth Filter Order and Half-Power Frequency

In filter design you will be given the passband (ωp , αp) and stopband (ωs , αs) parameters and you need to calculate, the order of the filter and the half-power cutoff frequency that meets these parameters.

Solve equation to determine upper limit for ωc

Solve equation to determine lower limit for ωc

Example Design the lowest-order Butterworth lowpass filter that meets the specifications ωp ≥ 10 rad/s, αp ≤ 2 dB, ωs ≤ 30 rad/s, and αs ≥ 20 dB. Plot the corresponding magnitude response to verify that design requirements are met.

Solution

Step 1: Determine the filter order K from equation

omegap = 10; alphap = 2; omegas = 30; alphas = 20;K = ceil(log((10^(alphas/10)-1)/(10^(alphap/10)-1))/(2*log(omegas/omegap)))

Step 2: Determine the range for the Half-Power cutoff frequency ωc

K = 3

omegac = [omegap/(10^(alphap/10)-1).^(1/(2*K)),omegas/(10^(alphas/10)-1).^(1/(2*K))]

ωc = 10.9350 13.9481 Choose ωc = 12.5

Continue Example

Step 3: Determine the filter transfer function H(s)

𝐻𝐻 𝑠𝑠 =1

𝑠𝑠3 + 2𝑠𝑠2 + 2𝑠𝑠 + 1

One way is to find the poles and then find H(s).

and then apply lowpass-lowpass frequency transformation, s s(1/12.5)

𝐻𝐻 𝑠𝑠 =1953.125

𝑠𝑠3 + 25𝑠𝑠2 + 312.5𝑠𝑠 + 1953.125

The second way it to start with the 3rd order prototype lowpass filter

omegac = 12.5; k = 1:K; A = poly(1j*omegac*exp(1j*pi/(2*K)*(2*k-1)))

A = 1.00 25.00 312.00 1953.125

Continue Example omega = linspace(0,35,1001); H = omegac^K./(polyval(A,1j*omega));plot(omega,abs(H));

Since the plot is not in dB, the attenuation parameters are converted, the passband floor is 1 − δp = 10−αp/20 = 0.7943,

and the stopband ceiling is δs = 10−αs/20 = 0.1.

The plot shows the Butterworth responses corresponding to the limiting values of ωc, theωc = 10.9350 curve exactly meets passband requirements and exceeds stopband requirements. Similarly, the ωc = 13.9481 curve exactly meets stopband requirements and exceeds passband requirements, ωc = 12.5 meets both requirements.

Example Design the lowest-order Butterworth bandpass filter that meets the specifications ωp1 ≤ 10 rad/s, ωp2 ≥ 20 rad/s, αp ≤ 2 dB, ωs1 ≥ 5 rad/s, ωs2 ≤ 35 rad/s, and αs ≥ 20 dB. Plot the corresponding magnitude response to verify that design requirements are met.

Step 1: Determine the Lowpass PrototypeSolution

First we normalize the passband frequency ωp = 1 and then we calculate the stopband frequency ωs for both transition bands of the bandpass filter and choose the most restrictive ωs.omegap1 = 10; omegap2 = 20; omegas1 = 5; omegas2 = 35; omegap = 1;omegas = abs([omegap*(omegas1^2-omegap1*omegap2)/(omegas1*(omegap2-omegap1)),...

omegap*(omegas2^2-omegap1*omegap2)/(omegas2*(omegap2-omegap1))])Omegas = 3.5000 2.9286

To meet the specifications of both transitions we choose the most restrictive which is the upper transition band, so ωs= 2.9286, this will lead to a higher order filter K.

Continue Example

K = 3

Next we determine the filter order K from equation

omegas = min(omegas); alphap = 2; alphas = 20;K = ceil(log((10^(alphas/10)-1)/(10^(alphap/10)-1))/(2*log(omegas/omegap)))

Next we calculate the range for the Half-Power cutoff frequency ωc

omegac = [omegap/(10^(alphap/10)-1).^(1/(2*K)),omegas/(10^(alphas/10)-1).^(1/(2*K))]

K = 3, ωp=1, ωs=2.9286, αp=2, αs=20

omegac = 1.0935 1.3616 Choose the average for ωc = 1.2276

𝐻𝐻 𝑠𝑠 =1

𝑠𝑠3 + 2𝑠𝑠2 + 2𝑠𝑠 + 1

Next we transform the normalized 3rd order prototype lowpass filter

s s(1/1.2276) 𝐻𝐻 𝑠𝑠 =1.85

𝑠𝑠3 + 2.455𝑠𝑠2 + 3.014𝑠𝑠 + 1.85

Continue Example

Step 2: We transform the prototype lowpass filter to the desired bandpass filter

𝐻𝐻 𝑠𝑠 =1.85

𝑠𝑠3 + 2.455𝑠𝑠2 + 3.014𝑠𝑠 + 1.85

𝑠𝑠 →𝑠𝑠2 + 20010𝑠𝑠

Continue Example

Example-Matlab Code

omegap1 = 10; omegap2 = 20; omegas1 = 5; omegas2 = 35; omegap = 1;omegas = abs([omegap*(omegas1^2-omegap1*omegap2)/(omegas1*(omegap2-omegap1)),...

omegap*(omegas2^2-omegap1*omegap2)/(omegas2*(omegap2-omegap1))]);omegas = min(omegas); alphap = 2; alphas = 20;K = ceil(log((10^(alphas/10)-1)/(10^(alphap/10)-1))/(2*log(omegas/omegap)));omegac = [omegap/(10^(alphap/10)-1).^(1/(2*K)),omegas/(10^(alphas/10)-1).^(1/(2*K))];omegac = mean(omegac); k = 1:K;pk = (1j*omegac*exp(1j*pi/(2*K)*(2*k-1))); A = poly(pk);a = 1; b = -pk*(omegap2-omegap1); c = omegap1*omegap2;pk = [(-b+sqrt(b.^2-4*a*c))./(2*a),(-b-sqrt(b.^2-4*a*c))./(2*a)];B = (omegac^K)*((omegap2-omegap1)^K)*poly(zeros(K,1)); A = poly(pk);omega = linspace(0,40,1001); H = polyval(B,1j*omega)./polyval(A,1j*omega);plot(omega,abs(H));

Comparing Chebyshev and Butterworth Response

The Chebyshev filter has a sharper cutoff (smaller transition band) than the same-order Butterworth filter, but this is achieved at the expense of inferior passband behavior (rippling instead of maximally flat). Chebyshev filters typically require lower order K than Butterworth filters to meet a given set of specifications.

Chebyshev Filters Type IThe magnitude of the K th-order Chebyshev polynomial filter is

K = 6

K = 7

𝛼𝛼𝑃𝑃 = 20log10 1 + 𝜖𝜖2

𝛼𝛼𝑃𝑃 = 10log10 1 + 𝜖𝜖2

𝜖𝜖2 = 10𝛼𝛼𝑃𝑃/10 − 1

Or

Determination of Chebyshev Filter Order

Replace ϵ and CK in the above equation using the equations below and solve for K

𝜖𝜖2 = 10𝛼𝛼𝑃𝑃/10 − 1

Note: Computing Chebyshev filter order is identical to the Butterworth case except that logarithms are replaced with inverse hyperbolic cosines.

Determination of Chebyshev Poles and H(s)the Chebyshev filter poles, and thus its transfer function, are found by squaring the magnitude response, substituting ω = s/j to yield H(s)H(−s), and then selecting the left half-plane denominator roots. Similar to how Butterworth poles lie on a semicircle, the poles of a Chebyshev lowpass filter lie on a semi-ellipse.

There is a close relationship between pole locations on a Chebyshev ellipse and those on Butterworth circles: the real part of the Chebyshev poles coincide with those of an equivalent-order Butterworth filter with radius ωpsinh[1/K sinh-1(1/ϵ)] and the imaginary part of the Chebyshev poles are equal to those of a Butterworth filter with radius ωpcosh[1/K sinh-1(1/ϵ)].

Coefficients of normalized Chebyshev denominator polynomials

Coefficients of normalized Chebyshev denominator polynomials

Example Design the lowest-order Chebyshev lowpass filter that meets the specifications ωp ≥ 10 rad/s, αp ≤ 2 dB, ωs ≤ 30 rad/s, and αs ≥ 20 dB. Plot the corresponding magnitude response to verify that design requirements are met.

Solution

Step 1: Determine the filter order K from equation

omegap = 10; alphap = 2; omegas = 30; alphas = 20; K = ceil(acosh(sqrt((10^(alphas/10)-1)/(10^(alphap/10)-1)))/acosh(omegas/omegap))

Step 2: Determine the range for the passband cutoff frequency ωP by setting ωs = 30 and K = 2 and solve the above equation for ωP.

K = 1.8479 K = 2

omegap = [omegap,omegas/cosh(acosh(sqrt((10^(alphas/10)-1)/(10^(alphap/10)-1)))/K)]

ωP = 10.00 11.3349 Choose the average so ωP = 10.6674

Continue Example

Step 3: Determine the prototype lowpass filter transfer function H(s)

𝐻𝐻 𝑠𝑠 =74.3963

𝑠𝑠2 + 8.5747𝑠𝑠 + 93.6594

First we find ϵ, |H(j0)|, poles, and then the lowpass prototype

epsilon = sqrt(10^(alphap/10)-1); k = 1:K; H0 = (mod(K,2)==1)+(mod(K,2)==0)/sqrt(1+epsilon^2);pk = -omegap*sinh(asinh(1/epsilon)/K)*sin(pi*(2*k-1)/(2*K))+...

1j*omegap*cosh(asinh(1/epsilon)/K)*cos(pi*(2*k-1)/(2*K));B = H0*prod(-pk), A = poly(pk)

|H(0)|= 0.7943 B = |H(0)|*p1*p2 =74.3963 A = [1.00 8.5747 93.6594]

Alternatively, the lowpass-to-lowpass transformation s → s/ωp = s/10.6674 applied to the appropriate coefficients of Table 2.4 yields the same denominator

Continue Example

𝐻𝐻 𝑠𝑠 =74.3963

𝑠𝑠2 + 8.5747𝑠𝑠 + 93.6594

omega = linspace(0,35,1001); H = B./(polyval(A,1j*omega)); plot(omega,abs(H));

Example Design the lowest-order Chebyshev bandstop filter that meets the specifications ωP1 ≥ 5 rad/s, ωP2 ≤ 35, αp ≤ 2 dB, ωs1 ≤ 10 rad/s, ωs2 ≥ 20 and αs ≥ 20 dB. Plot the corresponding magnitude response to verify that design requirements are met.

Solution

omegap1 = 5; omegap2 = 35; omegas1 = 10; omegas2 = 20; omegas = 1; omegap = abs([omegas*(omegap1*(omegas2-omegas1))/(-omegap1^2+omegas1*omegas2),...

omegas*(omegap2*(omegas2-omegas1))/(-omegap2^2+omegas1*omegas2)]) ωP = 0.2857 0.3415

Step 1: Determine the Lowpass PrototypeFirst we normalize the stopband frequency ωS = 1 and then we calculate the passband frequency ωP for both transition bands of the bandpass filter and choose the most restrictive ωP.

To meet the specifications of both transitions we choose the most restrictive which is the upper transition band, so ωP= 0.3415, this will lead to a higher order filter K.

Continue Example

K = 2

Next we determine the filter order K from equation

omegap = max(omegap); alphap = 2; alphas = 20; K = ceil(acosh(sqrt((10^(alphas/10)-1)/(10^(alphap/10)-1)))/acosh(omegas/omegap))

omegas = mean([omegas,...omegap*cosh(acosh(sqrt((10^(alphas/10)-1)/(10^(alphap/10)-1)))/K)])

omegas = 1 0.9038

Choose the average to provide both passband and stopband buffer zones, so ωS = 0.9519

Determine the range for the stopband cutoff frequency ωS by setting ωP = 0.3415 and K = 2 and solve the above equation for ωS.

Continue Example

Next, since we adjusted ωS we recalculate ωP using the equation

omegap = omegas/cosh(acosh(sqrt((10^(alphas/10)-1)/(10^(alphap/10)-1)))/K);

ωP = 0.3596omegap = 0.3596

Step 3: Determine the lowpass prototype filter transfer function H(s)

First we find ϵ, |H(j0)|, the poles, and then the lowpass prototype epsilon = sqrt(10^(alphap/10)-1); k = 1:K;pk = -omegap*sinh(asinh(1/epsilon)/K)*sin(pi*(2*k-1)/(2*K))+...

1j*omegap*cosh(asinh(1/epsilon)/K)*cos(pi*(2*k-1)/(2*K)); H0 = mod(K,2)+mod(K+1,2)/sqrt(1+epsilon^2); B = H0*prod(-pk), A = poly(pk)

|H(0)|= 0.7943 B = |H(0)|*p1*p2 =0.0846 A = [1.00 0.2891 0.1065]

𝐻𝐻 𝑠𝑠 =0.0846

𝑠𝑠2 + 0.2891𝑠𝑠 + 0.1065

Continue Example

𝐻𝐻 𝑠𝑠 =0.7943𝑠𝑠4 + 317.7313𝑠𝑠2 + 31773.1293

𝑠𝑠4 + 27.155𝑠𝑠3 + 1339.329𝑠𝑠2 + 5431.0𝑠𝑠 + 40000

Step 4: Determine the bandpass filter transfer function H(s) by transformation

𝐻𝐻 𝑠𝑠 =0.0846

𝑠𝑠2 + 0.2891𝑠𝑠 + 0.1065

𝑠𝑠 →10𝑠𝑠

𝑠𝑠2 + 200

Continue Example (Matlab Code) omegap1 = 5; omegap2 = 35; omegas1 = 10; omegas2 = 20; omegas = 1;omegap = abs([omegas*(omegap1*(omegas2-omegas1))/(- omegap1^2+omegas1*omegas2),...

omegas*(omegap2*(omegas2-omegas1))/(-omegap2^2+omegas1*omegas2)]) omegap = max(omegap); alphap = 2; alphas = 20;K = ceil(acosh(sqrt((10^(alphas/10)-1)/(10^(alphap/10)-1)))/acosh(omegas/omegap)) omegas = mean([omegas,...

omegap*cosh(acosh(sqrt((10^(alphas/10)-1)/(10^(alphap/10)-1)))/K)]) omegap = omegas/cosh(acosh(sqrt((10^(alphas/10)-1)/(10^(alphap/10)-1)))/K);epsilon = sqrt(10^(alphap/10)-1); k = 1:K;pk = -omegap*sinh(asinh(1/epsilon)/K)*sin(pi*(2*k-1)/(2*K))+...

1j*omegap*cosh(asinh(1/epsilon)/K)*cos(pi*(2*k-1)/(2*K)); H0 = mod(K,2)+mod(K+1,2)/sqrt(1+epsilon^2); B = H0*prod(-pk), A = poly(pk) a = 1; b = -(omegas2-omegas1)./pk; c = omegas1*omegas2;B = B/prod(-pk)*poly([1j*sqrt(c)*ones(K,1);-1j*sqrt(c)*ones(K,1)]) A = poly([(-b+sqrt(b.^2-4*a.*c))./(2*a),(-b-sqrt(b.^2-4*a.*c))./(2*a)]) omega = linspace(0,100,1001); H = polyval(B,1j*omega)./polyval(A,1j*omega); plot(omega,abs(H));

Inverse Chebyshev Filters Type IIThe magnitude response exhibits a maximally flat passband and an equiripple stopband. While Butterworth and Chebyshev type 1 filters have finite poles and no finite zeros, the inverse Chebyshev has both finite poles and zeros. The presence of zeros enhance the frequency response but the decay rate for ω > ωs is slower.

The inverse Chebyshev response can be obtained from a Chebyshevresponse in two steps. To begin, let |Hc(ω)| be a lowpass Chebyshevmagnitude response with passband edge ωp and ripple parameter ϵ, as shown in Fig. a. In the first step, we subtract |Hc(ω)|2 from 1 to obtain the response shown in Fig. b. In the second step, we interchange the stopband and passband with the lowpass-to-highpasstransformation of ω →−ωpωs/ω yields the inverse Chebyshevresponse, shown in Fig. c.

Elliptic Filters and Bessel-Thomson FiltersElliptic FiltersChebyshev filters have a smaller transition bands compared with that of a Butterworth filter because Chebyshev filters allow rippling in either the passband or stopband. Elliptic filter allows ripple in both the passband and the stopband, so it achieves a further reduction in the transition band. For a given transition band, an elliptic filter provides the largest ratio of the passband gain to stopband gain, or for a given ratio of passband to stopband gain, it requires the smallest transition band.

Bessel-Thomson FiltersUnlike the previous filters, where we approximate the magnitude response without paying attention to the phase response, the Bessel-Thomson filter is designed for maximally flat time delay over a bandwidth of interest. This means that the phase response is designed to be as linear as possible over a given bandwidth. Bessel-Thomson filters are superior to the filters discussed so far in terms of the phase linearity or flatness of the time delay. However, they do not fair so well in terms of the flatness of the magnitude response.