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Chapter 4Chapter 4

Digital Processing of Continuous-Time SignalsContinuous-Time Signals

清大電機系林嘉文

[email protected] 5731152

© The McGraw-Hill Companies, Inc., 2007Original PowerPoint slides prepared by S. K. Mitra 4-1-1

03-5731152

Digital Processing ofContinuous-Time Signals

Digital processing of a continuous time signal involves• Digital processing of a continuous-time signal involves the following basic steps:1 Conversion of the continuous time signal into a1. Conversion of the continuous-time signal into a

discrete-time signal2 Processing of the discrete-time signal2. Processing of the discrete time signal3. Conversion of the processed discrete-time signal

back into a continuous-time signalg• Conversion of a continuous-time signal into digital form

is carried out by an analog-to-digital (A/D) convertery g g ( )• The reverse operation of converting a digital signal into a

continuous-time signal is performed by a digital-to-


analog (D/A) converter


Since the A/D conversion takes a finite amount of time a• Since the A/D conversion takes a finite amount of time, a sample-and-hold (S/H) circuit is used to ensure that the input analog signal remains constant in amplitude until theinput analog signal remains constant in amplitude until the conversion is complete to minimize representation error

• To prevent aliasing, an analog anti-aliasing filter is p g, g gemployed before the S/H circuit

• To smooth the output signal of the D/A converter, which has a staircase-like waveform, an analog reconstruction filter is used



Both the anti aliasing filter and the reconstruction filter are• Both the anti-aliasing filter and the reconstruction filter are analog lowpass filters

• The most widely used IIR digital filter design method is• The most widely used IIR digital filter design method is based on the conversion of an analog lowpass prototype

• Discrete-time signals in many applications are generatedDiscrete time signals in many applications are generated by sampling continuous-time signals– There exists an infinite number of continuous-time signals, g ,

which when sampled lead to the same discrete-time signal– Under certain conditions, it is possible to relate a unique

ti ti i l t i di t ti i lcontinuous-time signal to a given discrete-time signal– If these conditions hold, then it is possible to recover the

original continuous-time signal from its sampled values


original continuous time signal from its sampled values

Effect of Sampling in the Frequency Domain

Let g (t) be a continuous time signal that is sampled• Let ga(t) be a continuous-time signal that is sampled uniformly at t = nT , generating the sequence g[n] where

g[n] = g (nT) ∞ < n < ∞g[n] = ga(nT), − ∞ < n < ∞

with T being the sampling period• The reciprocal of T is called the sampling frequency FT,

FT = 1/T• The frequency-domain representation of ga(t) is given by

its continuous-time Fourier transform (CTFT):

• The frequency-domain representation of g[n] is given by i DTFT


its DTFT:


To establish the relation between G (jΩ) and G(ejω) we• To establish the relation between Ga(jΩ) and G(ejω) , we treat the sampling operation mathematically as a multiplication of ga(t) by a periodic impulse train p(t)multiplication of ga(t) by a periodic impulse train p(t)

(t) i t f t i f id l i l ith i d T• p(t) consists of a train of ideal impulses with a period T as shown below

• The multiplication operation yields an impulse train:

( ) ( ) ( ) ( ) ( )t t t T t Tδ∞

∑


( ) ( ) ( ) ( ) ( )p a an

g t g t p t g nT t nTδ=−∞

= = −∑


• g (t) is a continuous-time signal consisting of a train of• gp(t) is a continuous-time signal consisting of a train of uniformly spaced impulses with the impulse at t = nTweighted by the sampled value ga(nT) of ga(t) at that instantg y p ga( ) ga( )

• There are two different forms of Gp(jΩ)• One form is given by the weighted sum of the CTFTs of• One form is given by the weighted sum of the CTFTs of



• To derive the second form we make use of the Poisson’s• To derive the second form, we make use of the Poisson s sum formula:

where ΩT = 2π/T, and Φ(jΩ) is the CTFT of ϕ(t)• For t = 0 the above equation reduces to• For t = 0, the above equation reduces to

• From the frequency shifting property of the CTFT, the CTFT of ga(t)e−jΨt is given by Ga(j(Ω + Ψ))S b tit ti ϕ(t) (t) jΨt i t th b ti• Substituting ϕ(t) = ga(t)e−jΨt into the above equation, we arrive at



• By replacing with Ω in the above equation we arrive at the• By replacing with Ω in the above equation we arrive at the alternative form of the CTFT Gp(jΩ) of gp(t), given by

• Therefore, Gp(jΩ) is a periodic function of Ω consisting of a , p(j ) p gsum of shifted and scaled replicas of Ga(jΩ), shifted by integer multiples of ΩT and scaled by 1/T

• The term on the RHS of the previous equation for k = 0 is the baseband portion of Gp(jΩ), and each of the remaining t th f t l t d ti f G (jΩ)terms are the frequency translated portions of Gp(jΩ)

• The frequency range is called the baseband or Nyquistband


band


• Assume g (t) is a band-limited signal with a CTFT G (jΩ) as• Assume ga(t) is a band-limited signal with a CTFT Ga(jΩ) as shown below

• The spectrum, P(jΩ) of p(t) having a sampling period T = 2π/ΩT is indicated below2π/ΩT is indicated below



• Two possible spectra of G (jΩ) are shown below• Two possible spectra of Gp(jΩ) are shown below

• If ΩT ≥ 2Ωm , there is no overlap between the shifted aliasing

T m preplicas of Ga(jΩ) generating Gp(jΩ)

• If ΩT < 2Ωm, there is an overlap of the spectra of the shifted


replicas of Ga(jΩ) generating Gp(jΩ) (Aliasing)


• If Ω > 2Ω g (t) can be recovered exactly from g (t) by• If ΩT > 2Ωm , ga(t) can be recovered exactly from gp(t) by passing it through an ideal lowpass filter Hr(jΩ) with a gain T and a cutoff frequency Ωc greater than Ωm and less than q y c g mΩc−Ωm as shown below



• The spectra of the filter and pertinent signals are as below• The spectra of the filter and pertinent signals are as below

• If ΩT < 2Ωm, due to the overlap of the shifted replicas of G (jΩ) the spectrum G (jΩ) cannot be separated by filteringGa(jΩ), the spectrum Ga(jΩ) cannot be separated by filtering to recover Ga(jΩ) because of the distortion caused by a part of the replicas immediately outside the baseband folded


p yback or aliased into the baseband

Sampling TheoremSampling Theorem• Sampling theorem - Let g (t) be a band-limited signal with• Sampling theorem - Let ga(t) be a band-limited signal with

CTFT Ga(jΩ) = 0 for |Ω| > Ωm. Then ga(t) is uniquely determined by its samples ga(nT), − ∞ ≤ n ≤ ∞ if y p ga( )

ΩT ≥ 2Ωmwhere Ω = 2π/Twhere ΩT = 2π/T

• The condition ΩT ≥ 2Ωm is often referred to as the Nyquistconditioncondition

• The frequency ΩT/2 is usually referred to as the folding frequencyfrequency


Sampling TheoremSampling Theorem• Given {g (nT)} we can recover exactly g (t) by generating• Given {ga(nT)}, we can recover exactly ga(t) by generating

an impulse train

and then passing it through an ideal lowpass filter Hr(jΩ)with a gain T and a cutoff frequency Ω satisfyingwith a gain T and a cutoff frequency Ωc satisfying

Ωm < Ωc < (ΩT − Ωm)• The highest frequency Ωm contained in ga(t) is usually

called the Nyquist frequency since it determines the minimum sampling frequency Ω = 2Ω that must be usedminimum sampling frequency ΩT = 2Ωm that must be used to fully recover ga(t) from its sampled version

• The frequency 2Ω is called the Nyquist rate


The frequency 2Ωm is called the Nyquist rate

Sampling TheoremSampling Theorem• Oversampling - The sampling frequency is higher than the O e sa p g e sa p g eque cy s g e t a t e

Nyquist rate• Undersampling - The sampling frequency is lower than the p g p g q y

Nyquist rate• Critical sampling - The sampling frequency is equal to the

Nyquist rate• In digital telephony, a 3.4 kHz signal bandwidth is

t bl f t l h tiacceptable for telephone conversation– a sampling rate of 8 kHz is used in telecommunication

In high quality analog music signal processing a bandwidth• In high-quality analog music signal processing, a bandwidth of 20 kHz has been determined to preserve the fidelity– a sampling rate of 44 1 kHz is used in CD music systems


a sampling rate of 44.1 kHz is used in CD music systems

Sampling TheoremSampling Theorem• Example - Consider the three sinusoidal signals:

g1(t) = cos(6πt)g2(t) = cos(14πt)

(t) (26 t)g3(t) = cos(26πt)• Their corresponding CTFTs are:

G ( jΩ) = π[δ(Ω − 6π) + δ(Ω + 6π)]G1( jΩ) = π[δ(Ω − 6π) + δ(Ω + 6π)]G2( jΩ) = π[δ(Ω − 14π) + δ(Ω + 14π)]G3( jΩ) = π[δ(Ω − 26π) + δ(Ω + 26π)]3( j ) [ ( ) ( )]


Sampling TheoremSampling Theorem• These continuous-time signals sampled at a rate of T = 0.1

sec, i.e., with a sampling frequency rad/sec • The sampling process generates the continuous-time

impulse trains g1p(t), g2p(t), and g3p(t)• Their corresponding CTFTs are given by

aliasing


aliasing

Sampling TheoremSampling Theorem• We now derive the relation between the DTFT of g[n] and

th CTFT f (t)the CTFT of gp(t)• To this end we compare

with

and make use of g[n] = ga(nT), − ∞ < n < ∞• Observation: We have

or, equivalently,


• As a result

Sampling TheoremSampling Theorem• We arrive at the desired result given by

• The relation derived above can be alternately expressed as

• From

• it follows that G(ejω) is obtained from Gp(jΩ) by applying the or from


( ) p(j ) y pp y gmapping Ω = ω/T (note, both G(ejω) are Gp(jΩ) are periodic)

Recovery of the Analog Signal (1/3)Recovery of the Analog Signal (1/3)• We now derive the expression for the output of the

id l l t ti filt H (jΩ) f ti f thideal lowpass reconstruction filter Hr(jΩ) as a function of the samples g[n] The imp lse response h (t) of the lo pass reconstr ction• The impulse response hr(t) of the lowpass reconstruction filter is obtained by taking the inverse DTFT of Hr(jΩ)

• Thus, the impulse response is given by, p p g y

• The input to the lowpass filter is the impulse train gp(t)


Recovery of the Analog Signal (2/3)Recovery of the Analog Signal (2/3)• Therefore, the output of the ideal lowpass filter is

i bgiven by:

• Substituting hr(t) = sin(Ωct) /(ΩTt/2) in the above and assuming for simplicity Ωc = ΩT/2 = π/T, we get

[ ]sin ( ) /t nT Tπ∞

• The ideal band-limited interpolation process is shown below

[ ]sin ( ) /( ) [ ]( ) /a n

t nT Tg t g n

t nT Tπ

π

∞

=−∞

−=

−∑The ideal band limited interpolation process is shown below

Ampl

itude

A

© The McGraw-Hill Companies, Inc., 2007Original PowerPoint slides prepared by S. K. Mitra 4-1-22Time

Recovery of the Analog Signal (3/3)Recovery of the Analog Signal (3/3)• It can be shown that when Ωc = ΩT/2 in

hr(0) =1 and hr(nT) = 0 for n ≠ 0r( ) r( )• As a result, from

we observe for all integer values of r in the range − ∞ < r < ∞

( ) [ ] ( )aag rT g r g rT= =

for all integer values of r in the range r • The above relation holds whether or not the condition of the

sampling theorem is satisfiedp g• However, for all values of t only if the

sampling frequency ΩT satisfies the condition of the


sampling theorem

Implication of the Sampling ProcessImplication of the Sampling Process• Consider again the three continuous-time signals: g1(t) =

(6 t) (t) (14 t) d (t) (26 t)cos(6πt), g2(t) = cos(14πt), and g3(t) = cos(26πt)• The plot of the CTFT G1p(jΩ) of g1p(t) is shown below

• It is apparent that we can recover any of its frequency-translated versions cos[(20k ± 6)πt] outside the basebandtranslated versions cos[(20k ± 6)πt] outside the baseband by passing g1p(t) through an ideal analog bandpass filter with a passband centered at Ω = (20k ± 6)π

• For example, to recover the signal cos(34πt), it will be necessary to use a bandpass filter with a frequency


response

Implication of the Sampling ProcessImplication of the Sampling Process• Likewise, we can recover the aliased baseband component

(6 t) f th l d i f ith (t) (t) bcos(6πt) from the sampled version of either g2p(t) or g3p(t) by passing it through an ideal lowpass filter with a frequency response:response:

Th i li i di t ti l th i i l• There is no aliasing distortion unless the original continuous-time signal also contains the component cos(6πt)cos(6πt)

• Similarly, from either g2p(t) or g3p(t) we can recover any one of the frequency translated versions, including the parentof the frequency translated versions, including the parent continuous-time signal g2(t) or g3(t) as the case may be, by employing suitable filters


Sampling of Bandpass Signals (1/3)Sampling of Bandpass Signals (1/3)• The conditions for the unique representation of a

ti ti i l b th di t ti i l bt i dcontinuous-time signal by the discrete-time signal obtained by uniform sampling assumed that the continuous-time signal is bandlimited in the frequency range from dc to somesignal is bandlimited in the frequency range from dc to some frequency Ωm

• Such a continuous-time signal is commonly referred to as aSuch a continuous time signal is commonly referred to as a lowpass signal

• There are applications where the continuous-time signal is pp gbandlimited to a higher frequency range ΩL ≤ |Ω| ≤ ΩH

• Such a signal is usually referred to as a bandpass signal• To prevent aliasing a bandpass signal can of course be

sampled at a rate greater than twice the highest frequency,


i.e. by ensuring ΩL ≥ 2ΩH

Sampling of Bandpass Signals (2/3)Sampling of Bandpass Signals (2/3)• The spectrum of the discrete-time signal obtained by

li b d i l ill h t l ithsampling a bandpass signal will have spectral gaps with no signal components present in these gaps Moreo er if Ω is er large the sampling rate also has to• Moreover, if ΩH is very large, the sampling rate also has to be very large which may not be practical in some situations

• A more practical approach is to use under sampling• A more practical approach is to use under sampling • Let ∆Ω = ΩH − ΩL define the bandwidth of the bandpass

signalsignal • Assume that the highest frequency contained in the signal is

an integer multiple of the bandwidth, i.e., ΩH = M(∆Ω)H• Choose the sampling frequency ΩT to satisfy the condition

ΩT = 2(∆Ω) = 2ΩH/M


T ( ) Hwhich is smaller than 2ΩH, the Nyquist rate

Sampling of Bandpass Signals (3/3)Sampling of Bandpass Signals (3/3)• Substitute the above expression for ΩT in

• This leads to

• As before, Gp(jΩ) consists of a sum of Ga(jΩ) and replicas of Ga(jΩ) shifted by integer multiples of twice the bandwidth ∆Ω d l d b 1/T∆Ω and scaled by 1/T

• The amount of shift for each value of k ensures that there will be no overlap between all shifted replicas no aliasingwill be no overlap between all shifted replicas no aliasing


Sampling of Bandpass SignalsSampling of Bandpass Signals

As shown above g (t) can be recovered from g (t) by• As shown above, ga(t) can be recovered from gp(t) by passing it through an ideal bandpass filter with a passband given by ΩL ≤ |Ω| ≤ ΩH and a gain of Tgiven by ΩL ≤ |Ω| ≤ ΩH and a gain of T

• Note: Any of the replicas in the lower frequency bands can be retained by passing gp(t) through bandpass filters with


y p g gp( ) g ppassbands ΩL − k(∆Ω) ≤ |Ω| ≤ ΩH − k(∆Ω), 1≤ k ≤ M −1

Analog Lowpass Filter SpecificationsAnalog Lowpass Filter Specifications• Typical magnitude response |Ha(jΩ)| of analog lowpass filter:

• In the passband, defined by 0 ≤ Ω ≤ Ωp , we require1− δp ≤ |Ha(jΩ)| ≤ 1+ δp, |Ω| ≤ Ωp , i.e., |Ha(jΩ)| approximates unity within

f δan error of ± δp• In the stopband, defined by Ωs ≤ Ω < ∞, we require

|H (jΩ)| ≤ δ Ω ≤ |Ω| < ∞ i e |H (jΩ)| approximates zero within an error


|Ha(jΩ)| ≤ δs, Ωs ≤ |Ω| < ∞ , i.e., |Ha(jΩ)| approximates zero within an error of δs

Analog Lowpass Filter SpecificationsAnalog Lowpass Filter Specifications• Ωp - passband edge frequency • Ωs - stopband edge frequency• δp - peak ripple value in the

passband• δs - peak ripple value in the

t b dstopband• Peak passband ripple

20l (1 δ )αp = −20log10(1 − δp )• Minimum stopband attenuation

( )αs = −20log10(δs )


Analog Lowpass Filter SpecificationsAnalog Lowpass Filter Specifications• Magnitude specifications may alternately be given in a

li d f i di t d b lnormalized form as indicated below

• Here, the maximum value of the magnitude in the passband assumed to be unityy

• - Maximum passband deviation, given by the minimum value of the magnitude in the passband

© The McGraw-Hill Companies, Inc., 2007Original PowerPoint slides prepared by S. K. Mitra 4-1-32• 1/A -Maximum stopband magnitude

Analog Lowpass Filter DesignAnalog Lowpass Filter Design• Two additional parameters are defined:

1) Transition ratio k = Ωp/Ωsfor a lowpass filter k < 1for a lowpass filter k 1

2) Discrimination parameter ll k 1usually k1

Butterworth Approximation (1/5)Butterworth Approximation (1/5)• The magnitude-square response of an N-th order analog

l B tt th filt i i blowpass Butterworth filter is given by

• First 2N −1 derivatives of |Ha(jΩ)|2 at Ω = 0 are equal to zero

• The Butterworth lowpass filter thus is said to have a maximally flat magnitude at Ω = 0maximally-flat magnitude at Ω = 0

• Gain in dB is G(Ω) = 10log10|Ha(jΩ)|2

As G(0) = 0 and G(Ω ) =10log (0 5) = 3 0103 3dB• As G(0) = 0 and G(Ωc) =10log10(0.5) = −3.0103 −3dB• Ωc is called the 3-dB cutoff frequency


Butterworth Approximation (2/5)Butterworth Approximation (2/5)• Typical magnitude responses with Ωc = 1

• Two parameters completely characterizing a Butterworth lowpass filter are Ωc and N

• These are determined from the specified banedges Ωp and Ωs, and minimum passband magnitude , and

i t b d i l 1/A


maximum stopband ripple 1/A

Butterworth Approximation (3/5)Butterworth Approximation (3/5)• Ωc and N are thus determined from

• Solving the above we get

• Since order N must be an integer, value obtained is rounded up to the next highest integerN is used next to determine Ω by satisfying either the• N is used next to determine Ωc by satisfying either the stopband edge or the passband edge spec exactly

• If the stopband edge spec is satisfied, then the passband


If the stopband edge spec is satisfied, then the passband edge spec is exceeded providing a safety margin

Butterworth Approximation (4/5)Butterworth Approximation (4/5)• Transfer function of an analog Butterworth lowpass filter is

i bgiven by

where

• Denominator DN(s) is known as the Butterworth polynomial of order N


Butterworth Approximation (5/5)Butterworth Approximation (5/5)• Example- Determine the lowest order of a Butterworth

l filt ith 1 dB t ff f t 1 kH dlowpass filter with a 1-dB cutoff frequency at 1 kHz and a minimum attenuation of 40 dB at 5 kHzNo• Now

hi h i ld 2 0 25895 d 10l 10(1/A2) 40which yields ε2 = 0.25895, and 10log10(1/A2) = −40 Which yields A2 = 10,000

• Therefore and

HenceHence

• We choose N = 4


We choose N 4

Chebyshev Approximation (1/3)Chebyshev Approximation (1/3)• The magnitude-square response of an N-th order analog

lowpass Type 1 Chebyshev filter is given bylowpass Type 1 Chebyshev filter is given by

where TN(Ω) is the Chebyshev polynomial of order N:

• Typical magnitude response plots of the analog lowpassTypical magnitude response plots of the analog lowpassType 1 Chebyshev filter are shown below


Chebyshev Approximation (2/3)Chebyshev Approximation (2/3)• If Ω = Ωs at the magnitude is equal to 1/A, then

• Solving the above we get

• Order N is chosen as the nearest integer greater than or g gequal to the above value

• The magnitude-square response of an N-th order analog lowpass Type 2 Chebyshev (also called inverse Chebyshev) filter is given by


where TN(Ω) is the Chebyshev polynomial of order N

Chebyshev Approximation (3/3)Chebyshev Approximation (3/3)• Typical response plots of Type 2 Chebyshev lowpass filter:

• The order N of the Type 2 Chebyshev filter is determined by• The order N of the Type 2 Chebyshev filter is determined by

E l D t i th l t d f Ch b h• Example - Determine the lowest order of a Chebyshevlowpass filter with an 1-dB cut-off frequency at 1 KHz and a maximum attenuation of 40 dB at 5 KHz


maximum attenuation of 40 dB at 5 KHz

Elliptic Approximation (1/2)Elliptic Approximation (1/2)• The square-magnitude response of an elliptic lowpass filter

is given byis given by

where R (Ω) is a rational function of order N satisfyingwhere RN(Ω) is a rational function of order N satisfying RN(1/Ω) = 1/RN(Ω), with the roots of its numerator lying in the interval 0 < Ω

Elliptic Approximation (2/2)Elliptic Approximation (2/2)• Example - Determine the lowest order of an Elliptic lowpass

filter with an 1 dB cut off frequency at 1 KHz and afilter with an 1-dB cut-off frequency at 1 KHz and a maximum attenuation of 40 dB at 5 KHzNote k = 0 2 and 1/ k =196 5134Note k = 0.2 and 1/ k1 =196.5134

• Substituting these values we getk’ = 0 979796 ρ = 0 00255135 and ρ = 0 0025513525k = 0.979796, ρ0 = 0.00255135, and ρ = 0.0025513525

• Hence N = 2.23308 (choosing N = 3)• Typical magnitude response plots with Ω =1 are shown• Typical magnitude response plots with Ωp =1 are shown

below


Analog Lowpass Filter DesignAnalog Lowpass Filter Design• Example - Determine the lowest order of an Elliptic lowpass

filter with an 1 dB cut off frequency at 1 KHz and afilter with an 1-dB cut-off frequency at 1 KHz and a maximum attenuation of 40 dB at 5 KHz

• Code fragments used• Code fragments used[N, Wn] = ellipord(Wp, Ws, Rp, Rs, ‘s’);[b, a] = ellip(N, Rp, Rs, Wn, ‘s’);[ , ] p( , p, , , );with Wp = 2*pi*1000;Ws = 2*pi*5000;Rp = 1;Rs = 40;


Design of Analog Highpass, Bandpass, and Bandstop Filters

• Steps involved in the design process:• Steps involved in the design process:Step 1 – Develop specifications of a prototype analog lowpass filter HLP(s) from specifications of desired analoglowpass filter HLP(s) from specifications of desired analog filter HD(s) using a frequency transformationStep 2 – Design the prototype analog lowpass filterp g p yp g pStep 3 – Determine the transfer function HD(s) of desired filter by applying inverse frequency transformation to HLP(s)

• Let s denote the Laplace transform variable of prototype analog lowpass filter HLP(s) and denote the Laplace transform variable ŝ of desired analog filter HD(ŝ)

• Then


Analog Highpass Filter Design (1/2)Analog Highpass Filter Design (1/2)• Spectral Transformation:

where Ωp is the passband edge frequency of HLP(s) andwhere Ωp is the passband edge frequency of HLP(s) and is the passband edge frequency of HHP(ŝ)

• On the imaginary axis the transformation isg y


Analog Highpass Filter Design (2/2)Analog Highpass Filter Design (2/2)• Example - Determine the lowest order of an Butterworth

lowpass filter with the specifications:

• Choose Ωp = 1, then

• Analog lowpass filter specifications:Ωp = 1, Ωs = 1, αp = 0.1 dB, αs = 40 dB, p , s , p , s ,

• Code fragments used[N, Wn] = buttord(1, 4, 0.1, 40, ‘s’);[ ] ( )[B, A] = butter(N, Wn, ‘s’);[num, den] = lp2hp(B, A, 2*pi*4000);


Analog Bandpass Filter Design (1/7)Analog Bandpass Filter Design (1/7)• Spectral Transformation:

where Ωp is the passband edge frequency of HLP(s), and are the lower and upper passband edge frequencies of desired bandpass filter H (ŝ)frequencies of desired bandpass filter HBP(ŝ)

• On the imaginary axis the transformation is


Analog Bandpass Filter Design (2/7)Analog Bandpass Filter Design (2/7)• Case 1:• To make we can either increase any one of

the stopband edges or decrease any one of the passbandedges as sho n beloedges as shown below

1) Decrease to⇒ larger passband and shorter leftmost transition bandg

2) Increase to⇒ No change in passband and shorter leftmost transition


g pband

Analog Bandpass Filter Design (3/7)Analog Bandpass Filter Design (3/7)• Note – the condition can also be satisfied by

d i hi h i t bl th b d idecreasing which is acceptable as the passband is reduced from the desired valueAlternatel the condition can be satisfied b increasing• Alternately, the condition can be satisfied by increasing which is not acceptable as the upper stop band is reduced from the desired valuefrom the desired value


Analog Bandpass Filter Design (4/7)Analog Bandpass Filter Design (4/7)• Case 2:• To make we can either increase any one of

the stopband edges or decrease any one of the passbandedges as sho n beloedges as shown below

1) Increase to⇒ larger passband and shorter rightmost transition bandg g

2) Decrease to⇒ No change in passband and shorter rightmost transition


g p gband

Analog Bandpass Filter Design (5/7)Analog Bandpass Filter Design (5/7)• Note – the condition can also be satisfied by

i i hi h i t bl th b d iincreasing which is acceptable as the passband is reduced from the desired valueAlternatel the condition can be satisfied b decreasing• Alternately, the condition can be satisfied by decreasing which is not acceptable as the lower stop band is reduced from the desired valuefrom the desired value


Analog Bandpass Filter Design (6/7)Analog Bandpass Filter Design (6/7)• Example - Determine the lowest order of an Elliptic

b d filt ith th ifi tibandpass filter with the specifications:

• Now and , since we choosewe choose

• We choose Ωp = 1• Hence• Hence

• Analog lowpass filter specifications: Ω = 1 Ω = 1 4 α = 1• Analog lowpass filter specifications: Ωp = 1, Ωs = 1.4, αp = 1 dB, αs = 22 dB


Analog Bandpass Filter Design (7/7)Analog Bandpass Filter Design (7/7)• Code fragments used:

[N, Wn] = ellipord(1, 1.4, 1, 22, ‘s’);[B, A] = ellip(N, 1, 22, Wn, ‘s’);[ d ] l 2b (B A 2* i*4 8989795 2* i*25/7)[num, den] = lp2bp(B, A, 2*pi*4.8989795, 2*pi*25/7);

• Gain plot:Gain plot:


Frequency KHz

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