5-transf analysis lti

Processamento Digital de SinaisEL66D - Engenharia Eletronica

Analise de Sistemas LIT por Transformadas

Prof. Daniel R. [email protected]

wikipipa.org

2014/05/18 Prof. Daniel R. Pipa 1/27

mailto:[email protected]

http://wikipipa.org

Analise de Sistemas LIT por Transformadas Resposta de sistmas LTI no domınio da frequencia

Senoide real

Supondo uma entrada senoidal real

xŒn� D Ax cos.!n C �x/ DAx

2e j�x e j!n

CAx

2e� j�x e� j!n

e sabendo que para sistemas LTI

H.e j!/xŒn� D e j!n yŒn� D H.e j!/e j!n

tem-se

yŒn� DAx

2

ˇH.e j!/

ˇe j.�xC†H.e j!//e j!n

CAx

2

ˇH.e j!/

ˇe� j.�xC†H.e j!//e� j!n

yŒn� D Ay cos.!n C �y/

comAy D

ˇH.e j!/

ˇAx �y D †H.e j!/ C �x .



Exemplo

Seja o sistemayŒn� D ayŒn � 1� C bxŒn�

com transf. ´ e DTFT

H.´/ DY.´/

X.´/D

b

1 � a´�1

´De j!

��! H.e j!/ Db

1 � ae� j!

tem-seˇH.e j!/

ˇD

jbj

j1 � a cos ! C ja sin !jD

jbjp

1 � 2a cos ! C a2

e†H.e j!/ D †b � tan�1 a sin !

1 � a cos !.



Exemplo cont.

Fazendo a D 0:8 eb D 0:2.

Para uma senoide com!0 D 2�=20 D 0:1� ,tem-seˇH.e j!0/

ˇD 0:58

†H.e j!0/ D

�0:255� . Logo, aamplitude decai para0:58 do valor deentrada e a fase eatrasada em 0:255� .Convertendo paraamostras,�0:255�=!0 D

�2:55 amostras.

205 5.1 Sinusoidal response of LTI systems

−p −p/2 0 p/2 p0

0.5

1

ω

−p/2

0 (rad

ians

)

0 10 20 30 40 50 60 70 80 90 100−1

−0.5

0

0.5

1

Am

plitu

de

n

Input

Output

Amplitude change

0.58

−0.26p

0.1p

2.55 samples

0.58

Time delay

p/2

Figure 5.1 Magnitude and phase response functions and input–output signals for the LTIsystem defined by (5.15). The higher frequency suffers more attenuation than the lowerfrequency (lowpass filter).

|H (ej0)| = |b|/(1−a) = 1. This yields b = ±(1−a). If a < 0, the maximum of |H (ejω)|occurs at ω = π . By requiring that |H (ejπ )| = |b|/(1 + a) = 1, we obtain b = ±(1 + a).Both cases can be satisfied by choosing

b = 1 − |a|, (5.20)

which implies |b| = 1 − |a| and ∠b = 0 because −1 < a < 1.Figure 5.1 shows plots of magnitude and phase response functions for a = 0.8 and

an input–output pair for the frequency ω = 2π/20. We can clearly see that sinusoidalinputs with frequencies close to ω = 0 pass with small attenuation; in contrast, sinu-soids with frequencies close to ω = π are severely attenuated. Since for a > 0,|H (ejω)|max

!|H (ejω)|min = (1 + a)/(1 − a), the peak of the magnitude response becomes

narrower as a approaches one. From the magnitude and phase response plots in Figure 5.1,the normalized gain at ω = 2π/20 is about 0.58 while the phase shift is about −0.26π

radians (or −0.26π/ω = −2.55 samples). These values are evident from the input–outputplots in Figure 5.1. "


Analise de Sistemas LIT por Transformadas Distorcao de sinais passando por sistemas LIT

Sistemas sem distorcaoPara nao haver distorcoes na “forma” do sinal

yŒn� D GxŒn � nd�,

onde G representa um ganho e nd um atraso. No domınio Fourier

Y.e j!/ D Ge� j!ndX.e j!/.

Portanto a resposta em frequencia e

H.e j!/ DY.e j!/

X.e j!/D Ge� j!nd .

ˇH.e j!/

ˇD G †H.e j!/ D �!nd

Para nao gerar distorcao, um filtro deve ter magnitude constante G efase linear em funcao de ! com inclinacao �nd.



Distorcao de faseQuando ˇ

H.e j!/ˇ

D G †H.e j!/ ¤ �!nd

ha apenas distorcao de fase.

Filtros passa-tudoSao chamados os sistemas sem distorcao de magnitude. Suascaracterısticas sao determinadas apenas pelas mudancas de fase.

UsoSao usados para corrigir distorcoes de fase gerados por outrossubsistemas.217 5.3 Distortion of signals passing through LTI systems

1

−1

1

−1

1

−1

1

−1

1

−1

1

−1

x[n]

y1[n]

y2[n]

y3[n]

y4[n]

y5[n]

n

n

n n

n

n

Original signal

Low-frequency attenuation

High-frequency attenuation

Constant phase shift

Linear-phase shift

Nonlinear-phase shift

(a) (b)

Figure 5.6 Magnitude (a) and phase (b) distortions. Clearly, it is difficult to distinguish theeffects of magnitude and phase distortion.

We note that if a system attenuates the low-frequency component c1 to 1/4, the resultingsignal y1[n] becomes “sharper.” In contrast, attenuating the high-frequency components iny2[n] results in a “smoother” signal. However, we cannot predict the extent of sharpeningor smoothing without computing the output signal.

Phase or delay distortion If the phase response is not a linear function of frequency,that is,

∠H (ejω) = −ωnd, (5.55)

the resulting distortion is known as phase or delay distortion.The phase response ∠H (ejω) gives the phase shift (in radians) experienced by each

sinusoidal component of the input signal. If we rewrite (5.12) as

y[n] = Ax|H (ejω)| cos[ωn + φx + ∠H (ejω)] (5.56)

= Ax|H (ejω)| cos!ω

"n + φx

ω+ ∠H (ejω)

ω

#$, (5.57)



Distorcao de magnitudeQuando ˇ

H.e j!/ˇ

¤ G †H.e j!/ D �!nd

ha apenas distorcao de magnitude.

Filtros de fase linearSao chamados os sistemas sem distorcao de fase. Suas caracterısticassao determinadas apenas pelas mudancas de magnitude.

UsoEm geral, visam-se filtros de fase linear, pois nao se deseja distorcoesdo sinal na banda de passagem.217 5.3 Distortion of signals passing through LTI systems

1

−1

1

−1

1

−1

1

−1

1

−1

1

−1

x[n]

y1[n]

y2[n]

y3[n]

y4[n]

y5[n]

n

n

n n

n

n

Original signal




Linear-phase shift


(a) (b)




∠H (ejω) = −ωnd, (5.55)





"n + φx

ω+ ∠H (ejω)

ω

#$, (5.57)

217 5.3 Distortion of signals passing through LTI systems

1

−1

1

−1

1

−1

1

−1

1

−1

1

−1

x[n]

y1[n]

y2[n]

y3[n]

y4[n]

y5[n]

n

n

n n

n

n

Original signal




Linear-phase shift


(a) (b)




∠H (ejω) = −ωnd, (5.55)





"n + φx

ω+ ∠H (ejω)

ω

#$, (5.57)


Analise de Sistemas LIT por Transformadas Filtros ideais e praticos

Filtros ideais

Filtros seletivos em frequenciaSao sistemas que bloqueiam frequencias na faixa de rejeicao edeixam passar sem distorcao frequencias na faixa de passagem.Portanto, devem ter fase linear na faixa de passagem.

Passa-baixas (low-pass)

Hlp.e j!/ D

(e� j!nd ; j!j < !c

0; !c < j!j � �

Passa-altas (hight-pass)

Hhp.e j!/ D

(0; j!j < !c

e� j!nd ; !c < j!j � �

Passa-faixa (band-pass)

Hbp.e j!/ D

(e� j!nd ; !l � j!j � !u

0; caso contrario

Rejeita-faixa (band-stop)

Hbs.e j!/ D

(0; !l � j!j � !u

e� j!nd ; caso contrario



Filtros ideais222 Transform analysis of LTI systems

0−p p w−wcwc

1

0−p p w−wl−wu wl wu

1

0−p p w−wcwc

1

0−p p w−wl−wu wl wu

1

(a) (b)

(c) (d)

Figure 5.9 Ideal frequency-selective filters: (a) lowpass filter, (b) bandpass filter, (c) highpassfilter, and (d) bandstop filter.

An ideal lowpass filter is defined by (5.67) with ωℓ = 0, whereas an ideal highpass filterhas ωu = π . Ideal bandstop filters have a distortionless response over all frequenciesexcept some stopband, ωℓ ≤ |ω| ≤ ωu, where H (ejω) = 0. We emphasize that the phaseresponse ∠H (ejω) is required to be linear only in the passband; there is no need for it to bedefined elsewhere because the response of the filter is zero. Figure 5.9 shows the frequencyresponses of four types of ideal filter.

To understand the implications of the “steep” transition from passband to stopband inideal filters, we consider an ideal lowpass filter with frequency response

Hlp(ejω) =!

e− jωnd , |ω| < ωc

0. ωc < |ω| ≤ π(5.69)

The impulse response corresponding to (5.69) is given by (see Example 4.13)

hlp[n] = sin ωc(n − nd)

π(n − nd). (5.70)

The impulse response and the step response of the ideal lowpass filter are illustrated inFigure 5.10 for nd = 0. We note that hlp[n] extends from −∞ to ∞; therefore we can-not compute the output of the ideal lowpass filter using a convolution sum. The impulseresponse hlp[n] has a DTFT Hlp(ejω) because it has finite energy. However, it should benoted that hlp[n] is not absolutely summable, that is,

∞"

n=−∞|hlp[n]| = ∞. (5.71)



Filtros praticosUm filtro passa-baixas ideal

Hlp.e j!/ D

(e� j!nd ; j!j < !c

0; !c < j!j � �

tem resposta ao impulso

hlpŒn� Dsin !c.n � nd/

�.n � nd/223 5.4 Ideal and practical filters

−20 −15 −10 −5 0 5 10 15 20−0.1

0

0.1

0.2

0.3

n

−20 −15 −10 −5 0 5 10 15 20

0

0.5

1

n

s[n]

h[n]

Figure 5.10 Impulse and step response sequences of the ideal lowpass filter.

Therefore, the ideal lowpass filter is unstable. Furthermore, since r−nhlp[n] is not abso-lutely summable for any value of r, the sequence hlp[n] does not have a z-transform.Since only systems with a rational system function can be computed recursively, wededuce that we cannot compute the output of the ideal lowpass filter either recursivelyor nonrecursively. In conclusion, the ideal lowpass filter is unstable and practicallyunrealizable.

The impulse response of the ideal bandpass filter can be obtained by modulating theimpulse response of an ideal lowpass filter with ωc = (ωu −ωℓ)/2 = #ω/2 using a carrierwith frequency ω0 = (ωu + ωℓ)/2. The result is

hbp[n] = 2sin ωc(n − nd)

π(n − nd)cos ω0n. (5.72)

The impulse responses of the ideal highpass and bandstop filters are given by

hhp[n] = δ[n] − hlp[n], (5.73)

hbs[n] = δ[n] − hbp[n], (5.74)

because Hhp(ejω) = 1 − Hlp(ejω) and Hbs(ejω) = 1 − Hbp(ejω). Therefore, all ideal filtersare unstable and unrealizable. Since all ideal filters can be expressed in terms of (5.69), werefer to Hlp(ejω) as the ideal lowpass prototype filter.

Ideal filters are used in the early stages of a design process to specify the modules in asignal processing system. However, since they are not realizable in practice, they must beapproximated by practical or nonideal filters. This is usually done by minimizing someapproximation error between the nonideal filter and a prototype ideal filter.

Freq. !c D �=4. Desloc. nd D 0.2014/05/18 Prof. Daniel R. Pipa 10/27


Filtros praticos

PoremhlpŒn� ¤ 0; n < 0 ) nao causal

1XnD�1

ˇhlpŒn�

ˇD 1 ) instavel

Filtros ideaisFiltros ideais nao sao realizaveis, pois sua resposta ao impulso e naocausal (depende de valores futuros) e instavel (nao e absolutamentesomavel).

Filtros praticosSerao necessarias aproximacoes que tornem hlpŒn� estavel e causal,para que o filtro possa ser usado na pratica.



Filtros praticosUma possıvel aproximacao e o simples truncamento de hlpŒn�

hlpŒn� D

8<:sin !c.n � nd/

�.n � nd/; 0 � n � M � 1

0; caso contrario.

Devido ao fenomeno de Gibbs, haverao oscilacoes (ripple) naresposta em frequencia.224 Transform analysis of LTI systems

1

0 π

Passband

Stopband

Transition-bandTransition-band

Stopband

Figure 5.11 Typical characteristics of a practical bandpass filter.

The design of practical filters that approach ideal behavior is the subject of Chapters 10and 11. To understand the nature of the approximations required to obtain a practical filterfrom an ideal filter, we note that we can obtain a causal FIR filter by truncating the impulseresponse of the ideal lowpass filter as follows

hlp[n] =

⎧⎨

⎩

sin ωc(n − nd)

π(n − nd), 0 ≤ n ≤ M − 1

0. otherwise(5.75)

As the delay nd and the length M of hlp[n] increase, the resulting filter Hlp(ejω) will be abetter approximation of the ideal lowpass filter.

A natural question arising at this point is how to evaluate the quality of a practical filter.Figure 5.11 shows the magnitude response of a typical practical bandpass filter. Comparedto the ideal bandpass filter in Figure 5.9, we observe a passband where |H (ejω)| fluctuatesabout one and stopbands where |H (ejω)| fluctuates close to zero. Between the passbandand stopbands are transition bands, where the filter neither passes nor rejects the inputfrequency components. A good filter should have only a small ripple in the passband,high attenuation in the stopband, and very narrow transition bands. In some applications,the specifications of phase characteristics or time-domain characteristics (for example, theovershoot of the step response) are also important. These issues, which are of fundamentalsignificance in filter design, are further investigated in Chapters 10 and 11.

5.5 Frequency response for rational system functions• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

In Section 3.6, we demonstrated that all LTI systems of practical interest are described bya difference equation of the form

y[n] = −N∑

k=1

aky[n − k] +M∑

k=0

bkx[n − k], (5.76)

Aproximacoes melhores serao vistas nas proximas aulas.2014/05/18 Prof. Daniel R. Pipa 12/27

Analise de Sistemas LIT por Transformadas Projeto de filtros por alocacao de polos e zeros

Dependencia dos polos e zeros na resposta em frequencia

Uma funcao de transferencia generica pode ser fatorada nos polos ezeros

H.´/ Db0 C b1´�1 C � � � C bM ´�M

1 C a1´�1 C � � � C aN ´�ND b0´N �M

QMkD1.´ � ´k/QNkD1.´ � pk/

e a DTFT se obtem fazendo ´ D e j!

H.e j!/ D b0e j!.N �M/

QMkD1.e j! � ´k/QNkD1.e j! � pk/

D b0e j!.N �M/

QMkD1

��!ZkZQN

kD1

��!PkZ

onde.e j!

� ´k/ D��!ZkZ .e j!

� pk/ D��!PkZ

sao vetores que iniciam em ´k e pk , respectivamente, e terminam nocırculo unitario e j! .



Dependencia dos polos e zeros na resposta em frequenciaMagnitude e fase da respo. em freq. do sistema

ˇH.e j!/

ˇD jb0j

QMkD1

ˇ��!ZkZ

ˇQN

kD1

ˇ��!PkZ

ˇ†H.e j!/ D !.M � N / C

MXkD1

†��!ZkZ �

MXkD1

†��!PkZ

Magnitude em uma certa freq. !

E o produtorio das distancias dos zeros sobre o produtorio dasdistancias dos polos.

Fase em uma certa freq. !

Somatorio dos angulos de��!ZkZ menos o somat. dos angulos de

��!PkZ.



Dependencia dos polos e zeros na resposta em frequencia

ZerosQuanto mais proximos docirc. unit. em !0, maiscontribuirao paraˇH.e j!0/

ˇ! 0.

PolosQuanto mais proximos docirc. unit. em !0, maiscontribuirao paraˇH.e j!0/

ˇ! 1.

231 5.6 Dependence of frequency response on poles and zeros

5.6 Dependence of frequency response on poles and zeros• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

The shape of the frequency response is determined by the impulse response or the coef-ficients of the difference equation. However, we cannot guess the shape of |H (ejω)| and∠H (ejω)| by inspecting the values of h[n] or {ak, bk}. In this section, we show that thereis a strong dependence of the shape of the frequency response on the location of poles andzeros of the system. We can use this dependence to (a) obtain a simple and intuitive proce-dure for determining quickly the magnitude and phase response, and (b) to gain physicalinsight into the filtering characteristics of LTI systems.

5.6.1 Geometrical evaluation of H (ejω) from poles and zeros

We start by noting that (5.79) can be equivalently written as

H (ejω) = b0

⎡

⎢⎢⎢⎢⎢⎣

M∏

k=1

(1 − zk e− jω)

N∏

k=1

(1 − pk e− jω)

⎤

⎥⎥⎥⎥⎥⎦= b0ejω(N−M)

⎡

⎢⎢⎢⎢⎢⎣

M∏

k=1

(ejω − zk

)

N∏

k=1

(ejω − pk

)

⎤

⎥⎥⎥⎥⎥⎦. (5.100)

This equation consists of factors of the form(ejω − zk

)and

(ejω − pk

). The factor

(ejω − zk

)is a complex number represented by a vector

−→ZkZ drawn from the point zk

(zero) to the point z = ejω in the complex plane, as illustrated in Figure 5.17. This complexnumber can be written in polar form as follows:

0

Z

PkZk

O

ω

βkαk

Φk Θk

1

zkpk θk

φk

Figure 5.17 The quantities required to compute the magnitude and phase response of a systemfrom the location of its poles and zeros.2014/05/18 Prof. Daniel R. Pipa 15/27


Aumento do ganho por um poloPara aumentar o ganho numa freq. !k , pode-se colocar um polodentro do cırculo unitario em pk D rke j! ; 0 < rk < 1. A freq. maisproxima do polo, !k , tera o maior aumento de ganhoˇ

Hnew.e j!k /ˇ

D1

1 � rk

ˇHold.e j!k /

ˇ

233 5.6 Dependence of frequency response on poles and zeros

can determine H (ejω) for every value of ω or equivalently any location of the point ejω

on the unit circle.

5.6.2 Significance of poles and zeros

To understand the effect of poles and zeros on the magnitude and phase responses, weseparately consider the case of a single pole and a single zero.

Gain enhancement by a pole Consider a pole pk = rk ejφk , as illustrated in Figure 5.18.To find the magnitude response |H (ejω)| for a certain value of ω, we connect the pole(point Pk) to the tip of vector z = ejω (point Z on the unit circle). If the length of this lineis Rk(ω), then

|H (ejω)| = κ

(PkZ)= κ

Rk(ω), (5.105)

where overbar denotes the length of a vector. The exact value of constant κ is not importantat this point. The line segment PkZ takes its minimum value 1 − rk at ω = φk, and itsmaximum value 1 + rk at ω = φk + π . Therefore, the length PkZ increases progressivelyas ω increases from φk to φk + π and then decreases continuously until ω approaches thevalue φk. Then, according to (5.105), |H (ejω)| decreases as ω goes from φk to φk + π andthen progressively increases as ω moves closer to φk (see Figure 5.18). We conclude thata pole pk = rk ejφk results in a frequency-selective response that enhances the gain aroundω = φk (angle of the pole) and attenuates the gain as we move away from φk. The dynamicrange of the magnitude response

|H (ejω)|max

|H (ejω)|min= 1 + rk

1 − rk(5.106)

00O

O

AA

φk φk

Pk Pk

ωk ωk

αk

(a) (b)

Z Z

11

Figure 5.18 Geometrical computation of magnitude (a), and phase (b), responses for the caseof a single pole.2014/05/18 Prof. Daniel R. Pipa 16/27


Aumento do ganho por um polo

A freq. mais distance do polo, � � !k , tera o menor aumento deganho ˇ

Hnew.e j.��!k//ˇ

D1

1 C rk

ˇHold.e j.��!k//

ˇSinais e sistemas reais no tempoPor que os sinais e sistemas reais no tempo, os polos sempre aparecemem complexos conjugados. Logo, deve-se adicionar dois polos

pk D rke j! e p�k D rke� j!



Aumento do ganho por um polo235 5.6 Dependence of frequency response on poles and zeros

0

Z

Z

Pk

O ω

−ω

φk

−φk

1r

Pk

Magnitude responses

Phase responses

pk

pk

pk*

pk*

Total

Total

−π −π/2 0 π/2 π0

5

10

ω

−π −π/2 0 π/2 π−π/2

0

π/2

ω

Figure 5.19 Geometrical explanation of the shape of magnitude and phase responsesgenerated by single and complex conjugate poles at pk = 0.9e±jπ/3. Only pole-zero patternswith mirror symmetry about the real axis, that is, real or complex-conjugate poles and zeros,result in magnitude responses with even symmetry and phase responses with odd symmetry.

the unit circle; hence, the system fully suppresses sinusoidal components with frequencyω = θk. Thus, we conclude that zeros have the opposite effect of poles. From (5.104)and the triangle OZZk in Figure 5.17, we can show that the phase response for a singlezero is

∠H (ejω) = −ω + $k(ω) = βk, (5.112)

and changes sign at ω = θk and ω = θk + π . Adding a complex-conjugate zero, to assurea system with real coefficients, makes the magnitude response function symmetric aboutω = 0. This can be proven using arguments similar to those used for complex conjugatepoles (see Problem 32).

Zeros outside the unit circle Although the poles of causal and stable systems should beinside the unit circle, their zeros can be anywhere in the z-plane. Moving a zero outside theunit circle, without changing its angle, has an interesting effect on its phase response. Thiseffect is illustrated using the geometrical construction in Figure 5.20. The phase responseof the zero inside the unit circle is equal to αin(ω) = $in(ω)−ω and continuously changessign at ω = 0. In contrast, for a zero outside the unit circle, the phase response changesfrom −π at ω = 0 − ϵ to π at ω = 0 + ϵ, where ϵ is an arbitrarily small positive number.Furthermore, since $out(ω) ≥ $in(ω) we have αout(ω) ≥ αin(ω). Thus, zeros outsidethe unit circle introduce larger phase shifts than zeros inside the unit circle. This topic isfurther discussed in Section 5.10. For a zero, z = rejθ , not on the real-line we can simply



Projeto de filtros por alocacao de polos e zeros

I Para se eliminar uma componente de freq. em ! D !0, aloca-seum zero com angulo !0 sobre o cırculo unitario.

I Para se amplificar uma componente de freq. em ! D !0,aloca-se um polo com angulo !0 perto, porem dentro, do cırculounitario. A amplificacao sera inversamente proporcional adistancia.

I Polos e zeros complexos devem aparecer em pares complexosconjugados, garantindo as simetrias em freq. e sinais reais notempo.

I Poles ou zeros na origem nao influenciam a magnitude, pois suadistancia ao cırculo unitario e sempre 1 para qualquer !.



ExemploProjetar um filtro IIR com as seguintes caracterısticas:

I Magnitude 0 em !1 D 0 e !3 D � .I Magnitude 1 em !2;4 D ˙�=5.I b0 D 0:5.

Solucao:I Dois zeros em ´ D e j0

D 1 e ´ D e j�D �1 para garantir

magnitude 0.I Dois polos em ´ D re˙ j�=5 para “puxar” ganho para cima.

Determinar r .

H.´/ D b0.1 � ´�1/.1 C ´�1/

.1 � re j�=5´�1/.1 � re� j�=5´�1/

D b01 � ´�2

.1 � 2r cos.�=5/´�1 C r2´�2/



Exemplo

Resposta em frequencia

H.e j!/ D b01 � e� j2!

.1 � 2r cos.�=5/e� j! C r2e� j2!/

Magnitude em ! D �=5 devido somente aos zerosˇH´.e j!/

ˇD 0:5

ˇ1 � e j2�=5

ˇD 0:5878

Precisa-se gerar um ganho de 1=0:5878 D 1:7 com os polos.Desprezando o polo mais distante

1

1 � rD 1:7 r D 0:4122



Filtros Notch

UsoSao filtros projetados para eliminar apenas uma frequencia especıfica,por exemplo, ruıdo de 60 Hz da rede.

Para se eliminar uma componente de freq. !0, podem-se alocar zerosem cima do cırculo unitario com angulos !0

H.´/ D .1�e j!0´�1/.1�e� j!0´�1/ D 1�.e j!0 Ce� j!0/´�1C´�2

Inserindo um ganho b0 para garantirˇH.e j!/

ˇmax D 1

H.´/ D b0

�1 � .2 cos !0/´�1

C ´�2�



Filtros Notch

ProblemaOs zeros tambem vao atenuar frequencias vizinhas, que nao deveriamser alteradas.

172 Fourier representation of signals

some operations on a signal or between two signals result in different operations betweentheir DTFTs.

4.5.1 Relationship to the z -transform and periodicity

The z-transform of a sequence x[n] was defined in Section 3.2 by

X(z) =∞!

n=−∞x[n]z−n. (4.101)

If the ROC of X(z) includes the unit circle, defined by z = ejω or equivalently |z| = 1, weobtain

X(z)|z=ejω =∞!

n=−∞x[n]e− jωn = X(ejω), (4.102)

that is, the z-transform reduces to the Fourier transform. The magnitude of DTFT isobtained by intersecting the surface |H(z)| with a vertical cylinder of radius one, centeredat z = 0. This is illustrated in Figure 4.26, which provides a clear demonstration of theperiodicity of DTFT. The radiant frequency ω is measured with respect to the positive realaxis and the unit circle is mapped on the linear frequency axis as shown in Figure 4.26.Multiple rotations around the unit circle create an inherent periodicity, with period 2π

ω

ω=0

ω=2pω= |H(e

jω)|

−1 −0.5 0 0.5 1−1

−0.5

0

0.5

1

Real Axis

Imag

inar

y A

xis

p

Figure 4.26 The relationship between the z-transform and the DTFT for a sequence with twocomplex-conjugate poles at z = 0.9ej±π/4 and two zeros at z = ±1.

EfeitoA banda do filtro e larga. Como diminuir a banda, ou seja, diminuir oefeito nas freq. vizinhas?



Filtros Notch

SolucaoColocam-se polos com mesmo angulo !0 bem perto dos zeros, ouseja, p D re j!0 ; r < 1. Os polos puxarao as freq. vizinhas para cima.

H.´/ D b0.´ � e j!0/.´ � e� j!0/

.´ � re j!0/.´ � re� j!0/D b0

1 � .2 cos !0/´�1 C ´�2

1 � .2r cos !0/´�1 C r2´�2241 5.7 Design of simple filters by pole-zero placement

−π −2π/5

−2π/5

0 π0

0.5

1

ω

−π 0 π−π

−π/2

0

π/2

π

ω

2π/5

2π/5

Figure 5.24 Magnitude and phase response of a second-order FIR notch filter (dashed line)and a second-order IIR notch filter with r = 0.9 and φ = 2π/5.

If we cascade several second-order sections, like (5.124), we can create FIR filters withbetter stopbands (see Problem 39).

If the zeros of (5.124) are placed on the unit circle by setting r = 1, the input fre-quency components at ω = ±φ are eliminated. Filters that have perfect nulls at certainfrequencies are known as notch filters. The second-order FIR notch filter has systemfunction

H(z) = b0[1 − (2 cos φ)z−1 + z−2]. (5.125)

The problem with FIR notch filters is that the bandwidth of the notches is large. A simpleway to create sharper notches is to place the zeros on the unit circle and two complex con-jugate poles at the same angle with the zeros, close to the zeros but inside the unit circle.The system function for the resulting notch filter is

G(z) = b01 − (2 cos φ)z−1 + z−2

1 − (2r cos φ)z−1 + r2z−2 . (5.126)

The creation of sharper notches is illustrated in Figure 5.24, which shows the magnituderesponses of (5.125) and (5.126) for r = 0.9 and φ = 0.4π radians. The gain b0 is chosenso that the maximum magnitude response is equal to one (see Problem 64).

Freq. !0 D 2�=5. Pontilhada: Filtro FIR, so zeros. Cheia: Sistmas IIR, com polos, r D 0:9.2014/05/18 Prof. Daniel R. Pipa 24/27

Analise de Sistemas LIT por Transformadas Conclusao

Sumario

O que foi visto hoje?

I Sistemas sem distorcao devem ter fase linear.I Filtros ideais sao instaveis e nao causais.I Aproximacoes sao necessarias. O simples truncamento de hŒn�

gera oscilacoes (ripple) devido ao fenomeno de Gibbs.I Zeros em sobre o cırculo unitario em e j!0 eliminam a

componente de freq.!0.I Polos perto do cırculo unitario em e j!0 amplificam a

componente de freq. !0.I Pode-se projetar filtros simples alocando polos e zeros. Tecnicas

mais avancadas serao vistas nos proximos capıtulos.



Exercıcios

Questoes: 5.12, 5.14, 5.17Problemas: 5.27



Respostas

Problemas:

I 5.27a: H.e j!/ D

0:5.1 � 0:45e j!C 0:05e j2!/=.1 � 0:325e j!

C 0:0225e j2!/.I 5.27b: Nao.I 5.27c: Existe, mas nao e unica.I 5.27d: Nao.


5-transf analysis lti

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