eec4-4a-ss-lecture-13.pdf

Signals & Systems FEEE, HCMUT

Ch-7: p ng tn s ca h thng LTI v thit k b lc tng t

Lecture-13

7.1. p ng tn s ca h thng LTI: biu Bode

7.2. Thit k b lc tng t


7.1. p ng tn s ca h thng LTI: biu Bode

7.1.1. p ng tn s ca h thng LTI

7.1.2. Biu Bode



p ng ca h thng LTI c hm truyn H(s) vi tn hiu est

st stf(t)=e y(t)=H(s)e

Khi h thng n nh v ROC cha trc o th ta c th thay s bi

j c c H(j )=H( ) p ng tn s, v ta c:

jt jt1 12 2

f(t)=cost y(t)= H(j)e + H( j)e

jtf(t)=cost y(t)=Re[H(j)e ]

f(t)=cost y(t)=|H(j)|cos t+ H(j)

V d:

Tng qut: f(t)=cos(t+) y(t)=|H(j)|cos t++ H(j)

, vi H(s) l bin i Laplace ca h(t)

jt jtf(t)=e y(t)=H(j)e



|H(j )| l t s bin ca ng ra vi ng vo li ca h

thng. Mt khc |H(j )| c gi tr khc nhau cc tn s khc

nhau p ng bin ca h thng

H(j ) l sai pha ca ng ra vi ng vo v H(j ) c gi tr khc

nhau cc tn s khc nhau p ng pha ca h thng

Vic v th ca p ng tn s l cn thit trong k thut!!!


7.1.2. Biu Bode

Xt h thng vi hm truyn: 1 22

1 2 3

K(s+a )(s+a )H(s)=

s(s+b )(s +b s+b )

1 2 1 2

2

1 3 1 3 2 3

Ka a (s/a +1)(s/a +1)H(s)=

b b s(s/b +1)(s /b +b s/b +1)

1 2 1 2

2

1 3 1 3 2 3

Ka a (j/a +1)(j/a +1)H(j)=

b b j(j/b +1)[ j /b +j b /b +1)

1 2 1 2

2

1 3 1 3 2 3

Ka a |j/a +1||j/a +1||H(j)|=

b b |j||j/b +1|| j /b +j b /b +1|2

2

1 2 1 3 3

j b a a b b b

H(j)= (j +1)+ (j +1) j (j +1) [ +j +1]


7.1.2. Biu Bode

Biu din p ng bin theo thang Logarit:

1 2

1 3 1 2

2

2

1 3 3

Ka a b b a a

j bb b b

20log|H(j)|=20log +20log|j +1|+20log|j +1|

20log|j| 20log|j +1| 20log| +j +1|

Th nguyn ca p ng bin theo thang Logarit l dB

Hng s: Ka1a2/b1b2: 20log [Ka1a2/b1b2]: hng, khng dch pha

Pole (hoc zero) ti gc: Pole: 20log|j|= 20log= 20u

u=log

Cn biu din trn thang tn s Logarit!!!

Zero: 20log|j|=20log=20u

0Pole: j=-90

0zero: j=90


7.1.2. Biu Bode

-30

0

-20

-10

10

20

30

0.01 0.1 1 10 100(u=-2) (u=-1) (u=0) (u=1) (u=2)

20lo

g|H

|,dB

1 decade

-20dB/decade

20dB/decade

Pole (hoc zero) ti gc:

Pole: 20log|j|= 20log= 20u

u=log

Zero: 20log|j|=20log=20u

Zero

Pole


7.1.2. Biu Bode

Pole (hoc zero) ti gc: 0Pole: j=-90

0zero: j=90

-150

0

-100

-50

50

100

150

0.01 0.1 1 10 100(u=-2) (u=-1) (u=0) (u=1) (u=2)

Phas

e, D

egre

es

-90

90

Pole

Zero


7.1.2. Biu Bode

Pole (hoc zero) bc 1: j ja aPole: -20log|1+ |; zero: 20log|1+ |

-18

0

-12

-6

6

12

18

0.01a 0.1a a 10a 100a

20lo

g|H

|,dB

20dB/decade

-20dB/decade3dB

1dB

1dB

0.1a 0.2a 0.5a a 2a 5a 10a

-3

-2.5

-2

-1.5

-1

-0.5

0

Err

or,

dB


7.1.2. Biu Bode

Pole (hoc zero) bc 1: j ja aPole: (1+ ); zero: (1+ )

-135

0

-90

-45

45

90

135

0.01a 0.1a a 10a 100a

Phas

e, D

egre

es

-450/decade

450/decade

0.01a 0.1a a 10a 100a

Ero

or,

Deg

rees

0.05a

-4

0

-2

2

4

6

-6


7.1.2. Biu Bode

Pole (hoc zero) bc 2: n n

2j

-20log|1+j2 + |

n n n n n n n

-40

-10

-30

-20

10

2020lo

g|H

|,dB

-40

dB

/dec

ade

0.1

0.2

0.3

0.5

0.7071

0

-5

10

Err

or,

dB

5

0.1

0.2

0.3

0.5

0.7071


7.1.2. Biu Bode

Pole (hoc zero) bc 2: n n

2j

[1+j2 + ]

0.1n 0.2 n 0.5 n n

2n 5 n n

0

-30

-60

-90

-120

-150

-180

Phas

e, D

egre

es

0.5

0.707

1

0.1

0.2

0.3


7.1.2. Biu Bode

V d 1: 20s(s+100)

H(s)=(s+2)(s+10)


7.1.2. Biu Bode

V d 2: 2

10(s+100)H(s)=

(s +2s+100)


7.2. Thit k b lc tng t

7.2.1. B lc thc t v cc yu cu thit k

7.2.2. S ph thuc ca p ng tn s theo poles v zeros


7.2.1. B lc thc t v cc yu cu thit k

|H(j )|

Gp

1

p s

Gs

0Passband Stopband

|H(j )|

Gp

1

ps

Gs

0PassbandStopband

|H(j )|

Gp

1

s1

Gs

0

Passband Stopband

p1 p2s2

|H(j )|

Gp

1

s1

Gs

0

Passband Stopband

p1 p2 s2

Stopband Passband

Lowpass

filter Highpass

filter

Bandpass

filter

Bandstop

filter



Xt h thng vi hm truyn H(s):

1 1 nn

1 1 n

(s-z )(s-z )...(s-z )P(s)H(s)= =b

Q(s) (s- )(s- )...(s- )

Kho st p ng tn s s=j :

1 2 nn

1 2 n

r r ...r|H(j)|=b

d d ...d

Im

Re

z1

z2

0

j

r1

r1

d1

d2

1 2 n

1 2 n

H(j)= + +...+

- - -...-



Tng li bng mt pole:

|H(j )| H(j )

Re

Im

0

j

-j



Tng li bng mt pole:

|H(j )| H(j )

Re

Im

0

j

-j

j



Tng li bng mt pole:

|H(j )| H(j )

Re

Im

0

j

-j



Tng li bng mt pole:

|H(j )| H(j )

Re

Im

0

j

-j

j



Gim li bng mt zero:

|H(j )|

H(j )

Re

Im

0

j

-j

j0



Gim li bng mt zero:

|H(j )|

H(j )

Re

Im

0

j

-j

j



Gim li bng mt zero:

|H(j )|

H(j )

Re

Im

0

j

-j



Gim li bng mt zero:

|H(j )|

H(j )

Re

Im

0

j

-j

j


B lc thng thp:



B lc thng cao:



B lc thng di:




B lc chn di:

eec4-4a-ss-lecture-13.pdf

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