주파수응답을이용한해석법...
TRANSCRIPT
![Page 1: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/1.jpg)
주파수 응답을 이용한 해석 법Continuous-Time
![Page 2: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/2.jpg)
주파수 응답의 정의
( ) ( )( )
Y s G sU s
= ( ) sinu t A tω=
( )( ) ( )1 2
( )( )n
b sG ss p s p s p
=− − −
2 2( ) ( ) ( ) ( ) AY s G s U s G ss
ωω
= =+
*0 01 2
1 2
( ) n
n
Y ss p s p s p s j s j
α α αα αω ω
= + + + + +− − − + −
1 2 *1 2 0 0( ) np tp t p t j t j t
ny t e e e e eω ωα α α α α−= + + + + +
![Page 3: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/3.jpg)
주파수 응답의 정의
*0 0( ) j t j ty t e eω ωα α−= +
( )0 2 2
( )
( ) ( )2
( )2
s j
j G j
A AG s s j G js j
A G j ej
ω
ω
ωα ω ωω
ω
=−
− ∠
= + = − −+
= −
( )*0 2 2
( )
( ) ( )2
( )2
s j
j G j
A AG s s j G js j
A G j ej
ω
ω
ωα ω ωω
ω
=
∠
= − =+
=
![Page 4: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/4.jpg)
주파수 응답의 정의
( ) ( )
( )
( ) ( )
( ) ( )
( ) ( ) ( )2 2
( )2 2
( ) sin ( )
j G j j t j G j j t
j t G j j t G j
A Ay t G j e e G j e ej j
e eA G jj j
A G j t G j
ω ω ω ω
ω ω ω ω
ω ω
ω
ω ω ω
− ∠ − ∠
− +∠ +∠
= − +
⎛ ⎞= − +⎜ ⎟⎜ ⎟
⎝ ⎠= +∠
![Page 5: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/5.jpg)
간단한 전기회로의 주파수 응답
dy u yCdt R
−=
( ) 1( )( ) 1
Y s G sU s RCs
= =+
( )2
1 1( )1 1
G jjRC RC
ωω ω
= =+ +
( )11( ) tan1
G j RCjRC
ω ωω
−⎛ ⎞∠ = ∠ = −⎜ ⎟+⎝ ⎠
![Page 6: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/6.jpg)
간단한 전기회로의 주파수 응답
![Page 7: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/7.jpg)
간단한 전기회로의 주파수 응답1f Hz=
![Page 8: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/8.jpg)
간단한 전기회로의 주파수 응답
3f Hz=
![Page 9: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/9.jpg)
주파수 응답과 극점의 관계
( ) ( )
2
22 2
1( )2 / 2 / 1
n
n n n n
G ss s s s
ωζω ω ω ζ ω
= =+ + + +
( ) ( )2
( )
1/ 2 / 1n n
G s
s sω ζ ω=
+ +
![Page 10: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/10.jpg)
표준형 2차 시스템의 극점
![Page 11: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/11.jpg)
표준형 2차 시스템의 ( )G s
0.3ζ = 0.5ζ =
![Page 12: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/12.jpg)
표준형 2차 시스템의 ( )G s
0.7ζ = 0.9ζ =
![Page 13: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/13.jpg)
보드 선도
10( ) 20log ( )dB
G j G jω ω=
![Page 14: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/14.jpg)
보드 선도
( )( )( )( )
1 2
1 2
( ) m
s z s zG s K
s s p s p− −
=− −
( )( )( ) ( )( )
1 2
1 2
( ) ( ) ms j
j z j zG j G s K
j j p j pω
ω ωω
ω ω ω=
− −= =
− −
( )( )( ) ( )( )0
1 2
1 1( )
1 1a b
m
j jG j K
j j jωτ ωτ
ωω ωτ ωτ
+ +=
+ +
![Page 15: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/15.jpg)
보드 선도
( )( )( ) ( )( )
( )
( )
10 10 01 2
10 0
1 2
10 0 10 10
10 10 1 10 2
1 1( ) 20log ( ) 20log
1 1
1 120log
1 1
20log 20log 1 20log 1
20log 20log 1 20log 1
a bmdB
a bm
a b
m
j jG j G j K
j j j
j jK
j j j
K j j
j j j
ωτ ωτω ω
ω ωτ ωτ
ωτ ωτ
ω ωτ ωτ
ωτ ωτ
ω ωτ ωτ
+ += =
+ +
+ +=
+ +
= + + + + +
− − + − + −
![Page 16: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/16.jpg)
보드 선도
( )( )( ) ( )( )
( ) ( )( ) ( ) ( )
01 2
0
1 2
1 1( )
1 1
1 1
1 1
a bm
a b
m
j jG j K
j j j
K j j
j j j
ωτ ωτω
ω ωτ ωτ
ωτ ωτ
ω ωτ ωτ
⎛ ⎞+ +∠ = ∠⎜ ⎟
⎜ ⎟+ +⎝ ⎠= ∠ +∠ + +∠ + +
−∠ −∠ + −∠ + −
![Page 17: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/17.jpg)
전달 함수의 기본 항
0mK s
( ) ( )1 ,1/ 1s sτ τ+ +
( ) ( ) ( ) ( )2 2/ 2 / 1 ,1/ / 2 / 1n n n ns s s sω ζ ω ω ζ ω⎡ ⎤ ⎡ ⎤+ + + +⎣ ⎦ ⎣ ⎦
![Page 18: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/18.jpg)
0mK s
( ) ( )10 0 10 0 10
10 0 10
20 log 20log 20log
20log 20 log
m mK j K j
K m
ω ω
ω
= +
= +
![Page 19: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/19.jpg)
• 위상
0mK s
( )( ) ( )( )
0 0
0 0 90
m mK j K j
K m j K m
ω ω
ω
∠ = ∠ +∠
= ∠ + ∠ = ∠ + × °
![Page 20: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/20.jpg)
( )1sτ +
10 10
10 10
1 20log 1 20log 1 0 1/1 20log 1 20log 1/
dB
dB
j jj jωτ ωτ ω τωτ ωτ ωτ ω τ
+ = + ≈ =+ = + ≈
( )210 101 20log 1 20log 1
dBj jωτ ωτ ωτ+ = + = +
![Page 21: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/21.jpg)
( )1sτ +
![Page 22: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/22.jpg)
( )1sτ +
( )( ) ( )( ) ( )
1 1 0 1/1 1 45 1/1 90 1/
jj jj j
ωτ ω τωτ ω τωτ ωτ ω τ
∠ + ≈ ∠ = °∠ + = ∠ + = ° =∠ + ≈ ∠ = °
![Page 23: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/23.jpg)
( )1sτ +
![Page 24: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/24.jpg)
( )1/ 1sτ +
10 10
10 10
1 20log 1 20log 1 0 1/1
1 20log 1 20log 1/1
dB
dB
jj
jj
ωτ ω τωτ
ωτ ωτ ω τωτ
= − + ≈ − =+
= − + ≈ −+
![Page 25: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/25.jpg)
( )1/ 1sτ +
![Page 26: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/26.jpg)
( )1/ 1sτ +
( )
( )
1 1 0 1/1
1 1 45 1/1
1 90 1/1
j
jj
jj
ω τωτ
ω τωτ
ωτ ω τωτ
⎛ ⎞∠ ≈ −∠ = °⎜ ⎟+⎝ ⎠⎛ ⎞
∠ = −∠ + = − ° =⎜ ⎟+⎝ ⎠⎛ ⎞
∠ ≈ −∠ = − °⎜ ⎟+⎝ ⎠
![Page 27: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/27.jpg)
( )1/ 1sτ +
![Page 28: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/28.jpg)
예제 6-1
( )1000( )
10G s
s s=
+ ( ) ( )1000 100( )
10 0.1 1s j
G js s j j
ω
ωω ω
=
= =+ +
10
10
10020log
20log 100 20log
40 20log
jωω
ω
= −
= −
![Page 29: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/29.jpg)
예제 6-1
![Page 30: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/30.jpg)
예제 6-1
![Page 31: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/31.jpg)
예제 6-2
( )( )( )
1000 1( )
10 100s
G ss s
+=
+ +
( )( ) ( )
( )( )( )
1000 1 1( )
10 /10 1 100 /100 1 0.1 1 0.01 1s j
s jG j
s s j jω
ωω
ω ω=
+ += =
+ + + +
![Page 32: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/32.jpg)
예제 6-2
![Page 33: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/33.jpg)
예제 6-2
![Page 34: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/34.jpg)
예제 6-2
![Page 35: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/35.jpg)
예제 6-2
![Page 36: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/36.jpg)
( ) ( )21/ / 2 / 1n ns sω ζ ω⎡ ⎤+ +⎣ ⎦
( ) ( )
( ) ( )( )
102
2102
1 20log 1 0/ 2 / 1
1 20log // 2 / 1
nn n dB
n nn n dB
j j
j j
ω ωω ω ς ω ω
ω ω ω ωω ω ς ω ω
≈ − =+ +
≈ −+ +
![Page 37: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/37.jpg)
( ) ( )21/ / 2 / 1n ns sω ζ ω⎡ ⎤+ +⎣ ⎦
![Page 38: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/38.jpg)
( ) ( )21/ / 2 / 1n ns sω ζ ω⎡ ⎤+ +⎣ ⎦
( ) ( )21 1
2/ 2 / 1n
n nj jω ω
ζω ω ζ ω ω=
=+ +
120log 20log 22
ζζ= −
![Page 39: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/39.jpg)
( ) ( )21/ / 2 / 1n ns sω ζ ω⎡ ⎤+ +⎣ ⎦
( ) ( )
( ) ( )( )
( ) ( )( )
2
2
2 22
1 1 0/ 2 / 1
1 2 90/ 2 / 1
1 / 180/ 2 / 1
nn n
nn n
n nn n
j j
jj j
j j
ω ωω ω ς ω ω
ς ω ωω ω ς ω ω
ω ω ω ωω ω ς ω ω
⎛ ⎞∠ ≈ −∠ = °⎜ ⎟⎜ ⎟+ +⎝ ⎠⎛ ⎞
∠ = −∠ = − ° =⎜ ⎟⎜ ⎟+ +⎝ ⎠⎛ ⎞
∠ ≈ −∠ − = − °⎜ ⎟⎜ ⎟+ +⎝ ⎠
![Page 40: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/40.jpg)
( ) ( )21/ / 2 / 1n ns sω ζ ω⎡ ⎤+ +⎣ ⎦
![Page 41: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/41.jpg)
( ) ( )21/ / 2 / 1n ns sω ζ ω⎡ ⎤+ +⎣ ⎦
![Page 42: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/42.jpg)
예제 6-3
( )2
10000( )2 100
G ss s s
=+ +
( )
( ) ( )( )
2
2
100( )/100 2 /100 1
100/10 0.2 /10 1
s j
G js s s
j j j
ω
ω
ω ω ω
=
=+ +
=+ +
![Page 43: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/43.jpg)
예제 6-3
![Page 44: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/44.jpg)
예제 6-3
10 10120log 20log 5 14
2dB
ζ= =
![Page 45: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/45.jpg)
예제 6-3
![Page 46: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/46.jpg)
안정도 여유
![Page 47: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/47.jpg)
안정도 여유
![Page 48: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/48.jpg)
안정도 여유
![Page 49: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/49.jpg)
안정도 여유
• 이득 여유
• 위상 여유
![Page 50: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/50.jpg)
안정도 여유
• 이득 여유
• 위상 여유
180 180
10 10 180 180
10180 180
0 ( ) ( )
20 log 1 20log ( ) ( )120log
( ) ( )
dBGM dB G j H j
G j H j
G j H j
ω ω
ω ω
ω ω
= −
= −
=
( )180 ( ) ( )c cPM G j H jω ω= °−∠
![Page 51: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/51.jpg)
안정도 여유
• 이득 여유
• 위상 여유
![Page 52: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/52.jpg)
안정도 여유
( )2( 1)( ) ( )0.1 1
sG s H ss s
+=
−
180 180
10 180 180 10
10 180 180
( ) ( ) 0
20log ( ) ( ) 20log 1
20log ( ) ( )
dBGM G j H j dB
G j H j
G j H j
ω ω
ω ω
ω ω
= −
= −
=
![Page 53: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/53.jpg)
주파수 영역에서의 제어기 설계Continuous-Time
![Page 54: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/54.jpg)
제어 시스템의 구조
![Page 55: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/55.jpg)
제어기 설계 과정
• 제어 시스템의 요구 사항 설정
• 센서와 구동기의 선정
• 제어 대상의 모델링
• 제어기의 설계
• 제어 시스템의 시뮬레이션
• 제어 시스템의 실험
![Page 56: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/56.jpg)
PD 제어기의 설계
( )( ) 1 dD s K T s= +
![Page 57: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/57.jpg)
예제 7-1
1( )( 1)
G ss s
=+
0lim
( 1)v s
KK s Ks s→
= =+
1 1ss
v
eK K
= =
( ) 100(1 0.1 )D s s= +
![Page 58: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/58.jpg)
예제 7-1
![Page 59: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/59.jpg)
예제 7-1
![Page 60: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/60.jpg)
진상 제어기의 설계
1( )1
TsD s KTsα+
=+
![Page 61: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/61.jpg)
진상 제어기의 설계
( ) ( )1 11 tan tan1
jT T Tj T
ωφ ω α ωα ω
− −⎛ ⎞+= ∠ = −⎜ ⎟+⎝ ⎠
10 max 10 101 1 1log log log2 T T
ωα
⎛ ⎞= +⎜ ⎟⎝ ⎠
max1
Tω
α=
1 1max
1tan tanφ αα
− −= −
max1tan2
αφα
−= max
1sin1
αφα
−=
+max
max
1 sin1 sin
φαφ
−=
+
![Page 62: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/62.jpg)
진상 제어기의 설계
10 max max 10120log ( ) ( ) 0.5 20logKG j H jω ωα
= − ×
max
1Tω α
=
![Page 63: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/63.jpg)
예제 7-2
( )100.5 20log 1/ 9dBα× =
max
1 1 0.1716.7 0.13
Tω α
= = =
1 0.17 1( ) 1001 0.13(0.17 ) 1
Ts sD s KTs sα+ +
= =+ +
max 50φ = ° 0.13α =
![Page 64: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/64.jpg)
예제 7-2
![Page 65: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/65.jpg)
예제 7-2
![Page 66: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/66.jpg)
예제 7-31( )
( 1)( / 5 1)G s
s s s=
+ +
0lim
( 1)( / 5 1)v s
KK s Ks s s→
= =+ +
1 1ss
v
eK K
= =
max 55φ = ° 0.1α = ( )100.5 20log 1/ 10dBα× =
max
1 1 0.674.73 0.1
Tω α
= = =
1 0.67 1( ) 101 0.1(0.67 ) 1
Ts sD s KTs sα+ +
= =+ +
![Page 67: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/67.jpg)
예제 7-3
![Page 68: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/68.jpg)
예제 7-3
![Page 69: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/69.jpg)
예제 7-3
max 32φ = ° 0.3α =
( )100.5 20log 1/ 5.2dBα× =
max
1 1 0.276.75 0.3
Tω α
= = =
0.67 1 0.27 1( ) 100.1(0.67 ) 1 0.3(0.27 ) 1
s sD ss s+ +
= ⋅+ +
![Page 70: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/70.jpg)
PI 제어기의 설계
1( ) 1i
D s KT s
⎛ ⎞= +⎜ ⎟
⎝ ⎠
![Page 71: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/71.jpg)
예제 7-4
1( )( 1)( / 5 1)
G ss s
=+ +
0.6( ) 10 1D ss
⎛ ⎞= +⎜ ⎟⎝ ⎠
3( ) 10 1D ss
⎛ ⎞= +⎜ ⎟⎝ ⎠
![Page 72: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/72.jpg)
예제 7-4
![Page 73: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/73.jpg)
예제 7-4
![Page 74: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/74.jpg)
지상 제어기의 설계
1( )1
TsD s KTsα+
=+
( ) ( )1 11 tan tan1
jT T Tj T
ωφ ω α ωα ω
− −⎛ ⎞+= ∠ = −⎜ ⎟+⎝ ⎠
![Page 75: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/75.jpg)
지상 제어기의 설계
10 1020 log ( ) ( ) 20lognew newc cKG j H jω ω α=
110
newc
Tω
=
![Page 76: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/76.jpg)
예제 7-5
1( )( 1)( / 5 1)
G ss s
=+ +
1018.8 20log α=
( )18.8/ 2010 8.7α = =
10 10 1.56.75new
c
Tω
= = =
1.5 1( ) 1008.7(1.5 ) 1
sD ss+
=+
![Page 77: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/77.jpg)
예제 7-5
![Page 78: 주파수응답을이용한해석법 Continuous-Timeccrs.hanyang.ac.kr/webpage_limdj/digitalcontrol/lecture6.pdf · 2019-12-02 · Microsoft PowerPoint - lecture6.ppt Author: limdj](https://reader034.vdocuments.net/reader034/viewer/2022042300/5ecad3ba27b32404c43c27ca/html5/thumbnails/78.jpg)
예제 7-5