Undergraduate
工程與科技

波德圖

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ページ1:

波德圖的定義:
「考慮下列系統的轉移函數(不失一般性) G(5)=
K(It Ts)
Wn
Wn
典型的
TE'S MAG & PHASE:
(S²係數工)
5+23 was twn
ef: 藍P133
Hag:200g(K) 12000g (1173) + 2000g(5)+20608 (11
25
wn
這很大條:注意到是在jw domain 不是 s domain
:
| G (jw) 18 = zales (6) 420 log ( (+3) - dolog (JW) -> log (1+ y
2
wn
wn²
<G(JW) = 2k+ < (+WT) Likw - 2 (1+
Wn
2330 - )
夠共軛極點/共軛薄點 (1452500元
大小與相位:
考慮共軛整桌
W
2
-20201
wn
-2008.
(4)+45=()
Wn
Wn
Complex poles & zeros is complex !! xD
X
eg: ples:
w<<Wh
: odB
w >>wn
al/Decade
很大的
but
wxwn 大小值隨阻尼号而有變化
阻尼↓,波德圖峰值
(理論上) 5=0.767峰值才消失
3.20.5 波德圖的峰值不明顯

ページ2:

波德圖
藍 9243
李16-1
jow
(5) = (5)| ==|9 (w) | < (u) = (w)
問:ex()這東西是是啥???哪来的??
(( the frequency response design method ))
frequency response (6.1)
a system
Y(S)
V(s) =
= G(s)
unit
step input
input (t) = A s (wot) 1 (t)
-u (s) =
W
Aw.
recall
Laplace
+
AW.
Y(5) = (5) 5²+w
=
s" two-
W
smut = swe
Coswt=
s'tw²
5584250
note: Yow 1:
Sin (1+0)=
5+w²
Awe
56050-WSMO
4(s).
G(S)->Y(s)
Cos(wt+6)=
e
=
AW
SA
5²+wo²
若G(S)的極點清楚(distinct)
Y(S) = 0 + sup
+...+
do
t
+ **
5-P2
5-Pn
S-.jwo
S+julo
Pzt
Pat
Juot
→y(t)=
*
-Jwot
負無窮大變零
+ne+doe
yss(t)
if Re (Pi) <o, Vi, then Yss (t) = αoedoe
+do e
* Juot
+
*
can be determined by
An
lim (5-3) G(S) (5-jo) (sejwo)
5570
M=1G (5)| = |G (-5), 6 = 2G(Sub)
A wo
23Wo
= - SA Me³Ø

ページ3:

100
李P6-7:
(RECALL)
* P6-7: for example:
mag. plot for
波德圖漸近線
first order term (jwc+1) dero
5202+1, 8=10
0
40dB
70dB
6dB
叫
1.0
100
1060
Pf (kind of)
01000/010/
45410+1
magnitude +1
weed, jwett~/dB
w2 331, jwc+1 & gue as we
☆轉折點觀念:在轉折拉着
趨近1,在轉折點後易
近線性
for 20 dB/ decade
Ĵwz+1 = jw10+1
for w>> 1
fake
10 and 100 for
example
JW10+1 2jw10
JW10/= w20
(101) = 40
solog(100x 10)=60
asymptotes:
W
wet
magnitude: 1
20dB / decade. !!!
Zods/ decade
wet line slope for
特别的

ページ4:

9. a certain
my system is represented by the asymptotic
bode diagram shown:
G
10
漸進的
w
10100
1000 10000
to a unit
find and Sketch the response of this system
input (assuming zero initial Conditions)
sol:
assume
all ple (HP
have a pole @ 10
(+1)
:10 第一個:他y軸不用p也是
7(5)=
☑(*)
夠奇葩了
10x10x
x(备+1)
stloo
5+10
(番+1) 盒)x10
* input unit step chimdien
Step 1
olts
X
Stiep
5($10)
反拉氏轉換不能直接
來,要部份分式展開

ページ5:

D
-9.
Stis
Conti (P)
94100
5(5+10)
=
S+100
5+ Sto
(Stra) S=0
100
=
10 = 10
Stro
be
S(Sho)
-(541) 5-10
-9
-10t
inverse laplane = lou(t) + (9) e ult)
起在10沒錯
扣掉也沒錯
X
但是e
起在1
to
f(t) = 10 -
9.1
=1
10
I start @ y=1
終值 exp項為零
全
y=10

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