EC1 - Ch2.8 1 2.8. Ñaùp öùng taàn soá (frequency response) 2.8.1 Haøm truyeàn ñaït mieàn taàn soá. 2.8.2 Ñôn vò decibel (dB) . 2.8.3 Ñaëc tuyeán taàn soá vaø Bode Plots . 2.8.4 Maïch coäng höôûng nối tiếp- song song. 2.8.5 Maïch loïc ñieän thuï ñoäng.
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EC1 - Ch2.8 1
2.8. ap ng tan so (frequency response)
2.8.1 Ham truyen at mien tan so.
2.8.2 n v decibel (dB) .
2.8.3 ac tuyen tan so va Bode Plots .
2.8.4 Mach cong hng ni tip- song song.
2.8.5 Mach loc ien thu ong.
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EC1 - Ch2.8 2
2.8.1 Ham truyen at mien tan so
ap ng tan so la s thay oi hanh vi cua mach theo tan sotn hieu. Khao sat ap ng tan so : da tren ham truyen mien tan so.
CircuitH(j)X() Y()
Y( ) OutputH(j )X( ) Input
= =
Ham truyen trong mien tan so H(j) hay H() nh ngha:
Trong o, in va out co the la dong hay ap. Thong dung nhat la t so: ap ra/ ap vao.
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EC1 - Ch2.8 3
Tm ham truyen at mien tan so:
i. Chuyen mach ve mien tan so:R R
L j LC
1j C
ii. Dung cac phng phap phan tch mach e thiet lap t so.
V du: Tm H(j) cua mach RC ? Chuyen mach ve mien tan so: Theo chia ap:
o
S
V 1V 1 jRCH(j) += =
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EC1 - Ch2.8 4
2.8.2 n v decibel (dB)
Trong ky thuat, o li G cua mach mo ta t so gia cong suat ngo ra va cong suat ngo vao, thng dung vi n v bel:
G = log10(P2/P1) [B]
(Alexander G. Bell (1847-1922), nha phat minh ien thoai)
c so cua no la dB c dung nhieu hn:G = 10log10(P2/P1) [dB]
Bieu dien theo ien ap va dong, ta co: G = 20log10(U2/U1) = 20log10(I2/I1) [dB]
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EC1 - Ch2.8 5
2.8.3 ac tuyen tan so va Bode Plots
Do la bieu thc phc nen ta co the bieu dien:
ac tuyen bien tan : bieu dien M() theo , khong n v.H(j ) M() () =
ac tuyen pha tan : Bieu dien () theo , n v deg. Thong dung hn ca la dung ac tuyen bien o logarithm:
G() = 20log10M() [dB] Ket luan: Ngi ta thng ve G() n v dB va () n v deg theo tan so khac vach LOG, vi cach bieu dien Bode Plots.
(Hendrik W. Bode (1905 1982)
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EC1 - Ch2.8 6
Giy v v thang chia LOG trc :
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EC1 - Ch2.8 7
Step1: Bieu dien H(j) dang chuan
( ) ( ) ( )( ) ( ) ( )
1 z z
1 p p
2j j j0 1z
2j j j
2p
K 1 2 1P(j)H(j )Q(j) j 1 2 1
+ + + = = + + +
T a thc theo j t va mau, ta a ve tha so bac 1 cho nghiem thc va bac 2 cho nghiem phc.
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EC1 - Ch2.8 8
Step2: Xap x bien tan bang oan thang
Plot 20log10K0 : horizontal line. Plot at = 1 : slope of 20dB/decade. Plot at = z1 : slope of 20dB/decade. Plot at = p1 : slope of 20dB/decade.
( ) ( ) ( )( ) ( ) ( )
1 z z
1 p p
2j j j0 1z
2j j j
2p
K 1 2 1H(j )
j 1 2 1
+ + + = + + +
Plot at = z : slope of 40dB/decade. Plot at = p : slope of 40dB/decade.
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EC1 - Ch2.8 9
Tong hp cach xap x:
Factor Amplitude Phase
20log10K0
0o
K0
20N dB/dec
90N o
[ j]N 1
- 20N dB/dec
- 90N o
1(j)N
1
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EC1 - Ch2.8 10
Summary of Bode Plots
Factor Amplitude Phase
Np
1(1 j )+ - 20N dB/dec
p0o
pp/10 10p
- 90N o
Nz(1 j )+
20N dB/dec
z 0o zz/10 10z
90N o
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EC1 - Ch2.8 11
Summary of Bode Plots
Factor Amplitude Phase
z z
j j 2 N [1 2 ( ) ( ) ]+ + 0o
zz/10 10z
180N o40N dB/dec
z
p p
j j 2 N
1[1 2 ( ) ( ) ]+ + 0o
pp/10 10p
- 180N o- 40N dB/dec
p
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EC1 - Ch2.8 12
Examples: The Bode PlotEx1: a) Find the transfer function
H(j) = V0()/Vs() ?b) Sketch the Bode plot ?c) Find v0 if vs = 10cos20t ?
( )( ) o01 j/16.7
( H(j ) ;v =4.17cos(20t-24.3 ) )1 j/5.: 5Ans 6+= +
Ex2: a) Find the transfer function H(j) = V0ut()/Vs() ?
b) Sketch the Bode plot ?c) Find v0ut if vs = 10cos(5000t+60o)V ?
( )( ) o00.011(j)( H(j ) ;v =2.2cos(5000t-77.54 ) )
1 j/100 1 jA /1000ns: = + +
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EC1 - Ch2.8 13
Examples: The Bode Plot
Ex3: a) Find the transfer function H(j) = V0()/Vi() ?
b) Sketch the Bode plot ?
( ) ( )2j j10 10Ans1( H(j ) )
[ 1]: =
+ +
Ex4: a) Find the transfer function H(j) = V0()/Vi() ?
b) Sketch the Bode plot ?
( ) ( )2
2j j10 10
(j )( H(j ) )100[ 1]
Ans: =+ +
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EC1 - Ch2.8 14
2.8.4 Hien tng cong hng nhanh : Nhanh xay ra cong hng ?
&U Z.I I YU = = T : Z hay Y : so thc .
Neu ap va dong cung pha.
U+
-
IZY
Xet mot nhanh (2 cc) cua mach :
Z thc(cong hng noi tiep)
Y thc(cong hng song song)
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EC1 - Ch2.8 15
2.8.5 Mach cong hng noi tiep :
Cho mach RLC noi tiep: ap vao u(t) co bien o co nh Um, tan so thay oi c.
1CZ R j( L )= +
a) Tr khang nhanh :
Tr khang nhanh thay oi theo tan so .
2 21C|Z| ( L )R = +
1CIm{Z} ( L )=
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EC1 - Ch2.8 16
b) Tan so cong hng noi tiep :
La tan so 0 thoa :
0
10 0 CIm{Z( )} ( L ) 0 = =
01LC
= 0 1f 2 LC=
Tai tan so cong hng : |Z| min = R va nhanh thuan tr.
Zmin
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EC1 - Ch2.8 17
c) Bang thong (BW) cua mach cong hng:
R m 2 21C
RU UR ( L )
= +
Dong qua nhanh ap tren R , co module:
R(max) mU U= Tan so cat : tan so ma
1R R(max)2
U U= 1 : tan so cat di. 2 : tan so cat tren.
Bang thong : 2 1BW = 2 1BW (f f ) Hz= hay
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EC1 - Ch2.8 18
Xac nh cac tan so cat :
1 2
mR( & ) m 2 21
C
R UU U2R ( L )
= =+ 2
1R R 12L 2L LC
= + +
RBWL
=
2
2R R 12L 2L LC
= + +
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EC1 - Ch2.8 19
d) He so pham chat :
max
T
WQ 2W
= Wmax : nang lng tch luy maxWT : nang lng tieu tan trong 1 chu ky mach cong hng noi tiep , ngi ta CM c :
21max L C m2W max(W W ) const LI= + = =
2 21 1T m m 02 2W RI T RI .2 / = =
0 0
0
L 1QR RC BW
= = =
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EC1 - Ch2.8 20
Tnh tan so cat theo he so pham chat : 2
1 0 20 0 0
R R 12 L 2 L LC
= + +
2
1 01 1 1
2Q 2Q
= + +
2
1 01 1 1
2Q 2Q
= + +
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EC1 - Ch2.8 21
e) o th vect tai cong hng :
I UR
ImIm
ReReU
UL
UC
mLm Cm 0 m
0
IU U LIC
= = = Do:
Cm 0 m 0Lm
m m m
U LI LU QU U RI R
= = = =
Cong hng noi tiep goi la cong hng ap v tai lan can tan so cong hng , ap tren cac phan t khang rat ln so vi tn hieu ap vao cua mach (Q lan) .
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EC1 - Ch2.8 22
V du1: Cong hng noi tiepTn hieu ra may phat song :
u(t) = 10cos(t) VTm : 0; BW; Q; Ulm vaUCm tai lan can 0?
Giai
0 -3 -5
1 2000 (rad/s)25.10 .10
= =3
0L 2000.25.10Q 25R 2 = = =
Lm Cm mU U Q.U 250 (V)= = =
3
R 2BW 80 (rad/s)L 25.10
= = =
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EC1 - Ch2.8 23
2.8.6 Mach cong hng song song : (oc)
1LY G j( C )= +
a) Dan nap nhanh :
Dan nap nhanh thay oi theo tan so .
2 21L|Y| G ( C )= +
1LIm{Y} ( C )=
Cho mach RLC song song : dong vao J(t) co bien o conh Jm, tan so thay oi c.
Ymin
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EC1 - Ch2.8 24
b) Tan so cong hng song song:
La tan so 0 thoa :
0
10 0 LIm{Y( )} ( C ) 0 = =
01LC
= 0 1f 2 LC=
Tai tan so cong hng : |Y| min = G va nhanh thuan tr.
Ymin
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EC1 - Ch2.8 25
c) Bang thong (BW) cua mach cong hng:
mLC 2 21
L
JUG ( C )
= +
Ap tren nhanh ap tren khung LC , co module: m
LC(max)JUG
= Tan so cat : tan so ma
1LC LC(max)2
U U= 1 : tan so cat di. 2 : tan so cat tren.
Bang thong : 2 1BW = 2 1BW (f f ) Hz= hay
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EC1 - Ch2.8 26
Xac nh cac tan so cat :
1 2
m mLC( & ) 2 21
L
J JU2GG ( C )
= =+ 2
1G G 1( )2C LC2C
= + +
GBWC
=
21
G G 1( )2C LC2C = + +
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EC1 - Ch2.8 27
d) He so pham chat :
mach cong hng song song , ngi ta CM c : 21
max L C LCm2W max(W W ) const CU= + = =2 21 1
T LCm LCm 02 2W GU T GU .2 / = =0 0
0
C 1QG LG BW
= = =
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EC1 - Ch2.8 28
e) o th vect tai cong hng :
LCmLm Cm 0 LCm
0
UI I CUL
= = = Do:
Cm 0 LCm 0Lm
m m LCm
I CU CI QJ J GU G
= = = =
Cong hng song song goi la cong hng dong v tai lan can tan so cong hng , bien o dong qua cac phan t khang rat ln so vi bien o tn hieu dong a vao mach (Q lan) .
ULC IR
ImIm
ReReJ
ICIL
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EC1 - Ch2.8 29
V du: Mach cong hng
VD1: Find 0 , Q and BW ?
(Ans:a) 1.5811 rad/s; 0.1976;
8 rad/s b) 5 krad/s; 20; 250 rad/s )
VD2: Find 0 ?
(a)
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EC1 - Ch2.8 30
V du: Mach cong hng
VD4: Tm tan so cong hng ?
(Ans: 12,9 Mrad/s )
VD5: Tm L e mach cong hng tan so 100 rad/s ?
(Ans: 5 mH )
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EC1 - Ch2.8 31
V du: Mach cong hng
VD6: Determine the value of so that i(t) is in phase with e(t) ? Find i(t) if E = 60 V ?
(Ans: 100 rad/s ; 5cos(100t) A )
VD7: Determine R1 and L1 in terms of R2, L2 and ? Find R1and L1 when R2 = 5 k , L2 = 1.25 H at 4 krad/s ?
(Ans: 2.5 k ; 625 mH )
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EC1 - Ch2.8 32
2.8.7 Mach loc ien thu ong
a) nh ngha mch lc in: H(j)X() Y()
b) Phn loi mch lc in:
Thu ong: ch gom cac phan t R, L, C.
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EC1 - Ch2.8 33
V d: Mch lc in
Tm hm truyn H(j) = V0/Vi , v c tuyn bin tn v suy ra loi mch lc ? Tnh tn s ct c ti |H(j)| = 0,707 ? Tnh H(j3c) ? Xc nh v0(t) khi vi(t) = 50cos(3ct 10o ) V ?
p ng tn s: H(j) = 6000/(6000 + j). Tn s ct: c = 6000. Ta c H(j3c) = 0,3162-71,57o. vo(t) = 15,81cos(18000t 81,57o).
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