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Transfer Func,ons and Frequency Response Part 2
ESC201A : Introduc,on to Electronics
rhegde Dept. of Electrical Engineering
IIT Kanpur
Sketching of Transfer function: Bode Magnitude Plot
3 4
10 1 ( )1 1
10 10
Hj j
ωω ω
= ×+ +
ω
( ) ( )H dBω
20
310
-20dB/decade
410
-20dB/decade
-40dB/decade
2 210 10 103 4 20Log ( ( ) ) 20 20 (1 ( ) ) 20 (1 ( ) )
10 10H Log Logω ωω = − + − +
-20dB/decade
2
Sketching of Transfer function Bode Magnitude Plot
3 4 5
10 1 1 ( )1 1 1
10 10 10
Hj j j
ωω ω ω
= × ×+ + +
ω
( ) ( )H dBω
20
310-20dB/decade
410
-20dB/decade
-40dB/decade
2 2 210 10 10 103 4 5 20Log ( ( ) ) 20 20 (1 ( ) ) 20 (1 ( ) ) 20 (1 ( ) )
10 10 10H Log Log Logω ω ωω = − + − + − +
-60dB/decade
510-20dB/decade
3
( ) ( )NH jω ω=
10 10 20Log ( ( ) ) 20 ( )H N Logω ω= ×
1
20N dB/decade
( ) ( )H dBω
ω
4
Bode Magnitude Plot
Determine transfer function?
vS
vO(t)
R
C ( ) ( )( )
O
S
VHV
ωω
ω=
( )1j CRHj CRω
ωω
=+
1 j Cω
3
3
( / ) ( )1 ( / )
dB
dB
jHjω ω
ωω ω
=+ 3dB 3dB
1 1= ; f =RC 2 RC
ωπ
210 10 10
3 3
20Log ( ( ) ) 20 ( ) 20 (1 ( ) )dB dB
H log logω ωω
ω ω= − +
5
ω3dB
20 dB/decade
( ) ( )H dBω
ω
210 10 10
3 3
20Log ( ( ) ) 20 ( ) 20 (1 ( ) )dB dB
H log logω ωω
ω ω= − +
-20 dB/decade
High Pass Filter 6
Bode Magnitude Plot
33dB
1 f 102
HzRCπ
= =
High Pass Filter 7
Bode Plot segments
31 2
1 1 ( ) ( / ) {1 ( / )}( / ) {1 ( / )}
n to r mH K j j
j jω ω ω ω ω
ω ω ω ω= × × × +
+
( ) ( )H dBω
ω
20log(K)
20n dB/decade
oω2ω
-20m dB/decade
-20r dB/decade
1ω
20t dB/decade 3ω
8
Example: 1 1 ( ) 20010 100
H jj j
ω ωω ω
= × × ×+ +
1 1 ( ) 0.21 1
10 100
H jj j
ω ωω ω
= × × ×+ +
( ) ( )H dBω
ω-14
10
-20 dB/decade
100
-20 dB/decade
-20dB/decade
1 20 dB/decade
9
0.1+6 dB
-34
Filter -pass a band of frequency and reject the remaining
f
f f
f
|H(f)| |H(f)|
|H(f)||H(f)|
Low pass High pass
Band pass Band Stop
10
R =1K
C 1µFvS
vO
vS
vO(t)
R
C
11
R =1K
C 1µFvS
vO
3dB Frequency of single capacitor filters
33dB
1= 10 /RC
rad sω =
C 1µFvS
vO
R1 =1K
R2 =1K
3dB1 2
1=R R C
ω
CLinear Circuit 3dB
eq
1 1=R C
ωτ=
12
One can often tell the type of filter by looking at behavior at very low and very high frequencies and keeping in mind that capacitor offers very high impedance at low frequencies and very low impedance at high frequencies.
Bandpass filter
Low f High f
Vo ~0 Vo ~0
13
Bandpass Filter
f2 f1>f2
vS R1
C1
C2
vO(t)R2
2 11 1 2 2
1 1 f ; f2 2RC R Cπ π
≅ ≅14
Example: Band Pass filter
15
Bandstop Filter
f1 f2>f1
Will this work?
16
What does this circuit do?
Low f High f
Vo ~Vin Vo ~Vin 17
Bandstop or Notch Filter
What does this circuit do?
18
R-L Circuits (Filters)
R
vS
vO(t)
L
( ) ( )( )
O
S
VHV
ωω
ω=
3
3
( / ) ( )1 ( / )
dB
dB
jj LHR j L j
ω ωωω
ω ω ω= =
+ +j Lω
3dBRL
ω =
High pass filter
19
R-L Circuits
vS
vO(t)L
R
( ) ( )( )
O
S
VHV
ωω
ω=
3
1 ( )1 ( / )dB
RHR j L j
ωω ω ω
= =+ +
j Lω
3dBRL
ω =
Low pass filter
20
Amplitude Modulated (AM) Radio
Different radio channels are separated by very narrow frequency interval.
For example, one may want to receive a 450KHz signal but reject 460KHz or 440KHz
( ) ( )H dBω
ω
450KHz
460KHz
-60dB
This implies an attenuation of -6000 dB/decade !!
2460log( ) 10450
decades−≅
21
Resonance
Galloping Gertie, the first Tacoma Narrows Bridge in Tacoma Washington, United States. The bridge opened on July 1, 1940 and from the start became notorious for its movement during windy days, earning the nickname "Galloping Gertie". The wind-induced collapse occurred on November 7, 1940, due partially to a physical phenomenon known as mechanical resonance…..wikepedia
22
Nuclear magnetic resonance
23
Resonance
A small disturbance leads to oscillatory behavior 24
T = 1.1µs 25
T = 0.9µs 26
T = 1µs
The amplitude is 10 times larger even though input magnitude is same ! 27
Series Resonant Circuit
vS Ci(t)
LR
Resonance is a condition in which capacitive and inductive reactance cancel each other to give rise to a purely resistive circuit
1eqZ R j L j
Cω
ω= + −
Resonant frequency: 1 10O OO
j L jC LC
ω ωω
− = ⇒ =
12Of LCπ
=eqZ R=
Current and voltage are in phase (power factor is unity) and current is maximum ! 28
vS Ci(t)
LR
2 2
( )1( )
mVIR L
C
ωω
ω
=
+ −
ω1 and ω2 are called half-power frequencies since P ∝ I2
29
12 2
11
( )1 2( )
m mV VIRR L
C
ωω
ω
= =
+ −
22 2
22
( )1 2( )
m mV VIRR L
C
ωω
ω
= =
+ −ω1 and ω2 are called half-power frequencies since P ∝ I2
30
2 2
( )1( )
mVIR L
C
ωω
ω
=
+ −
1 2Oω ωω=2 1
RBL
ω ω= − =
Quality (Q) factor: Sharpness of resonance Peak Stored Energy2
Energy dissipated in one period at resonanceQ π=
2
2
122 12
mO
m O
L I LQRI R T
ωπ
×= × =
×
1 1O
O
QCRLC
ωω
= ⇒ =
31
vS C
LR
vS C
LR0.1K j0.9K
-j1.1K Z=0.1K-j0.2K
j1K
-j1K
0.1K
Z=0.1K
Not very large change in impedance as we approach resonance ! 32
Impedance is in kΩ
Impedance is in kΩ
vS C
LR
vS C
LR0.1K j0.9meg
-j1.1meg
Z=0.1K-j0.2meg
j1meg
-j1meg
0.1K Z=0.1K
very large change in impedance as we approach resonance ! Implying high quality factor 33
Impedance is in MΩ
Impedance is in kΩ
Quality factor Q
vS C
LR j1K
-j1K
0.1K
Z=0.1K
vS C
LR j1meg
-j1meg
0.1K
Z=0.1K
1O OL CQ or QR R
ω ω= =
34
2 1RBL
ω ω= − =OLQR
ω=
O OQBω ω
ω= =
ΔHence Q represents sharpness of resonance
For high Q circuits: 35
R-L-C filters
vS C
vO(t)LR
j Lω
j Cω−vS
vO(t)R
C
Lj Lω
j Cω−
36
vO(t)L
R
C
vSj Lωj Cω−
vS
RvO(t)
L
C
j Lω
j Cω−
37
vO(t)L
R
C
vSj Lωj Cω−
How much Q do we need to pass 450KHz but reject 460KHz by 60dB?
2 2
( )( )( ) 1( )
O
IN
V RHV
R LC
ωω
ωω
ω
= =
+ −
Assuming VIN = 1V and noting that Q = ωOL/R
22 2
2
1( )1 ( 1)
O
O
VQ
ωωω
=
+ −
For ω=ωO, VO = 1 so the signal simply passes through !
3 62 450 10 2.8 10 /O rad sω π= × × × = ×38
For an attenuation of -60dB or 10-3 at ω: Q=23,000
Such a large value of Q is not easy to get !
22 2
2
1( )1 (1 )
O
O
VQ
ωωω
=
+ −
sradsrad
/1089.2104602
/10827.210450263
630
×=××=
×=××=
πω
πω
39
Example: for Q = 104 at 450KHz
3Suppose 10 ; 0.28 ; 125L H R C pF−= ⇒ = Ω ⇒ =oLQRω
=
Suppose 0.1 ; 28 ; 1.25L H R C pF= ⇒ = Ω ⇒ =
Parallel Resonance
C LIM 0 R
C LvS
R
1 1eqY j C j
R Lω
ω= + −
Resonant frequency: 1 10O OO
j C jL LC
ω ωω
− = ⇒ =
12Of LCπ
=eqZ R=
40
LIM 0 R+V- C
2 22
2
( )11 ( )
mI RVR C LL C
ω
ωω
=
+ −
41
2 22
2
( )11 ( )
mI RVR C LL C
ω
ωω
=
+ −
For high Q: 42
What is the resonant frequency ?
43 2o
Ofωπ
=