Post on 16-Dec-2015
ELE1110C Tutorial 3ELE1110C Tutorial 3
27/09/200627/09/2006Cathy, KAI CaihongCathy, KAI Caihong
OutlineOutline
• Lecture Reviews
• Examples
How to represent a sine wave How to represent a sine wave (sinusoid)?(sinusoid)?
• Complex number notation (L2.137)– a+jb
Represent a vector with a magnitude and phase
– Magnitude =
– Phase =
– Unit vector =
a
jb22 ba
ab
tan 1
cos sinj 1
,1sinjcose j
• Rotation: multiplied by j (rotate anti-clockwise)
22 2 , cos sin cos sin , . .
j tj tLet t then t j t t j t i e e e
• => Period = 2π/w, Frequency = 1/Period = w/2π• => w : Angular Frequency (i.e. w = 1000 x 2π
for a 1000Hz sine wave)
Impedance CalculationImpedance Calculation• Capacitor:
( . . )
/ *
/ / 1/
j t
j t
c
Let V Ae i e V A
I CdV dt C j Ae j C V
Z V I V j CV j C
Impedance CalculationImpedance Calculation• Impedance of capacitor:
• Magnitude of impedance :
• Low frequency => ZC is large => open circuit
• High frequency => ZC is small => short circuit
1 C
VZ
I jwC
1CZ Cw
Circuit AnalysisCircuit Analysis
• If V1 is an varying voltage, what is V2?
R
C V2V1
Capacitor – Capacitor – Transfer FunctionTransfer Function• Definition
– Impedance of capacitor :
• By KVL
• So that
• Hence
2C
VZ
I
1 2 CV IR V IR IZ
1
1
C
I VR Z
2 1 1 1
1/ 1
1/ 1
C
CC
Z j CV IZ V V V
R Z R j C j CR
Capacitor – Capacitor – Transfer FunctionTransfer Function• Transfer function T(w)= Output / Input
=
• Note that the output response depends on the frequency of the input sinusoid
• High frequency ( ),– High frequency signals are cut
• Low frequency ( ), – Low frequency signals are passed – The circuit is a low pass filter
2
1
( ) 1
( ) 1
V w
V w jwCR
w 2 ( ) 0V w
0w 2 ( ) 1V w
• Usually we focus more on power ration in filter design:
2 2
1 1
1 ( )1out
in
P
P CRj CR
• Therefore, 3dB point (half-power, or cut-off point):
120.5 / 1 ( )out inP P CR
• At half power point,
=> CCR = 1
=> C = 1/RC
• since C = 2f , therefore fC = 1/2RC
2
1 1 1
2 21 ( )OUT
IN
V
V wCR
33dB point of a LPFdB point of a LPF
Magnitude and phase response of a low pass filter Magnitude and phase response of a low pass filter (LPF)(LPF)
– Magnitude response =
– Phase response =
2IN
OUT
)wCR(1
1VV
)wCR(tanVV 1
IN
OUT
• dB is the logarithm of a ratio between powers
– 3dB => POUT is 2 times PIN
– 10dB => POUT is 10 times PIN
– 20dB => POUT is 100 times PIN
– -3dB => POUT is half of PIN , and so on
IN
OUT
P
Plog10dB
Examples from Problem Sheet 3
• Example 1:• Find the transfer function of the following circuit
VOUTC
R1
VIN
R2
• Let Y = (1/ jwC)
• As no current will flow into R2, V1 = Vout, I2 = 0• I0 = I1 = Vin/(R1 + Y)• Vout = V1 = I0 * Y =YVin/(R1+Y)• Then you can find Vout/Vin
Example 2:
• Find the transfer function of the circuit below.
• Let Y1 = (1/ jwC1), Y2 = (1/ jwC2)
• V1 = Vin*(Y1//(Y2+R2)) / ( (Y1//(Y2+R2)) + R1)
• Vout = V1*(R2/(R2+Y2))
• Therefore, Vout = [Vin*(Y1//(Y2+R2)) / ( (Y1//(Y2+R2)) + R1) ] *(R2/(R2+Y2))
• (Vout/Vin )= (Y1//(Y2+R2)) *(R2/(R2+Y2)) / ( (Y1//(Y2+R2)) + R1) *(R2/(R2+Y2))
Example 3:
• The following diagram shows an elevator call button - an example of proximity switch. The button has two sides separated by insulator. One side is a plate conductor and the other side is a ring conductor. The button behaves as a capacitor (C1). When a finger is placed somewhere near the middle of the ring, the finger acts like an earth as the body can absorb a fair amount of charges quickly. The capacitance between the ring and the finger is C2 and the capacitance between the plate and the finger is C3. Also shown are the equivalent circuits for the two states.
• The circuit for the elevator call button is as follows.
• Suppose C1=30pF, and C2=C3=10pF, find V with i) no finger present, and ii) finger present.
• As Q = CV, we can deduce that the ratio of voltage across 2 capacitors in series is:
1)
, A BA A B B
B A
V CC V C V
V C
301, ,
30
2
AA B s
B
sA B
VV V V
V
VV V
• 2) As no current will flow into the earth point (finger), the earthing has no effect to this circuit.
11 1
30, 30 30 5 35,10 10
35, ,
30
6 7,
13 13
B A
BA B S
A
A S B s
C C
VV V V
V
V V V V