clock degradation and ac steady state analysis · state analysis ch. 8 ac steady state analysis of...
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Clock Degradation and AC Steady State Analysis
Ch. 8
AC steady state analysis of RLC circuitsCkts are driven by sinusoidal sources (sin(ωt), cos(ωt))In steady-state, the output is also sinusoidalWe will introduce techniques that allow to solve for steady-state behavior without using DE’sThese techniques resemble ordinary DC analysis, but use complex arithmetic
Also introduce RMS value of signals (§ 9.4)
§ 9.4
Application: clock-signal degradation by interconnect network
Use superposition and Fourier series (Ch. 7)
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Sinusoidal Signals
§ 8.2.1
( )ϕω += tVtv cos)(
Sinusoidal signals are fully specified with
Phase angleϕ
Radian frequency (T = Periodicity)ω=2π/TAmplitudeV
cos –sin
2π
Many convenient relations. Example:
( ) ( )tt ωω sinsin −=−
( ) ( )tt ωω coscos =−
( )tt ωπω sin2
cos −=⎟⎠⎞
⎜⎝⎛ +
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Phase Relations
( ) ( )Rtitv =
R
+ -v(t)
i(t)
Are voltage and current in phase ?
( ) ( )ϕω += tIti cos ( ) ( )ϕω += tRItv cos YES for a resistor
L
v(t)
i(t)
+ -( ) ( )
dttdiLtv =
( ) ( )ϕω += tIti cos ( ) ( )ϕωω +−= tLItv sin ⎟⎠⎞
⎜⎝⎛ ++=
2cos πϕωω tLI
NO for an inductor
( )tt ωπω sin2
cos −=⎟⎠⎞
⎜⎝⎛ +( ) ( )tt
dtd ωωω sincos −= ( ) ⎟
⎠⎞
⎜⎝⎛ +=
2coscos πωωω tt
dtd
Expl 8.1
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C
+ -v(t)
i(t)( ) ( )
dttdvCti =
( ) ( )ϕω += tVtv cos ( ) ⎟⎠⎞
⎜⎝⎛ ++=
2cos πϕωω tCVti
NO for a capacitor
Capacitor Phase RelationsAre voltage and current in phase ?
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Phase Relations Summary
I = ωCV cos (ωt + ϕ + π/2)V(cos (ωt + ϕ)C
V = ωLI cos (ωt + ϕ + π/2)I(cos (ωt + ϕ)L
V = RI cos (ωt + ϕ)I(cos (ωt + ϕ)R
Response (result quantity)Forcing function (applied quantity)
Resistor current and voltage are in phase
Inductor voltage leads (voorlopen) induction current by 90°
Capacitor current leads capacitor voltage by 90°
a leads b b lags a (achterlopen)
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Combination of Sinusoidal Sources
( )3020 −tcos ω ( )6040 +tcos ω ( ) ( )43.33cos72.44 +ω= ttv
=+ +– – +–
Result of adding two sinusoidal signals with same frequency is again a sinusoid with same frequency but with different phase
0 5 10 15-50
0
50v1v2v1+v2
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Proof using Trigonometry (2)
( ) ( ) ( )60cos4030cos20 +ω+−ω= tttv
30sinsin2030coscos20 tt ω+ω= 60sinsin4060coscos40 tt ω−ω+
( ) ( ) tt ω−+ω+= sin60sin4030sin20cos60cos4030cos20tt ω−ω= sin64.24cos32.37
20 cos (ωt – 30) + 40 cos (ωt + 60) = 44.72 cos (ωt + 33.43)
Proof:
( ) vuvuvu sinsincoscoscos +=−
( ) uu sinsin −=−
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Proof using Trigonometry (3)
24.64
37.32
44.72
43.3372.4432.37acos =⎟
⎠⎞
⎜⎝⎛
43.3372.4464.24asin =⎟
⎠⎞
⎜⎝⎛ ( )43.33sin
72.4464.24
=⇔
( ) tttv ωω sin64.24cos32.37 −=
( ) ⎟⎠⎞
⎜⎝⎛ −×= tttv ωω sin
72.4464.24cos
72.4432.3772.44
( )tt ω−ω= sin43.33sincos43.33cos72.44( ) vuvuvu sinsincoscoscos +=−
( ) ( )43.33cos72.44 += ttv ω
( )43.33cos72.4432.37
=⇔
( ) uu sinsin −=−
33.43
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Proof using Trigonometry (4)
( ) ( ) ( )60cos4030cos20 +ω+−ω= tttv
30sinsin2030coscos20 tt ω+ω= 60sinsin4060coscos40 tt ω−ω+
( ) ( ) tt ω−+ω+= sin60sin4030sin20cos60cos4030cos20tt ω−ω= sin64.24cos32.37
( )43.33cos72.44 +ω= t
20 cos (ωt – 30) + 40 cos (ωt + 60) = 44.72 cos (ωt + 33.43)
Proof:
0 5 10 15-50
0
50v1v2v1+v2
Q.E.D.
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Review
Result of adding two sinusoidal signals with same frequency is again a sinusoid with same frequency but with different phase
Procedure of adding sinusoidal signal using trigonometrypossible, but requires some ingenuity
Will not elegantly solve our general problem
First, we will review polar form of complex numbers
Then we are ready for phasor concept
20 cos (ωt – 30) + 40 cos (ωt + 60) = 44.72 cos (ωt + 33.43)
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Polar Form and Angle Notation
jb
Im
Reaθ
r
A = a + jb θ = angle, or argument= arg A
r = amplitude, or magnitude= |A|
22 bar +=
abatan=θ
a = r cosθ
b = r sinθ
A = a + jb = r cosθ + jr sinθ = r (cosθ + j sinθ )= r ejθ (Eulers Identity)
A = a + jb = r ejθ ≡ r∠θ
Polar form Angle notation
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Only Magnitude and Phase20 cos (ωt – 30) + 40 cos (ωt + 60) = 44.72 cos (ωt + 33.43)
Addition of sinusoidal signals with same frequency is again a sinusoid with same frequency but with different phase
same frequency but different phase and amplitude
Conclusion: for analysis of time-invariant linear systems driven by sinusoidal signals of a given frequency ω, we only need to explicitly work with amplitude and phase
In general: a linear time invariant network can’t change the frequency of a signal, only the phase and amplitude
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Phasors
amplitude and phase differenceare the principal concernsconvenient representation:vectors rotating atconstant angular velocity ω
lengths = amplitudesangles = phase differences
conventions:counterclockwise rotationprojection on x-axis
phasor diagram issnapshot of the rotatingconfiguration
From wisselstroomtheorie (Otten)
( )tjIe ω
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From wisselstroomtheorie (Otten)
Also see Figure 8.3
( )tjIe ω
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PhasorsThus: we only need to focus on magnitude and phase
These can be represented as one complex numberIm
Re
magnitude V
phase ϕ
( ) ( )ϕω += tVtv cos
( )ϕω= jtj VeeRe
( )ϕ∠= ω Ve tjRe
( ) ϕϕω ∠⇒+ VtcosV Phasor Transform
( ) ( )ϕωϕ ωϕ +=⇒∠ tVeVeV tjj cosRe Inverse Phasor transform
Because frequency does not change, we can eliminate it from the notation, using it only implicitly and put it back later after all calculations ( ) ( ) ϕϕϕω ω ∠⇒∠=+ VVetV tjRecos
Rotating vector
( )( )ϕω += tjVeRe
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Phasor vs Polar Form and Complex Arithmetic
A phasor V∠ϕ is like a polar form number A∠θ, but is implicitly associated with a frequency ω
Otherwise, it is a complex number and obeys all rules of complex arithmetic. Examples:
( )ϕθϕθ +∠⋅=∠⋅∠ baba
( )ϕθϕθ −∠=∠∠ b/ab/a
=∠+∠ ϕθ ba( )ϕθϕθ sinsincoscos bajba +++=
( ) ( )22 sinsincoscos ϕθϕθ baba +++= ⎟⎠
⎞⎜⎝
⎛++
∠ϕθϕθ
coscossinsinatan
bba
ϕϕθθ sincossincos jbbjaa +++
Addition requires rectangular coordinatesr∠θ = a + jb22 bar +=
abatan=θ
22 bar +=
abatan=θ
a = r cosθ
b = r sinθ
a = r cosθ
b = r sinθ
1.
2.
3.
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Example1 : Proof using Phasors
( ) ( ) ( )60cos4030cos20 +ω+−ω= tttv
20 cos (ωt – 30) + 40 cos (ωt + 60) = 44.72 cos (ωt + 33.43)
Proof using phasors (see example 8.3):
to frequency domain
60403020 ∠+−∠=V
( ) ( ) 60sin4060cos4030sin2030cos20 jj ++−+−=
jj 64.34201032.17 ++−=
43.3372.44 ∠=
( ) ( )tjj eetv ω43.3372.44Re=
back to time domain
( )43.33cos72.44 += tω
j.. 64243237 +=
3.(Use from previous slide)
...=V extra bold font for phasors...=V extra bold font for phasors
Q.E.D.
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Example 2: Phase Relations Using Phasors
Multiplication by j is rotation in complex plane by 2π
(because ( ) ( ) ( )2121 argargarg zzzz +=× )
V Iπ/2 Current lags voltage
Voltage leads current
⇒ Voltage leads current by π/2
Three ‘phasor’ ways to illustrate I-V phase relation (of inductor)
tjj eIedtdL
dtdL ωϕ== IV ILjω=
( ) ( )ϕω += tIti cos
ϕωϕ ∠== IeIe tjjI
Inductance:
1.
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ϕπω jj IeLe ⋅= 2
Inverse Phasor Transform (back to time domain)
( )⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛=
⎥⎦⎤
⎢⎣⎡ +⎟
⎠⎞
⎜⎝⎛ + tj
LIetvωπϕ
ω 2Re
⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛ +++⎟
⎠⎞
⎜⎝⎛ ++=
2sin
2cosRe πϕωπϕωω tjtLI
= ωLI cos (ωt + ϕ + π/2)
IV Ljω=
⎟⎠⎞
⎜⎝⎛ +
⋅= 2πϕ
ωj
IeL ⎟⎠⎞
⎜⎝⎛ +∠=
2πϕωLI
2.
3.
⇒ Voltage leads current by π/2
Example 2: Phase Relations Using Phasors
⇒ Voltage leads current by π/2
ϕϕ jIeI =∠=I ⇒
ϕωθ jIeLjV ⋅=∠
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Impedance
Ohms law defines resistance( ) ( )Rtitv =
Capacitor relates current and voltage through differential equation
( ) ( )dt
tdvCti =
In the frequency domain, we will find
IV Z=Z is called the impedance of the capacitor
Impedance is defined as the ratio of the voltage and current phasors
(Complex) impedances and phasors allow frequency domain solution of RLC circuits as if they are R circuits, by using complex arithmetic. Avoid D.E.’s
§8.3
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Impedance of a Capacitor
tjj eVedtdC ωϕ= tjj eVeCj ωϕω= VCjω=
The (complex) impedance of a capacitor is given byCj
CjZ
ω−
=ω
=1
VIdtdC=
IVCjω
=1⇒
CjZ
ω=
1⇒
How can we compute Z?tjj eVeV ωϕϕ =∠=V
C
I+
-V
ϕ∠= VV
IV Z=
frequency domain
θ∠= II
V
C
i(t)+
-v(t)
( ) ( )ϕω += tVtv cos
( ) ( )dt
tdvCti =
time domain
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Impedance of an Inductor
I+
-V
ϕ∠= II
i(t)+
-v(t)
( ) ( )ϕω += tIti cos
tjj eIedtdL
dtdL ωϕ== IV
ILjω=
The impedance of an inductor is given by Z = jωL
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Impedance, Resistance, Reactance
C1j
Cj1Z
LjZ RZ
ω−=
ω=
ω==
jXRZ +=R is the resistance
X>0 : X is the inductive reactance
X<0 : X is the capacitive reactance
impedance is a complex number equal to the ratio of the phasor voltage and the phasor currentmeasured at the same port (terminal pair)
X is the reactance
From wisselstroomtheorie (Otten)
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Admittance, Conductance, Susceptanceadmittance is reciprocal impedance: a complex number equal to the ratio of the phasor current and the phasor voltage measured at the same port (terminal pair)
L1j
Lj1Y
CjY R1Y
ω−=
ω=
ω=
=
jBGY +=G is the conductance
X is the susceptanceX>0 : X is the capacitive susceptance
X<0 : X is the inductive susceptance
From wisselstroomtheorie (Otten)
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Circuit Simplification
Now we know impedance of R, C, L
Can we also determine impedance of series and parallel connections?
YES: circuit simplification works!
KVL and KCL are also true in frequency domain
V1 V2+ +- -I
+ -
VZ1 Z2
V+ -
Zeq
I
V1 = I Z1 V2 = I Z2 V = I Zeq Zeq = Z1 + Z2
§8.3.2-8.3.3
(Complex) impedances and phasors allow frequency domain solution of RLC circuits as if they are R circuits, by using complex arithmetic. Avoid D.E.’s
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Example: Simplification of Series RC
+ -
Ri(t)C
v(t)
R
+ -
V
I Z
+ -V
I
IV ⎟⎠
⎞⎜⎝
⎛+=
CjR
ω1
RCCR
ω−
∠ω
+=1atan1
222
Cj
ω−
R
Cjω1
I⎟⎠⎞
⎜⎝⎛
ω−
+=CjR
z
IZ=
CjRZ
ω−
+= ⎟⎠⎞
⎜⎝⎛ −
=RCω
θ 1atan
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Impedance is not a Phasor
RCCR
ω−
∠ω
+=1atan1
222
Z has a magnitude and angle (argument)Can also be written in polar form and with angle notationBut it is NOT A PHASOR --- just a complex quantityBecause there is not an implicit frequency (angle is not phase)Even if Z is frequency dependent
CjRZ
ω−
+= θ∠= Z
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Simplification of Parallel Circuits
R L
+
-V
I
R jωL
+
-V
I
Z1
+
-V
I
Z2
1ZVI1 =
22 Z
VI =
eqZV
=
∑=i ieq ZZ
11∑=i
ieq YY
Zeq
+
-V
I
RVI1 =
Ljω=
VI2
LjR ω+=
VVIeqZV
=
LjRZY
eqeq ω
+==111
LjRLRjZeq ω+
ω=
21 ZZVVIII 21 +=+=
KCL
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Determine vc(t) when vs(t) = V cosωt ⇔ V∠0
Plan: determine I and multiply by impedance of C
τ
⎟⎠⎞
⎜⎝⎛ωτ−
∠ω
+ω
π−∠=
1atan120
222
CRC
V
+
-vs C
R
vc
+
-vs C
R
vc
+
-V C
R
VcI
+
-V C
R
VcI
+
-V C
R
VcI
Frequency Domain Circuit Analysis Example
211 π
−∠ω⋅=
ω⋅=
CCjIIVc
⎟⎠⎞
⎜⎝⎛ω−
∠ω
+
∠==
RCCR
V1atan1
0
222Z
VIZ
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Back to Time Domain
⎟⎠⎞
⎜⎝⎛ωτ−
−π
−∠+τω
=1atan
2122V
( ) ( )θωτω
++
= tVtvc cos122 2
1atan π−⎟
⎠⎞
⎜⎝⎛ωτ
=θ
Amplitude depends on frequency
(Phase-shift too)
More on this later...
⎟⎠⎞
⎜⎝⎛ωτ−
∠ω
+ω
π−∠=
1atan12
222
CRC
VCV ( ) ( )αα −−= atanatan
atan has odd symmetry
21atan
122π
−⎟⎠⎞
⎜⎝⎛ωτ
∠+τω
=V
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Intermezzo
+
-vs C
R
vc
+
-vs C
R
vc
+
-V C
R
VcI
+
-V C
R
VcI
+
-V C
R
VcI
We have
21atan
122π
−⎟⎠⎞
⎜⎝⎛ωτ
∠+τω
=VVc
Book (example 8.6) has
ωττω
atan122
−∠+
=VVc
Hence -atanωτ should equal2
1atan π−⎟
⎠⎞
⎜⎝⎛ωτ
Proof on next slide
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Proof
Show that 2
1atanatan π−=−
xx
x1tan =ϕ ϕ=
x1atan
x−=⎟⎠⎞
⎜⎝⎛ π
−ϕ2
tan
( )2
atanatan πϕ −=−=− xx
Q.E.D.
21atan π−=
x
ϕ
x
1
1
-x
π/2
( )xx −=− atanatan
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V
R VC ?I
Cjω1
V
R VC ?I
Cjω1
AC steady state
VC?
V
I
Resistive Circuits versus Phasor CircuitsDC
( )21 RRVI +=
21
221 RR
RVRIV+
⋅=⋅=
⎟⎠
⎞⎜⎝
⎛+=
CjR
ω1VI
RCjRCj Cj
Cjω+
=+
⋅=ω
⋅=ω
ω
11
1
1VVIVc
You should really see clearly that procedure from previous slides (e.g. determine I and multiply by impedance) is just the same as shown here --- just use complex arithmeticAnd you should be able to apply such techniques on the exam(with numbers it might be easier)
V
R1 V1 ?
R2
I
V
R1 V1 ?
R2R2
I
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Example of Circuit Simplification (1)
+
-40∠0
-j3
9
j3
-j19
10
1 0.2 j0.6 V0
Determine V0 phasor, using source transformation techniques
Frequency not explicitly specified, but is implicitly given in impedance of elements
Plan:
1. Eliminate left-most loop by replacing it by Thévenin equivalent
2. Compute current through circuit
3. Multiply current by impedance of right-most branch
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+
-40∠0
-j3
9
j3
-j19
10
1 0.2 j0.6 V0
3140
jI
+=
Determine V0 phasor, using source transformation techniques
22))((
))(())((
dcdjcbja
djcdjcdjcbja
djcbja
+
−+=
−+−+
=++
Example of Circuit Simplification (2)
124)31(1040 jj −=−=
4 - -j3
9j3
-j19
10
1
0.2 j0.6
j12
V0
I
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4 - -j3
9j3
-j19
10
1
0.2 j0.6
j12
V0
10)39)(31( jjZ −+
=
4 - 1.8+j2.4-j19
10
0.2 j0.6
j12
V0
Z
241892739)39)(31( 2 jjjjjj +=−+−=−+
Example of Circuit Simplification (3)
4.28.1 j+=
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4 - -j3
9j3
-j19
10
1
0.2 j0.6
j12
V0
10)39)(31( jjZ −+
=
4 --j19
10
0.2 j0.6
j12
V0
Z
241892739)39)(31( 2 jjjjjj +=−+−=−+
Example of Circuit Simplification (3)
4.28.1 j+=
1.8+j2.4
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4 - 1.8+j2.4-j19
10
0.2 j0.6
j12
V0
08.156.125
2739)43(4)13(12
16121236 jj
jj
jjI +=
+=
−−
=−−
=
1236)4.28.1)(124( jjjV −=+−=
84.1812.36)1910)(08.156.1(0 jjjV −=−+=
36-j12
1.8+j2.4
-j19
10
0.2 j0.6
+
-
V0
IV
74.4084.1812.36 22 =+ °−=− 55.27
12.3684.18atan
°−∠= 55.2774.40
Example of Circuit Simplification (4)
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Note
+
-40∠0
-j3
9
j31
36-j12
1.8+j2.4
+
-
We had replaced circuit on the left by equivalent on the right
This is only valid for a specific frequency
For that frequency, we cannot distinguish circuits by looking (measuring) at its ports
Other frequencies would require/produce other equivalent circuits!
But calculation procedure remains the same!
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Resonance°−∠= 55.2774.400V
+
-40∠0
-j3
9
j3
-j19
10
1 0.2 j0.6 V0
Note: |V0| > 40Seems impossible: output voltage exceeds source voltageBut it is correct, this phenomenon is called resonance(R)LC circuits driven by sinusoidal signals can exhibit resonance: Output voltage can peak because of interaction between energy stored and released in L and C elements when damping factor is low.See § 8.6 (and Figure 8.23 in particular)
Resonance! (§ 8.6)
RLC-exercises can have answers with voltages exceeding those of any (or even the sum) of the sources!
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Complete response = transient + steady-statetransient (natural response) always dies outsteady-state (forced response) is what remains
AC Steady-state solution is our current goalAlways has same frequency as source signalSinusoidal sources => sinusoidal responses with same frequency ( ) ( )xxx tVtv ϕω += cos ( ) ( )yyy tIti ϕω += cosω is known from the start, need only compute magnitudes and phases (e.g. Vx and ϕx) Phasors
Analyze circuits with one or more sinusoidal sources
( ) ( )ϕω += tVtv ms cos [V]Vm = amplitude, or magnitude [V]ω = 2πf = frequency [rad/s] ϕ = phase angle [degree]
Frequency Domain Analysis Summary
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Phasors are complex numbers representing magnitude and phase, implicitly associated with (source) frequency
Phasors allow frequency domain representation and analysis of circuit
R R C Cj
Cj ω−
=ω1
L Ljω
(Time domain) differential equations become (frequency domain) algebraic equationsSolve RLC circuits like resistive circuits (but use complex arithmetic)
(use either technique: node voltage method, mesh current, voltage and current division source transformation, series/parallel simplification , Thévenin/Norton equivalents, superposition, and so on)
Return to time domain by using inverse phasor transform
( ) ( )xxtj Qetq ϕ∠= ωRe (q(t) is unknown quantity, current
or voltage)
Frequency Domain Analysis Summary
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Effective ‘value’ of Time-Varying Signals
Peak – to – peak value 2V?No: does not distinguish between waveforms (same for all those above)Average value? No: is zero for symmetric waveforms like these.
§9.4
What is a good measure (maat) for voltage / strength / size / effectiveness of time varying signal?
-V
V
t-V
V
t
-V
V
t
V
-Vt
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Root–Mean–Square Value
Effective voltage relative to power delivered to a resistive load
DC: V = I R RIR
VVIP 22===
Time - Varying
( ) ( )∫=T
dttitvT
P0
1 Average Power
( )∫=T
dttvTR 0
211
2EFFV
RVP EFF
2= ( )∫=≡
TRMSEFF dttv
TVV
0
21
The Root-Mean-Square (RMS) value of a signal is that value that would deliver the same power to a resistive load as a DC signal with same voltage
root
meansquare
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Root – Mean – Square Value
The RMS value of a signal (time varying voltage or current) is with respect to dissipation equivalent to the same DC value
RVP RMS
2= ( )∫=
TRMS dttv
TV
0
21
RIP RMS2= ( )∫=
TRMS dtti
TI
0
21
Integrate over representative time interval
One period (or integer multiple)
Or ‘sufficiently long’
A multi-meter shows RMS value
Mains electricity is 220V RMS
RMS is standard way of specifying AC signal/sources (unless otherwise noted)
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RMS Example
( )⎥⎥
⎦
⎤
⎢⎢
⎣
⎡∫ −+∫=T
T
T
RMS dtVdtVT
V2
22
0
22 1
VVRMS =
( )∫=2
0
22 21TRMS dtV
TV
-V
+V
0 T t
VVRMS 2=
Compute VRMS
22V=[ ]TTV T 2
0
24=∫=∫=
TTdt
TVdtV
T 0
2
0
2 11 [ ] 20
2VT
TV T ==
2xV
0 T t
Same signal, but shifted along V-axis. Do you expect VRMS to be equal, larger or smaller than on the left?
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RMS Value of Sinusoidal SourcesV
-V
VVRMS 21
= (See book)
The RMS value of a sinusoidal source is 0.707 times the amplitude
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RMS Value of Fourier Series Sources( )[ ]∫ ∑ θ+ω+=
TnnRMS dttnAA
TV
0
20 cos1
( ) ( )( )...2cos1cos 210 +θ+ω+θ+ω+ nn tAtAA
( ) ( )( )...2cos1cos 210 +θ+ω+θ+ω+× nn tAtAA
Again a signal with period T, integral will vanish (cosα cosβ =...)
Several kinds of terms
A20
When integrated, gives constant contribution
( )nn tnAA θ+ω× cos0
( ) ( )kknn tkAtnA θ+ω×θ+ω coscos
Integral of cosine over (multiple) complete period(s) will vanish
‘Rectified’ cosine wave, integral yields An
2/2( ) ( )nnnn tnAtnA θ+ω×θ+ω coscos
∑+=∞
=1
220 2n
nAA
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Filtering
Amplitude and phase depend on frequencyCircuit principles like these are fundamental for filteringFiltering: selection of frequency rangesExamples: tuning of radio signals, filters in loudspeaker boxes(to separate signals for woofer and tweeter), ...Very central to Signal Processing, Radio Reception, …But currently very often done digitally (if frequency low enough)
( ) ( )θωτω
++
= tVtvc cos122 +
-V C
R
VcI
+
-V C
R
VcI
+
-V C
R
VcI
21atan πωτ
θ +⎟⎠⎞
⎜⎝⎛=
Next slides will peak-ahead to associated topics
Naturally, in the frequency domain§8.5
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Transfer Function and Bode DiagramTransfer function ≡ Ratio of output phasor to input phasor
( )ωτ+
=≡ωjV
VcH1
1 ( ) ( ) ( )ωτ+−=ω≡ω jHH dB 1log20log20
Decibel: 20 times common logarithm (base 10) of signal strength
-20 dB
10 x frequency change
0-3
log(ω)
|H(ω)|DB
10
-10
-30
-20
-40-50
τ101
τ1
τ10
τ100
τ1000
high frequency asymptote
low frequency asymptote
half-power point
corner frequency
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Transfer Function and Bode DiagramTransfer function ≡ Ratio of output phasor to input phasorDecibel: 20 times common logarithm (base 10) of signal strength
Also a standard plot for phase
We are not going to study this any further
But particularly good example of power of phasor analysis
0-3
log(ω)
|H(ω)|DB
10
-10
-30
-20
-40-50
τ101
τ1
τ10
τ100
τ1000
0-3
log(ω)
|H(ω)|DB
10
-10
-30
-20
-40-50
τ101
τ1
τ10
τ100
τ1000
0-3
log(ω)
|H(ω)|DB
10
-10
-30
-20
-40-50
τ101
τ1
τ10
τ100
τ1000
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Overview/Conclusion
Phasor/frequency domain analysis (§8.1- §8.3)
Clock system analysis (§8.4) skipped, but we covered the tools
Preview of filtering and transfer functions (§8.5)
RLC Circuits in frequency domain skipped (§8.6)
Complex Fourier Series skipped (§8.7)
Root-Mean-Square values (§9.4)
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ExamWe have covered an introduction to Circuit Analysis
Exam will be on Fr 19/11/04, 2pm-5pm
All material (see next slides)
Indien het eind-tentamen >= 4.0:
Eindcijfer = max (eindtentamen, gemiddelde v. deeltentamen en eindtentamen)
If you have questions during preparation, you may contact me e.g. by e-mail: [email protected]
Keep on watching web site, I might put extra info on it. E.g. as a result of questions by some of you.
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Contents
Fanout and capacitive couplingmore examples of circuits → ODE → solution
C4
Package inductance and RLC circuit analysis coupled ODE’swith non-real natural frequencies
C5
Interconnect and RC ladder circuitsCoupled ODE’s, natural frequencies
C3
Gate delay and RC circuitsFirst order RC circuits, linear first order ODETime response, energy, powe
C2
Fundamental conceptsCharge, current, voltage, powerCircuits, circuit elementsKirchoff laws
C1
Please make VERY sure that you can really fluently apply this!
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Contents (ctd)
Thévenin and Norton equivalent modelsControlled sourcesRoot-mean-square values Nodal analysis
S 9.1 – 9.5.1
Clock degradation and ac steady state analysisAlgebraic solution of circuitsPhasor representation(complex) impedance and admittance
S 8.1 – 8.3
Clock skew and signal representation Time domain vs frequency domain analysisSuperposition(7.6 only briefly)
S 7.1 – 7.7
Only one solution: invest your time and practice – practice – practice
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The End
Good luck, and thanks for your attention