Power dissipation.
Filters. Transmission lines.
Eugeniy E. Mikhailov
The College of William & Mary
Week 4
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Power dissipation
Recall that power dissipated by element is
P = VI
where V and I are real.
Since we use a substitute
V cos(ωt) → Veiωt and I cos(ωt) → Ieiωt ,
we need to write
P = Re(V )Re(I)
Recall the Ohm’s law
V = ZI
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Power dissipation by a reactive element
Theorem
Average power dissipated by a reactive element (C or L) is 0
Lets use as example an inductor.
π
ZL = iωL = ei 2 ωL, IL = Ip eiωt
π
π
VL = ZL IL = ei 2 ωLIL = ωLIp ei(ωt+ 2 )
Re(IL ) = Ip cos(ωt), Re(VL ) = −ωIp L sin(ωt)
Thus average power dissipated by the inductor
Z
T
Z
0
P=
T
Re(IL )Re(VL )dt = −
P=
−ωIp2 L
Ip cos(ωt)ωIp L sin(ωt)dt
0
Z
T
cos(ωt) sin(ωt)dt =
0
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ωIp2 L
Z
0
T
1
sin(2ωt)dt = 0
2
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Fourier transform
If function f (t) goes to zero at ±∞ then f̂ (w) exists such as
Z ∞
f (t) =
f̂ (ω)eiωt dω
−∞
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Transfer function
Time domain
Vin (t)
Frequency domain
H(t)
Z
Vout(t)
Vin(ω)
t
−∞
Vout(ω)
Vout (ω) = G(ω)Vin (ω)
H(t − τ )Vin (τ )dτ
Vout (t) =
G(ω)
Where G is complex transfer
function or gain.
Definition
G(ω) =
Vout (ω)
= |G(ω)|eiφ(ω)
Vin (ω)
Often used values of G in dB
dB = 20 log10 (|G(w)|)
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Simple example: RC low-pass filter
R
Vin(ω)
Vout(ω)
C
G(ω) =
1
Vout (ω)
1
1
1
= iωC 1 =
=
1
Vin (ω)
iωRC 1 + iωRC
1 + iωRC
R + iωC
defining ω3dB =
1
RC
G(ω) =
1
1
ω = r
1 + i ω3dB
1+
eiφ , φ = atan(−
ω2
2
ω3dB
ω
)
ω3dB
Note
|G(ω = ω3dB )| = 20 log10
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1
1+1
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= 20 log10
1
√
2
= −3dB
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Bode plots
Definition
|G|, dB
Bode plot: plots of magnitude and phase of the transfer function,
where |G| is often plotted in dB
R
Vin(ω)
Vout(ω)
0
-5
-10
-15
-20
-25
-30
-35
-40
-45
0.01
0.1
C
1
ω/ω3db
10
100
1
ω/ω3db
10
100
G(ω) =
1
ω
1 + i ω3dB
Phase, rad
0
-π/4
-π/2
0.01
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|G|, dB
RC high-pass filter
C
with ω3dB =
1
ω/ω3db
10
100
0.1
1
ω/ω3db
10
100
π/4
0
0.01
G(ω) =
0.1
π/2
Vout(ω)
R
Phase, rad
Vin(ω)
0
-5
-10
-15
-20
-25
-30
-35
-40
-45
0.01
ω
i ω3dB
Vout (ω)
R
iωRC
=
=
=
ω
1
Vin (ω)
1 + iωRC
1 + i ω3dB
R + iωC
1
RC
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RL filters
RL high-pass filter
RL low-pass filter
L
Vin(ω)
0
-5
-10
-15
-20
-25
-30
-35
-40
-45
0.01
Vout(ω)
R
Vin(ω)
R
R
, ω3dB =
R + iωL
L
0.1
1
ω/ω3db
10
100
iωL
R + iωL
0.1
1
ω/ω3db
10
100
0.1
1
ω/ω3db
10
100
π/2
Phase, rad
0
Phase, rad
0
-5
-10
-15
-20
-25
-30
-35
-40
-45
0.01
Vout(ω)
L
G(ω) =
|G|, dB
|G|, dB
G(ω) =
R
-π/4
-π/2
0.01
0.1
1
ω/ω3db
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10
100
π/4
0
0.01
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Filters chain
Vin(ω)
G1(ω)
G2(ω)
G3(ω)
Gn(ω)
Vout(ω)
Technically next stage loads the previous and it is quite hard to
calculate total transfer function.
However if we use rule of 10 to avoid overloading the previous filter.
Every next stage resistor Ri+1 > 10Ri we can approximate
Gt (ω) ≈ G1 (ω)G2 (ω)G3 (ω) · · · Gn (ω)
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Example band pass filter
R1
Vin(ω)
C2
C1
i ω2ω
3dB
1 + i ω2ω
For R1 = 1k Ω, R2 = 100k Ω,
C1 = C2 = .01µF
0
-5
-10
-15
-20
-25
-30
-35
-40
-45
100
1000
10000
100000
1e+06
1e+07
100000
1e+06
1e+07
ω, s-1
π/2
3dB
π/4
Phase, rad
3dB
|G|, dB
Gt (ω) ≈ G1 (ω)G2 (ω)
1
Gt (ω) ≈
1 + i ω1ω
Vout(ω)
R2
0
-π/4
-π/2
100
1000
10000
ω, s-1
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|G|, dB
Notch filter - Band stop filter
R
C
Vin(ω)
Vout(ω)
L
0
-10
-20
-30
-40
-50
-60
-70
-80
-90
0.01
R=1
R=100
0.1
1
ω/ω0
10
100
π/2
1
ω0 = √
LC
Phase, rad
π/4
0
-π/4
R=1
R=100
-π/2
0.01
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1
ω/ω0
10
100
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