W A T K I N S - J O H N S O N C O M P A N Y Semiconductor
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Transcript W A T K I N S - J O H N S O N C O M P A N Y Semiconductor
Engineering 43
Fourier
Transfer Fcn
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx
Fourier Transform
A Fourier Transform A Conceptual Example
is an integral
• This Irregular Signal
transform that reexpresses a function
in terms of different
Sine/Cosine waves
• Is the SAME as the
of varying
Sum of these Sinusoids
amplitudes,
wavelengths, and
phases.
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
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Fourier Transform
John Baptiste
Joseph Fourier
investigated Time
Varying Heat-Flow
in a Metal Bar
His great Insight:
ANY Periodic
Function Could be
Expressed as the
sum of Sinusoidal
Functions
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For a Given,
arbitrary Periodic
Function, f(t), The
Fourier Equivalents
f t b 0
B
m
cos m 0 t m
m 1
f t d 0
D
n
sin n 0 t n
n 1
Bruce Mayer, PE
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Example: Square Wave
The SquareWave Shown at Bottom-Lt
can be described by a sum-of-sines
v sq
4A
sin 0 t
4A
3
sin 3 0 t
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4A
5
sin 5 0 t
4A
7
sin 7 0 t
4A
9
sin 9 0 t
Bruce Mayer, PE
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Transfer Fuction, H(f)
iin
Consider a “Black Box”
v in
that takes Input Power,
vin & iin Transforms this
Power into an Output, vout & iout
iout
v out
• A typical transformation would be to “FilterOut” certain electrical frequencies.
For Phasor Voltages
V out
Vin & Vout Define the
Hf
voltage
V in
Transfer Function as
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx
Transfer Function
Hf
V out
V in
Note that the Transfer Function
• Is a Function of FREQENCY ONLY
• Can Change (and usually does change)
the Magnitude and Phase-Angle of many
of the incoming, frequency-dependent,
electrical signals
Measuring an Unknown “Black Box”
Apply Sinusoidal Vin
(Vin0°), Measure Vout
(Voutφ°) and Plot:
Vout / Vin and φ
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
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Hf
Example Transfer Function
f Hz
H f
f Hz
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx
Example Transfer Function
Find vout for vin = 1.35Vcos(40∙2πt+65°)
H f
Hf
−25
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f Hz
Bruce Mayer, PE
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Example Transfer Function
Then at 40 Hz
(40∙2π rads/sec)
H 40 Hz 25 150
V out
V in
Using the Values
Taken from the H(f)
Mag & Phase
Graphs
Recall vin
V out 25 150 1 . 35 V 65
In Phasor for
Or in the Time
Domain
v in 1 . 35 V cos 40 2 t 65
V in 1 . 35 V 65
Thus
V out 33 . 75 V 85
v out t 33 . 75 V cos 40 2 t 85
V out H 40 Hz V in
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx
MultiFrequency Example 6.2
Note the THREE
Frequencies
• 0 Hz
• 1000 Hz
– 1000∙2π
rad/sec
• 2000 Hz
– 2000∙2π
rad/sec
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx
Ex6.2 Transfer Function
Apply to vin the Transfer Function
From the Transfer Function find
H 0 4 0 H 1000 3 30 H 2000 2 60
• Apply To components of vin
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx
Example 6.2
Using This H(f) Set find
H 0 4 0 H 1000 3 30
H 2000
2 60
V out 1 H 0 V in1 4 0 3 0 12 0 12
V out 2 H 1000 V in 2 3 30 2 0 6 30
V out 3 H 2000 Vin 3 2 60 1 70 2 10
Note that the above Phasors CanNOT
be added as they have DIFFERENT
Frequencies.
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx
Example 6.2
Because of Differing Frequencies
MUST add TIME-DOMAIN Voltages
V out 1 12
v out 1 t 12
V out 2 6 30
v out 2 t 6 cos 1000 2 30
V out 3 2 10
v out 3 t 2 cos 2000 2 10
Then vout(t)
is simply
the SUM
of the above
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx
1st Order Lo-Pass Filter
Consider the RC Ckt
Shown below
ZC
1
j C
1
j 2 f C
Notice the Limits of
Behavior
V in
V out
lim Z C lim
f 0
f 0
lim Z C lim
In the Frequency
Domain the Cap
Impedance, Zc
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f
f
1
j 2 f C
1
j 2 f C
0
A cap is
• OPEN to Low-Freq
• SHORT to Hi-Freq
Bruce Mayer, PE
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1st Order Lo-Pass Filter
Thus the Behavior
of a Cap-Based
Impedance
• At LO-Frequencies a
Cap acts as an
OPEN Circuit
V in
• At HI-Frequencies a
Cap Acts as a
SHORT Circuit
ZC
Now use Phasor VDivider on RC ckt
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ZR R
ZC
V out
Z tot
V in
V out
1
j 2 f C
1 j 2 f C
R 1 j 2 f C
Multiplying Top&Bot
by j2πfC
V out
1
1 j 2 f RC
V in
Bruce Mayer, PE
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V in
1st Order Lo-Pass Filter
Then the Transfer
Function
Hf
V out
V in
1
1 j 2 f RC
ReWriting
Hf
Hf
1
1 jf 2 RC
1 j f
1 f
Note The Mag & Ph
of H(f) in terms of fB :
jf 1 f B
1
1
fB
fB
Where
Hf
2
fB
1
2 RC
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fB is the “Break
point” Frequency at
which H(f) falls to
70.7% of its Original
Magnitude Value.
1
1 f
H f arctan
fB
f
2
fB
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx
Lo-Pass Filter
V in
V out
The LoPass Filter Transfer Function
fB : is also call the Half-Power-Frequency
• Recall Full Power to a Resistor:
I R
or
• Then HALF Power: I
V
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R
2
2
or
2
V
2
2
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx
R
2
R
LR (LowPass) Filter
Find the Transfer
Function for LR Ckt
Z L j 2 fL
V in
V out
I
Use Ohm Find The
Single Loop Current
I
V in
ZL R
V in
j 2 fL R
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Then also by Ohm
1
V in
V in
V out I R
j 2 fL R 1
j 2 fL R 1
R
ReWriting
V out
V in
j 2 fL R 1
V in
j
f
R
2 fL
1
V in
1 j
f
fB
Arrive at Xfer Fcn very
similar to RC Ckt
Hf
V out
V in
where : f B R
1
1 jf
fB
2 L
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx
The deciBel (dB)
Named after
Alexander Graham
Bell, the deciBel
(dB) relates two
Power Levels
L dB 10 log
P2
P1
SomeTimes The
Power Level is
Referenced to a
Standard Value, P0
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In this case
L dB 10 log
P
P0
ReCall a Current or
Voltage delivering
Power to a Resistor
Pv V
2
R
Pi I R
2
Then the dB in
Current or Voltage
Ratios
Bruce Mayer, PE
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The deciBel (dB)
dB In Terms of
Voltage Ratios
L dB
V 22 R
P2
10 log
10 log 2
P1
V1 R
V 22
10 log 2
V1
2
V
V
10 log 2 20 log 2
V
V
1
1
Or dB for Currents
L dB
I 12 R
10 log
10 log 2
P1
I2 R
P2
2
I
I
I
10 log 2 20 log 2
10 log
I
I
I
1
1
2
2
2
1
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Now we Defined
Hf
V out
V in
Hf
V out V in
2
Since |H(f)| is a
Voltage Ratio,
define
Hf
dB
20 log H f
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx
dB Plots (SemiLog) Plot
Plotting H(f) on the logarithmic dB Scale
makes it easier to distinguish Very
Large (104 vs 105) or Very Small (10−4
vs 10−5) Points on the Plots
85 db 20 log H f
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Hf
10
85 20
10
4 . 25
0 . 0000562
Bruce Mayer, PE
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Cascaded NetWork Gain
Consider the
Transfer Function of
the “BlackBox” at
V out
Right
Hf
V in
Looking inside the
BlackBox find
V out
V in
V out 2
V in 1
V out 2
V in 1
1
V out 2
V in 1
Note that with
Vout1 = Vin2
Engineering-43: Engineering Circuit Analysis
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V out 1
V out 1
V out
Hf
V out 2
V in
V in 1
V out 1
so : H f
V out 1
V out 2
V out 1
V out 2
V in 1
V in 2
V in 1
V out 1
or
V in 2
H 1 f H 2 f
Or in dB form
Hf
H 1 f H 2 f
20 log H f
20 log H 1 f H 2 f
20 log H f
20 log H 1 f
Hf
dB
H1 f
dB
20 log
H 2 f
dB
Bruce Mayer, PE
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H 2 f
Reading Logarithmic Scales
Tools Needed
• Ruler
• Scientific Calculator
To Find a Value of a Pt
Between Decades m & n
• Use Ruler to Measure
– Decade Distance, dd
– Distance from Pt to Lower
Decade (Decade m), dp
• Then The Value at the Pt
V 10
d p dd
10 m
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx
10
10
-30
-31
d d 21.1 mm
10
-32
V 10
15.4 21.1
10
d p 15.4 mm
10
32
10
0.730
10
32
5.37 10
32
-33
400
405
410
415
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420
425
x
430
435
440
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx
445
450
Octave
An octave is the interval between two
points where the frequency at the
second point is twice the frequency of
the first.
Given Frequencies f1 & f2
N oct
OR
N oct
f1
log 2
f
2
log f 1 f 2
log 2
Engineering-43: Engineering Circuit Analysis
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MUSICAL Octaves
Octave
1
2
3
4
5
6
7
8
Frequency (Hz)
63
125
250
500
1k
2k
4k
8k
Wavelength in air
(70oF, 21oC) (ft)
17.92
9.03
4.52
2.26
1.129
0.56
0.28
0.14
Wavelength in air
(70oF, 21oC) (m)
5.46
2.75
1.38
0.69
0.34
0.17
0.085
0.043
Bruce Mayer, PE
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WhiteBoard Work
Let’s This Nice
Problem
v in t
v out t
Find the OutPut
Voltage for For
this Input
v in t 17 V 23 V cos 1000 2 t 31V cos 12000 2 t
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx
All Done for Today
79.5 MHz
Notch
Filter
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx
Engineering 43
Appendix
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx
Logarithm Change
of Base Proof
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
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Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx
White Board RL Filter Problem
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
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LR Filter Transfer Function
1
0.9
f = 0:10:20e3
HfB = 1./sqrt(1+(f/fB).^2);
plot(f,HfB,'LineWidth',3), grid, xlabel('f
(Hz)'), ylabel('|H(f)')
disp('showing fB plot - hit ANY KEY to
continue')
pause
fB = 2700/(2*pi*68e-3)
Hf = abs(2700./(2700 + j*2*pi*f*68e-3));
plot(f,Hf,'LineWidth',3), grid, xlabel('f
(Hz)'), ylabel('|H(f)')
0.8
|H(f)
0.7
0.6
0.5
0.4
0
0.2
0.4
0.6
0.8
1
f (Hz)
1.2
Engineering-43: Engineering Circuit Analysis
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1.4
1.6
1.8
2
4
x 10
Bruce Mayer, PE
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P5.57 Graphics
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx
P5.81 Graphics
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-06a_Fourier_XferFcn.pptx