High and Low pass Fillter
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Transcript High and Low pass Fillter
Lecture 23
Filters
Hung-yi Lee
Filter Types
Lowpass filter
Bandpass filter
wco : cutoff frequency
Bandwidth B = wu - wl
Highpass filter
Notch filter
Real World
Ideal filter
Transfer Function – Rules
• Filter is characterized by its transfer function
H s =
N s
D s
=K
s z1 s z 2 s z m
s p1 s p 2 s p n
The poles should be at the
left half of the s-plane.
We only consider
stable filter.
Given a complex pole or zero, its complex
conjugate is also pole or zero.
Transfer Function – Rules
• Filter is characterized by its transfer function
H s =
N s
D s
=K
s z1 s z 2 s z m
s p1 s p 2 s p n
n m :improper filter
As the frequency increase,
the output will become
infinity.
n m :proper filter
We only consider proper
filer.
The filters consider have more poles than zeros.
Filter Order
H s =
N s
D s
=K
s z1 s z 2 s z m
s p1 s p 2 s p n
Order = n
The order of the denominator is the order of the filter.
order=1
order=4
order=4
Outline
• Textbook: Chapter 11.2
Second-order Filter
First-order Filters
Lowpass
Filter
Highpass
Filter
Lowpass
Filter
Highpass
Filter
Bandpss
Filter
Notch
Filter
First-order Filters
Firsr-order Filters
H s =
N s
zero or first order
first order
D s
0 or 1 zero
1 pole
Case 1:
Case 2:
1 pole, 0 zero
1 pole, 1 zero
H s = K
1
s p
H s = K
sz
s p
Firsr-order Filters - Case 1
Lowpass filter
H s = K
w
1
s p
As ω increases
Magnitude decrease
Phase decrease
Pole p is on the
negative real axis
Firsr-order Filters - Case 1
• Amplitude of the transfer function of the first-order
low pass filter
Ideal
Lowpass
filter
First-order
Lowpass
filter
Firsr-order Filters - Case 1
• Find cut-off frequency ωco of the first-order low
pass filter
w
At DC
Lowpass filter
H s = K
0 = K
1
s p
1
| p|
Find cut-off frequency ωco
such that
| p|
w co = K
1
| p|
1
2
w co = | p |
Firsr-order Filters - Case 2
H s = K
sz
s p
w
Zero can be
positive or
negative
|z|
| p|
Case 2-1: Absolute value of zero is
smaller than pole
Magnitude is proportional to the
length of green line divided by the
length of the blue line
Low frequency ≈ |z|/|p|
Because |z|<|p|
The low frequency signal will
be attenuated
If z=0, the low frequency can
be completely block
Not a low pass
Firsr-order Filters - Case 2
H s = K
sz
s p
w
Case 2-1: Absolute value of zero is
smaller than pole
Magnitude is proportional to the
length of green line divided by the
length of the blue line
High frequency
The high frequency signal
will pass
|z|
| p|
If z=0 (completely block low
frequency)
H s = K
s
s p
High pass
First-order Filters - Case 2
H s = K
s
s p
• Find cut-off frequency ωco of the first-order high
pass filter
w = K
| jw |
w co
| jw p |
w |p|
2
2
co
= K
w co = K
2
=
1
2
w co = | p |
| j w co |
| j w co - p |
=
K
2
(the same as low
pass filter)
First-order Filters - Case 2
H s = K
sz
s p
w
Case 2-2: Absolute value of zero is
larger than pole
Low frequency ≈ |z|/|p|
Because |z|>|p|
The low frequency signal will
be enhanced.
High frequency: magnitude is 1
The high frequency signal
will pass.
Neither high pass
nor low pass
First-order Filters
vh
1
H lp s =
1
sC
R
=
1
1
sRC 1
sC
1
vl
1
H lp j w =
w =
RC
jw
1
Lowpass filter
If vh is output
Highpass filter
0 = 1
= 0
RC
1
2
w
RC
RC
If vl is output
RC
1
s
RC
1
sC
Consider vin as input
=
w co =
1
2
(pole)
RC
s
H hp s = 1 H lp s =
s
1
RC
0 = 0
= 1
w co =
1
RC
(pole)
First-order Filters
H lp s =
vh
R
R
sL R
=
L
s
R
L
R
R
H lp j w =
vl
L
jw
R
w =
L
R
w
L
2
L
0 = 1 = 0 w co =
H hp s = 1 H lp s =
2
s
s
R
L
R
L
(pole)
Cascading Two Lowpass Filters
V out
Vx
V in
H lp 1 s =
H lp s =
Vx
V in
H lp 2 s =
H lp 1 s
V out
Vx
H lp 2 s
Cascading Two Lowpass Filters
V out
Vx
V in
1
1
H lp s =
=
sC 1
R1
1
sC 1
sC 2
R2
=
1
1
sC 1 R1 1
sC 2
1
1 s C 1 R 1 C 2 R 2 s C 1 C 2 R 1 R 2
2
1
sC 2 R 2 1
Cascading Two Lowpass Filters
V out
Vx
V in
Z eq s
The first low pass filter is influenced by the
second low pass filter!
H lp 1 s =
Vx
V in
=
Z eq s
Z eq s R1
Cascading Two Lowpass Filters
V out
Vx
V in
1
Z eq s =
|| R 2
sC 1
sC 2
1
1
R2
sC 1
sC 2
=
1
1
R 2
sC 1
sC 2
Z eq s
1
2
s C 1C 2
2
s C 1C 2
=
sC 2 R 2 1
s C 1 C 2 s C 1 C 2 R 2
2
Cascading Two Lowpass Filters
V out
Vx
V in
Z eq s
H lp s = H lp 1 s H lp 2 s =
=
1
Z eq s
Z eq s R1
sC 2
1
sC 2
1
1 s R 1C 1 C 2 R 2 R 1C 2 s C 1C 2 R 1 R 2
2
R2
Second-order Filters
Second-order Filter
H s =
N s
0, 1 or 2 zeros
D s
Second order
Must having
two poles
2 poles
Case 1:
No zeros
Case 2:
One zeros
Case 3:
Two zeros
Second-order Filter – Case 1
Case 1-1
Case 1-2
w
w
p1
p1
p2
p2
Second-order Filter – Case 1
Case 1-1
Real Poles
w
The magnitude is
l1
The magnitude
monotonically decreases.
l1
p1
p2
l1l 2
As ω increases
l2
l2
K
Decrease faster than first
order low pass
Second-order Filter – Case 1
Case 1-2
Complex Poles
w
The magnitude is
l1
As ω increases,
l2
p1
l1
l2
p2
K
l1l 2
l1 decrease first and then increase.
l2 always increase
What will happen to magnitude?
1. Increase
2. Decrease
3. Increase, then decrease
4. Decrease, then increase
Second-order Filter – Case 1
Case 1-2
Complex Poles
w
If ω > ωd
l1
l2
p1
wd
wd
p2
l1 and l2 both increase.
The magnitude must decrease.
What will happen to magnitude?
1. Increase
2. Decrease
3. Increase, then decrease
4. Decrease, then increase
Second-order Filter – Case 1
Case 1-2
Complex Poles
l1
p1
wd
wd
p2
When ω < ωd
w
w
l2
l1 =
w d w
l2 =
w d w
2
2
2
2
Maximize the magnitude
K
l1l 2
Minimize
l1l 2 =
2
wd w
2
2
wd w
2
Second-order Filter – Case 1
Minimize l l = w w w w
Minimize f w = w w w w
df w
= 2 w w 1 w w 2 w w w
dw
w w 1 w w w w w w = 0
2
2
1 2
2
2
d
d
2
2
2
2
d
d
2
2
d
2
2
2 w w d
w d w
2
2
w
d
= 0
w wd w
2
= wd w
2
=0
d
w w d w 2w = 0
2
w
2
2
d
w wd w
d
2
d
2 w w d
d
2
d
2
d
2
w wd
2
= 0
w d w w d w = 0
2
w =
wd
2
2
(maximize)
Second-order Filter – Case 1
w =
wd
2
2
Lead to maximum
The maxima exists when w d
w
Peaking
w
p1
p1
wd
wd
wd
wd
p2
wd
p2
No Peaking
wd
Peaking
Second-order Filter – Case 1
w =
wd
2
2
Lead to maximum
The maxima exists when w d
wd
l1
w
wd
p2
w =
w
p1
=K
l2
Peaking
K
l1l 2
1
w d w
2
2
w d w
2
2
Assume w = w d2 2
w = K
1
2w
0 = K
d
1
wd
2
2
Second-order Filter – Case 1
K
H s =
s
2
w0
Q
sw
2
0
w0
Q
p1 , p 2 =
w0
Q
2
4 w 02
2
For complex poles
p1
w0
wd
wd
4 w 02 0
=
w0
wd = w0
2
Q
wd = w0 1
2Q
w0
p2
w0
Q
2
2
1
2
1
4Q
2
Second-order Filter – Case 1
H s =
s
2
w0
Q
sw
jw
2
0
w = w0
=
Q times
p1
K
H jw =
K
w
w0
wd
wd
w0
Q
K
w
2
jw w 0
2
j
w0
w
Q
K
H jw 0 =
w = w0
w0
2
j
w0
p2
2
0
2
Q
w 0 =
K
w
2
0
Q
0 =
K
w0
2
Q times of DC gain
Second-order Filter – Case 1
w =
w
2
d
2
K
H s =
s
2
w0
w = K
Lead to maximum
sw
Q
For complex poles
2
0
w0
Q
=
p1 , p 2 =
w0
Q
w0
Q
2
2
w0
2Q
wd = w0 1
2w
4 w 02
2
4 w 02 0
1
1
Q
2
1
4Q
2
d
Second-order Filter – Case 1
w =
w
2
d
2
H s =
s
2
w0
=
s w0
2
Q
w = w0 1
w = K
Lead to maximum
K
1
2Q
2
w0
wd = w0 1
2Q
1
2w
d
1
4Q
2
Lead to maximum
The maximum value is K
The maximum exist when Q
1
2
1
2w
= 0 . 707
= K
d
1
Q
w0
2
1
1
4Q
2
Second-order Filter – Case 1
Case 1-1
Real Poles
Case 1-2
Complex Poles
w
w
p1
p1
p2
wd
wd
Q 0 .5
(No Peaking)
p2
0 . 707 Q 0 . 5
Which one is considered as closer
to ideal low pass filter?
K
H s =
s
2
w0
Q
Q 0 .5
s w0
2
Q =
1
Complex
poles
= 0 . 707
2 (Butterworth filter)
Q 0 . 707
Peaking
Butterworth – Cut-off Frequency
K
H s =
s
2
w0
s w0
H jw =
=
0 =
K
s
2
= 0 . 707
2
2
Q
=
Q =
1
2w 0 s w 0
2
K
w0
2
w co =
K
j w 2
w
2w 0 jw w 0
2
K
2
0
w
2
j 2w 0 w
1
K
2 w0
2
w co = w 0
ω0 is the cut-off frequency
for the second-order
lowpass butterworth filter
(Go to the next lecture first)
Second-order Filter – Case 2
Case 2: 2 poles and 1 zero
Case 2-1: 2 real poles and 1 zero
w
z1
p 2 p1
w
p2
z1 p1
w
p 2 p1
z1
Second-order Filter – Case 2
Case 2: 2 poles and 1 zero
Case 2-1: 2 real poles and 1 zero
w
p 2 p1
z1
flat
Bandpass Filter
Second-order Filter – Case 2
Case 2-2: 2 complex poles and 1 zero
w
Two
Complex
Poles
p1
-40dB
w0
w0
+
z1
Zero
+20dB
p2
| z1 |
Second-order Filter – Case 2
Case 2-2: 2 complex poles and 1 zero
| z 1 | w 0
-40dB
-20dB
Two
Complex
Poles
w 0 | z1 |
-40dB
w0
+
+20dB
-20dB
Zero
+20dB
| z 1 | w 0
| z1 |
w0
| z1 |
Second-order Filter – Case 2
Case 2-2: 2 complex poles and 1 zero
Highly Selective
+20dB
-20dB
| z 1 |= 0
Two
Complex
Poles
-40dB
w0
+
w0
Zero
+20dB
Bandpass Filter
| z1 |
Bandpass Filter
• Bandpass filter: 2 poles and zero at original point
H bp ( s ) = K
w
s
s w 0 Q s w 0
2
2
Find the frequency for the
maximum amplitude
bandpass filter
ω0?
Bandpass Filter
• Find the frequency for the maximum amplitude
H bp ( s ) = K
H bp ( j w ) =
=
s
s w 0 Q s w
2
2
0
K w 0 Q j w
jw
2
w 0 Q j w w
K
jw Q
w0
1
w0Q
jw
=
2
0
=
=
w 0
K 2
s w 0
w0
Q s w 0
2
K
j w 2
w 0 Q j w
K
jw Q
Q s
1
jw 0 Q
w
=
w0
2
1
w 0
Q jw
K
w
w0
1 jQ
w
w0
Bandpass Filter
• Find the frequency for the maximum amplitude
H bp ( j w ) =
K
w
w0
1 jQ
w
w
0
K
a bp (w ) =
w
w0
1 Q
w
w
0
2
a bp (w ) is maximized
when w = w 0
(Center frequency)
The maximum value is K’.
2
a bp ( 0 ) = 0
a bp ( ) = 0
(Bandpass filter)
Bandpass Filter
a bp (w ) is maximized
K
a bp (w ) =
w0
2 w
1 Q
w
w
0
w
when w = w 0
2
The maximum value is K’.
bandpass filter
K
K / 2
B
wl
w0
Bandwidth B
= ωr - ωl
wr
Bandpass Filter - Bandwidth B
K
a bp (w ) =
=
w0
2 w
1 Q
w
w
0
2
K
2
2
2
w
w0
1
w w = Q2
0
1
2
2
w w 0w w 0 = 0
Q
w
w0
=2
1 Q
w
w
0
w0
w
1
=
w0 w
Q
2
2
w =
1
Q
w0
1
2
w 0 4 w 0
Q
2
Four answers?
Pick the two positive
ones as ωl or ωr
Bandpass Filter - Bandwidth B
2
w =
1
w0
Q
1
2
w 0 4 w 0
Q
2
2
1
wr =
Q
w0
1
2
w 0 4 w 0
Q
2
B = wr wl =
w0
Q
w = w rw l
2
0
Q measure the narrowness
of the pass band
Q is called quality factor
2
wl =
1
Q
w0
1
2
w 0 4 w 0
Q
2
H bp ( s ) = K
= K
s
s w 0 Q s w 0
2
2
s
s Bs w0
2
2
Bandpass Filter
H bp ( s ) = K
= K
s
s
2
w 0
2
Q s w0
s
s Bs w0
2
2
Usually require a specific bandwidth
The value of Q determines the
bandwidth.
When Q is small, the
transition would not be
sharp.
Stagger-tuned Bandpass Filter
Stagger-tuned Bandpass Filter
- Exercise 11.64
s
H s = K
s
2
w1
K = 1600
s
sw
Q
Bandpass Filter
Center frequency: 10Hz
2
1
s
2
w2
sw
Q
w 1 = 10
w 2 = 40
2
2
Bandpass Filter
Center frequency: 40Hz
We want flat passband.
dB
Tune the value of
Q to achieve that
10 Hz
40 Hz
Stagger-tuned Bandpass Filter
- Exercise 11.64
H s = K
s
2
s
s
w1
w2
Q
s w1 s
2
2
Q
K = 1600
s w2
2
w 1 = 10
w 2 = 40
Test Different Q
Q=3
Q=1
Q=0.5
Second-order Filter – Case 3
• Case 3: Two poles, Two zeros
Case 3-1: Two real zeros
Two real
poles
Two
Complex
poles
w
w
p1
z1
z1
p2
p1
z2
z2
p2
High-pass
Second-order Filter – Case 3
Case 3: Two poles, Two zeros
Case 3-2: Two Complex zeros
s
p 1 s p 2 = s
2
w0
Q
w0
p1
w
w
s w0
2
z1
Fix ω0
Larger Q
s z1 s z 2 =
Larger θ
s
2
w
Q
s w
2
z2
Fix ωβ
Larger Q β
Larger θ β
p2
Second-order Filter – Case 3
Case 3: Two poles, Two zeros
Case 3-2: Two Complex zeros
w0 w
w0
p1
w
w
z1
Two
poles
w0
-40dB
Two
zeros
z2
w
+40dB
p2
Second-order Filter – Case 3
Case 3: Two poles, Two zeros
Case 3-2: Two Complex zeros
w0 w
s
2
w
Q
H s =
s
2
w0
Q
w0
p1
w
w
s w
2
z1
s w0
2
w 0 = 10, w = 1, Q = Q = 10
z2
High-pass
Notch
p2
Second-order Filter – Case 3
Case 3: Two poles, Two zeros
Case 3-2: Two Complex zeros
w
w0 w
w0
w
z1
p1
Two
poles
w0
-40dB
Two
zeros
w
+40dB
p2
z2
Second-order Filter – Case 3
Case 3: Two poles, Two zeros
Case 3-2: Two Complex zeros
w0 w
s
2
w
Q
H s =
s
2
w0
Q
s w
w
w
w0
2
z1
p1
s w0
2
w 0 = 1, w = 10, Q = Q = 10
Low-pass
Notch
p2
z2
Second-order Filter – Case 3
Case 3: Two poles, Two zeros
Case 3-2: Two Complex zeros
w0 = w
Q Q
w0 = w
w
p1
z1
Large Q
Two
poles
Two
zeros
w0
-40dB
z2
w
+40dB
small Qβ
p2
Second-order Filter – Case 3
Case 3: Two poles, Two zeros
Case 3-2: Two Complex zeros
w0 = w
Q Q
w0 = w
w
p1
z1
w 0 = w = 10
Q = 100, Q = 10
z2
p2
Second-order Filter – Case 3
Case 3: Two poles, Two zeros
Case 3-2: Two Complex zeros
w0 = w
Two
poles
Two
zeros
Q Q
w0 = w
w
p1
z1
small Q
w0
w
-40dB
+40dB
Larger Qβ
p2
z2
Second-order Filter – Case 3
Case 3: Two poles, Two zeros
Case 3-2: Two Complex zeros
w0 = w
Q Q
w0 = w
w
p1
z1
w 0 = w = 10
Q = 10, Q = 100
Standard
Notch
Filter
p2
z2
Second-order Filter – Case 3
Case 3: Two poles, Two zeros
Case 3-2: Two Complex zeros
w0 = w
s
w0
Q
H s =
s
2
w0
Q
w 0 = w z1
Q Q
If the two zeros
are on the ω axis
The notch filter will
completely block the
frequency ω0
2
w
p1
Q =
s w0
2
s w0
2
=
sw
2
0
s
2
w0
Q
2
sw
2
0
p2
z2
Notch Filter
s
2
w0
Q
H no s =
s
2
w0
Q
w
no w =
w
2
0
2
0
s w0
2
The extreme value
is at ω= ω0
s w0
2
no w 0 =
w w
0
Q
2
w 0w
Q
2
w
2 2
w
2 2
no 0 = 1
Q
Q
no = 1
(Notch filter)
Second-order RLC Filters
A
v in
C
v in
v out
v out
B
v out
v in
D
v in
v out
RLC series circuit can implement high-pass, lowpass, band-pass and notch filter.
Second-order RLC Filters
A
v in
v out
DC (O)
Infinity (X)
Low-pass Filter
B
v out
v in
DC (X)
Infinity (O)
High-pass Filter
Second-order RLC Filters
v out
w
C
p1
v in
z1
p2
H s =
R
R sL
1
=
sCR
sC R s CL 1
2
s
=
s s
2
R
L
R
L
sC
Band-pass Filter
1
LC
Second-order RLC Filters
– Band-pass
C
v out
v in
40pF to 360pF
L=240μH, R=12Ω
s
H s =
s s
2
R
L
R
L
w0
Q
Frequency range
Center frequency: w 0 = 1
1
LC
w
2
0
f0 =
Max: 1.6MHz
w0
2
LC
= 1 2
LC
min: 0.54MHz
Second-order RLC Filters
– Band-pass
C
v out
v in
40pF to 360pF
L=240μH, R=12Ω
s
H s =
s s
2
Frequency range 0.54MHz ~ 1.6MHz
R
L
R
L
w0
Q
w0 = 1
1
LC
w
2
0
Q =
LC
w 0L
R
w0 Q = R L
Q =
1
L
R
C
Q is 68 to 204.
Band-pass
1
s
v in
1
s
Z eq
v out
s
1
s
s 1
s
=
s 1
s s 2
s
V out = V in
1
1 s 1
= || 1 = ||
s
s
s s
1
1 s 1
=
Z eq
1 Z eq
s 1
1
1
1
s
s
s s 2
= V in
s 1 s 1
1
s s 2
= V in
s
s 3s 1
2
Band-pass
1
s
v in
1
v out
s
H s =
s
s 3s 1
2
Band-pass Filter
V out = V in
H s =
V out
V in
=
s
s 3s 1
2
s
s 3s 1
2
w =1
2
0
w0 = 1
B=
w0
Q
Q =
1
3
=3
Second-order RLC Filters
C
v out
D
v out
v in
v in
s
H bp s =
s s
2
H s = 1 - H bp s
R
L
R
L
LC
1
s
2
1
LC
=
s s
2
R
L
Notch Filter
1
LC
Active Filter
V in = i Z i
Basic Active Filter
V out = i Z f
-i
H s =
i
0
0
V in
V out
V out
V in
=
Z
f
Zi
First-order Low-pass Filter
Z
R
f
H s =
f
=
V out
=
||
f
sC
=
Ri
Zi
R
Zi
V in
Z
1
R iR
1
f
sC
f
R
f
1
sC f
f
R i 1 sR f C f
w co = 1 R f C f
f
First-order High-pass Filter
Z
H s =
f
Zi
=
Z
f
Zi
R
Ri
V in
V out
=
f
1
sC i
sC i R
f
sC i R i 1
w co = 1 R i C i
Active Band-pass Filter
Band-pass Filter
Active Band-pass Filter
H 1 s =
=
R
R i 1 sR f C f
R
1 sR C 1
H 2 s =
=
R
H bp s = H 1 s H 2 s =
1 sRC 1
f
sC i R
f
sC i R i 1
sR C 2
sR C 2 1
sR C 2
sR C 1
2
?
Loading
H s =
v out
Z
The loading Z will change
the transfer function of
passive filters.
v in
Z
f
Zi
v out
The loading Z will NOT
change the transfer
function of the active filter.
Z
Cascading Filters
If there is no loading
H s =
The transfer function is H(s).
V in
H s V in
One Filter Stage Model
V out
V in
V out
Cascading Filters
V1
H 1 s V1
1st Filter with
transfer function H1(s)
H 2 s V 2
V2
2st Filter with
transfer function H2(s)
Overall Transfer Function: H s =
V3
V1
H 1 s H 2 s
V3
Cascading Filters
H 1 s V1
V1
1st Filter with
transfer function H1(s)
V 2 = H 1 s V1
V 3 = H 2 s V 2
Z i2
Z o1 Z i 2
H 2 s V 2
V2
V3
2st Filter with
transfer function H1(s)
V 3 = H 2 s H 1 s V1
H s =
V3
V1
Z i2
Z o1 Z i 2
= H 2 s H 1 s
Z i2
Z o1 Z i 2
H s = H 2 s H 1 s
Cascading Filters
V1
H 1 s V1
1st Filter with
transfer function H1(s)
Z i2
Z o1 Z i 2
H 2 s V 2
V2
2st Filter with
transfer function H1(s)
H s = H 2 s H 1 s
If zero output impedance (Zo1=0)
or If infinite input impedance (Zi2=∞)
V3
Cascading Filters
– Input & Output Impedance
Z o s
IT
VT
Z i s
Z i s =
H s V i
VT
IT
Z o s
Vi = 0
Z i s
H s V i V T
IT
Z o s =
VT
IT
H s = H 2 s H 1 s
If zero output
impedance (Zo1=0)
or If infinite input
impedance (Zi2=∞)
Cascading Filters
– Basic Active Filter
-i =0
i =0
Z o s =
0
0
0
V in
=0
IT
VT
VT
IT
=0
Active Notch Filter
A
B
Which one is correct?
Active Notch Filter
Low-pass
Filter
High-pass
Filter
Add
Together
Homework
• 11.19
Thank you!
Answer
• 11.19: Ra=7.96kΩ, Rb= 796Ω,
va(t)=8.57cos(0.6ω1t-31。)
+0.83cos(1.2ω2t-85。)
vb(t)=0.60cos(0.6ω1t+87。)
+7.86cos(1.2ω2t+40。)
(ω1 and ω2 are 2πf1 and 2πf2 respectively)
• 11.22: x=0.14, ωco=0.374/RC
• 11.26(refer to P494): ω0=2π X 6 X 10^4, B= ω0=2π X 5 X
10^4, Q=1.2, R=45.2Ω, C=70.4nF
• 11.28(refer to P494): C=0.25μF, Qpar=100, Rpar=4kΩ,
R||Rpar=2kΩ, R=4kΩ
Acknowledgement
• 感謝 江貫榮(b02)
• 上課時指出投影片的錯誤
• 感謝 徐瑞陽(b02)
• 上課時糾正老師板書的錯誤
Appendix
Aliasing
Sampling
Wrong
Interpolation
Actual
signal
High frequency becomes low frequency
Phase
cos 10 t
cos 20 t
filter w = 1 w = 90
cos 10 t
2
cos 20 t
2
Table 11.3 Simple Filter
Type
Lowpass
Highpass
Bandpass
Notch
Transfer Function
H (s) =
H (s) =
H (s) =
H (s) =
Properties
a (0 ) = K
K w co
s w co
a (w co ) = K
a ( ) = K
Ks
s w co
a (w co ) = K
K w 0 Q s
2
K s 2s w0
2
2
2
B = w0 Q
a (w 0 ) = KQ w 0
s w 0 Q s w 0
2
2
a (w 0 ) = K
s w 0 Q s w 0
2
2
B = w0 Q
98
Loudspeaker for home usage with three
types of dynamic drivers
1. Mid-range driver
2. Tweeter
3. Woofers
https://www.youtube.com/watch?v=3I62Xfhts9k
From Wiki
• Butterworth filter – maximally flat in passband and
stopband for the given order
• Chebyshev filter (Type I) – maximally flat in stopband,
sharper cutoff than Butterworth of same order
• Chebyshev filter (Type II) – maximally flat in passband,
sharper cutoff than Butterworth of same order
• Bessel filter – best pulse response for a given order
because it has no group delay ripple
• Elliptic filter – sharpest cutoff (narrowest transition
between pass band and stop band) for the given order
• Gaussian filter – minimum group delay; gives no
overshoot to a step function.
Link
• http://www.ti.com/lsds/ti/analog/webench/weben
ch-filters.page
• http://www.analog.com/designtools/en/filterwizar
d/#/type
Suppose this band-stop filter were to
suddenly start acting as a high-pass
filter. Identify a single component
failure that could cause this problem
to occur:
If resistor R3 failed open, it would cause
this problem. However, this is not
the only failure that could cause the same
type of problem!