Review
n
Phasors
¨ Source
n
vs. Impedence representation
First Order Circuits
¨ Initial and Final conditions
n
Second Order Circuits
¨ Solution
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
1
Lecture #7
OUTLINE
n
n
n
n
n
Decibels
Transfer function
First-order lowpass filter
Cascade connection and Logarithmic frequency
scales
Bode Plots
Reading
n
Chap 6
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
2
1
Bel and Decibel (dB)
n
A bel (symbol B) is a unit of measure of ratios of power
levels, i.e. relative power levels.
¨ The name was coined in the early 20th century in honor of
Alexander Graham Bell, a telecommunications pioneer.
¨ The bel is a logarithmic measure. The number of bels for a given
ratio of power levels is calculated by taking the logarithm, to the
base 10, of the ratio.
¨ one bel corresponds to a ratio of 10:1.
¨ B = log10(P1/P2) where P1 and P2 are power levels.
n
The bel is too large for everyday use, so the decibel
(dB), equal to 0.1B, is more commonly used.
¨ 1dB = 10 log10(P1/P2)
n
dB are used to measure
¨ Electric power, Gain or loss of amplifiers, Insertion loss of filters.
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
3
Logarithmic Measure for Power
n
n
To express a power in terms of decibels, one starts by
choosing a reference power, Preference, and writing
Power P in decibels = 10 log10(P/Preference)
Exercise:
¨ Express a power of 50 mW in decibels relative to 1 watt.
¨ P (dB) =10 log10 (50 x 10-3) = - 13 dB
n
Exercise:
¨ Express a power of 50 mW in decibels relative to 1 mW.
¨ P (dB) =10 log10 (50) = 17 dB.
n
dBm to express absolute values of power relative to a
milliwatt.
¨ dBm = 10 log10 (power in milliwatts / 1 milliwatt)
¨ 100 mW = 20 dBm
¨ 10 mW = 10 dBm
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
4
2
Power in 1st and 2nd Order Circuits
n
n
n
When dealing with 1st or 2nd order cirduits it is
convenient to refer to the frequency difference
between points at which the power from the circuit is
half that at the peak of resonance.
Such frequencies are known as “half-power
frequencies”, and the power output there referred to
the peak power (at the resonant frequency) is
10log10(Phalf-power/Ppeak) = 10log10(1/2) = -3 dB.
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
5
Logarithmic Measures for Voltage or Current
From the expression for power ratios in decibels, we can
readily derive the corresponding expressions for voltage
or current ratios.
Suppose that the voltage V (or current I) appears across
(or flows in) a resistor whose resistance is R. The
corresponding power dissipated, P, is V2/R (or I2R). We
can similarly relate the reference voltage or current to the
reference power, as
Preference = (Vreference)2/R or Preference= (Ireference)2R.
Hence,
Voltage, V in decibels = 20log10(V/Vreference)
Current, I, in decibels = 20log10(I/Ireference)
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
6
3
Note that the voltage and current expressions are just
like the power expression except that they have 20 as
the multiplier instead of 10 because power is
proportional to the square of the voltage or current.
Exercise: How many decibels larger is the voltage of a
9-volt transistor battery than that of a 1.5-volt AA
battery? Let Vreference = 1.5. The ratio in decibels is
20 log10(9/1.5) = 20 log10(6) = 16 dB.
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
7
Transfer Function
n
Transfer function is a function of frequency
¨ Complex quantity
¨ Both magnitude and phase are function of
frequency
Vin
H (w ) =
EE40 Summer 2006: Lecture 7
Two Port
filter network
Vout
Vout
Ð(q out - q in )
Vin
Instructor: Octavian Florescu
8
4
Filters
n
Circuit designed to retain a certain
frequency range and discard others
Low-pass: pass low frequencies and reject high
frequencies
High-pass: pass high frequencies and reject low
frequencies
Band-pass: pass some particular range of
frequencies, reject other frequencies outside
that band
Notch: reject a range of frequencies and pass
all other frequencies
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
9
Common Filter Transfer Function vs.
Freq
H( f)
H( f)
Low Pass
High Pass
Frequency
Frequency
H( f)
H( f)
Band Pass
Frequency
EE40 Summer 2006: Lecture 7
Band Reject
Frequency
Instructor: Octavian Florescu
10
5
First-Order Lowpass Filter
H ( jw ) =
VC
1 /( jwC )
1
1
=
=
=
Ð - tan -1 (wRC )
2
V 1 /( jwC ) + R 1 + jwRC
1 + (wRC )
Letw B = 1 / RC
H ( jw ) =
and
1
w
1+ j
wB
f B = 1 / 2pRC
1
=
1+ (
w 2
)
wB
Ð - tan -1 (
w
)
wB
1
H ( jw B ) =
Ð - tan -1 (1)
2
+
V
-
R
+
VC
C
-
æ | H ( jw B ) | ö
æ 1 ö
÷÷ = 20 log10 ç
20 log10 çç
÷ = -3dB
è 2ø
è | H (0) | ø
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
11
First-Order Highpass Filter
H ( jw ) =
VR
R
jwRC
wRC
ép
ù
=
=
=
Ð - ê - tan -1 (wRC )ú
2
V 1 /( jwC ) + R 1 + jwRC
ë2
û
1 + (wRC )
Letw B = 1 / RC
and
f B = 1 / 2pRC
w
wB
w
ép
æ w öù
wB
÷÷ú
H ( jw ) =
=
Ð - ê - tan -1 çç
w
2
w
è wB øû
ë
1+ j
1 + ( )2
wB
wB
H ( jw B ) =
1
Ð - tan -1 (1)
2
æ | H ( jw B ) | ö
æ 1 ö
÷÷ = 20 log10 ç
20 log10 çç
÷ = -3dB
è 2ø
è | H ( 0) | ø
EE40 Summer 2006: Lecture 7
VR
+
V
-
Instructor: Octavian Florescu
R
+
VC
C
-
12
6
First-Order Lowpass Filter
H ( jw ) =
VR
=K
V
VR
+
V
-
EE40 Summer 2006: Lecture 7
R
+
VL
L
-
Instructor: Octavian Florescu
13
First-Order Highpass Filter
H ( jw ) =
VL
=K
V
VR
+
V
-
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
R
+
VL
L
-
14
7
First-Order Filter Circuits
High Pass
VS
+
–
R
C
Low Pass
Low
Pass
VS
+
–
R
L
High
Pass
HR = R / (R + 1/jwC)
HR = R / (R + jw L)
HC = (1/jw C) / (R + 1/jwC)
HL = jw L / (R + jwL)
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
15
Gain or Loss Expressed in Decibels
The gain produced by an amplifier or the loss of a filter
is often specified in decibels.
The input voltage (current, or power) is taken as the
reference value of voltage (current, or power) in the
decibel defining expression:
Voltage gain in dB = 20 log10(Voutput/Vinput)
Current gain in dB = 20 log10(Ioutput/Iinput
Power gain in dB = 10 log10(Poutput/Pinput)
Example: The voltage gain of an amplifier whose input
is 0.2 mV and whose output is 0.5 V is
20log10(0.5/0.2x10-3) = 68 dB.
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
16
8
Change of Voltage or Current
with a Change of Frequency
One may wish to specify the change of a quantity
such as the output voltage of a filter when the
frequency changes by a factor of 2 (an octave) or 10
(a decade).
For example, a single-stage RC low-pass filter has at
frequencies above w = 1/RC an output that changes
at the rate -20dB per decade.
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
17
Bode Plot
n
Plot of magnitude of transfer function vs.
frequency
¨ y-axis
is the 20log of the magnitude of the
transfer function in dB and x-axis is ω
¨ Both x and y axes are in log scale
n
Plot of phase of transfer function vs.
frequency
¨ y-axis
is the phase of transfer function in
degrees and x-axis is ω
¨ y-axis is linear and x-axis is in log scale
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
18
9
Bode Plot Example
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
19
Magnitude Plot of a Low Pass Filter
H ( jw ) =
1
1+ (
w 2
)
wB
Ð - tan -1 (
w
)
wB
Note that :
| H (0) |= 1 and
1
| H ( jw B ) |=
2
lim(| H ( jw ) |) = 0
¥
ω ->
wB
w
For w << w B , | H ( jw ) |» 1
For w >> w B , | H ( jw ) |»
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
20
10
Magnitude Plot…
Remember we are plotting 20log(|H(jω )|)
20 log(| H ( jw B ) |) = -3dB
For w << w B , | H ( jw ) |» 1
so 20 log(| H ( jw ) |) = 0dB
w
For w >> w B , | H ( jw ) |» B
w
so 20 log(| H ( jw ) |) = 20 log(w B ) - 20 log(w )
constant
EE40 Summer 2006: Lecture 7
Corresponds to a slope
of -20dB/decade
Instructor: Octavian Florescu
21
Magnitude Plot
20log(|H(jω )|)
ω
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
22
11
Phase Plot of a Low Pass Filter
1
H ( jw ) =
1+ (
w 2
)
wB
Ð - tan -1 (
w
)
wB
Note that :
ÐH (0) = 0 o
ÐH ( jw B ) = -45o
lim ÐH ( jw ) = -90 o
ω ->
¥
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
23
Magnitude Plot
H(jω )
ω
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
24
12
Bode Plot of High Pass Filter…
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
25
Second Order Circuits
1
L
R
w02 =remember
z ,0 = w2 that,=
z
Also
C
EE40 Summer 2006: Lecture 7
L
R
C
2L
Instructor: Octavian Florescu
26
13
Magnitude Response For ζ < 1
1
1 - w LC + jwRC
1
1
| H ( jw ) |=
=
2
2
2
2
2
1 - w LC + (wRC )
æ w2 ö æ
w ö
ç1 - 2 ÷ + çç 2z
÷
÷
ç w ÷
è wo ø
o ø
è
H ( jw ) =
2
(
)
2
w
1
= o
w 2 LC w 2
so 20 log(| H ( jw ) |) = 40 log(w B ) - 40 log(w )
For w >> wo , | H ( jw ) |»
This corresponds to a
slope of -40dB/decade
For w << wo , | H ( jw ) |» 1
so 20 log(| H ( jw ) |) = 0dB
Note that :
| H ( jwo ) |=
1
2z
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
27
Magnitude Plot For ζ < 1
H(jω )
ω
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
28
14
Phase Response For ζ < 1
2
w
1
= o2
2
w LC w
For w << wo , | H (w ) |» 1
For w >> wo , | H (w ) |»
æ
w
ç 2z
wo
æ wRC ö
-1 ç
ÐH ( jw ) = - tan -1 ç
÷ = - tan ç
2
w2
è 1 - w LC ø
çç 1 - 2
è wo
Note that :
ÐH (0) = 0o
and
ö
÷
÷
÷
÷÷
ø
lim(ÐH ( jw )) = -180o
ω ->
ÐH ( jwo ) = -90 o
¥
Approximations for z £ 1 :
ÐH ( j 0.1wo ) @ 0o
and
EE40 Summer 2006: Lecture 7
ÐH ( j10wo ) @ -180 o
Instructor: Octavian Florescu
29
Phase Response For ζ < 1 Continued
H(jω )
ω
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
30
15
Bode Plot For ζ >= 1
n
For ζ >= 1, we can know that we factor H(jω ) in the
following form:
H ( jw ) =
1
æ w öæ
w ö
çç1 + ÷÷çç1 + ÷÷
è w1 øè w 2 ø
where ω 1 = ω 2 if ζ = 1
n
We shall find the bode plot of this transfer function
when we talk about higher order filters.
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
31
Higher Order Filters
n
Higher order filters can be factorized into a product
of 1st and 2nd order filters
n
How do we account for the complete response?
¨
Simply add all individual responses to get the overall
response
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
32
16
Addition of responses
H ( jw ) =
2
æ jw öæ
jw öæç
jw æ jw ö ö÷
÷÷ K
çç
÷÷çç1 +
÷÷ 1 + 2z
+ çç
wo1 è wo1 ø ÷
è wB1 øè wB 2 øçè
ø
2
æ jw öæ
jw öæç
jw æ jw ö ö÷
çç
÷÷çç1 +
÷÷ K
÷÷ 1 + 2z
+ çç
ç
wo 2 è wo 2 ø ÷
è wB 3 øè wB 4 øè
ø
2
æ
æ jw ö
æ
jw ö
jw æ jw ö ö÷
÷÷ |
÷÷ | +20 log | çç1 +
÷÷ | +20 log | ç1 + 2z
20 log(| H ( jw ) |) = 20 log | çç
+ çç
ç
wo1 è wo1 ø ÷
è wB1 ø
è wB 2 ø
è
ø
2
æ
æ jw ö
æ
jw ö
jw æ jw ö ö÷
÷÷ | -20 log | çç1 +
÷÷
÷÷ | -20 log | ç1 + 2z
- 20 log | çç
+ çç
ç
wo 2 è wo 2 ø ÷
è wB 4 ø
è wB 3 ø
è
ø
2
æ
æ jw ö
æ
jw ö
jw æ jw ö ö÷
÷÷
÷÷ + Ðçç1 +
÷÷ + Ðç1 + 2z
ÐH ( jw ) = Ðçç
+ çç
ç
wo1 è wo1 ø ÷
è wB1 ø
è wB 2 ø
è
ø
2
æ
æ jw ö
æ
jw ö
jw æ jw ö ö÷
÷÷ - Ðçç1 +
÷÷
÷÷ - Ðç1 + 2z
- Ðçç
+ çç
ç
wo 2 è wo 2 ø ÷
è wB 4 ø
è wB 3 ø
è
ø
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
33
Magnitude Plot
20log(|H(jω )|)
ω
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
34
17
Magnitude Plot
H(jω )
ω
EE40 Summer 2006: Lecture 7
Instructor: Octavian Florescu
35
More on Second-Order Filter Circuits
Band Pass
R
VS
+
–
Z = R + 1/jwC + jw L
Low
Pass
C
High
Pass
L
EE40 Summer 2006: Lecture 7
HBP = R / Z
Band HLP = (1/jwC) / Z
Reject
HHP = jw L / Z
HBR = HLP + HHP
Instructor: Octavian Florescu
36
18
Ideal Filters
H( f)
H( f)
Low Pass
Frequency
High Pass
Frequency
H( f)
H( f)
Band Pass
Frequency
EE40 Summer 2006: Lecture 7
Band Reject
Frequency
Instructor: Octavian Florescu
37
19