Transcript Lecture 4 - University of California, Berkeley

EECS 105 Fall 2003, Lecture 4
Lecture 4: Resonance
Prof. Niknejad
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Lecture Outline


Some comments/questions about Bode plots
Second order circuits:
–
–
–
Series impedance and resonance
Voltage transfer function (bandpass filter)
Bode plots for second order circuits
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Good Questions…

Why does the Bode plot of a simple pole or zero
always have a slope of 20 dB/dec regardless of the
break frequency?

You’ve been sloppy with signs, what’s the deal?

Why does the arctangent plot look so funny?

Why do we factor the transfer function into terms
involving jω?
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Bode Plot Question #1

Note that the slope of a pole or zero is independent
of the break point:
20 log 1  j  20 log   20 log   20 log 
This term is a constant

On a log scale, this term has
a fixed slope of 20 dB/decade
On a log-log scale, all straight lines have the same
slope … the slope gets translated into an intercept
shift!
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Bode Plot Question #2

Why do you sometimes use a positive sign and
other times a negative sign in the transfer function?
(1  j z1 )(1  j z 2 )(1  j zn )
H ( )  G0 ( j )
(1  j p 2 )(1  j p 2 ) (1  j pm )
K
Which one is right?



The plus sign is right! For “passive” circuits, the
poles are all in the LHP (left-half plane). A simple
RC circuit has:   RC  0
Otherwise the circuit has a negative resistor!
We can synthesize negative resistance with active
circuits…
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Bode Plot Question #3

I know what an arctan looks like and it looks
nothing like what you showed us!
Linear Scale
Department of EECS
Log Scale
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Log Scales

Log scales move forward non-uniformly…
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Why jω?


When we factor our transfer functions, why do we
always like to put things in terms of jω, as opposed
to say ω?
Recall that we are trying to find the response of a
system to an exponential with imaginary argument:
e


j t
LTI System
H
H ( j) e j (t  )
Real sinusoidal steady-state requires the argument
to be imaginary. We must therefore only consider
the transfer functions for such values …
If this doesn’t make sense, hang in there!
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Second Order Circuits

The series resonant circuit is one of the most
important elementary circuits:

The physics describes not only physical LCR
circuits, but also approximates mechanical
resonance (mass-spring, pendulum, molecular
resonance, microwave cavities, transmission lines,
buildings, bridges, …)
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Series LCR Impedance

With phasor analysis, this circuit is readily
analyzed
Z
Z  jL 
1
j C
R
1 

Z  jL 
 R  R  jL1  2

j C
  LC 
1
1 

Im[ Z ]  L1  2
0
  LC 
Department of EECS
1
 
LC
2
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Resonance



Resonance occurs when the circuit impedance is
purely real
Imaginary components of impedance cancel out
For a series resonant circuit, the current is
maximum at resonance
+ VL – + VC –
VL
+
Vs
+
VR
−
−
VC
VR
Vs
  0
Department of EECS
VL
VL
Vs
VR
Vs
VR
VC
VC
  0
  0
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Series Resonance Voltage Gain

Note that at resonance, the voltage across the
inductor and capacitor can be larger than the
voltage in the resistor:
+ VL –
VL  I j0 L 
+ VC –
+
VR
−
Vs
V
j 0 L  s j 0 L
Z (0 )
R
 jQ  Vs
Vs 0 L
Vs
VC  I

  j 0 L
j0C Z (0 ) j
R
1
  jQ  Vs
0 L
Department of EECS
1 1
LC 1
L 1 Z0
Q




R
0 C R
C R
C R R
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Second Order Transfer Function

So we have:
+
Vo
−

H ( j ) 
V0
R

1
Vs
jL 
R
j C
To find the poles/zeros, let’s put the H in canonical
form:
V0
j CR
H ( j ) 

Vs 1   2 LC  j RC

One zero at DC frequency  can’t conduct DC due
to capacitor
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Poles of 2nd Order Transfer Function

Denominator is a quadratic polynomial:
R
j
L
V0
j CR
H ( j ) 


2
1
R
Vs 1   LC  j RC
2
 ( j )  j
LC
L
R
j
1
2
L


H ( j ) 
0
R
LC
02  ( j ) 2  j
L
j
H ( j ) 
Q
  ( j )  j
2
0
Department of EECS
 0
2
0
Q
Q
0 L
R
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Finding the poles…

Let’s factor the denominator:
( j )  j
2
0
02
0
Q
 02  0
0
1


  
 j0 1 

2Q
4Q
2Q
4Q 



2
0
Poles are complex conjugate frequencies
The Q parameter is called the
“quality-factor” or Q-factor
This is an important parameter:
Im
Re
0
Q R
 
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Resonance without Loss

The transfer function can be parameterized in terms
Im
of loss. First, take the lossless case, R=0:
2
 


2
0
0
  

 0 
  j0

 2Q

4
Q

 Q 



Re
When the circuit is lossless, the poles are at real
frequencies, so the transfer function blows up!
At this resonance frequency, the circuit has zero
imaginary impedance and thus zero total impedance
Even if we set the source equal to zero, the circuit
can have a steady-state response
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Magnitude Response

The response peakiness depends on Q
0 R

j 0
0 L
Q
H ( j ) 

 R

02   2  j 0
02   2  j 0
0 L
Q
j
H ( j0 )  1
Q 1

j
H ( 0)  0
Q  10
H ( j 0 ) 
02
Q
02  02  j0
0
1
Q
Q  100
0
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
How Peaky is it?

Let’s find the points when the transfer function
squared has dropped in half:
2
 0 
  
Q
1
2

H ( j ) 

2
2
 0 
2
2 2
0      
 Q
1
1
2
H ( j ) 

2
2
 02   2 

  1
 0 / Q 


2
  

  1
 0 / Q 
2
0
Department of EECS
2
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Half Power Frequencies (Bandwidth)

We have the following:
2
  

  1
 0 / Q 
2
0
2
 
2
0
02   2
 1
0 / Q
0
Q
  0
ab  0
a b  0
2
 0 
  02   a  b

 
2Q
 4Q 
Take positive frequencies:
     
Department of EECS
Four solutions!
2
0
0
Q
ba
ab  0
a b  0

1

0 Q
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
More “Notation”

Often a second-order transfer function is
characterized by the “damping” factor as opposed
to the “Quality” factor
  ( j )  j
2
0
2
0
1  ( j )  j
2
Q

Q
0

0
1
0
1  ( j ) 2  ( j )2  0
1
Q
2
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Second Order Circuit Bode Plot

Quadratic poles or zeros have the following form:
( j ) 2  ( j )2  1  0
damping ratio

The roots can be parameterized in terms of the
damping ratio:
 1 
( j ) 2  ( j )2  1  (1  j ) 2
Two equal poles
 1 
( j ) 2  ( j )2  1  (1  j 1 )(1  j 2 )
j     2  1
Two real poles
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Bode Plot: Damped Case

The case of ζ >1 and ζ =1 is a simple generalization
of simple poles (zeros). In the case that ζ >1, the
poles (zeros) are at distinct frequencies. For ζ =1,
the poles are at the same real frequency:
 1 
( j ) 2  ( j )2  1  (1  j ) 2
(1  j )  1  j
2
2
Asymptotic
Slope is 40 dB/dec
20 log 1  j  40 log 1  j
2
(1  j ) 2  1  j   1  j   21  j 
Asymptotic Phase Shift is 180°
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Underdamped Case

For ζ <1, the poles are complex conjugates:
( j ) 2  ( j )2  1  0
j     2  1    j 1   2


For ωτ << 1, this quadratic is negligible (0dB)
For ωτ >> 1, we can simplify:
20 log ( j ) 2  ( j )2  1  20 log ( j ) 2  40 log 

In the transition region ωτ ~ 1, things are tricky!
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Underdamped Mag Plot
ζ=0.01
ζ=0.1
ζ=0.2
ζ=0.4
ζ=0.6
ζ=0.8
ζ=1
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Underdamped Phase

The phase for the quadratic factor is given by:
 2 

 ( j )  ( j )2  1  tan 
2 
 1  ( ) 





2

1
For ωτ < 1, the phase shift is less than 90°
For ωτ = 1, the phase shift is exactly 90°
For ωτ > 1, the argument is negative so the phase
shift is above 90° and approaches 180°
Key point: argument shifts sign around resonance
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Phase Bode Plot
ζ=1
Department of EECS
ζ=0.01
0.1
0.2
0.4
0.6
0.8
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Bode Plot Guidelines

In the transition region, note that at the breakpoint:
1
( j )  ( j )2  1  ( j )  ( j )2  1  2 
Q
From this you can estimate the peakiness in the magnitude
response
Example: for ζ=0.1, the Bode magnitude plot peaks by 20
log(5) ~14 dB
The phase is much more difficult. Note for ζ=0, the phase
response is a step function
For ζ=1, the phase is two real poles at a fixed frequency
For 0<ζ<1, the plot should go somewhere in between!
2





Department of EECS
2
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Energy Storage in “Tank”

At resonance, the energy stored in the inductor and
capacitor are
1
1 2
2
wL  L(i (t ))  LI M cos 2 0t
2
2
2
1
1 1

2
wC  C (v(t ))  C   i ( )d 
2
2 C

2
1
I M2
1
I
2
M
 C 2 2 sin 2 0t 
sin
0 t
2
2 0 C
2 0 C
1 2
1
1 2
2
2
ws  wL  wC  I M ( L cos 0t  2 sin 0t )  I M L
2
0 C
2
1
WL ,max  WS  I M2 L
2
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Energy Dissipation in Tank

Energy dissipated per cycle:
wD  P  T 

1 2
2
IM R 
2
0
The ratio of the energy stored to the energy
dissipated is thus:
1 2
LI M
wS
0 L 1
Q
2



wD 1 I 2 R  2
R 2 2
M
2
0
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
Physical Interpretation of Q-Factor

For the series resonant circuit we have related the Q factor
to very fundamental properties of the tank:
wS
Q  2
wD



The tank quality factor relates how much energy is stored in
a tank to how much energy loss is occurring.
If Q >> 1, then the tank pretty much runs itself … even if
you turn off the source, the tank will continue to oscillate
for several cycles (on the order of Q cycles)
Mechanical resonators can be fabricated with extremely
high Q
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 4
Prof. A. Niknejad
thin-Film Bulk Acoustic Resonator (FBAR)
RF MEMS



Agilent Technologies(IEEE ISSCC 2001)
Q > 1000
Pad
Resonates at 1.9 GHz
Thin Piezoelectric Film

Can use it to build low power oscillator
C1
C0
Cx
Department of EECS
R0
Rx
C2
Lx
University of California, Berkeley