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Basics of Fourier transform
)()( Ttftf Periodic function
T
dttf0
)(
T
k
T
k
T
kkkk
dttktfT
b
dttktfT
a
dttfT
a
tkbtkaatf
0
0
0
0
0
0
01
01
0
)sin()(2
;)cos()(2
;)(1
);sin()cos()(
0)2
sin()2
cos(
)2
sin()2
sin(2
)2
cos()2
cos(2
0
0
0
dttT
mtT
k
dttT
mtT
kT
dttT
mtT
kT
T
km
T
km
T
T
2 0
k
kk
kkk
kkk
a
btg
bac
tkcatf
)(
);cos()(
22
01
0
The ‘vectors’ are good orthonormal base for every finite energy signal
)( dttf
)()(2
1)),((
2 ;)()(
1- tfdtet
dtetf
tj
tj
)sin()cos( tjte tj
f(t) must be limited: physical signals carrylimited amounts of energy
FOURIER TRANSFORM of a non-periodic function
dtttOjdtttE )sin()()cos()()(
By definition of Dirac’s delta dtee tjtj )'( '
tje
f(t) must have a finite number of discontinuities
f(t) can be split in its even E(t) and odd O(t) components:
2
)()()(
;2
)()()(
tftftO
tftftE
Hp.
Fourier transform
InverseFourier transform
Periodic vs non- periodic functions: Fourier spectrum
Time domain Frequency domain
Periodic
Not periodic
2 4 6 8 10
0 .05
0 .10
0 .15
0 .20
0 .25
0 .30
0 .35
||tet
4 2 0 2 4 6 8 10
0 .1
0 .2
0 .3
0 .4
|F|
ω
0 2 4 6 8 10
0 .5
1 .0
1 .5
2 .0
2 1 1 2
1 .0
0 .5
0 .5
1 .0
|F|
)()( ])([
;)()(
)()(
)()(])([
)(
)(
gfgf
tdgedfe
dttgedfe
dtdtgfegf
jj
tjj
ttj
)(g
It is much easier to describe the transfer function of sequential ‘blocks’ in term of frequency response: only at the last step the time behavior time domain is inferred
Convolution and Fourier transform
ω→t
dtgftgf )()()]([
Time domain Frequency domain
The integral in time domain turns intosimpler algebraic product in the Fourier frequency domain.
)( f))](()([ fg
dtfgtfg )()()]([
Frequency domain Time domain
So, if this means a 3dB gain, if the attenuation is -20dB in the signal
Logarithmic attenuation and gain ratios
Attenuation and gains relative to voltage (V), current (I), power (P), but also pressureand other physical quantities are usually measured as adimensional ratios towards areference value of the measured quantity, let us say a voltage V0, current I0, power P0
By definition:
010log20
V
VdBX
20
V
V1.0
0
V
V
20 dB every decade of gain
-20 dB every decade of attenuation
2
1
0
V
VOtherwise, a -6dB attenuation means 20
dB)(in
0
10X
V
V
If we deal with power 22 or IVP
0102
0
2
100
10 log20log10log10 V
V
V
V
P
PdBY m
Bode’s diagrams
If we want to study the behavior of a ‘block’ in the frequency domain responding to sinusoidal stimulus we define:
)(
)0(0
VV
VV
RCjarctg
RCjj
V
V
in
out
)(
)(1
12
Example: RC networkR
C
VinVout
RC Time constant (TC)
ω→0
0.707
0 .1 1 10 10 0 10 00
0 .005
0 .010
0 .050
0 .100
0 .500
1 .000
→20 dB
1 decade
-3 dB
0 .1 1 10 10 0
80
60
40
20
0
ωωt ωt
ωt: roll-off freq
BW
IDEAL Frequency/istantaneous signal mixer
)(tS
)(tR
)()()( tRtStM
Signal
‘Reference’
Some simplifications without loss of generality:
•S(t) and R(t) can be choosen periodic, even with different pulsations ωs and ωr
• we can choose <S(t)>=0,<R(t)>=0 (Fourier expansion coefficients a0=0)
)sin()( rrtRtR )sin()( sstStS
LOCK IN COMPONENTS: SIGNAL MIXER
ak=0 k>0, b1=R ak=0 k>0, b1=S
bk=0 n>1 bk=0 n>1
Mixer output )()()( tRtStM
)sin()sin()( ssrr ttRStM
))cos((2
1))cos((
2
1)( rsrsrsrs tRStRStM
In our particular case
s
r
rs
0 )2cos(2
1cos
2
1)( tRSRStMHp.
DC component AC component
nnn ;22
If ωs≠ ωr the DC component vanishes, no matter about phase lags between the two signals
LOCK IN COMPONENTS: SIGNAL MIXER Frequency Content
The IDEAL low-pass filter (infinite roll-off)
))(())(())(())(()),(( rsrsrsrsRStM
)),(( tM
|F|
ωωrωr-ωs ωs ωr+ωs
ω2ω
0
0 0 2ω
|F||F|
ωt: rolloff frequency
ONLY DC component pass
LP
ωt ω
|A(ω)|
1
cos2
1)( RSM
Provided that φ could be regulated we have singled out the weight of the component in the signal S(t) at the reference frequency ωr
LOCK IN COMPONENTS: LOW-PASS FILTER
After the LP filter only the RMS amplitude of the signal will be extracted, acting like a demodulator
LOCK IN COMPONENTS: Low-pass filter propertiesLOCK IN COMPONENTS:
The LP filter behaves as an integrator for spectral components in the signal with pulsations larger than ωt: this is equivalent to an integration performed up to nTC (up to infinite if roll-off or ‘order’ n of the filter is infinite).The istantaneous mixer output is so integrated to yield by definition self-correlation between the reference and the input signal
dtttnTC
RSR
nTC
nTC
0
)sin()sin(1
2 lim)(
2pA
Signal switched on at t=0
0 .5 1 .0 1 .5 2 .0
0 .5
1 .0
1 .5
2 .0pA
C R
Rectifier
TC=RC
LOCK IN COMPONENTS: The phase-sensitive detector (PSD)
LP
ωt ω
|A(ω)|
1
)(tS
)(tR cos2
1)( RSM r
Given a reference R(t) with proper pulsation ωr, whatever the form of the input signalS(t), the DC output component of this block will depend only on the weight of thespectral component of the signal at ωr, apart relative phase lag φ.
RMS value of the stimulus is known, then φ has to be tuned to find the max rms value of the signal , in this respect this block is phase-sensitive
2
R
2
S
0Q
A zero frequency (very narrow) band-pass filter is obtained with:
or S/N ratio (typ. ωr ≈ 1kHz, Δω ≈ 0.01Hz)
0Q
A zero frequency (very narrow) band-pass filter is obtained with:
or S/N ratio (typ. ωr ≈ 1kHz, Δω ≈ 0.01Hz)
Lock-in building blocks: The phase shifter and PLL
LP
ωt ω
|A(ω)|
1
)(tS
)sin()( ttR r
)cos()(' ttR r
Φ
A phase shifting network (Φ) has to be applied to the reference (most common) to maximize output Mx
cos22
SRM X
sin22
SRMY
Mixer 1
Mixer 2
reference
Reference inphase quadrature
X
Y
YX
M
Marctg
MMM
22
Phase of the output signal can be fed back to drive the phase shifter:auto-phase-locking through a Phase Locked Loop (PLL) block
)sin()( ttR r
sin
cos
Mφ
From ideal to a real lockin amplifier
IDEAL mixer carries only ωs± ωr
frequency content
REAL mixer carries ωs± ωr AND ωs, ωr AND image frequencies ωs± 2ωr, ωr± 2ωs
But this is the least problem, due to the following LP filter, cutting off/integratinghigh frequency band of the mixer ouput.
IDEAL LP filter has a cutoff frequency ωt =0 and infiniteroll-off
REAL LP cannot have ωt =0, but only ωt→0, AND roll-off must be finite
LP filter output can NOT be noise immune if ouput spectral BW is zero (ωt =0 and infinite roll-off) also output power is zero!
ωt =0 means infinite integration time or TC: definitely we won’t wait the eternity to read the output
A REAL measuring device will be affected by noise from several sources and in any circuit block: in our case the worst one is the noise at the amplifier input stage
Noise sources
Intrinsic : mainly noise of the input terminals of the amplifier stages
Noise amplitude vs frequency
log(
Vno
ise)
log(f )
1/f noise
0
White noise
0.1 1 10 100 1kHz
Johnson noise
Shot noise
•1/f: ensemble of excitation-deexcitation processes in semi-conductors into environmentalthermal bath, almost independent from amplifier input bandwith BW
•White noise
BWTRkV Bnoise
RMS 4 BWeII RMSnoiseRMS 2
•Johnson noise •Shot noise
Thermal fluctuationsof electron density inany resistor R
Thermal/quantumfluctuations of discretenumber of chargecarriers: e=1.6·10-19C Extrinsic: MUCH more complex
• RF/EMI interferences
• Mains supply lines radiating at 50/100 Hz
• Capacitive/Inductive coupling with surrounding devices
• Ground loopsAre only most common source of external noise
Spectral density hasto be recognized forevery particular set-upof experiment
Lock-in I/O Signal and noise power spectral densities2
log(
Vno
ise)
log(f )0
0.1 1 10 100 1kHz
Input BW
• Typical power spectral density at the input in a ideally good case (no ext noise): colored areas proportional to the power of noise and signal
Noise power
Signal power(at ωr)
SignalBURIEDin noise
• After PSD detector/integrator
2 lo
g(V
nois
e)
log(f )0
0.1 1 10 100 1kHz
LP filterbandwidth
Signal
Noise
LesserLP filter BWHigher TC
HigherS/N ratio
ωt
BUT
If TC is too large (ωt→0)not only noise, but also signal power will be lost !
S/N ratios at low frequencies will be poorer in any case: choose reference/stimulus in the 100 Hz-10 kHz freq. range
Input BW
Shifted to DC
SystemThe ‘classic’ lock-in setup
Signal
ωt ω
|A(ω)|
1
Band-pass FilterAC amplifier
+
Noise
Gac
Mixer
LP / integrator
Φ
Ref. generator
Gdc
DC Output
ABB
C D
Question: how will an unknown system respond to an external harmonic stimulus?
The reference can be generated:
• internally: a built-in oscillator excites the system directly or through transducers
• externally: further device excites the system and a PLL circuit has to drive the built-in oscillator to desired stimulus frequency and compensate phase lags
Phase shifter PLLtracks φ,ωr
0 .2 0 .4 0 .6 0 .8 1 .0
1 .0
0 .5
0 .5
1 .0 ωr
0 .2 0 .4 0 .6 0 .8 1 .0
1 .0
0 .5
0 .5
1 .0 ωr
A B
C D
Hzr 200
r
60 Hz supply noise
Spectral transfer functions of lock-in blocks
The response function A(λ) could not be trivially linear (a), and typically is NOT linear (b),or even resonant (c)
Response of a physical system to a periodic stimulus I
0 .2 0 .4 0 .6 0 .8 1 .0
1 .0
0 .5
0 .5
1 .0 ωr λ
)sin(0 tUU r )sin(),( 0 tt rr
....)sin()sin()()),(( 0 krk
krr tkAtAtA
A(λ)System
A will be modulated with respectto the harmonic stimulus at ωr
BUT
Linear/harmonicterm
non-linear distortion higher order harmonics
Anharmonic terms
(c)
Response of a physical system to a periodic stimulus II
The modulated λ itself could be periodic but not harmonic, like square wave atpulsation ωr: typical of laser beam intercepted by a mechanical chopper wheel
)sin( ;)sin()(0
krkk
krk tkstkt
)(
00 !
1 ;)(
!
1)( m
m
m
m
mm
m
AA
m
A
mA
m
m kkk
m sAm
tA
0 1
)(
!
1))((
Taylor series
Fourier series
problem:this full series expansion is almost unmanageable !
Hp:
•Δλ<<λ0
• λ1 (and λ2) >> λk for k>1(2)
• λ1>λ2
•| A(k)|<<|A(1)|
Small λ modulation
First harmonic ωr (and first overtone 2ωr) dominant
Only lower order terms of Taylorseries will contribute
Discussion of ))(( tA
........)2(2
1 ....)()( 2211
21
21
)2(2211
)1()0( sssAssAAtA
2)()sin(()
2)()sin((
2
1
)sin()sin(
qprqpr
qpqrpr
tqptqp
sstqtp
Remember that
....)22(2
12121212111
2111
21
)2( ssssA
The k-th order spectral weight will be: T
krk dttAtkT
S0
))(()sin(1
........)~~~(2
1
...)~~~~(~))(()sin(
112
1112
1112
1)2(
22221111)1()0(
kkk
kkkkkkr
sssA
ssssAsAtAtk
)21(
2)()21(sin~
2121
krk tksEx.:
Which of these terms will give an output after the PSD/integrator?
Lock-in output vs. mixer harmonics
Most generally N
NrNkr tNCtAtk )sin())(()sin(
ppkN
ppk
m
N kkNm
AC
2
;2!
)(
N: integer sum of harmonics
ΔφN: sum of phases
At the output of PSD/demodulator we have:
Only surviving terms are for N=0 p
pkk
When the base (k=1) harmonic is fed into the reference input mixer weobtain an estimate of the spectral weight S1 at the lock-in output
Spectral weights Sk vs signal derivatives
k=1 The dominant contribution comes from:initial hypotheses discard higher ordercontributions
22)1( ~
nsA
11)1( ~
nsA
112
1)2( ~
2
1
2
1nsA
The output is proportional to the first derivative of the input signal,again provided that modulation of the parameter Δλ<<λ0
k=2
2 terms:
Experiment is modulated at ωr but reference input of the mixer is drivenat 2ωr by means of a frequency doubler/multiplier:
2f detection
Which one will prevail?
1)2()1( AA
21
In most general case the answer is not unique!
Not necessarily S2 is univocally proportional to second derivative only
Peculiar cases of λ coupling
Linear coupling of λ and A:good approx. For Δλ→0
1 , 0 ;)( )()1()0( kAAAA m
)cos(~ )1()1(kkkkkk AsAS
)2sin(422
1
)cos(~
1
21)2(
21)2(
2
11)1(
11)1(
1
AAS
AsAS n
Only first derivative will be detected
Linear coupling of λ and excitation U
....)~~~~(22
1)~~(~
11111111
21)2(
111)1()0( kkkkkkkk ssssAssAsAS
Sk term is proportional mainlyto the k-th derivative
Notice that 2f detection of 2ndderivative requires a -π/2 phase shift with respect to usual1f detection
)sin()( 10 tt r
0 .5 1 .0 1 .5 2 .0 2 .5 3 .0
0 .2
0 .4
0 .6
0 .8
1 .0
Non-linear resonant coupling with λ
Only 2ω component
For Δλ<Γ
)(A)sin()( 10 tt r
22
21
)(
21
1)(
res
A
full width half maximum (FWHM)
)1()0()( AAA
res
The RMS value of the modulation of A(λ), versus amplitude (AM) or frequency (FM) modulation of λ, yields a term proportional to the first derivative of A(λ), if Δλ<Γ
2 4 6 8 10
0 .4
0 .2
0 .2
0 .4
Γ
Example: STS point spectroscopy
Tip
sample
The tip is held at fixeddistance from the sample
Tip or sample bias is scannedwith a linear ramp and I(V) isacquired
Numerical derivative ‘Lock-in’ derivative
MUCH betterS/N ratio!
The bias is modulatedwith a small amplitudevoltage (some mV)
0 .2 0 .4 0 .6 0 .8 1 .0
1 .0
0 .5
0 .5
1 .0