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