Fundamentals of Radio Interferometry

Adapted from NRAO Synthesis Imaging Summer School Lectures By Rick Perley

Interferometry in Action

Interferometry in Action cont…

Michelson Interferometer

•  The movable mirror is used to adjust the extra path-length so that the split beam is either in or out of phase with the original beam

•  The fringe pattern is

the resulting interference pattern for a monochromatic signal

Beam Pattern Origin

•  Antenna’s response results from coherent phase summation of the electric field at the focus.

•  First null will occur at angle where one extra wavelength of path is added across the full width of the aperture:

θ ~ λ/D

On-axis incidence

Off-axis incidence

The Standard Parabolic Antenna Response

The power response of a uniformly illuminated circular parabolic antenna of 25-meter diameter, at a frequency of 1 GHz.

Getting Better Resolution

•  The 25-meter aperture of a VLA antenna provides insufficient resolution for modern astronomy. –  30 arcminutes at 1.4 GHz, when we want 1 arcsecond or

better! •  Building a bigger telescope is not practical – 1 arcsecond

resolution at λ = 20 cm requires a 40 km aperture. –  The world’s largest fully steerable antenna (operated by the

NRAO at Green Bank, WV) has an aperture of only 100 meters ⇒ 4 times better resolution than a VLA antenna.

•  Building a 40 km wide antenna isn’t feasible so synthesis of the equivalent aperture through combinations of elements is needed

•  ‘Aperture synthesis’, was developed in the 1950s in England and Australia. Martin Ryle (University of Cambridge) earned a Nobel Prize for his contributions.

If source emission is unchanging, there is no need to collect all incoming rays at one time. Imagine sequentially combining pairs of signals. Breaking aperture into N sub-apertures, there will be N(N-1)/2 pairs to combine. This approach is the basis of aperture synthesis.

Aperture Synthesis – Basic Concept

The Stationary, Monochromatic Interferometer

A small (but finite) frequency width, and no motion. Consider radiation from a small solid angle dΩ, from direction s.

X

s s

An antenna b €

τg = b⋅ s /c

)cos(2 tVV ω=])(cos[1 gtVV τω −=

2/])2cos()[cos(21 gg tVV ωτωωτ −+

2/)]/2cos([2/])cos([ 2121 cVVVVR gc sb ⋅== πυωτ

multiply

average

V is the voltage response to the source

τg is the lag

In Phase: τg = nλ/c

Quadrature Phase: τg = (2n+1)λ/4c

Anti-Phase: τg = (2n+1)λ/2c

Examples of the Signal Multiplications The two input signals are shown in red and blue.

The desired coherence is the average of the product (black trace)

Signal Multiplication, cont. •  Averaged signal is independent of time t but dependent on

lag, τg – a function of direction, and hence on the distribution of the brightness.

•  V denotes the voltage of the signal which depends upon the

source intensity by:

so the term V1V2 is proportional to source intensity, Iν. (measured in Watts.m-2.Hz-2.ster-2).

•  The product depends on antenna areas and electronic gains, but these factors can be calibrated for.

•  To determine the dependence of the response over an

extended object, we integrate over solid angle. •  Provided there is no spatial coherence between emission from

different directions, this integration gives a simple result.

€

V ∝ E ∝ I

The ‘Cosine’ Correlator Response •  The response from an extended source is obtained by

integrating the response over the solid angle of the sky: for no frequency dependence and no spatial coherence. Key point: the vector s is a function of direction, so the phase in the cosine is dependent on the angle of arrival. This expression links what we want – the source brightness on the sky) (Iν(s)) – to something we can measure (RC, the interferometer response).

Ω⋅= ∫∫ dcIRC )/2cos()( sbs πνν

A Schematic Illustration

The COS correlator ‘casts’ a sinusoidal fringe pattern of angular scale λ/b (b is antenna separation) radians, onto the sky. The correlator multiplies the source brightness by this wave pattern, and integrates (adds) the result over the sky.

Orientation set by baseline geometry. Fringe separation set by baseline length and wavelength.

- + - + - + - Fringe Sign

λ/B rad.

Source brightness

Odd and Even Functions

•  But Rc, is insufficient – it is only sensitive to the ‘even’ part of the brightness, IE(s).

•  Any real function, I(x,y), can be expressed as the sum of two real functions which have specific symmetries:

An even part: IE(x,y) = (I(x,y) + I(-x,-y))/2 = IE(-x,-y) An odd part: IO(x,y) = (I(x,y) – I(-x,-y))/2 = -IO(-x,-y)

= + I IE IO

Recovering the ‘Odd’ Part: The SIN Correlator

The integration of the cosine response, Rc, over the source brightness is sensitive to only the even part of the brightness:

since the integral of an odd function (IO) with an even function (cos

x) is zero. To recover the ‘odd’ part of the intensity, IO, we need an ‘odd’

coherence pattern. Let us replace the ‘cos’ with ‘sin’ in the integral:

since the integral of an even times an odd function is zero. To

obtain this necessary component, we must make a ‘sine’ pattern.

Ω⋅=Ω⋅= ∫∫∫∫ dcIdcIR EC )/2cos()()/2cos()( sbssbs πνπν

Ω⋅=Ω⋅= ∫∫∫∫ dcIdcIR OS )/2(sin)s()/2(sin)( sbsbs πνπν

Making a SIN Correlator

•  We generate the ‘sine’ pattern by inserting a 90 degree phase shift in one of the signal paths.

X

s s

An antenna b

cg /sb ⋅=τ

)cos(2 tVV ω=])(cos[1 gtVV τω −=

2/])2sin()[sin(21 gg tVV ωτωωτ −+

2/)]/2sin([2/])sin([ 2121 cVVVVR gs sb ⋅== πυωτ

multiply

average

90o

Define the Complex Visibility

We now DEFINE a complex function, V, to be the complex sum of the two independent correlator outputs:

where This gives us a beautiful and useful relationship between the

source brightness, and the response of an interferometer: Although it may not be obvious (yet), this expression can be

inverted to recover I(s) from V(b).

φiSC AeiRRV −=−=

!!"

#$$%

&=

+=

−

C

S

SC

RR

RRA

1

22

tanφ

Ω=−= ⋅−∫∫ dsIiRRV ciSC e /2)()( sbb νπ

ν

Picturing the Visibility

•  The intensity, Iν, is in black, the ‘fringes’ in red. The visibility is the net dark green area.

RC RS

Long Baseline

Short Baseline

•  The Visibility is function of source structure and interferometer baseline.

•  The Visibility is NOT a function of absolute position of antennas

(provided emission is time-invariant, and is located in far field). •  The Visibility is Hermitian: V(u,v) = V*(-u,-v). This is a

consequence of the intensity being a real quantity. •  There is a unique relation between any source brightness

function, and the visibility function. •  Each observation of the source with a given baseline length

provides one measure of the visibility. •  Sufficient knowledge of visibility function (as derived from an

interferometer) will provide estimate of the source brightness.

Comments on the Visibility

Examples of Visibility Functions •  Top row: 1-dimensional even brightness distributions. •  Bottom row: The corresponding real, even, visibility functions.

Geometry – the perfect, and not-so-perfect

To give better understanding, we now specify the geometry. Case A: A 2-dimensional measurement plane. Imagine the measurements of Vν(b) taken entirely on a plane. A considerable simplification occurs if we arrange the coordinate system so one axis is normal to this plane. Let (u,v,w) be coordinate axes, with w normal to plane and distances measured in wavelengths. The components of unit direction vector, s, are: and for the solid angle

€

s = l,m,n( ) = l,m, 1− l2 −m2( )221 mldldmd −−=Ω

Direction Cosines

The unit direction vector s is defined by its projections on the (u,v,w) axes. These components are called the Direction Cosines.

221)cos(

)cos()cos(

mln

ml

−−==

=

=

γ

β

α

The baseline vector b is specified by its coordinates (u,v,w) (measured in wavelengths).

)0,,(),,( vuwvu λλλλλ ==b

u

v

w

s

α β

γ

l m b

n

The 2-d Fourier Transform Relation

Then, νb.s/c = ul + vm + wn = ul + vm, from which we find,

which is a 2-dimensional Fourier transform between the projected

brightness: and the spatial coherence function (visibility): Vν(u,v). And we can now rely on a century of effort by mathematicians on how

to invert this equation, and how much information we need to obtain an image of sufficient quality. Formally,

With enough measures of V, we can derive I.

dldmemlmlIvuV vmuli∫∫ +−

−−= )(2

221),(),( πυ

ν

)cos(/ γνI

dvduevuVmlI vmuli∫∫ += )(2),()cos(),( πνν γ

•  Which interferometers can use this special geometry? a) Those whose baselines, over time, lie on a plane (any plane).

All E-W interferometers are in this group. For these, the w-coordinate points to the NCP.

– WSRT (Westerbork Synthesis Radio Telescope) – ATCA (Australia Telescope Compact Array) – Cambridge 5km telescope (almost).

b) Any coplanar array, at a single instance of time. – VLA or GMRT in snapshot (single short observation)

mode. •  What's the ‘downside’ of this geometry?

–  Full resolution is obtained only for observations that are in the w-direction. Observations at other directions lose resolution.

•  E-W interferometers have no N-S resolution for observations at the celestial equator!!!

•  A VLA snapshot of a source will have no ‘vertical’ resolution for objects on the horizon.

Interferometers with 2-d Geometry

VLA Coordinate System

w points to the source, u towards the east, and v towards the NCP. The direction cosines l and m then increase to the east and north, respectively.

b s0 s0

22 vu +w

v

dldmemlmlIvuV vmuli∫∫ +−

−−= )(2

221),(),( πυ

ν

We have shown that under certain (and attainable) assumptions about electronic linearity and narrow bandwidth, a complex interferometer measures the visibility, or complex coherence:

(u,v) are the projected baseline coordinates, measured in wavelengths, on a plane oriented facing the phase center, and (l,m) are the sines of the angles between the phase center and the emission, in the EW and NS directions, respectively.

Making Images

vu)vu,()cos() )vu(2 ddeV(l,mI mli∫∫ ++= πν γ

This is a Fourier transform relation, and it can be in general be solved, to give:

This relationship presumes knowledge of V(u,v) for all values of u and v. In fact, we have a finite number, N, measures of the visibility, so to obtain an image, the integrals are replaced with a sum:

vu)]vu(2exp[)v,u(V1

)1

n ΔΔ+= ∑=

mliN

(l,mI nn

N

nnn πν

If we have Nv visibilities, and Nm cells in the image, we have ~NvNm calculations to perform – a number that can exceed 1012!

Making Images

•  The sum on the last page is in general complex, while the sky brightness is real. What’s wrong?

•  In fact, each measured visibility represents two visibilities, since V(-u,-v) = V*(u,v).

•  This is because interchanging two antennas leaves Rc unchanged, but changes the sign of Rs.

•  Mathematically, as the sky is real, the visibility must be Hermitian.

•  So we can modify the sum to read:

vu])vu(2cos[A1

)1

ΔΔ++= ∑=

nnn

N

nn ml

N(l,mI φπν

But Images are Real

•  Each cosine represents a two-dimensional sinusoidal function in the image, with unit amplitude, and orientation given by: α = tan-1(u/v).

•  The cosinusoidal sea on the image plane is multiplied by the visibility amplitude An, and a shifted by the visibility phase φn.

•  Each individual measurement adds a (shifted and amplified) cosinusoid to the image.

•  The basic (raw, or dirty) map is the result of this summation process.

•  The actual imaging is performed using FFTs

Interpretation

+

-

+ - +

l

m

The rectangle below represents a piece of sky. The solid red lines are the maxima of the sinusoids, the dashed lines their minima. Two visibilities are shown, each with phase zero.

α

A simple example

For a 1 Jy point source, all visibility amplitudes are 1 Jy, and all phases are zero. The lower panel shows the response when visibilities from 21 equally-spaced baselines are added.

The individual visibilities are shown in the top panel. Their (incremental) sums are shown in the lower panel.

1-d Example: Point-Source

•  For a centered square object, the visibility amplitudes decline with increasing baseline, and the phases are all zero or 180.

•  Again, 21 baselines are included.

Example 2: Square Source

Real interferometers must accept a range of frequencies (among other things, there is no power in an infinitesimal bandwidth)! So consider the response of an interferometer over frequency.

First define the frequency response functions, G(ν), as amplitude and phase variation of the signals paths over frequency.

Then integrate:

The Effect of Bandwidth.

νν

ντπνν

ννν devGvGIV gi2*

21

2

2

)()()(1∫Δ+

Δ−Δ= s

G

ν ν0

Δν

The Effect of Bandwidth

If the source intensity does not vary over frequency width, we get

where the G(ν) are assumed to be square, real, and of

width Δν. The frequency ν0 is the mean frequency within the bandwidth.

The fringe attenuation function, sinc(x), is defined as:

ΩΔ= −

∫∫ dIV gig e τνπ

ν ντ 02)(csin)(s

6)(1

)sin()(csin

2xxxx

πππ

−≈

=

for x << 1

The Bandwidth/FOV limit

Source emission is attenuated by function sinc(x), known as ‘fringe-washing’ function. Noting that τg ~ (b/c) sin(θ) ~ bθ/λν ~ (θ/θres)/ν, attenuation is small when

The ratio Δν/ν is the fractional bandwidth. The ratio θ/θres is the source offset in units of the fringe separation, λ/b. In words: the attenuation is small if the fractional bandwidth times the angular offset in resolution units is less than unity. Significant attenuation of the measured visibility is to be expected if the source offset is comparable to the interferometer resolution divided by the fractional bandwidth.

1<<Δ

resθθ

νν

Bandwidth Effect Example Finite Bandwidth causes loss of coherence at large angles,

because the amplitude of the interferometer fringes are reduced with increasing angle from the delay center.

1/

=Δ

Bλθ

νν

•  The trivial solution is to avoid observing large objects! (Not helpful).

•  Although there are computational methods which allow recovery of the lost amplitude, the loss in SNR is unavoidable.

•  The simple solution is to observe with a small bandwidth. But this causes loss of sensitivity.

•  So, the best (but not cheapest!) solution is to observe with LOTS of narrow channels.

•  Modern correlators will provide tens to hundreds of thousands of channels of appropriate width.

Avoiding Bandwidth Losses

There are still more details…

•  Another consequence of observing with finite bandwidth is that sensitivity of the interferometer is not uniform over the sky.

•  Because internal electronics of interferometer don’t work at the observed frequency, other complications must be worked out.

•  We have assumed everywhere that the values of the visibility are obtained ‘instantaneously’. This is of course not reasonable, for we must average over a finite time interval.

•  Time averaging, if continued too long, will cause loss of measured coherence which is analogous to bandwidth smearing.

Ninth Synthesis Imaging Summer School Socorro, June 15-22, 2004

ALMA: Atacama Large Millimeter Array ALMA will bring to millimeter and sub-millimeter astronomy the aperture synthesis techniques of

radio astronomy enabling precision imaging on sub-arcsecond angular scales. The emission at

millimeter wave-lengths is provided by thermal emission from cool gas, dust, and solid bodies, the

same material that shines brightly at far infrared wavelengths. Presently, such natural cosmic

emission can be studied only from space with the coarse angular resolution and limited sensitivity.

ALMA will image at 1 mm wavelength with a

resolution of 0.01" the same that will be achieved by the Next Generation Space Telescope. It will

complement VLT providing the same image detail and clarity. The reconfigurability of ALMA antennas

gives ALMA a zoom-lens capability allowing it to make high-fidelity images of large regions of the sky.

ALMA

•  Up to 80 high-precision antennas, in the Chilean Andes District of San Pedro de Atacama, 5000 m above sea level. "

•  Wavelengths of 0.3 to 9.6 millimeters, where the Earth’s atmosphere above a high, dry site is largely transparent.

•  12 m antennas with reconfigurable baselines ranging from 15 m to 18 km.

•  Resolutions as fine as 0.005" will be achieved at the highest frequencies, a factor of ten better than the Hubble Space Telescope.

“What are the basic properties of the fundamental particles and forces?”

Neutrinos, Magnetic Fields, Gravity, Gravitational Waves, Dark Energy

“What constitutes the missing mass of the Universe?” Cold Dark Matter (e.g. via lensing), Dark Energy, Hot Dark Matter (neutrinos)

“What is the origin of the Universe and the observed structure and how did it evolve?”

Atomic hydrogen, epoch of reionization, magnetic fields, star-formation history……

“How do planetary systems form and evolve?”

Movies of Planet Formation, Astrobiology, Radio flares from exo-planets……

“Has life existed elsewhere in the Universe, and does it exist elsewhere now?”

SETI

CORNERSTONE OBSERVATORIES: ALMA, JWST, SKA, AND ELT

SKA: Fundamental questions

Top priorities for a new generation radio telescope

•  Detect and image neutral hydrogen in the very early phases of the universe when the first stars and galaxies appeared

– “epoch of re-ionisation” •  Locate 1 billion galaxies via their neutral hydrogen

signature and measure their distribution in space – “dark energy”

•  Origin and evolution of cosmic magnetic fields

– “the magnetic universe” •  Time pulsars to test description of gravity in the strong

field case (pulsar-Black Hole binaries), and to detect gravitational waves

•  Planet formation – image Earth-sized gaps in proto-

planetary disks

What instrument do we need? A radio telescope with •  sensitivity to detect and image atomic hydrogen at the

edge of the universe à very large collecting area •  fast surveying capability over the whole sky à very large

angle field of view •  capability for detailed imaging of the structures of the

planetary gaps and how they change à large physical extent

•  a wide frequency range to handle the science priorities

SQUARE KILOMETRE ARRAY

Square Kilometer Array

Ninth Synthesis Imaging Summer School Socorro, June 15-22, 2004

Optical Astronomy Challenges

•  Always aiming to increase sensitivity and resolution

•  Interferometry can be done in the optical but it is hard

•  Atmosphere poses challenge

•  Time domain astronomy

Radio vs. Optical

•  VLA – 27 antennae Bmax ~ 5.2 Mλ at 44 GHz

•  NPOI (Navy prototype optical interferometer) – 6 antennae Bmax ~ 967 Mλ at 667 THz

Radio vs. Optical

P.Napier, Synthesis Summer School, 18 June 2002 3

eg. VLA observing at 4.8 GHz (C band)

Interferometer Block Diagram

Antenna

Front End

IF

Back End

Correlator

Radio vs. Optical

•  Baseline ~ 3x104 m •  Wavelength ~ 1x10-2 m •  Integration time ~ 6x102 s •  Spatial coherence scale ~ 3x106 waves

•  Baseline ~ 3x102 m •  Wavelength ~ 1x10-6 m •  Integration time ~ 1x10-2 s •  Spatial coherence scale ~ 1x105 waves

Coherence Volume r02 t0:

Radio: 5.4x1015 Optical: 1x108 (normalized)

Factor of ~5x107 advantage for radio over optical interferometry

Fundamental Differences – Radio & Optical

•  Temporal coherence of atmosphere – t0 –  Minutes (radio) vs. milliseconds (optical)

•  Spatial coherence of atmosphere – r0 –  Kilometers (radio) vs. centimeters (optical)

•  Coherence function of the fields –  Radio -- Direct measurement of amplitude and phase –  Optical -- No direct measurement of either