7/30/2019 604-tutorial2

1/4

Astronomy Statistics and Measurement - Part II: Measurement

Tutorial question sheet 2

Adam L. Woodcraft

Home page:http://woodcraft.lowtemp.org

1. a. You are probably familiar with images like this from the Hubble Wide Field/Planetary

camera. The reason for the odd shape is that the instrument is made from a mosaic of 4

CCDs, each with the same number of pixels, but a rather complex optical system arranged

so that one of them (the planetary array) has a different plate scale to the other three (the

wide field arrays), and therefore has a higher resolution. What compromise must have

been made with the wide field CCDs? In what way do they fail to fully take advantage of

the abilities of the telescope?

Image courtesy of NASA

b. The planetary array has a focal ratio of f/28.3. What is the focal length corresponding

to this focal ratio? (Primary mirror diameter is 2.4 m.) The spacecraft is 13 m long. Is

there a problem?

c. Each pixel subtends 46 milliarcseconds on the sky. How big is a pixel?

d. To make an image which is limited by the telescope resolution, it is in fact necessary

to have two pixel widths within an Airy disc. This is called Nyquist sampling. Are the

pixels in the planetary camera sufficiently small for operation at a wavelength of 500 nm?

e. The focal ratio for the wide field arrays is 12.9, and the pixels have the same di-

mensions as the planetary array. Show that the pixels would be correct size for Nyquist

sampling at a wavelength of approximately 2 m? Could the arrays be used at that wave-

length?

1

7/30/2019 604-tutorial2

2/4

2. In the lectures, I claimed that the reason that the squared term appears in the definition

DQE = (S/N)out(S/N)in

2

, (1)

is that the time t taken to achieve a given signal to noise ratio increases as t2. Show thatthis is true. (Hint: look in your notes from the statistics part of the course)

3. For a long enough exposure, dark current will saturate the wells in a CCD (i.e. fill

them with electrons so there is no space for any more). One method of avoiding this would

be to cool the detector. What else could we do to stop this happening? Why wouldnt we

do this with a camera used for astronomy?

4. a. Modern CCDs can store 100 000 electrons in each well. Taking this to be the

maximum signal that can stored, at what percentage of this maximum signal does the

readout noise become equal to the photon noise? Assume an RQE of 80% and a read-outnoise of 4 electrons.

b. Suppose the CCD has a dark current of 100 electrons/sec at room temperature (300 K).

Using the equation

q exp

Eg2kbT

(2)

from the lecture notes, calculate the number of electrons generated by dark current in a 1

hour exposure if the array is cooled to 200 K. Take kb as 8.62105eV/K, and ignore the

fact that the energy gap will change slightly with temperature. How does the dark noise

compare to the readout noise?

5. a. In an intrinsic semiconductor, photons with energy greater than the energy gap

create electron-hole pairs. Does the same thing happen in an extrinsic semiconductor?

b. In a photoconductor made from a highly doped extrinsic semiconductor, dark current

can flow in the impurity band without electrons having to be raised to the conduction band

by thermal excitation (this is why blocked impurity band detectors are used). Does this

mean that the dark current is independent of temperature? (Hint: look in the bolometer

section of your notes).

6. a. The NEP of a perfect photoconductor (i.e. in which the noise is only due to photon

noise) is given byNEPblip = 2h

Np, (3)

where h is Plancks constant (6.626 1034 J s), is the frequency of the radiation andNp is the number of photons per second. Derive this result from the definition of NEP:

NEP =InoiseS

(4)

(you need to include a factor of 2 which comes from randomness in the generation of

charge carriers).

b. Show that NEPblip can also be expressed as

NEPblip = 2

hcQ

, (5)

2

7/30/2019 604-tutorial2

3/4

where Q is incident power, c is the speed of light and is the wavelength.

c. Tests on a Ge:Be photoconductor show a responsivity S = 10A W1 at a wavelength

of 42 m. An NEP of1.7 1016WHZ1 was measured for a photon flux of7.1 1013

W. How close does this come to being background limited?

7. The bolometers in the SPIRE instrument on the Herschel Space Telescope are NTD

germanium composite spiderweb bolometers, and they have the following properties (ap-

proximately; they vary from pixel to pixel):

Thermal conductivity: G = G0T, where G = 150 pW K1 at 300 mK, and =1.5.

Heat capacity: C= C0T, where C= 0.6 pJK1 at 300 mK, and is assumed to be 1.

a. For the array operating at 360 m, the expected background load (mostly from the tele-

scope) is 3 pW. The heat sink temperature T0 is expected to be 300 mK. What temperatureis the bolometer absorber expected to operate at (neglect the power from current through

the resistor)? (Hint: The power across a thermal link with conductance G(T) is given by

P =TT0

G(T)dT, (6)

where T is the bolometer absorber temperature.)

b. What is the time constant at this temperature? (Assume = C/G)

c. The resistance of the thermistor is given by

R(T) = R0 expTgT

12

, (7)

where R0=100 and Tg=40. What is the thermometer resistance at this temperature?

d. Suppose we instead operated the array bolometers at a heat sink temperature of 100 mK.

Show that the resistance would become approximately 500 M. This is too large for con-venient operation. What would we change about the thermistor to reduce the resistance,

without changing its size?

e. The arrays operating at different wavelengths in the instrument have different back-

ground loadings. What would we change in the design of the different arrays to ensure

that they all operate at a similar absorber temperature?

8. Starting from Plancks law:

B(, T) =2h3

c21

eh

kt 1, (8)

show that ifh kbT this reduces to the Rayleigh-Jeans approximation,

B =2kbT

2

c2. (9)

3

7/30/2019 604-tutorial2

4/4

How can you tell that this is a classical approximation?

9. By expanding equation (75) in the lecture notes, show that the only term corresponding

to a frequency less than so, and thus the only useful term for heterodyne detection, isgiven by equation (81).

10. Consider a heterodyne receiver for radio astronomy. The upper sideband corresponds

to signals at an original frequency of 115 GHz and the lower one to 107 GHz.

a. Calculate the frequency of the local oscillator and the intermediate frequency.

b. A spectral line is observed. To determine which sideband it originates from, the ob-

server increases the local oscillator frequency by 100 Hz. The signal moves to a lower

frequency. Which sideband is it from?

11. a. Since 1960, there have been on average around 40 large radio telescopesoperating. Assuming that the average power received by each telescope is 1016 W (areasonable value if we exclude observations of our sun), then what will the total amount

of energy received by the end of this year be?

b. In gamma ray astronomy, photons are observed with energies of 100 Tev. How many

such photons would it take to equal the value you derived above? (1 electron volt (eV) is

1.6 1019 J)

c. Photons with this energy are detected by using purpose-built optical telescopes to ob-

serve the Cerenkov radiation that they produce in the atmosphere. One group has built a

telescope with a 17 m primary diameter. Why do you think that gamma ray astromomerscan afford such a telescope when optical astronomers currently have to make do with 10 m

at best? (Hint: they dont have lots more money)

4