# Qs re aspects of the Holmdel Horn Antenna used to find the CMB

• I
Gold Member
I have been trying to understand some trechnical aspects of the Holmdel Horn Antenna used to first confirm CMB. My references are:

Regarding antenna gain

Quote from (1)
Antenna gain is usually defined as the ratio of the power produced by the antenna from a far-field source on the antenna's beam axis to the power produced by a hypothetical lossless isotropic antenna, which is equally sensitive to signals from all directions. Usually this ratio is expressed in decibels, and these units are referred to as "decibels-isotropic" (dBi). An alternative definition compares the received power to the power received by a lossless half-wave dipole antenna, in which case the units are written as dBd. Since a lossless dipole antenna has a gain of 2.15 dBi . . .
(Underlining mine)

Quote from (2)
An isotropic radiator is a theoretical point source of electromagnetic or sound waves which radiates the same intensity of radiation in all directions.

Q1: Can someone please explain: “antenna from a far-field source on the antenna's beam axis”?

Q2: Does the 2.15 dBi mean that a dipole antenna emits or receives
1/102.15 = 0.00708
times the energy that an isotropic radiator antenna would emit or receive?

Regarding technical aspects

Quotes from (3)
In 1964, Arno Penzias and Robert Woodrow Wilson at the Crawford Hill location of Bell Telephone Laboratories in nearby Holmdel Township, New Jersey had built a Dicke radiometer that they intended to use for radio astronomy and satellite communication experiments. On 20 May 1964 they made their first measurement clearly showing the presence of the microwave background, with their instrument having an excess 4.2K antenna temperature which they could not account for.
...
The CMB has a thermal black body spectrum at a temperature of 2.72548±0.00057 K.

Quotes from (4)
Holmdel Photo (1962)

This type of antenna ... consists of a flaring metal horn with a curved reflecting surface mounted in its mouth, at a 45° angle to the long axis of the horn. The reflector is a segment of a parabolic reflector, so the antenna is really a parabolic antenna which is fed off-axis. ... It is extremely broad-band, has calculable aperture efficiency, and the walls of the horn shield it from radiation coming from angles outside the main beam axis. The back and side lobes are therefore so minimal that scarcely any thermal energy is received from the ground. The antenna has a gain of about 43.3 dBi and a beamwidth of about 1.5° at 2.39 GHz and an aperture efficiency of 76%.

Q3: Does the gain of 43.3 dBi (41.41 dBi more than a dipole antenna with 2.15 dBi gain) mean that antenna receives 1/1041.41 = 3.69 x 10-42 times the power of a dipole antenna? I am thinking my guess about this must be wrong because this is such an infinitesimal amount of reception.

Q4: How is it possible that “scarcely any thermal energy is received from the ground”? The ambiant temperature should cause the material of the horn to have the same temperature. Ambiant temperature in NJ should be about 290 K. This is over 100 times as hot as the CMB. Also, the receptive solid angle of the dipole to the CMB (assuming that the receiving antenna is a dipole) is much smaller than the about 4π solid angle from the surrounding horn.

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stefan r
...
Q3: Does the gain of 43.3 dBi (41.41 dBi more than a dipole antenna with 2.15 dBi gain) mean that antenna receives 1/1041.41 = 3.69 x 10-42 times the power of a dipole antenna? I am thinking my guess about this must be wrong because this is such an infinitesimal amount of reception.

You left out the "deci" part of "decibel". 104.141 = 1.38 x 104. Around fourteen thousand times the power when aimed at the source.

Gold Member
Hi stefan:

So I made the same error in Q2.
The 2.15 dBi means that a dipole antenna emits or receives
1/100.215 = 0.785
times the energy that an isotropic radiator antenna would emit or receive. Is that correct?

Regards,
Buzz

davenn
Gold Member
Q1: Can someone please explain: “antenna from a far-field source on the antenna's beam axis”?

google near field and far field attributes of an antenna

here's an easy starting point ....
https://www.bing.com/search?q=near+...&src=IE-SearchBox&FORM=IENTTR&conversationid=

Q2: Does the 2.15 dBi mean that a dipole antenna emits or receives
1/102.15 = 0.00708

a dipole has 2.15 dB gain over an isotropic radiator, almost double, ( 3.0 dB is double) and it is effective for both receive and transmit

Q3: Does the gain of 43.3 dBi (41.41 dBi more than a dipole antenna with 2.15 dBi gain) mean that antenna receives 1/1041.41 = 3.69 x 10-42 times the power of a dipole antenna? I am thinking my guess about this must be wrong because this is such an infinitesimal amount of reception.

you are messing up your numbers .... you are adding "-" into your exponentials eg 10-42 ... dunno why you are doing that it's wrong

it has 43.3 dB gain over a Isotropic radiator which has 0dB gain or 41.15 dB ( NOT dBi) over a dipole.
every 3 db the power is doubled a quick remembering

3 dB = x2
10 dB = x 10
20 dB = x 100
30 dB = x 1000
40 dB = x 10,000

so for rounded figured, that huge horn antenna has a bit less than 10,000 x the gain of a dipole

Hi stefan:

So I made the same error in Q2.
The 2.15 dBi means that a dipole antenna emits or receives
1/100.215 = 0.785
times the energy that an isotropic radiator antenna would emit or receive. Is that correct?

Regards,
Buzz

no, your numbers are still messed up

Dave

Buzz Bloom
Gold Member
a dipole has 2.15 dB gain over an isotropic radiator, almost double
Hi Dave:

Thank you for explaining the dB nomenclature. I use to know this five decades ago, but I have over time mis-remembered it.

So, re Q2: the gain of the dipole of 2.15 dBi means the dipole has reception/radiation of
100.215 = 1.64​
times that of the of the isotropic antenna.

Re Q3: The gain of the Holmdel antenna of 41.41 dB more than the dipole antenna means that the Holmdel antenna receives
104.141 = 13,836​
times the reception/radiation of the dipole antenna.

What now seems confusing is: Geometrically why does a dipole antenna have less reception/radiation than an isotropic antenna? I am thinking in terms of receiving radiation from a wide area, like for example, the CMB. Why wouldn't receiving from all directions acquire more radiation than just from the non-isotropic dipole?

Regards,
Buzz

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Gold Member
I much appreciate the posts re Q2 and Q3 from @stefan r and @davenn. After some thought, I think I now also understand Q1.

The quote “antenna from a far-field source on the antenna's beam axis” means that the dipole antenna has a preferred direction, "the beam axis", while the isotropic antenna does not. Therefore the dBi advantage of the dipole is comparing the radiation/reception from/to a localized source along the beam axis to radiation/reception uniformly with respect all directions.

Now, if you, or anyone, could help me with Q4, I would very much appreciate it.

Regards,
Buzz

davenn
Gold Member
So, re Q2: the gain of the dipole of 2.15 dBi means the dipole has reception/radiation of
100.215 = 1.64times that of the of the isotropic antenna.

yes

Re Q3: The gain of the Holmdel antenna of 41.41 dB more than the dipole antenna means that the Holmdel antenna receives
104.141 = 13,836times the reception/radiation of the dipole antenna.

that seems incorrect
as with the table in my last post, it should be a little shy of 10,000 x that of a dipole

What now seems confusing is: Geometrically why does a dipole antenna have less reception/radiation than an isotropic antenna? I am thinking in terms of receiving radiation from a wide area, like for example, the CMB. Why wouldn't receiving from all directions acquire more radiation than just from the non-isotropic dipole?

because it is not radiating is all directions equally, only an isotropic radiator ( a point source ) can do that, and we cannot make one of them.

A radiation pattern of a dipole is like a doughnut ... there is no signal received/radiated off the end tips

the Z axis is where the dipole is ... that is the dipole is vertical inside the doughnut pattern

Q4: How is it possible that “scarcely any thermal energy is received from the ground”? The ambient temperature should cause the material of the horn to have the same temperature. Ambient temperature in NJ should be about 290 K. This is over 100 times as hot as the CMB. Also, the receptive solid angle of the dipole to the CMB (assuming that the receiving antenna is a dipole) is much smaller than the about 4π solid angle from the surrounding horn.

the shape and angle of the horn means that the receiving antenna down at the narrow end of the horn is well shielded from the ground by the horn.
The actual physical temperature of the antenna isn't an issue in the way you suggest. It adds very little to the overall thermal noise coming into the front of the antenna from the environment

from Wiki
In telecommunication, antenna noise temperature is the temperature of a hypothetical resistor at the input of an ideal noise-free receiver that would generate the same output noise power per unit bandwidth as that at the antenna output at a specified frequency. In other words, antenna noise temperature is a parameter that describes how much noise an antenna produces in a given environment. This temperature is not the physical temperature of the antenna. Moreover, an antenna does not have an intrinsic "antenna temperature" associated with it; rather the temperature depends on its gain pattern and the thermal environment that it is placed in.

Antenna noise temperature has contributions from several sources:

Galactic noise is high below 1000 MHz. At around 150 MHz, it is approximately 1000K. At 2500 MHz, it has leveled off to around 10K.

Earth has an accepted standard temperature of 290K.

The level of the sun's contribution depends on the solar flux. It is given by

T A = 3.468 F λ 2 10 G / 10 {\displaystyle T_{A}=3.468F{{\lambda }^{2}}10^{G/10}}

where F {\displaystyle F}
is the solar flux,
λ {\displaystyle \lambda }
is the wavelength,
and G {\displaystyle G}
is the gain of the antenna in decibels.
The antenna noise temperature depends on antenna coupling to all noise sources in its environment as well as on noise generated within the antenna. That is, in a directional antenna, the portion of the noise source that the antenna's main and side lobes intersect contribute proportionally.

For example, a satellite antenna may not receive noise contribution from the earth in its main lobe, but sidelobes will contribute a portion of the 290K earth noise to its overall noise temperature.

Noise from the ground and environment, in general, is easily dealt with. Most of these radio astronomy systems use a switch between the receiver and the antenna. It measures the noise level coming in the antenna and then switches to a cryogenically cooled dummy load (using nitrogen etc) that is producing maybe only a few Kelvin of noise level. The difference in the noise level between the two is then removed from the recorded data when the antenna is pointing at a galactic or further distance noise source that is being studied. This then gives a true noise level for the object being studied, since all unwanted noise is removed

Dave

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Buzz Bloom
Gold Member
Hi Dave:
that seems incorrect
as with the table in my last post, it should be a little shy of 10,000 x that of a dipole
You quoted my:
Re Q3: The gain of the Holmdel antenna of 41.41 dB more than the dipole antenna means that the Holmdel antenna receives
104.141 = 13,836 times the reception/radiation of the dipole antenna.​
I think you must have confused yourself. The dB value of 4.141 is greater than the 4 value corresponding to 10000 in the table in your post #4.

In particular I am interested in understanding how the horn shields the "Earth heating" noise.
the shape and angle of the horn means that the receiving antenna down at the narrow end of the horn is well shielded from the ground by the horn.
The article
https://treelinebackpacker.com/2013/05/06/calculate-temperature-change-with-elevation/
says
You will lose an average 3.5 degrees Fahrenheit for every 1000 feet of elevation you gain.​
I think this means that the exterior of the horn is being radiated by a 290 K source from all directions. Therefore the horn material will acquire an equilibrium temperature of 270 K. Therefore, the antenna inside the horn will be radiated from all directions with 290 K radiation unless the antenna is protected by a screen maintained at a much lower temperature.

Here is a quote from another article:
https://www.nps.gov/parkhistory/online_books/butowsky5/astro4k.htm
They removed the effects of radar and radio broadcasting, and suppressed interference from the heart in the receiver itself by cooling it with liquid helium to -269°C, only 4° above absolute zero--the temperature at which all motion in atoms and molecules stops.​
What my limited search skills have not been able to find is a description of the geometry of exactly how the liquid helium was configured so that the 290 K radiation became effectively screened from the receiving antenna. Can you help me with this?

Noise from the ground and environment, in general, is easily dealt with. Most of these radio astronomy systems use a switch between the receiver and the antenna. It measures the noise level coming in the antenna and then switches to a cryogenically cooled dummy load (using nitrogen etc) that is producing maybe only a few Kelvin of noise level. The difference in the noise level between the two is then removed from the recorded data when the antenna is pointing at a galactic or further distance noise source that is being studied. This then gives a true noise level for the object being studied, since all unwanted noise is removed
I am not sure I correctly understand this description. This is what I think it means.

(b) A switch is used so that two different radiation signals are measured non-concurrently: (i) The ambient noise alone, and (ii) the ambient noise together with the CMB radiation.
(c) What is measured is the power (watts) of these two signals. The difference between (i) and (ii) is the power of the CMB radiation.
(d) The ambient noise consists of (i) the radiation of the liquid helium, and (ii) the radiation of the external sources not completely screened by the liquid helium.
What is missing from the story is that the antenna must be attached to a tuned circuit so that the power only within a particular bandwidth near a particular frequency is measured. Presumably the choice of peak frequency and the bandwidth would be adjustable.
If this is correct, I would like to understand the geometry.

Regards,
Buzz

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davenn
Gold Member
You quoted my:
Re Q3: The gain of the Holmdel antenna of 41.41 dB more than the dipole antenna means that the Holmdel antenna receives
104.141 = 13,836 times the reception/radiation of the dipole antenna.I think you must have confused yourself. The dB value of 4.141 is greater than the 4 value corresponding to 10000 on the table in your post #4.

sorry my mistake ... I read 13.863 x not 3,863 x period instead of a comma. My eyes aint as good as they used to be

In particular I am interested in understanding how the horn shields the "Earth heating" noise.
In particular I am interested in understanding how the horn shields the "Earth heating" noise. The Earth radiates at a temperature of about 290 K.
The article
https://treelinebackpacker.com/2013/05/06/calculate-temperature-change-with-elevation/
says
You will lose an average 3.5 degrees Fahrenheit for every 1000 feet of elevation you gain.I think this means that the exterior of the horn is being radiated by a 290 K source from all directions. Therefore the horn material will acquire an equilibrium temperature of 270 K. Therefore, the antenna inside the horn will be radiated from all directions with 290 K radiation unless the antenna is protected by a screen maintained at a much lower temperature.

you are still missing the point of my previous post .... this isn't a PHYSICAL temperature thing
so that temp Vs height info is irrelevant .....

here again is my quote from wiki
carefully note the bit bolded in red

In telecommunication, antenna noise temperature is the temperature of a hypothetical resistor at the input of an ideal noise-free receiver that would generate the same output noise power per unit bandwidth as that at the antenna output at a specified frequency. In other words, antenna noise temperature is a parameter that describes how much noise an antenna produces in a given environment. This temperature is not the physical temperature of the antenna. Moreover, an antenna does not have an intrinsic "antenna temperature" associated with it; rather the temperature depends on its gain pattern and the thermal environment that it is placed in.

Antenna noise temperature has contributions from several sources:

Galactic noise is high below 1000 MHz. At around 150 MHz, it is approximately 1000K. At 2500 MHz, it has leveled off to around 10K.

Earth has an accepted standard temperature of 290K.

The level of the sun's contribution depends on the solar flux. It is given by

T A = 3.468 F λ 2 10 G / 10 {\displaystyle T_{A}=3.468F{{\lambda }^{2}}10^{G/10}}

where F {\displaystyle F}
is the solar flux,
λ {\displaystyle \lambda }
is the wavelength,
and G {\displaystyle G}
is the gain of the antenna in decibels.
The antenna noise temperature depends on antenna coupling to all noise sources in its environment as well as on noise generated within the antenna. That is, in a directional antenna, the portion of the noise source that the antenna's main and side lobes intersect contribute proportionally.

For example, a satellite antenna may not receive noise contribution from the earth in its main lobe, but sidelobes will contribute a portion of the 290K earth noise to its overall noise temperature.

when the open end of the horn antenna is pointing at the ground, it will receive LOTS of ground noise but not all of it because the ground is radiating across a very wide spectrum of frequencies and mostly infra-red. Whilst the antenna is designed for a relatively narrow bandwidth of frequencies a LONG WAY from the IR freq range. When the horn is pointing at the sky, it will receive virtually zero ground noise but rather it will receive lots of sky, sun ( if the sun is near the aperture), and galactic noise and of course signal from the sources you are studying.

Here is a quote from another article:
https://www.nps.gov/parkhistory/online_books/butowsky5/astro4k.htm
They removed the effects of radar and radio broadcasting, and suppressed interference from the heart in the receiver itself by cooling it with liquid helium to -269°C, only 4° above absolute zero--the temperature at which all motion in atoms and molecules stops.What my limited search skills have not been able to find is a description of the geometry of exactly how the liquid helium was configured so that the 290 K radiation became effectively screened from the receiving antenna. Can you help me with this?

OK there are 2 sections there which really should have had better separation

this part for a start ....

is NOT done by cooling the receiver .... it's done with physical electronics filtering

this part goes together
and suppressed interference from the heart in the receiver itself by cooling it with liquid helium to -269°C, only 4° above absolute zero

special note on this bit I have bolded to see the relevance
electronics produce noise, you reduce get rid of hat noise by cooling the electronics. This is standard practice for high gain very low noise satellite receiving systems

What my limited search skills have not been able to find is a description of the geometry of exactly how the liquid helium was configured so that the 290 K radiation became effectively screened from the receiving antenna. Can you help me with this?

again, the cooling by the liquid helium DOESNT remove the ground noise. The ground noise is removed by making the antenna very directional and not having it point at the ground. The receiving antenna is physically shielded from the ground by the horn

I am not sure I correctly understand this description. This is what I think it means.

OK there is a dipole/monopole ( usually a 1/4 wave monopole) antenna at the narrow end of the horn
here's a typical cross-section I have drawn

(b) A switch is used so that two different radiation signals are measured non-concurrently: (i) The ambient noise alone, and (ii) the ambient noise together with the CMB radiation.

Not quite .... change to this to.....
(i)
a ... noise from a cooled dummy load resistor across receiver front end. This gives a receiver noise figure
b ... then, depending on the installation, the antenna is usually aimed at a "cool part of the sky" and a signal noise level is taken
c ... then received signal from the object/area of the sky of interest is measured

Then a and b are subtracted from c to leave just the signal from the region/object of interest

(c) What is measured is the power (watts) of these two signals. The difference between (i) and (ii) is the power of the CMB radiation.

The receiver produces a voltage level, a very small one, uV = microvolts of each of the signals in the a, b and c and then the maths is done manually or by computer to get the end result signal level

(d) The ambient noise consists of (i) the radiation of the liquid helium, and (ii) the radiation of the external sources not completely screened by the liquid helium.

scrap that and go with what I have written above ( remember I commented that the helium doesn't do any screening)

cheers
Dave

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Gold Member
Hi Dave:

I am trying to understand something here that I have never studied before, and I much appreciate your patience with me.

I would like to focus now on the mechanisms that prevents that part of the 290 K black body radiation that is in the range of the 2.7 K CMB from being too noisy with respect to the CMB signal. The are two contributions that make this noise small.
(1) The actual power ratio within a chosen receiver bandwidth for both noise and target signal, and
(2) Assuming (1) is not sufficient for a desired signal to noise ratio, the noise shielding effect of the horn.

I intend as a first step to calculate using Plank's Law the ratio of (1) the noise power from a 1 m2 black body source at 290 K to (2) the target power of a 1 m2 2.7 K source. The calculation will be with respect to the total received power for both (1) and (2) within a frequency range centered on the peak frequency μpeak of 2.7 K black body radiation and with a bandwidth of some reasonable size (e.g.,0.9 μpeak to 1.1 μpeak). Then I may need to seek your help in adjusting this ratio to take into account the appropriate source areas for each of the two sources, and also to perhaps choose a different band width.

I expect this step 1 calculation will take me a while to complete.

Regards,
Buzz

sophiecentaur
Gold Member
2020 Award
@Buzz Bloom Do you have a text book as a main source of information about how noise turns up in receiving equipment? I think you need a bit of structure in your learning of this topic. If you don't have a text book then there are many links that Google can give you. This could be more useful than producing a list of questions and learning from the answers you get because you may not be asking all the important questions. You have the problem of jumping into the middle of a very difficult subject in which there are many different facets, each of which is dead hard!
I would like to focus now on the mechanisms that prevents that part of the 290 K black body radiation that is in the range of the 2.7 K CMB from being too noisy with respect to the CMB signal. The are two contributions that make this noise small.
(1) The actual power ratio within a chosen receiver bandwidth for both noise and target signal, and
(2) Assuming (1) is not sufficient for a desired signal to noise ratio, the noise shielding effect of the horn.
Noise Power in all resistors at a given temperature is the same but the actual resistance of the resistor determines the Noise Voltage. That's important to bear in mind
When a signal arrives from deep space we can assume, for simplicity, that there is no other unwanted noise introduced on the way. The signal in this case is the 2.7K CMB radiation. The signal enters the antenna (dish, horn or anything else. If the antenna is directive enough, it will not 'see' the hot local ground so we can ignore that for a start. But there is noise due to the temperature of the metalwork of the antenna; currents are running all over the antenna and it is hot. But the resistance of each element of the reflecting surface is low (much lower than the (say) 50Ω of the receiver input load. One way of looking at this is to say there is the receiver resistance (a Cold 50Ω) in series with a hot antenna resistance of a small fraction of an Ohm. So the noise current is Dominated by what's generated in the Cold resistance.

The receiver input stage will introduce some Extra noise power (Noise Figure) and the 50Ω load can be cooled to liquid He temperatures so the noise power will be very low (noise temperature of around 4K, say) The low RMS noise voltage due to the antenna metal is very much 'diluted' by the resistance ratio of RAntenna and 50Ω of the receiver load. So it's the cold front end noise that counts. The 40dB or so of gain that the antenna gives you just enough margin to allow you to measure the 2.5K CMB signal in the presence of a 'hot' antenna.
It is harder than that, of course and any inadequacies in the directivity of the receiving antenna can introduce side lobes in the pattern and those side lobes can actually be looking at off axis hot sources, such as the ground. A Horn antenna is not so very highly directional but I believe the radiation pattern is good.

There is another problem with looking at the CMB radiation and that is it's thermal noise. So, unlike a CW signal, you can't improve your Carrier to Noise Ratio by filtering. You are stuck with the same ratio for whatever bandwidth you use. On the other hand, you can swing round and pick the very best direction for reception because the source is very diffuse.

Buzz Bloom
Gold Member
o you have a text book as a main source of information about how noise turns up in receiving equipment?
Hi Sophie:

Thanks for very much your post. I do not have any text book on this topic. I am using what I can find on the Internet, mostly from Wikipedia.

I believe I understand the nature of the information I am seeking, but I think my problem is using the correct vocabulary to express it. I am making progress with the math for the "first step", and I have in mind a thought experiment to capture what in particular puzzles me.

I expect to post my thought experiment with the associated math within a few days. (What slows me down are the careless mistakes I make in my calculations.)

The thought experiment specifically explores the ratio of received power from two radiating black body sources through a filter that passes EM frequencies within a specific range. Each source is a disk with an area of 1 m2. One disk has a fixed temperature of 2.7255 K and the other of 290 K. The peak frequencies of the two sources are respectively 160.2 GHz and 17.05 THz. The filter range is 53.41 GHz to 320.5 Ghz.

Regards,
Buzz

Gold Member
Hi @sophiecentaur, @davenn, @stefan r:

I have finally complected the math for the thought experiment I want to ask about.

(1) Imagine a spherical shell S containing a vacuum environment encased in a larger shell maintained at the temperature of liquid helium. One-half of the inner shell surface H1 is maintained at 290 K, approximately the ambient temperature of Earth's environment. The other half of the inner shell H2 is maintained at 2.7 K, the approximate temperature of the CMB radiation.

(2) At the center of S is an an antenna oriented to receive some fraction Fbb of the H1 black body EM radiation, and the same fraction Fbb of the H2 radiation.

(3) The antenna is connected to a filter with a flat bandwidth between 80% and 120% of the peak frequency of 2.7 K radiation: approximately 127 GHz to 190 Ghz. (I understand that in practice it is more likely a tuned RLC circuit would be used, but a flat bandwidth filter simplifies the math.) I am assuming there is a resistor of the filter circuit across which voltage measurements are made.

(4) The measured voltage will consist of two parts, one part V1 from the H1 radiation, and one part V2 from the H2 radiation.

(5) V1 = P * 2904 * Ff,1 and V2 = P * 2.74 * Ff,2. P is a constant independent of which hemisphere is considered. Ff,i is the fraction of the EM power from Hi between the frequencies of the filter bandwidth, the index i having values 1 and 2. I calculated values for Ff,1 and Ff,2 by calculating the definite integral below with several pairs of limits, and adjusting for the temperature involved.
ab x3 / (ex-1) dx​
The bottom line is:
signal (2.7 K) voltage to noise (290 K) voltage ratio = 0.0164.​

My big area of confusion about this is that this ratio (or some other similar ratio based on (1) a different band width, and (2) different antenna geometry) seems likely to be inadequate for the successful detection of the CMB signals which led to achieving a good estimate for the CMB temperature. I hope someone will clarify this for me.

This conceptual spherical configuration is intended to very roughly simulate what the Holmdel antenna did. The external horn material must have been at the ambient temperature, and an opening through which CMB EM would reach the antenna would correspond to the H2 surface. The actual configuration would have a much larger area ratio between H1 to H2, and this would make the signal to noise ratio worse.

Regards,
Buzz

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sophiecentaur
Gold Member
2020 Award
One-half of the inner shell surface H1 is maintained at 290 K, approximately the ambient temperature of Earth's environment.
The temperature of the hot half of the sphere is not a primary issue. The antenna reduces the contribution of energy from that direction to a very low value. Given infinitely conducting metal work and a radiation pattern that rejects signals from the hot half, there will be no contribution at all at the receiver feed point. I already made the point that low resistivity metal work, the contribution of thermal noise power from that source will be 'diluted'. The general principle is that a good reflector is a poor absorber so it is also a poor emitter of the thermal energy due to its temperature (290K say). This link has a graph of the effective noise temperature of a receiving dish antenna ( which varies with the elevation because of the presence of the hot ground)

Gold Member
The general principle is that a good reflector is a poor absorber so it is also a poor emitter of the thermal energy due to its temperature (290K say).
Hi sophie:

Thank you for your response. If you made this point before, I apologize for missing it.

This is my interpretation of your response. The horn is not a black body because it has high reflectance, like a very shiny metal. Therefore, if it is a very good reflective material at 290 K, only a small fraction of its black body power would be radiated.

On page 2 of a technical guide about: Reflective Materials and Coatings:
the figure about Spectraflect reflectance coating gives a value of about 95% reflectivity in the wavelength range of 300 nm to 2400 nm which respectviely correspond to the peak wavelengths for temperatures of 9700 K and 1200 K. This raises some uncertainty about the reflectivity of black body radiation at 290 K.

I am having great difficulty finding any references about reflectivity with respect black body radiation at 290 K. Can you help me?

I found
The article starts on page 197. Figure II on page 201 seems to be useful. Column shows wavelength in units of 0.001 nm = 10 μm. The row for 1 unit (10 μm) corresponds to 290 K. The column for silver gives for this row a reflectivity of 96.4%. This improves the signal to noise ratio by dividing it by 0.036. Thus the ratio of 0.00164 improves to 0.0454.

Do you think this ratio is sufficient to detect the CMB and determine a good estimate for its temperature? If not, then some additional explanation is needed.

The antenna reduces the contribution of energy from that direction to a very low value. Given infinitely conducting metal work and a radiation pattern that rejects signals from the hot half, there will be no contribution at all at the receiver feed point.
The following
"The antenna reduces the contribution of energy from that direction to a very low value,"​
seems to mean the geometry is the mechanism for accomplishing this. Can you explain his?
Also, can you explain,
"...a radiation pattern that rejects signals from the hot half...?"​
The horn, which is the source of the 290 K radiation, seems to (almost?) completely surround the antenna.

Regards,
Buzz

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davenn
Gold Member
The horn, which is the source of the 290 K radiation, seems to (almost?) completely surround the antenna.

you need to read the second half of my post #7 again it seems it still hasn't clicked with you

I see I also requoted it in post #9 and hilited it in red

Dave

sophiecentaur
Gold Member
2020 Award
The horn is not a black body because it has high reflectance, like a very shiny metal.
There is no reason why it should be a black body so - no problem.
[EDIT: You use shiny foil to wrap a to chicken in. That's to stop it radiating so much energy. - Same thing]

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Gold Member
I see I also requoted it in post #9 and hilited it in red
Hi Dave:

Is the following the quote you are referring to?
This temperature is not the physical temperature of the antenna. Moreover, an antenna does not have an intrinsic "antenna temperature" associated with it; rather the temperature depends on its gain pattern and the thermal environment that it is placed in.​
I think you misunderstand the temperature I am talking about in the thought experiment of my post #13. The 290 K temperature in my thought experiment is the temperature of one half of the spherical shell, not the antenna at the center. This 290 K half sphere is by analogy corresponding to the temperature of the exterior horn material of the Holmdel device. With this understanding, is the quote still relevant?

Regards,
Buzz

Gold Member
You use shiny foil to wrap a to chicken in. That's to stop it radiating so much energy.
What I edited in (under ADDED) at the end of my post #15 was making the point that even if the Holmdel horn material was highly reflective, like silver, the signal to noise ratio (for the thought experiment) was still small: 0.0454. In applying this result to the actual Holmdel apparatus, the shell almost completely surrounds the antenna, unlike the equal areas of the 290 K and 2.7 K areas on the hypothetical sphere. Therefore the ratio of the radiating 290 K solid angle is much large than the solid angle through which the CMB radiation enters the apparatus. The estimate the s/n ratio of the Holmdel apparatus, the 0.0454 s/n ratio would be divided by the solid angle ratio (from the perspective of the antenna) between (a) the Holmdel horn and (b) the opening through the CMB enters the apparatus. My question about this is:
How is a such low s/n ratio compatible with the success of the Holmdel apparatus?​

Do you disagree with my discussion regarding the low s/n ratio?

Regards,
Buzz

sophiecentaur
Gold Member
2020 Award
Therefore the ratio of the radiating 290 K solid angle is much large than the solid angle through which the CMB radiation enters the apparatus.
You don't seem to be taking into account the sensitivity pattern of the horn. Where are you suggesting would be the source of 290K noise be. A solid angle of around 90° and a clear sky would not see any hot sources.
The reflectivity of the horn material would only need to apply at the microwave frequencies that were received. I think you are assuming that the horn is just a 'hole' through which the signal enters from a hemisphere. In fact it is a directive antenna which rejects the ground radiation and provides virtually none of its own thermal noise.
It would probably help if you accepted that the system actually worked and to find reasons for that, rather than finding reasons for it NOT to have worked. Your hemispheres at different temperatures are not relevant. Are you familiar with the concept of antenna gain?

davenn
Gold Member
I think you misunderstand the temperature I am talking about in the thought experiment of my post #13. The 290 K temperature in my thought experiment is the temperature of one half of the spherical shell, not the antenna at the center. This 290 K half sphere is by analogy corresponding to the temperature of the exterior horn material of the Holmdel device. With this understanding, is the quote still relevant?

No, you are still misunderstanding how irrelevantly small the 290K from the ground is at the antenna because of how much the horn shields the antenna from that noise

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Gold Member
Where are you suggesting would be the source of 290K noise be.
Hi Sophie:

I just noticed another of my careless errors in my calculations. I will update the s/n ratio value when I complete the corrections.

Regarding your question in the above quote:
The source I have in mind is the material used to make the horn shaped structure that has an antenna somewhere inside this structure at its "core". This structure will be in thermal equilibrium with the air outside, about 290 K.It will radiate a portion of the black body radiation for 290 K. The portion depends of the reflectivity of the material.

Regards,
Buzz

Gold Member
No, you are still misunderstanding how irrelevantly small the 290K from the ground is at the antenna because of how much the horn shields the antenna from that temperature
Hi Dave:

Regards,
Buzz

Gold Member
Hi Sophie and Dave:

My mistake did not effect the s/n value (0.00164). The error related to misreading the wavelengths in the Table II reference. The following is from the new reference I found cited below.
The materials are listed alphabetically. The entry for silver includes the following line.
Plate (0.0005 Ni) 93-371 C .06-.07​
I interpret this to mean the following.
(a) The material is called "Plate silver".
(b) It includes 0.05% Ni as a component.
(c) The temperature range for which the emissivity range is valid is 93 C to 371 C (366 K to 644 K).
(d) The emissivity range corresponding to (c) is 6% to 7%.
(e) Since the lower temperature limit is somewhat greater than 290 K, I am assuming that the emissivity for 290 K would be somewhat less than 6%.

Using the 6% emissivity value would raise the s/n ratio to 0.00164/0.06 = 0.044. The previous value 0.0454 was based on an emissivity assumption of about 3.6%. In either case, the resulting s/n ratio seems too small for a successful discovery of CMB and it's temperature.

The smallest emissivity value (2%) I found in the following table.

The 2% emissivity would increase the s/n ratio to 0.082. Would that be good enough?

Regards,
Buzz

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sophiecentaur
Gold Member
2020 Award
The smallest emissivity value (2%) I found in the following table.
Is this the emissivity at the relevant microwave frequency? It never needs to be 'shiny' and I have seen film of the horn experiment. The horn was just a regular painted grey / green, afair. It was not silver or gold but a (cheap) surplus horn from some other application. There were pigeons nesting deep inside the horn at one stage, so the story goes. They really did upset the SNR!
Just take a look at photos of state of the art radio telescopes. They are not shiny and often they will be covered in a light layer dust and bird droppings. You are not taking a realistic approach to this.
I haven't looked at your calculations but the snr you are quoting is incredibly low ( starting with -27dB??) Could they really dredge up a noise like signal that low down in the mush?
Did you look at the graph of antenna noise temperature in the link I gave you? At an elevation of around 40 degrees, it's not much more than a couple of K. That's the reason the measurements were possible. If you are interested, you could see what calculation you should be doing in order to get that value.
Therefore the ratio of the radiating 290 K solid angle is much large than the solid angle through which the CMB radiation enters the apparatus.

Do you disagree with my discussion regarding the low s/n ratio?
I disagree with your reasoning. There is virtually no thermal noise arriving due to the hot metal compared with the received CMBR. The noise is predominantly from the front end stage of the receiver and they went to a lot of trouble there, using a helium cooled parametric amp, I think. If what you say is true, there would have been no point, would there?