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

[Buzz Bloom]

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

(1) https://en.wikipedia.org/wiki/Antenna_gain
(2) https://en.wikipedia.org/wiki/Isotropic_radiator
(3) https://en.wikipedia.org/wiki/Cosmic_microwave_background#History
(4) https://en.wikipedia.org/wiki/Holmdel_Horn_Antenna#Technical​

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
1/10^2.15 = 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 1/10^41.41 = 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.

 

[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 1/10^41.41 = 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". 10^4.141 = 1.38 x 10⁴. Around fourteen thousand times the power when aimed at the source.

 

[Buzz Bloom]

Hi stefan:

So I made the same error in Q2.
The 2.15 dBi means that a dipole antenna emits or receives
1/10^0.215 = 0.785
times the energy that an isotropic radiator antenna would emit or receive. Is that correct?

Regards,
Buzz

 

[davenn]

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 ... don't know 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
1/10^0.215 = 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

see my above comments

Dave

 

[Buzz Bloom]

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

10^0.215 = 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

10^4.141 = 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

 

[Buzz Bloom]

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]

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 radiation
• Earth heating
• The sun
• Electrical devices
• The antenna itself

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 removedDave

 

[Buzz Bloom]

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
10^4.141 = 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 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.

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.

(a) The antenna receives radiation at the core of the horn.
(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

 

[davenn]

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:

• https://en.wikipedia.org/w/index.php?title=Galactic_radiation&action=edit&redlink=1
• Earth heating
• The sun
• Electrical devices
• The antenna itself

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 ...

They removed the effects of radar and radio broadcasting

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.

(a) The antenna receives radiation at the core of the horn.

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

 

[Buzz Bloom]

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 μ_peak of 2.7 K black body radiation and with a bandwidth of some reasonable size (e.g.,0.9 μ_peak to 1.1 μ_peak). 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]

@Buzz Bloom Do you have a textbook 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 textbook 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 R_Antenna 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]

o you have a textbook 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 textbook 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

 

[Buzz Bloom]

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 F_bb of the H₁ black body EM radiation, and the same fraction F_bb 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 * F_f,1 and V₂ = P * 2.7⁴ * F_f,2. P is a constant independent of which hemisphere is considered. F_f,i is the fraction of the EM power from H_i between the frequencies of the filter bandwidth, the index i having values 1 and 2. I calculated values for F_f,1 and F_f,2 by calculating the definite integral below with several pairs of limits, and adjusting for the temperature involved.

∫_a^b x³ / (e^x-1) dx​

The bottom line is:

signal (2.7 K) voltage to noise (290 K) voltage ratio = 0.0164.​

I will post details about this calculation if anyone wants to see it.

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

 

[sophiecentaur]

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)

 

[Buzz Bloom]

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:

https://www.labsphere.com/site/assets/files/2553/a-guide-to-reflectance-materials-and-coatings.pdf​

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?

ADDED
I found

https://nvlpubs.nist.gov/nistpubs/bulletin/07/nbsbulletinv7n2p197_A2b.pdf .​

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

 

[davenn]

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 redDave

 

[sophiecentaur]

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]

 

[Buzz Bloom]

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

 

[Buzz Bloom]

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]

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]

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

 

[Buzz Bloom]

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

 

[Buzz Bloom]

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:

Please look at my post #22 for my thinking about the 290 K radiation source.

Regards,
Buzz

 

[Buzz Bloom]

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.

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

https://www.omega.com/literature/transactions/volume1/emissivitya.html​

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

Regards,
Buzz

 

[sophiecentaur]

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?

 

[Buzz Bloom]

Are you familiar with the concept of antenna gain?

Hi Sophie:

Yes, but I do not understand the geometry of the Holmdel apparatus.The diagram in Dave's post #9 doesn't seem to match the picture of the Holmdel.

Do you know in this picture where the antenna is, and what direction it's directionality is pointing? Is the horn supposed to reflect the CMB radiation to the antenna? Do you know what material the horn is made of? Is this material highly reflective at the 159 GHz peak frequency for 2.7 K?


Regards,
Buzz

 

[Buzz Bloom]

Is this the emissivity at the relevant microwave frequency?

Hi Sophie:

The referenced table is generally about emissivity at higher temperature than that corresponding to microwaves. The 290 K radiation is generally far infrared.

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.

You are convincing me that the very low snr value is an artifact of an error in my calculations which I have not so far been able to find. If I posted my calculations, would you examine them and try to find my error?

Regards,
Buzz

 

[sophiecentaur]

If I posted my calculations, would you examine them and try to find my error?

There is no need to post your calculations. I think your error is in assuming far more radiation from the metal of the antenna than there is. As the link I have mentioned says, it's in the order of just a few (thickness of the line on that graph)K.
Let's face it, the receiver feed is buried down inside that horn and it is receiving some radiation from almost every direction except the square hole up at the end, I guess that, if you made the horn out of resistive material (377Ω or what they coat stealth aircraft with) then the noise temperature might indeed be 290K'

I found this link after not much of an effort and that has a lot of data about the horn, which is actually part of an offset parabolic reflector. Its gain is around 40dB. A big devil and a narrow beam!

 

[Buzz Bloom]

that has a lot of data about the horn, which is actually part of an offset parabolic reflector. Its gain is around 40dB. A big devil and a narrow beam!

Hi Sophie:

Thanks for the reference about the horn. I am guessing that the angle selectivity of the parabolic reflector may be the mechanism that improves the snr to a usable level. It would help if I could see an actual construction diagram of the Holmdel apparatus. The article is full of vocabulary and concepts new to me, so it will take me a while to get my head around it. Thank you very much for your patience with me.

Regards,
Buzz

 

[sophiecentaur]

I am guessing that the angle selectivity of the parabolic reflector may be the mechanism that improves the snr to a usable level

Of course. That's what reflector antennae are all about. If you can't find much info about "Offset Paraboic Reflectors" then look up "Parabolic Reflectors" to find out the sort of beam patterns they produce and the design criteria.
I have frequently made this point to people who post on PF but you seem to have jumped in far too deep, too fast into this topic for you to expect to get a sensible result straight away. You need to read up more about the basics of communications theory. There is loads of stuff at all levels available. Q and A has its limits and can be pretty inefficient as a way of learning about a wide topic like this one.

 

[Buzz Bloom]

Q and A has its limits and can be pretty inefficient as a way of learning about a wide topic like this one.

Hi Sophie:

Since you have so very helpful to me, I feel it to awkward for me to disagree with you. However, I have learned what I was seeking to learn here on the PF in about 10 days. If I had found textbooks as a source and tried to learn from them, I am pretty sure it would have taken me months to get to this point.

Thank you a gain for your help.

Regards,
Buzz

 

[davenn]

Yes, but I do not understand the geometry of the Holmdel apparatus.The diagram in Dave's post #9 doesn't seem to match the picture of the Holmdel.

It's just a different shaped horn ... it's just a convenience thing for ease of pointing at the sky whilst being able to rotate it in 2 directions...
don't get hung up on it ... it's irrelevent

regardless of if it is the simple horn I posted in #9 or the one the topic is about. If the horn is pointing at the sky, then there is VERY LOW (close to zero) level of ground (290K) noise entering the opening of the horn. This is what makes dish and horn antennas so very effective. Their directionality is VERY SHARP.

Do you know in this picture where the antenna is, and what direction it's directionality is pointing? Is the horn supposed to reflect the CMB radiation to the antenna? Do you know what material the horn is made of? Is this material highly reflective at the 159 GHz peak frequency for 2.7 K?

swing the opening arrow towards you a little
The antenna is right at the apex of the horn right in front of the receiver
I'm sure that article you originally posted a link to commented on the structure
it is aluminium ... and you can observe the inside surface is made of unpolished panels

Dave

 

[sophiecentaur]

It would help if I could see an actual construction diagram of the Holmdel apparatus.

I guess the search for that would be up to you, I'm afraid. That article tells us that the horn has a section of parabolic form for its reflector. Your response rather goes to show that you really need to approach this along a tried and tested path of learning, which has facts and ideas presented in a logical order, not leaving out the important bits on the way. Your attempt to get anywhere by just asking questions has left you without the tools to understand the whole system - rather proving my point.
How would you possibly know the right questions to ask if you do it your way? It would be Brownian Motion, which never gets anywhere fast.
Remember, there is nothing fundamentally better about a Holmdel horn, any more than the receiving equipment was better than we use these days. It's an interesting historical bit of kit of a design which is seldom used these days. Better to look into the sort of set ups that are used in conventional comms reflectors and radio telescopes. There is far more written about them and that will help you far more than one fuzzy photo and a very limited paper.
The CMBR is observed every day by modern equipment.

 

[Buzz Bloom]

Remember, there is nothing fundamentally better about a Holmdel horn, any more than the receiving equipment was better than we use these days.

Hi Sophie:

BTW: Although the material I read did not describe the specific characteristics of the Holmdel parabolic reflector, I was able to make a calculation from the data I did find that the reflector diameter was about 1 m.

The reason I chose to learn about the Holmdel rather than current technology was mostly a historical-technical interest.

Regards,
Buzz

 

[sophiecentaur]

I did find that the reflector diameter was about 1 m.

This doesn't seem right. The Photograph shows a rectangular shaped aperture which is four or five human heights in each axis. Are you talking of a secondary reflector inside the horn? How is this relevant to the basic gain of the antenna?
D'you know, I think we have reached the end of the line with this thread. You keep introducing ideas and questions that are only serving to impede your learning about this topic. There is no alternative but for you to go through the topic of microwave receiving antennae from square one.

 

[Buzz Bloom]

Are you talking of a secondary reflector inside the horn?

Hi Sophie:

Yes. Based on the previous discussion in the thread, the internal parabolic antenna has a gain of 43.3 dBi. It is the directionality of this component that makes it possible for the apparatus as a whole to avoid the thermal noise from the 290 K radiation from the aluminum of the horn. This gain corresponds to that provided by a 1 m parabolic reflector with a simple antenna at its focus.

Solving this equation for for D gives 1 m, with G = 43.3, k = 76%, and
λ_2.7K = c/ν = 1.889 cm, where
ν = 158.7 GHz corresponding to the peak frequency at 2.73 K.

Regards,
Buzz

 

[davenn]

I did find that the reflector diameter was about 1 m.

Don't know where you got that from ??
It is very much bigger than that and it doesn't have a diameter as such ... it is basically rectangular with a slightly curved surface 6.1m at it's longest length

The antenna is 50 feet (15 m) in length with a radiating aperture of 20 by 20 feet (6 by 6 m) and is constructed of aluminium.

all this info was in the links in your original post

Hi Sophie:

Yes. Based on the previous discussion in the thread, the internal parabolic antenna has a gain of 43.3 dBi. It is the directionality of this component that makes it possible for the apparatus as a whole to avoid the thermal noise from the 290 K radiation from the aluminium of the horn. This gain corresponds to that provided by a 1 m parabolic reflector with a simple antenna at its focus.
View attachment 221888
Solving this equation for for D gives 1 m, with G = 43.3, k = 76%, and
λ_2.7K = c/ν = 1.889 cm, where
ν = 158.7 GHz corresponding to the peak frequency at 2.73 K.

Regards,
Buzz

As noted by Sophi and myself , that is pretty much all incorrect

It is the directionality of this component that makes it possible for the apparatus as a whole to avoid the thermal noise from the 290 K radiation from the aluminum of the horn

please read carefully ... getting frustrated having to constantly repeat myself
Thermal radiation from the horn material is so low that it is basically irrelevant .. I have given you links to that several times
The metal sides of the horn SHIELD THE ANTENNA from the thermal radiation from the groundDave

 

[Buzz Bloom]

Hi @davenn and @sophiecentaur:

I am sorry that I misunderstand what I have read.

https://www.revolvy.com/main/index.php?s=Holmdel Horn Antenna&item_type=topic

The reflector is a segment of a parabolic reflector, so the antenna is really a parabolic antenna which is fed off-axis. A Hogg horn combines several characteristics useful for radio astronomy. 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. Consequently, it is an ideal radio telescope for accurate measurements of low levels of weak background radiation. 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%.​

The underlining is mine.

From this text I decided that I could determine the radius of the "parabolic antenna". In my post #36 I showed the calculation of a parabolic telescope diameter (1 m) from the other data I had. I also used another formula for determining the beamwidth.

Using the value ψ=1.5^o I calculated D = 88 cm.

I am completely baffled about why this is wrong.

Regards,
Buzz

 

[davenn]

I am completely baffled about why this is wrong.

because you used 1 metre instead of 6 metres

The reflector is a segment of a parabolic reflector, so the antenna is really a parabolic antenna which is fed off-axis. A Hogg horn combines several characteristics useful for radio astronomy. 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. Consequently, it is an ideal radio telescope for accurate measurements of low levels of weak background radiation. 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%.

The underlining is mine.

and you didn't underline the important sentence that tells you what I have been trying to impress upon you several times
read the bolded bit ... it gives you the result of the effectiveness of the horn as commented on in the previous sentence that you did underline

D

 

[Buzz Bloom]

because you used 1 metre instead of 6 metres

Hi Dave:

ADDED
I have just again looked over my spreadsheet I used to do calculations and found another error. I suggest we postpone any further discussion until I figure out how to fix the error.

PREVIOUS POST VERSION

Two more things I don't understand.
(1) Why do you say I "used" 1 meter?
The 1 m was a calculation based on the other data:
G = 43.3, k = 76%, and λ_2.7K = c/ν = 1.889 cm.
(2 Why doesn't

a radiating aperture of 20 by 20 feet (6 by 6 m)​

mean an opening in the horn, rather than a parabolic reflector.
In yout post #36, you indicate that the "antenna" is in the "shed" attached to the small left end of the horn. Perhaps the shed also houses the 1 m diameter parabolic reflector. If not, what is the means by which the CMB signal reaches the "antenna" if there is a 6 m by 6 m reflector?

https://en.wikipedia.org/wiki/Antenna_aperture

In electromagnetics and antenna theory, antenna aperture, effective area, or receiving cross section, is a measure of how effective an antenna is at receiving the power of electromagnetic radiation (such as radio waves). The aperture is defined as the area, oriented perpendicular to the direction of an incoming electromagnetic wave, which would intercept the same amount of power from that wave as is produced by the antenna receiving it. At any point, a beam of electromagnetic radiation has an irradiance or power flux density (PFD) which is the amount of energy passing through a unit area of one square meter. If an antenna delivers Po watts to the load connected to its output terminals (e.g. the receiver) when irradiated by a uniform field of power density PFD watts per square meter, the antenna's aperture Aeff in square meters is given by:

A_eff = P_o/PFD.​

This concept seems to me to mean something other than πD²/4 where D is the diameter of the open circular boundary of a parabolic reflector. A 6 m by 6 m square area doesn't seem to me to have any relationship to the parabolic reflector described in the quote in my post #38.

One more reference
https://en.mimi.hu/astronomy/aperture.html
Aperture

The aperture of a telescope is the diameter of the light collecting region, assuming that the light collecting region has a circular geometry .
...the size of the opening through which light passes in an optical instrument such as a camera or telescope. A higher number represents a smaller opening while a lower number represents a larger opening.​

In addition to the 2 confusions I mentioned above, how do you explain in the absence of a parabolic reflector the avoidance of the 290 K noise radiation emitted by the walls of the horn, and which then is received by the ultimate receiver antenna within the horn, and which (both entirely and also within the frequencies of interest) is much greater than that of the CMB radiation entering through the 36 m² aperture?
It is much greater for two reasons:
(a) the 290 K radiation is much greater per m² than the CMB radiation;
(b) the area of the 290 K radiation is much greater than the (aperture) 36 m² area.

Regards,
Buzz

 

[sophiecentaur]

290 K radiation from the aluminum of the horn

What do you mean by that? If the aluminium of the rest of the horn is radiating the full black body radiation then how does a standard paraboloid antenna mange to look at weak signals? What is different about the aluminium face of the 'reflector?

I am completely baffled about why this is wrong.

If you look at the patterns of a range of antennae of the same basic aperture, you will get a wide variation. This is because the basic aperture is affixed by the 'illumination' of the reflector by the feed antenna. You can choose a narrower beam with poor side lobe performance and that will give you maximum gain but wouldn't;t be suitable for low signal reception. You can also choose a wider beam with good side lobe performance and that will have lower gain but be much more immune to well-off-axis sources. The formulae that you are using are based on rule of thumb and 'typical' antennae.
Yet again, I am surprised that you are surprised that you keep coming across things that don't make sense to you. You just don't know enough about the topic. That's not a problem but the only way round it is to learn a lot more and then revisit the topic. It will make more sense then. This is yet another example of not asking the 'right' question because you are not coming to the topic from the right direction.

If I had found textbooks as a source and tried to learn from them, I am pretty sure it would have taken me months to get to this point.

I just read this again. The fact is that you don't know what point you are 'at'. How do you know that, if you stray away from this particular place in your knowledge space, that you will be in a position to answer another closely related question? This is why people need to sit Degree and Masters Courses etc. before they are considered to be good candidates for Engineering jobs.

 

[Buzz Bloom]

What do you mean by that? If the aluminium of the rest of the horn is radiating the full black body radiation then how does a standard paraboloid antenna mange to look at weak signals? What is different about the aluminium face of the 'reflector?

My conceptual model of the Holmdel, which you have convinced me is wrong, is that there is a parabolic reflector which focuses the target radiation onto the end-point antenna (epa). Almost all of the target radiation which enters the horn through the aperture, gets focused onto this epa. The radiation from the horn only send a small part of its radiation onto the epa. I am not sure how to properly calculate this "small part" of the radiation, but I will give it a try. (See (9) below.) The final signal (filtered 2.73 K radiation) to noise (filtered 290 K radiation) calculation is

SNR = 16.5​

Suppose we assume that the epa has a functional area which "sees" the radiation from the walls of the horn. Given an assumed parabolic reflector of about 1 m diameter (I now have come to understand that 1 m is a wrong calculation), assume the epa is a straight rod of length about 5 cm, and about a 1 mm radius. The area of the rod is then
(1) A_rod = π × 0.001 × 0.05 ~= 0.00015 m².
There are other factors:
(2) The power ratio corresponding to the temperatures:

R_power = (290/2.73)⁴ ~= 1.3 × 10⁸​

(3) The 290 K radiating area

A₂₉₀ ~= (1/2) * (4*6 m) * 16 m =~= 200 m²​

(4) The 2.73 K radiating area

A_2.73 = (1/4) × π × 36 m² ~= 12 m²​

Assume all of the (4) area is focused onto the epa.
(5) The fraction of the 290 K wall area radiation that hits the epa
Assume an average effective distance from a random wall point to the epa is 2 m.
Each point on the wall radiates on the average onto a hemisphere of

2 π 4 m².​

The fraction of the radiation that hits the epa is

R₂₉₀ ~= 0.00015 m² / 2 π 4 m² ~= 0.000006​

(5) The ratio of 290 K power hitting the epa to the 2.73 K power hitting the epa is

R_power = 1.3 × 10⁸ × 0.000006 × 200 m² / 12 m² ~= 13,000​

(6) A useful black body radiation integral

f(x) = x³/(e^x-1)
F(x) = ∫₀^x f(x) dx
F(∞) = π⁴ / 15 ~= 6.494​

(7) The fraction of 2.73 K radiation between 90% and 110% of the peak 2.73 K frequency.

F_2.73 = (F(β - F(α))/F(∞) ~= 0.24
α = 2.257
β = 3.386​

(8) The fraction of 290 K radiation between 90% and 110% of the peak 2.73 K frequency.

F₂₉₀ = (F(β* - F(α*))/F(∞) ~= 1.12 × 10^-6.
α* = (2.73/290) × α = 0.0210
β* = (2.73/290) × β = 0.0315​

(9) The final calculation! The signal to noise ratio (SNR) = ratio of the 2.73 K to 290 K power received by the epa and after filtering with a flat frequency range between 90% and 110% of the peak 2.73 K frequency.

SNR = F_2.73 / (F₂₉₀ × R_power)

~= 0.24 / (13,000 × 1.12 × 10^-6)~= 16.5​

BTW Sophie, I like the bit you have at the end of your posts.

Q. Can you play the piano?
A. Dunno, I have never tried.​

I am not sure what you think it means, but I think it means that the answerer is naive in thinking that if he had previously tried, he would know. I have also decided that the answerer is probably a male.

Regards,
Buzz

 

[sophiecentaur]

I am not sure what you think it means, but I think it means that the answerer is naive in thinking that if he had previously tried, he would know. I have also decided that the answerer is probably a male.

Male and totally self confident. (Or is that tautology?)
I actually don't understand whether or not your analysis makes sense. I'm afraid.
Why not try following through the process with a simple circular paraboloid with a simple Horn feed? The feed can give a range of illumination percentages and a range of side and back lobe values. You can pick and mix in a very few steps. I still don't know why you are considering the 290K aluminium when what counts is the Noise Temperature of the reflecting surface. The hot bit that counts is the part of the ground etc that the feed sees directly (feed horns have some side sensitivity and are not normally buried down inside the paraboloid.

 

[Buzz Bloom]

The feed can give a range of illumination percentages and a range of side and back lobe values. You can pick and mix in a very few steps.

Hi Sophie:

Can you recommend a reference that explains how to work with "illumination percentages" and "side and back lobe values"? These are unfamiliar concepts to me, although there is a sort of hint in Post #7.

I still don't know why you are considering the 290K aluminium when what counts is the Noise Temperature of the reflecting surface. The hot bit that counts is the part of the ground etc that the feed sees directly (feed horns have some side sensitivity and are not normally buried down inside the paraboloid.

Generally, the temperature of the horn will be in equilibrium with the air temperature. The radiation from the ground, and from any other solid areas outside the horn, will not ever directly reach the epa, except for the possibility that the source is on a direct (or reflected) line of sight to the epa. This exception will not happen if the aperture points towards the sky. Instead this radiation from outside solids will hit the horn wall and then will raise the wall's temperature a small amount before the wall re-radiates enough to reestablish equilibrium with the air temperature.

The "Noise Temperature of the reflecting surface" can be considered as an additional 290 K area nearer the epa than the wall. (I assume this is the parabolic surface.) My intuition is that this is a smaller amount of radiation hitting the epa than the wall radiation. To take this into account, I would need to know (or guess) (a) the area of the parabolic reflector, and (b) the average distance (actually the average inverse distance squared) from a random reflector point to the epa at the focus point.

How do you know that, if you stray away from this particular place in your knowledge space, that you will be in a position to answer another closely related question? This is why people need to sit Degree and Masters Courses etc. before they are considered to be good candidates for Engineering jobs.

I am an elderly retired person. My participation in the PFs is a hobby, not related to any job, not am I seeking any job. This thread was started because I was curious about how the Holmdel was able to do what it did that made it famous. In post #1 I asked 4 questions. The first 3 were quickly answered, but the 4th has continues to be a confusing conversation for me, in part due to my careless errors in calculations.
The most confusing part of the conversation I think is what I perceive to be the ambiguous usage of he term "antenna".
(a) The entire big Holmdel hon is called an antenna.
(b) A parabolic reflector with a limited directionality antenna at the focus is an antenna.
(c) The limited directionality antenna at the focus of a parabolic reflector is an antenna.is an antenna.
I have tried in my questions to be specific about which of these three things I am talking about, but in both responses and reference quotes, there is some ambiguity.

One aspect of my question (4) for post #1 that seems still unresolved in my mind is whether or not the Holmdel apparatus contains inside the pyramidal horn exterior a parabolic reflector component. In post #1 I quoted from Wikipedia:

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.

I interpreted the Wikipedia tesxt as follows: the text "a segment of" means "a partial", and "mounted in its mouth" means "inside the pyramidal horn exterior".

Male and totally self confident. (Or is that tautology?)

Just probably a tautology.

Regards,
Buzz

 

[davenn]

Two more things I don't understand.
(1) Why do you say I "used" 1 meter?
The 1 m was a calculation based on the other data:
G = 43.3, k = 76%, and λ2.7K = c/ν = 1.889 cm.

which is obviously a wrong result since you can see the reflector is substantially larger than 1m and the text indicates that as well

(2 Why doesn't
a radiating aperture of 20 by 20 feet (6 by 6 m)mean an opening in the horn, rather than a parabolic reflector.

in this case, it is the same thing

In yout post #36, you indicate that the "antenna" is in the "shed" attached to the small left end of the horn. Perhaps the shed also houses the 1 m diameter parabolic reflector. If not, what is the means by which the CMB signal reaches the "antenna" if there is a 6 m by 6 m reflector?

absolutely and utterly definitely NOT

it reflects off the reflector that you can see at the opening and is funnelled down the horn to the antenna element

it really is as simple as that ... I showed you a basic horn antenna way earlier on in the thread
stick a reflector at around 45 deg angle at the open end of the horn

This concept seems to me to mean something other than πD2/4 where D is the diameter of the open circular boundary of a parabolic reflector. A 6 m by 6 m square area doesn't seem to me to have any relationship to the parabolic reflector described in the quote in my post #38.

what don't you understand about it ?

One more reference
https://en.mimi.hu/astronomy/aperture.html
Aperture
The aperture of a telescope is the diameter of the light collecting region, assuming that the light collecting region has a circular geometry .
...the size of the opening through which light passes in an optical instrument such as a camera or telescope. A higher number represents a smaller opening while a lower number represents a larger opening.

that's all true and that is specifically referring to an optical situation telescope or camera lens ... no problems there

Aperture is just the size of the opening regardless of its shape ...
and in the case of the of the antenna being discussed, it is roughly 6m x 6m according to supplied info

In addition to the 2 confusions I mentioned above, how do you explain in the absence of a parabolic reflector the avoidance of the 290 K noise radiation emitted by the walls of the horn, and which then is received by the ultimate receiver antenna within the horn, and which (both entirely and also within the frequencies of interest) is much greater than that of the CMB radiation entering through the 36 m2 aperture
It is much greater for two reasons:
(a) the 290 K radiation is much greater per m2 than the CMB radiation;
(b) the area of the 290 K radiation is much greater than the (aperture) 36 m2 area.

As I have previously stated, neither of which are relevant

Do I really have to repeat previous answers for a 4th time ??
I'm begging you ... PLEASE reread them and let it sink in

As Sophi said many posts ago
because you haven't done any basic study on antenna systems and theory. You continue to ask all these Q's in a haphazard way
when with a bit of good background study, you would already have the answers Dave

 

[Buzz Bloom]

you can see the reflector is substantially larger than 1m and the text indicates that as well

Hi Dave:

Can you highlight where in the Holmdel picture I can see a greater than 1 M reflector?
Where does the text say that there is a greater than 1 M reflector?

in this case, it is the same thing

Please explain to me how an opening can be the same thing as a reflector.

Your post #45 has more in it that convinces me that I still do not understand your concept of what is the reflector in the Holmdel apparatus. Please explain the concept.
(a) Is it parabolic?
(b) Is it the same thing as the opening in the horn?
(c) Does the red arrow on the right end of the edited photo in your post #32 point to it? Is it a part of the exterior wall of the horn? If not, please describe in detail exactly where it is.

I showed you a basic horn antenna way earlier on in the thread

Do you mean the diagram in your post #9? If so, this diagram has nothing that looks like a parabolic reflector. Please describe what part of this diagram you intend to represent the reflector.Regards,
Buzz

 

[Buzz Bloom]

THIS POST WAS AN ACCIDENT

 

[davenn]

Can you highlight where in the Holmdel picture I can see a greater than 1 M reflector?

BUZZ ... seriously !

LOOK at my post of the antenna with the red arrows on it. the blunt end of the right arrow ends on the reflector

Where does the text say that there is a greater than 1 M reflector?

you even quoted the correct text in a post a couple of posts ago
it approximates both measurements ...

Buzz Bloom said: ↑
(2 Why doesn't
a radiating aperture of 20 by 20 feet (6 by 6 m)mean an opening in the horn, rather than a parabolic reflector.

(b) A parabolic reflector with a limited directionality antenna at the focus is an antenna.

this also from another of your posts is wrong ... the reflector is only part of the antenna system ... it ISNT the actual antenna

Please explain to me how an opening can be the same thing as a reflector.

Your post #45 has more in it that convinces me that I still do not understand your concept of what is the reflector in the Holmdel apparatus. Please explain the concept.
(a) Is it parabolic?

the opening size is the APERTURE

The reflector is according to the text parabolic

(b) Is it the same thing as the opening in the horn?

the reflector is the reflector
the opening is the aperture
the measurement of both is the same

(c) Does the red arrow on the right end of the edited photo in your post #32 point to it? Is it a part of the exterior wall of the horn? If not, please describe in detail exactly where it is.

read first comment

I will do a cross section side view in my next post

Dave

 

[davenn]

basic cross section drawing of the Holmdel antenna ( not to scale)

the actual horn and reflector showing the reflector surface ... lines are dotted where the surface is hidden by the front part of the horn panelling
Look at the shaded/outlined area compared to the size of the two guys ... the reflector area is HUGE ... it ISNT 1m x 1m

Dave

 

[Buzz Bloom]

I will do a cross section side view in my next post

H i Dave:

I apologize for my being dense and not previously understanding your concept of the parabolic reflector. I think I got it now.

The reflector is part of one of the four walls of the Holmdel horn which has a truncated pyramid shape with the exception of one of the reflector wall which is described later. The particular reflector wall is either (1) attached to the side of the truncated pyramid opposite the side with the large aperture/opening, or (2) physically a part of that wall. The reflector is geometrically a curved shape positioned at a 45^o angle to the 6m by 6m base of the pyramid. At this angle incoming radiation will be reflected to pass in the direction of the central axis of the pyramid toward the narrow end. The radiation will also be focused by the curved shape onto the "antenna" at the end of the apparatus near or in the shed.

If this is correct, then I thank you for being patient with me, and I am please that I finally understand your concept. I will now have to do some thinking to calculate an approximation for the implied SNR.

A few minor questions?
(a) Is the diameter of the parabolic reflector 6m or something different?
(b) Is the reflector part of the wall or attached to it? (Your diagram in post #49, which I saw after stating this post, seems to be saying "part of the wall")
(c) Do you have an estimate for the geometry and size of the antenna at the left of the apparatus?

Regards,
Buzz