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

  1. Feb 28, 2018 #1

    Buzz Bloom

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    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)
    Holmdel.png
    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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  3. Feb 28, 2018 #2

    stefan r

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    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.
     
  4. Feb 28, 2018 #3

    Buzz Bloom

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    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
     
  5. Feb 28, 2018 #4

    davenn

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    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=

    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

    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


    no, your numbers are still messed up :wink:

    see my above comments

    Dave
     
  6. Mar 1, 2018 #5

    Buzz Bloom

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    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
     
    Last edited: Mar 1, 2018
  7. Mar 1, 2018 #6

    Buzz Bloom

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    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
     
  8. Mar 1, 2018 #7

    davenn

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    yes

    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


    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

    norm3D1lam.jpg

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


    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
    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
     
  9. Mar 2, 2018 #8

    Buzz Bloom

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    Hi Dave:
    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 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?

    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
     
    Last edited: Mar 2, 2018
  10. Mar 2, 2018 #9

    davenn

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    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 :rolleyes::rolleyes:


    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


    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.


    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
    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


    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


    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

    horn crosssection.GIF



    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


    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


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


    cheers
    Dave
     
  11. Mar 2, 2018 #10

    Buzz Bloom

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    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
     
  12. Mar 5, 2018 #11

    sophiecentaur

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    @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!
    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.
     
  13. Mar 6, 2018 #12

    Buzz Bloom

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    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
     
  14. Mar 8, 2018 #13

    Buzz Bloom

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    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.​
    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
     
    Last edited: Mar 8, 2018
  15. Mar 8, 2018 #14

    sophiecentaur

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    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)
     
  16. Mar 8, 2018 #15

    Buzz Bloom

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    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?

    ADDED
    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 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
     
    Last edited: Mar 8, 2018
  17. Mar 8, 2018 #16

    davenn

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

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


    Dave
     
  18. Mar 9, 2018 #17

    sophiecentaur

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    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]
     
    Last edited: Mar 9, 2018
  19. Mar 9, 2018 #18

    Buzz Bloom

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    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
     
  20. Mar 9, 2018 #19

    Buzz Bloom

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    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
     
  21. Mar 9, 2018 #20

    sophiecentaur

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    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?
     
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