Antenna Temperature of Microwave Radiometer

Hi..

I have designed microstrip antenna in 11.4 GHz for radiometer system and it has 60° of beamwidth. The detected target has diameter of 1 cm and its temperature is 80°C. The distance between the target and the antenna is 10cm. I need your help for calculating the antenna temperature of radiometer.

Thanks in Advanced
Hadi

When the target would fill the viewing field of the radiometer completely, the radiometer should read the temperature of the target (after correction for own noise contribution).

The viewing field of the radiometer is about 0.9 sr. The target covers about 8m sr. So the contribution of the target to the noise temperature will be

(273+80)*8m/0.9 = 3K.

Note that the target also covers some part of the background radiation, so actual increase due to this target depends on the background temperature. If the background temperature is also 80 degrees, you will not notice any difference.

The ratio of (target solid angle)/(viewing solid angle of sensor) is so low, that the overall noise temperature will very likely be dominated by the noise temperature of the background (and don't forget the resistive loss in the microstrip).

The above is a simple reasoning valid for optical behavior. Your target is not large with respect to lambda, so due to resonance you can have significant difference between actual contribution and this simple model result.

Something that may make life easier: If you transmit with your patch antenna, and whatever structure absorbs Z percent of the transmitter power, that structure will introduce T*Z/100 Kelvin to the noise temperature. T is the temperature of the structure in kelvin.

WimRFP said:

When the target would fill the viewing field of the radiometer completely, the radiometer should read the temperature of the target (after correction for own noise contribution).

The viewing field of the radiometer is about 0.9 sr. The target covers about 8m sr. So the contribution of the target to the noise temperature will be

(273+80)*8m/0.9 = 3K.

Note that the target also covers some part of the background radiation, so actual increase due to this target depends on the background temperature. If the background temperature is also 80 degrees, you will not notice any difference.

The ratio of (target solid angle)/(viewing solid angle of sensor) is so low, that the overall noise temperature will very likely be dominated by the noise temperature of the background (and don't forget the resistive loss in the microstrip).

The above is a simple reasoning valid for optical behavior. Your target is not large with respect to lambda, so due to resonance you can have significant difference between actual contribution and this simple model result.

Something that may make life easier: If you transmit with your patch antenna, and whatever structure absorbs Z percent of the transmitter power, that structure will introduce T*Z/100 Kelvin to the noise temperature. T is the temperature of the structure in kelvin.

Click to expand...

Thank you so much.........could you please give me the formulas did you used for calculations of antenna temperature????. The 80° of target is apparent temperature and Background has 40° temperature.

I used these formulas

1) I calculated the solid angle of target and it is Omega S=At/R^2 (sr.) where At=2pi*(radius of target)^2 .... and R is distance between the target and antenna.

2) Later I calculated the solid angle of antenna radiation pattern which it is Omega A= (lambda)^2/Ar.....Ar is the aperture of antenna

3) the antenna temperature =Ts*(Omega S/Omega A).....where Ts is temperature of target (80°)

Did you used these formulas???

Hello,

I used same formulas for the solid angle (as under 1). The formula of At is valid for small diameter w.r.t. distance as it assumes the disk is flat.

For the sensor itself, the above isn't met, so this formula will introduce some error. Please visit Wikipedia (solid angle) for the exact formulas for the solid angle of a cone.

Formula 2 I don't understand, I can't see that lambda is required for calculating the viewing solid angle for the antenna as the beam width is given.

When doing noise temperature calculations, you should work in K (not degr. Celsius).

For your case; first calculate the solid angle for the background minus the solid angle of the target (as that shadows the radiation from the background). Then calculate the noise temperature contributed by the background. It will be just below the background temperature.

Second, calculate the noise temperature for the target and add this to background noise temperature, this gives the total antenna noise temperature (assuming no loss).

Note that the formula under 3 is only valid when the gain is constant within the viewing angle. This isn't true for your antenna, so in real world you have to include the actual gain of the antenna. But I think this isn't required for your exercise.

Also try to understand the matter (that is why the formulas are as they are).

WimRFP said:

Hello,

I used same formulas for the solid angle (as under 1). The formula of At is valid for small diameter w.r.t. distance as it assumes the disk is flat.

For the sensor itself, the above isn't met, so this formula will introduce some error. Please visit Wikipedia (solid angle) for the exact formulas for the solid angle of a cone.

Formula 2 I don't understand, I can't see that lambda is required for calculating the viewing solid angle for the antenna as the beam width is given.

When doing noise temperature calculations, you should work in K (not degr. Celsius).

For your case; first calculate the solid angle for the background minus the solid angle of the target (as that shadows the radiation from the background). Then calculate the noise temperature contributed by the background. It will be just below the background temperature.

Second, calculate the noise temperature for the target and add this to background noise temperature, this gives the total antenna noise temperature (assuming no loss).

Note that the formula under 3 is only valid when the gain is constant within the viewing angle. This isn't true for your antenna, so in real world you have to include the actual gain of the antenna. But I think this isn't required for your exercise.

Also try to understand the matter (that is why the formulas are as they are).

Click to expand...

Could you please give me some references to understand the matter of FORMULAS?......Thanks

Hello,

You may try Antennas (for all applications), Kraus, Marhefka. This has a section on remote sensing / Radio Astronomy. You may also look into basic documents on Radio Astronomy. When I needed it, I had to figure it out myself based on basic physical and electrical principles.

If you keep in mind the basic principle I mentioned in my first reply (regarding seeing the antenna as a transmit antenna), it is not that hard to understand. Same principle you can use on loss in the antenna (or feed cable).

Dear Raed:

While WimRFP's answer looks correct, the approach is not.
Any microwave radiometer connected to an antenna responds not only to an object you mentioned, but to the complex surrounding scene of objects and media.
To use WimRFP's explanation, you would have to locate your patch antenna in a free space where you could assume zero other contributions. Only then (and still the galactic background counts by ~3 Kelvins) his calculation can be true.
Another problem is that any antenna, including your patch, has the main lobe and side lobes. The side lobes usually are pointed to ground or surrounding objects, so the ambient temperature of such objects (290-300 Kelvins) must be added to antenna output.

To make the situation simpler, your radiometer should have a voltage output proportional to antenna input. To calibrate the radiometer, you should connect one-two-three "standard" noise temperature sources to radiometer input, best chosen to be close to the expected antenna temperature. Usual choices are 290 K ambient, 350-400 K for medical objects, etc.
Then you could set the output voltmeter to "zero" for the ambient noise source; with the antenna pointed to the tested object, your voltmeter will read the difference between the ambient and unknown temperature, including the 290 K background, emitted from the ground, walls, etc.
Real experiment will show you best what to do. Calculations are only estimates- and you will be surprised when you touch the real world! I promise!

Jiripolivka: What is wrong with my approach?

When I reread my postings, I took the background into account (but without a calculation example) and mentioned gain depencence.

I know in real world that you need to take into account G(rho, theta) and T(rho, theta), but even without using this it is clear that the 1 cm small target doesn't contribute much to the overall antenna noise temperature.

Dear WimRFP:

your approach is not wrong but not complete. As I wrote, the complex scene is often overlooked if a response of angularly small object is to be detected.
This is why I recommend to run a real test; I have exchanged e-mail with the requester earlier. He attempts to design a medical radiometer. With this type of application, the radiometric equations you used in your calculation will also be affected by the dispersion in the tissue, attenuated by the air/tissue boundary, and masked by local-oscillator return if a superheterodyne receiver is used.
People are used to see a scene in the visible light with a considerable contrast due to various objects. The same scene at microwaves has often a smaller contrast while unnoticed background becomes more important.

From the first posting I thought it was more like a home work question then a real application. I agree that this application involves more the we discussed here.

As far as I know, the contribution to the total noise temperature of Ttarget*0.01*(percentage of dissipated transmitting power) does hold always (when you would use the receiver antenna as a transmitting antenna).

In my opinion, it will be almost impossible to detect the actual temperature of a 10 mm diameter target at 100mm distance from an antenna with -3 dB BW of about 60 degrees.

I do agree that the small target will be almost undetectable under given conditions.

In radio astronomy, this suggests why large antennas compared to wave length are needed. Then the best for distant objects with a small viewing angle, VLBI offers the best approach.

In medical applications of radiometry it is always almost impossible to model a real main lobe in lossy and complex tissue. Instead of antennas applicators are used (small cavities with one side open) often filled with alumina powder to improve the transition through the skin boundary. Still a reliable temperature measurement of internal objects is almost impossible.

Dr. Jiripolivka............Thank you so much for your help... ........I'm designing two radiometers at the same time....one is for medical application (Near field) and the second is for industrial applications (Far field)........I was asking about the temperature antenna for far field application.

Regards

In both cases, you should consider the COMPLETE SCENE what your radiometer antenna is "looking at".
In medical application, the antenna is named "applicator" as its aperture is applied to tissue surface. Due to tissue complex impedance, the impedance matching between the applicator and tissue boundary is one of the problems; the other is that the applicator radiation pattern is significantly deformed due to tissue impedance and inner structure. Often one cannot know what "hot spot" inside of the tissue is "seen" by the applicator.
In the industrial application, the complete scene consists of the surrounding objects like walls and other structures which can or cannot emit the ambient temperature (absorbing walls often do but metal objects have a low emissivity, they rather reflect other objects' emission). Important interference comes from windows (low temperature from the sky), from fluorescent lights (typically 10 00 Kelvins), persons moving around (35 deg.C ~ 330 Kelvins), etc. On such background it may be (and is) difficult to detect small objects like yours.
In microwave radiometry, one advice is to use "matched" antennas- their radiation patterns (main beam angle) should be approximately equal or not too larger than the viewing angles of the observed object. Your planar antenna is apparently quite unsuitable due to its too wide beam; try to use a better-matched antenna.

---------- Post added at 19:32 ---------- Previous post was at 19:31 ----------

correction: fluorescent lights emit rather 10 000 Kelvins approx. over 2....20 GHz

jiripolivka said:

In both cases, you should consider the COMPLETE SCENE what your radiometer antenna is "looking at".
In medical application, the antenna is named "applicator" as its aperture is applied to tissue surface. Due to tissue complex impedance, the impedance matching between the applicator and tissue boundary is one of the problems; the other is that the applicator radiation pattern is significantly deformed due to tissue impedance and inner structure. Often one cannot know what "hot spot" inside of the tissue is "seen" by the applicator.
In the industrial application, the complete scene consists of the surrounding objects like walls and other structures which can or cannot emit the ambient temperature (absorbing walls often do but metal objects have a low emissivity, they rather reflect other objects' emission). Important interference comes from windows (low temperature from the sky), from fluorescent lights (typically 10 00 Kelvins), persons moving around (35 deg.C ~ 330 Kelvins), etc. On such background it may be (and is) difficult to detect small objects like yours.
In microwave radiometry, one advice is to use "matched" antennas- their radiation patterns (main beam angle) should be approximately equal or not too larger than the viewing angles of the observed object. Your planar antenna is apparently quite unsuitable due to its too wide beam; try to use a better-matched antenna.

---------- Post added at 19:32 ---------- Previous post was at 19:31 ----------

correction: fluorescent lights emit rather 10 000 Kelvins approx. over 2....20 GHz

Click to expand...

Thank you so much for helping me..... I have already designed and fabricated a small horn antenna (3 cm X 3 cm) and it has 10 dBi gain and beamwidth about 50°.........I will build a box of absorbing matrial and I will put the antenna and the target (heated resistor) inside this box and I will see the output voltage of radiometer.

Regards

Your heated resistor is small with respect to its wavelength (1cm versus 3cm). So (anti) resonance may affect the actual solid viewing angle as seen by the radiometer's antenna. Also the ohmic value of the resistor may have affect.

Even the orientation of the "cold" wires that supply DC power to the resistor will affect your measurement as they provide a coupling path from the antenna to your 1cm large object. So heating the resistor by other means (with wires removed), may show other result. You may search for "mie scattering". It is not directly appricable to your case, but it may give you some idea of relation between size and scattering area.

Dear Raed:

In microwave radiometry it is good to look at your system as a photographic camera. Your system consists of a lens (antenna) and a radiometer (a CCD or film).
To observe various scenes, your antenna or lens should be adjusted to bring an observed target into focus, or, to get a maximum radiated energy into the sensor. In radiometry the approach is called "matching the antenna to a target".
This means that if you know the vision angle of a target, the antenna beam width should be as close to it if possible.
If the antenna beam is wider that the target vision angle, you can get from the antenna only a fraction of target radiated energy.
We measure target temperature over a distance. The antenna output temperature then is:

Ta = Tt x {vision angle squared}/{beam width squared}.

The squared angles are used here to replace the solid angles; the equation is only valid for circularly symmetrical target projections as well as antenna conical beams.
So far you have made and used too small antennas with too wide beams; then the radiometer cannot resolve the tiny temperature step over background. Try a larger antenna; you will see the difference.

Let us assume that Hadi wants to measure the temperature of his 1cm large target at 10cm distance. This target has a viewing angle, as seen from the detector (antenna), of 5.7 degrees.

Let us try to make an antenna with 5.7 degrees beam width.
The beam width is limited by the diffraction limit (57*lambda/size). This means the size (for example diameter of reflector) should be in the range of 0.3m (assuming good illumination). Such an antenna would have a gain in the range of about 28 dBi.

The far field distance (Fraunhofer distance) for a dish with D=0.3m and lambda=0.03m is about 3 m. His target is at 0.1m distance (that is in the Fresnel diffraction region). The beam at a distance of 0.1m will be about 0.3m wide (meters, not degrees). In fact it is a collimated beam at this distance.

All antenna design software that outputs gain or directivity assumes that the target is in the far field zone (Fraunhofer region), so you can't design a short sighted antenna that has to focus at short distance by using gain or directivity approach.

I think (with my limited experience with short sighted antennas) that with an aperture diameter of 0.1m, a spot size at 0.1m from the antenna of about 0.04m would be possible. If so, his target covers about 6% of the antenna's spot size.

A communication antenna with size of about 0.1m (think of a flat panel array) would have a directivity of about 18dBi. However as mentioned before it is useless as its focus is at infinity.