Physics 728 

Prof. Dale E. Gary
NJIT 
Front End Receiving System
Power vs. Temperature
The power level of the radiation (W) can be traced from its reception by the feed, through the receiving system. The "signal" is generally noiselike (white noise, containing all frequencies in the band). For convenience, which will become clear, we often consider the equivalent noise temperature corresponding to the power levelP = kT Δν
although we also refer to the power level in decibel milliwatts [dBm].We can consider the power received by the antenna,
P_{a} = kT_{a}Δν, (1)
where T_{a} is the antenna temperature, and the output power of the receiver asNoise From The Source (i.e. the Signal) T_{a}.P_{tot} = P_{a} + P_{sys} => T_{tot} = T_{a} + T_{sys },where T_{sys} is the system temperature, and represents the added noise of the system. It is a figure of merit, and should be kept as low as possible. We can break the system temperature into several contributions:T_{sys} = T_{bg} + T_{sky} + T_{spill} + T_{loss} + T_{cal} + T_{rx },where
lump into T_{a} lump into T_{rx}T_{bg} = noise contribution from microwave and galactic backgroundsTo give you an appreciation for the magnitude of these contributions, the following table gives the VLA performance as an example:
T_{sky} = noise contribution from atmospheric emission
T_{spill} = noise contribution due to ground radiation (spillover and scattering)
T_{loss} = noise contribution due to losses in feed
T_{cal} = noise contribution due to injected noise
T_{rx} = receiver noise temperature
λ T_{bg} T_{sky} T_{spill} T_{loss} T_{cal} T_{rx} T_{sys} 92 cm 25 3 15 7 5 70 125 20 cm 3 3 14 8 2 30 60 6 cm 3 3 7 5 2 30 50 3.6 cm 3 2 5 2 2 16 30 2 cm 3 8 6 13 6 80 116 1.3 cm 3 17 6 21 7 100 154
 Note: T_{bg }, T_{sky }, and T_{spill} vary with position on the sky.
 T_{bg } is the 3 K cosmic background (leftover emission from the Big Bang), except at the lowest frequency where radio emission from the galaxy becomes important.
 T_{sky} = T_{sky}(1 − e^{−τ}o^{csc E}), where τ_{o} = zenith opacity, E = elevation angle. The value of T_{sky} is generally small except at high frequency where the atmosphere beocmes more opaque.
To see how important the unwanted T_{sys} is, let's compare it with a typical signal. Say we have a point source of flux density 1 Jy [= 10^{−26} W m^{−2} Hz^{−1}]. If observed with a radio telescope of 10 m diameter, what is T_{a}?Sensitivity of a Single DishS = 2kT_{a}ν^{2}/c^{2}Δν . (2)Note that ΔΩ > ΔΩ_{source} so use ΔΩ_{beam}. From θ_{FWHM} ~ λ/DΔΩ_{beam} = π/4 (θ_{FWHM})^{2} = π/4 (c^{2}/D^{2}ν^{2}) (3)Insert (3) into (2) and solve for T_{a},T_{a} = Sc^{2}/2kν^{2} 4/π (D^{2}ν^{2}/c^{2}) = 4SD^{2}/2πk = 4(10^{−26})(10^{2})/2π(1.38 x 10^{−23}) = 0.04 K!So in order to measure a 1 Jy source, we have to measure 0.04 K against of order 100 K of system noise. If we make the antenna more directive (D larger) then the situation improves. The 100 m Bonn telescope, or the GBT antenna would see 4 K from a 1 Jy source. The 1000 ft Arecibo dish would see ~40 K. However, recall from last time that the effective area of an antenna is reduced by the efficiency, A = ηA_{o}, with η ~ 0.5, so the problem is even worse. The appropriate antenna temperature expression for a source of flux density S isT_{a} = SηA/2k. (4)so Bonn has 1.5 K/Jy, Arecibo: 15 K/Jy. Clearly we are measuring very small signals.
We can take advantage of the fact that the signal will be correlated from one sample to the next, while the noise will not be, and "beat down the noise" by making many sample measurements and adding them up. According to the Nyquist theorem, a time series of measurements of signal of bandwidth Δν, of duration τ, will contain 2Δντ independent samples, so the noise should go down by the squareroot of the number of samples, orSources of Noise in Receiversthe Front EndΔT = T (2Δντ)^{−1/2 }, (5)where T is the equivalent temperature of the signal plus noise. A more general expression, from Crane and Napier (1989) and Anantharamaiah (1989) [an earlier version of the NRAO Summer School book] is
ΔT =[ T_{a}^{2} + T_{a}T_{sys} + T_{sys}^{2}/2 ] ^{1/2} (6) Δντ Normally when observing cosmic sources, we can ignore T_{a} relative to T_{sys }, which reduces to equation (5) with T = T_{sys}. However, what happens when we observe a strong source such as the Sun? The brightness temperature of the quiet Sun is about 10,000 K at, say, 10 GHz (recall homework problem set #1), so if we observe the Sun with a 25 m antenna with aperture efficiency ~ 0.5, the 10,000 K source will fill the beam and the antenna temperature will be 5,000 K. This is now much larger than the typical system temperature, so we say that the source dominates the noise, and the sensitivity of the antenna is now less than before. We will revisit this when we examine the sensitivity of an interferometer, and discuss the minimum noise level in radio images.
Let us now examine the signal path from the feed to the receiver. This chain is called the front end, and its characteristics dominate the noise of the system. A typical front end consists of the feed, some connecting cables, perhaps with filters or couplers, then a first stage lownoise amplifier (LNA), then typically a second stage LNA, followed by the receiver itself, as shown in the schematic below:A front end that I built for Lucent Technologies (the Solar Radio SpectroPolarimeter) is shown in the figure below left, while the EOVSA front end is shown below right: Before considering individual elements, we develop the concept of a general 2port device, as shown by the rectangle in the schematic below:
The input to this device is some hot resistor R, which generates noise P_{G} = kT_{o}Δν by virtue of its temperature T_{o} = ambient (~290 K). The device has some internal resistance giving added noise P_{N}. We can describe the gain of the device as G =P_{GN} /P_{G},so that the output power isP_{GN} = GkT_{o}Δν.
Now replace the 2port device with an ideal (lossless) device and adjust the input to give an output P_{N} (not P_{GN}). The temperature T of the external resistor required to attain output power P_{N} of the original device is called the noise temperature of the device. If the original device has no internal noise, then P_{N} = 0 and T = 0.We define the noise figure, NF, as NF = (P_{GN} + P_{N}) / P_{GN} (= 1 if no loss)
orSaturationNF = 1 + P_{N} / P_{GN} = 1 + GkTΔν/ GkT_{o}Δν = 1 + T / T_{o} . (6)So the noise figure is related to noise temperature byT = (NF − 1) T_{o} [T_{o} = ambient temperature, usually considered to be 290 K]Often, NF is given in dB:NF_{dB} = 10 log NFExample: Miteq LNA noise figure.Attenuators as 2Port
One type of 2port device is a "passive" attenuator. Transmission lines and cables between components have losses, and can be considered as an attenuator. The "gain" of an attenuator is G = ε < 1, and associated with this gain is the loss factor L = 1/ε.As an example, a 3 dB attenuator has
L_{dB} = 3 dB => L = 2 => ε = 0.5.As before, but replacing G with εP_{GN} = εkT_{o}Δν= kT_{o}Δν/L.The noise output of an attenuator isP_{N} = (1 − ε ) kT_{Phys}Δν,where T_{Phys}= physical temperature of cable or device.Then from (6), the noise figure of an attenuator is
NF = 1 + P_{N} / P_{GN} = 1 + (1 −ε) kT_{Phys}Dn/ e kT_{o}Dn= 1 + (L  1)T_{Phys} / T_{o} .But recall that the noise temperature isT = (NF − 1) T_{o} = (L − 1) T_{Phys} .The meaning of this is that a 3 dB attenuator (loss factor L = 2) will contribute T = T_{Phys} to the noise temperature of the system, i.e. about 290 K, if it is put in the front end (before the first amplifier). Thus, the attenuator cuts the input signal by 3 dB (factor of 2), but at the same time it also introduces 290 K of noise into the systema double whammy. But lossy components are sometimes used in low noise front ends, so how to we get away with it? To see that, we have to examine a series of 2port devices.Total System Temperature of a Series of 2Port Devices
An actual system is just a linear chain of 2port devices, as shown in the figure below:where the input (i.e. from the feed) is shown as T_{o}, and each twoport device is labeled with its noise temperature and gain. Some of the gains may be less than one (i.e. a lossy cable or attenuator). The power output of the whole system will be P = G_{1}G_{2}G_{3}...G_{n}kT_{o}Δν + G_{1}G_{2}G_{3}...G_{n}kT_{1}Δν + G_{2}G_{3}...G_{n}kT_{2}Δν + G_{3}...G_{n}kT_{3}Δν + ...and the corresponding system temperature isT = T_{o} + T_{1} + T_{2 }/ G_{1} + T_{3 }/ G_{1}G_{2} + ... + T_{n}_{ }/ G_{1}G_{2}...G_{n}_{−1} .You can see that the external temperature (the antenna temperature) just gets added to by all of the noise temperatures of the following devices, but each stage after the first stage gets divided by the total gains of the preceeding stages. This makes the first amplifier stage allimportant.Let's apply this to a real system, according to the block diagram below:
This is basically the block diagram for the system discussed at the beginning of this section. The total temperature of the system (from our introductory remarks above) is T_{sys} = T_{bg} + T_{sky} + T_{spill} + T_{loss} + T_{cal} + T_{rx },where now we can identify the receiver temperature as
lump into T_{a} lump into T_{rx}T_{rx} = (L− 1) T_{Phys} + LT_{2} + LT_{3 }/ G_{2} + ... .To be even more specific, say we have a system in which the cabling and coupler loss is 0.4 dB, => L = 1.10. Further, say the noise figure of the first amplifier is 2.5 dB =>T_{2} = (NF − 1) T_{Phys} = (1.778 − 1) 290 K = 225 K, with gain G_{2}= 25 dB = 316. Finally, say the noise figure of the second amplifier is 8 dB => T_{2} = (NF − 1) T_{Phys} = (6.31 − 1) 290 K = 1539 K. Plugging in these numbers (which are similar to the actual numbers for OVSA):T_{rx} = (L− 1) T_{Phys} + LT_{2} + LT_{3 }/ G_{2} = (1.10 − 1) 290 K + (1.10) 225 K + (1.10)1539 K/ 316Notice that the noise temperature of the first stage is all important, and as long as it has a lot of gain, following noise temperatures are not so important. Notice also that the loss term ahead of the first amplifier contributes to every following stage, not just the first, so if it is not small it will dominate. For example, there are two feeds used on the OVSA antennas, one of which has an intrinsic loss of 3 dB, while the other has very little loss. For homework, you will recalculate the receiver temperature using a loss of 3.4 dB.
= 29 K + 247.5 K + 5.3 K = 282 KNote that the noise figure, and noise temperature, are defined relative to the ambient temperature. What if we cool the front end (the lossy parts and the first stage amplifier) to, say, 30 K? Then the above receiver temperature would become:
T_{rx} = (1.10 − 1) 30 K + (1.10) 23 K + (1.10)1539 K_{ }/ 316so for the greatest sensitivity we will want to cool the receiver.
= 3 K + 25.3 K + 5.3 K = 33.6 KWe repeat, however, that if we are observing a bright source like the Sun, the Sun itself sets the system temperature because the receiver temperature is small compared to the antenna temperature. So it does no good to cool the receivers for a solar radiotelescope, except in so far as it is needed for calibration on cosmic sources.
Another issue for solar work is the large and rapid changes in total flux density from the Sun. The largest flare might be of order 10^{5} sfu, so let us see what this does to the signal out of the first amplifier of the OVSA system. The OVSA system is a broadband system, covering 118 GHz. First, what is the input power, P_{a}, due to a 10^{5} sfu flare, observed with one of OVSA's 27 m antennas? We insert (4) into (1) to getReceiverP_{a} = SηAΔν/2 ,where S = 10^{5}10^{−22} W m^{−2} Hz^{−1}= 10^{−17} W m^{−2} Hz^{−1}, η = 0.5, A = 572 m^{2}, and Δν = 1.7 x 10^{10} HzP_{a} = (10^{−17}) (0.5) (572) (1.7 x 10^{10})/2 = 2.43 x 10^{−5} W = 2.43 x 10^{−2} mW.Let's convert this to dBm, and use dB everywhere to do the calculation. To get dBm, take the log of the power in mW and multiply by 10, so 2.43 x 10^{−2} mW becomes −16.1 dBm. The cable/coupler loss (for the linear feed) is −0.4 dB, the first stage amplifier gain is about 25 dB, so the output of the first stage amplifier is −16.1 −0.4 + 25 dBm = 8.5 dBm. The first stage amplifier will have saturation problems at 10 dBm, so this is getting close to a problem with saturation. The second stage amplifier also has 25 dB of gain, but it is a "mixer/preamp" that combines the signal with a local oscillator signal and has a bandwidth of only 100 MHz. For this stage, the bandwidth is reduced by a factor of 17 GHz/100 MHz = 170, which is −22 dB, and it has a gain of 25 dB, for a net gain of 3 dB. Thus, the signal out of the mixer/preamp is 11.5 dBm, while saturation occurs at only 5 dBm. We conclude that the second stage will saturate badly on large flares! How do we solve this? We simply put a 10 dB attenuator between the first and second stage amplifiers. How much affect will this have on the receiver temperature?
−16.1 dBm −0.4 dB 25 dB 3 dB Net gain of each stage −16.1 dBm −16.5 dBm 8.5 dBm 11.5 dBm Output power
of each stage
To end this lecture, let us look at what happens to the signal after it leaves the front end and enters the receiver. The figure below is the block diagram for a downconverter module in the Expanded Owens Valley Solar Array.
This is a dualchannel receiver, with H pol and V pol signals being received separately on optical fiber. The signals enter on the left and exit on the right. The signals are converted to electrical form, go through two stages of frequency conversion, which selects a 400 MHz intermediate frequency (IF) portion of the incoming 118 GHz radio frequency (RF), adjusts its power level by means of amplifiers and attenuators, and then provided this clean 400 MHz IF to the digitizers in the correlator chassis (not shown). We will discuss more about receivers and following electronics when we get to correlators.