RF systems: noise and linearity

Welcome to part two, the part with the most exam points. Last year, three of the ten problems came from this lecture, a noise-figure cascade, a cascade intercept point and a compression and noise-floor question, and almost every other problem needs the decibel conversions we start with. We start with decibels, then noise and the noise figure, then non-linearity and intercept points, and we finish with sensitivity and dynamic range, which combine everything.

Radio engineers work in decibels because gains multiply and signal levels span many orders of magnitude. A power ratio in decibels is 10 times the logarithm of the ratio, so doubling the power adds 3.01 dB and multiplying by ten adds 10 dB. An absolute power needs a reference: dBW is relative to one watt, and dBm to one milliwatt, so the dBm value is always the dBW value plus 30. Voltages use 20 times the logarithm, because power goes with the square of the voltage, and the simple formula holds only when both voltages are across equal impedances. Voltage levels are given in dBV, relative to one volt, or dBµV, relative to one microvolt, and the dBµV value is the dBV value plus 120. Keep the rule in mind: 10 log for powers, 20 log for voltages, and a plain dB is always a ratio, never a level.

Let us do the three conversions from the slides. Ten times the logarithm of 0.05 is minus 13.01, so 0.05 watts is minus 13.01 dBW. Adding 30 gives 16.99 dBm, which you can check directly: 50 milliwatts over one milliwatt, ten log of 50, is 16.99. For the voltage, 20 times the logarithm of ten to the minus 3 is minus 60 dBV, and adding 120 gives 60 dBµV. Going back from minus 70 dBm, divide by ten and raise ten to that power: ten to the minus 7 milliwatts, which is 0.1 nanowatts, a typical received signal.

Decibels make cascades easy, but there is one classic trap. Gains in dB add, and a level in dBm plus a gain in dB gives a level in dBm, so 30 dBm through a 10 dB amplifier is 40 dBm; but two levels in dBm cannot be added: 30 dBm plus 30 dBm is one watt plus one watt, two watts, which is 33 dBm, not 60. For part a, 200 milliwatts out for one milliwatt in is a gain of 200, which is 23.01 dB. For part b, the attenuator halves the power, a loss of 2, which is 3.01 dB, or a gain of minus 3.01 dB. For part c, convert each level to milliwatts: one, plus 1.995, plus 0.251, which is 3.246 milliwatts, and ten log of that is 5.11 dBm: barely more than the largest term, as always when adding powers. And for part d, the gains in dB simply add, 3 times 23 minus 3 is 66 dB, a linear gain of almost four million.

Noise sets the smallest signal a receiver can detect, so it limits the range of every radio link. The basic source is thermal noise: in any resistor above absolute zero, the random agitation of charges produces a noise voltage whose r m s value is the square root of 4 k T B R. With Boltzmann's constant, 1.38 times ten to the minus 23 joules per kelvin, a 50 ohm resistor at 290 kelvin in a 1 megahertz bandwidth gives 0.89 microvolts. The power that this resistor delivers to a matched load is simply k T B, which does not depend on the resistance at all. At the reference temperature T zero of 290 kelvin, k T zero is 4 times ten to the minus 21 watts per hertz, which is minus 204 dBW per hertz, or minus 174 dBm per hertz: memorise this number, it appears in every noise problem. Devices add other noise as well: shot noise from the discrete nature of the current, and flicker noise, which grows as one over the frequency and matters mostly at low frequencies.

The quality of the received signal is measured by the signal-to-noise ratio, the signal power over the noise power, usually in decibels. Every real component adds noise of its own, so the SNR gets worse from input to output, and the noise figure measures by how much: the input SNR over the output SNR, at least one, or in decibels the input SNR minus the output SNR. It is defined with a matched noise source at the reference temperature of 290 kelvin at the input. Equivalently, the component can be described by its equivalent noise temperature T e: the output noise is k B G times T zero plus T e, so the noise factor is one plus T e over T zero, and T e is T zero times F minus one. A practical warning: the formulas use the linear factor F, while datasheets give the noise figure in decibels, so always convert before using the cascade formula.

When several stages are cascaded, their noise combines according to the Friis formula. The total noise factor is the first factor, plus the second factor minus one divided by the first gain, plus the third factor minus one divided by the product of the first two gains, and so on, all in linear units. Each stage's own noise is divided by the gain in front of it, because the signal has already been amplified when it gets there. So, provided its gain is high, the first stage dominates, and that is why every receiver starts with a low-noise amplifier with a high gain: the stages after it then hardly matter; with a lossy or low-gain first stage, compare the terms instead. A passive lossy component, such as a filter, a cable or a passive mixer at room temperature, has a gain of one over its loss and a noise factor equal to its loss, which is why a lossy cable in front of the LNA is so harmful. And the total gain in dB is just the sum of the gains minus the sum of the losses.

This is the worked example from the slides. First convert everything to linear: gains of 10, 0.794 and 0.501, and noise factors of 1.585, 1.259 and 2.512. Friis gives 1.585, plus 0.259 over 10, plus 1.512 over 10 times 0.794, which is 1.80, or 2.56 dB: notice how little the filter and the mixer add after the 10 dB amplifier. The gain is 10 minus 1 minus 3, which is 6 dB, a factor of 3.98. The equivalent noise temperature is 290 times 0.80, 232 kelvin. The antenna is colder than T zero, 150 kelvin, so the output noise is k B G times 150 plus 232: 1.38 times ten to the minus 23, times ten to the 7, times 3.98, times 382, which is 2.1 times ten to the minus 13 watts, minus 96.8 dBm; the slide rounds it to 2.08 times ten to the minus 13 watts with the same dBm value.

This was the first problem of last year's exam, and it is pure Friis. Two dB is a factor of 1.585, so the total factor is 1.585 plus 0.585 over 10, which is 1.643. In decibels that is 2.16 dB: the second amplifier adds only 0.16 dB. The equivalent noise temperature is 290 times 0.643, 186.6 kelvin; one solved copy wrote 185.6 because it rounded F to 1.64. For the output noise, add the decibels: minus 174 dBm per hertz, plus the noise figure, plus ten log of the bandwidth, plus the gain. That is minus 174 plus 2.16 plus 64.77 plus 20, which is minus 87.1 dBm, exactly as in both solved copies.

In decibels the noise calculations become a sum, which the bar chart builds step by step. Start from minus 174 dBm per hertz, add ten log of the bandwidth, add the noise figure, and add the gain if you want the noise at the output: that is the noise floor. Without the gain, the same sum is the noise referred to the input, k T zero B F. The sensitivity, the weakest signal that still gives the required output SNR, is the input noise plus that minimum SNR. With a 1 megahertz bandwidth, a 5 dB noise figure and a 10 dB minimum SNR, the bars give minus 174 plus 60 plus 5 plus 10, which is minus 99 dBm. Notice that each decade of bandwidth costs 10 dB, so a narrow-band receiver is much more sensitive than a wide-band one.

Every real amplifier is non-linear, and we model its output as a power series of the input: alpha zero plus alpha one times the input plus alpha two times its square plus alpha three times its cube. With one tone at the input, the squared and cubed terms generate a DC term and harmonics at twice and three times the frequency; they are far from the signal, so a filter removes them. The cubic term also changes the fundamental: its gain becomes alpha one plus three quarters alpha three A squared, and since alpha three is negative in practical amplifiers, the gain falls as the amplitude grows, which is gain compression. The real problem appears with two close tones: the cubic term produces third-order intermodulation products at twice omega one minus omega two and twice omega two minus omega one, right next to the wanted signals, in red, where no filter can remove them.

On a plot of output power against input power, both in dBm, the fundamental is a straight line of slope one until it starts to compress; the 1 dB compression point is where it falls 1 dB below the ideal line. The third-order products grow with the cube of the input, so on the decibel plot their line has a slope of 3: every extra dB of input adds 3 dB of intermodulation. Extending both straight lines, they meet at the third-order intercept point; its input coordinate is the IIP3, and its output coordinate the OIP3. The two are related by the gain, OIP3 equals IIP3 plus G, and for a cubic non-linearity the 1 dB compression point lies about 9.6 dB below the IIP3. A useful extra: the input-referred intermodulation level is 3 times the input power minus twice the IIP3, so the carrier-to-intermodulation ratio is twice the distance between the IIP3 and the input power.

For linearity the cascade formula works the other way round from Friis: the reciprocal of the total IIP3 is the sum of each stage's reciprocal IIP3, multiplied by the gain in front of it, so a stage weighs more the more gain sits in front of it and the lower its own IIP3: later stages are often the bottleneck, but not always. Convert to linear: gains of 1.585, 0.316 and 100, and intercepts of 10, 39.8 and 100 milliwatts. The sum is one over 10, plus 1.585 over 39.8, plus 1.585 times 0.316 over 100, which is 0.1448 per milliwatt; here the first stage contributes 69 %, the second 27 % and the last only 3.5 %, because the attenuator in front of it cuts its gain, so compare the terms instead of trusting a rule. Its inverse is 6.91 milliwatts, 8.4 dBm. The total gain is 2 minus 5 plus 20, 17 dB, so the OIP3 is 25.4 dBm. And the 1 dB compression point is about 9.6 dB below the IIP3, minus 1.2 dBm at the input.

This budget problem combines all three cascade calculations. The gain is the easy part: minus 2, plus 10, minus 4, plus 20, which is 24 dB. For the noise figure, the figure gives no noise figure for the filter and the mixer, so take each passive part's noise figure equal to its loss: 2 dB and 4 dB. Friis then gives 1.585 plus 0.927 plus 0.240 plus 0.396, which is 3.148, or 4.98 dB; the filter in front costs a full 2 dB. For the intercept point, the figure just says IP3; in Pozar, where this figure comes from, it means the output intercept, so subtract each stage's gain: the RF amplifier has an IIP3 of 0 dBm, the mixer 15 plus 4, 19 dBm, and the IF amplifier 20 minus 20, 0 dBm. The cascade sum is 0.631, plus 6.31 over 79.4, plus 2.51, which is 3.22 per milliwatt, so the input IP3 is minus 5.08 dBm: the IF amplifier, with the full gain in front of it, dominates. If you read the IP3 values as input intercepts instead, the answer is plus 5.41 dBm, so on the exam always write down which reading you used.

Last year's exam had a cascade intercept question: one solved copy crossed it out, and the other, where it is problem 1, got it wrong, so let us do it properly; the third block's gain is hidden on the copy, but it hardly matters. The datasheet values are output intercepts, so first subtract each stage's gain: the LNA has an IIP3 of 10 minus 20, minus 10 dBm, and the mixer 5 minus 15, also minus 10 dBm. In milliwatts, the sum is one over 0.1, plus 100 over 0.1, plus a negligible term for the very linear filter, about 1 010 per milliwatt. So the system IIP3 is about one microwatt, minus 30 dBm, set almost entirely by the mixer, which sees the full 20 dB of LNA gain; assuming the third block is a 2.5 dB loss, the net gain is 32.5 dB and the OIP3 is 2.5 dBm. With the given input of minus 40 dBm, the intermodulation products sit twice 10, that is 20 dB, below the carriers, and since all the noise figures are given, Friis gives a total noise factor of 1.615, a noise figure of 2.08 dB. The classmate's copy put the output intercepts straight into the formula as input intercepts and got minus 15 dBm, which was marked incorrect: always convert OIP3 to IIP3 first.

The sensitivity is the minimum input signal for which the receiver still delivers an acceptable output SNR. Start from the definition of the noise factor with the output SNR at its minimum allowed value and the input noise equal to k T zero B. Solving for the signal, the minimum power is k T zero, times the bandwidth, times the noise factor, times the minimum SNR. In decibels, the sensitivity in dBm is minus 174, plus the noise figure, plus ten log of the bandwidth, plus the minimum SNR. For a digital link the minimum SNR comes from a bit-error-rate curve: the SNR that gives the required BER. And the formula tells you how to improve a receiver: a better LNA, a narrower bandwidth, or a modulation that needs less SNR.

The satellite problem runs the sensitivity formula backwards. First convert the BER: 0.001 % is ten to the minus 5, and on the slide's curve that level is reached at an SNR of about 17 dB. Then solve the sensitivity formula for the noise figure: the sensitivity, plus 174, minus ten log of the bandwidth, minus the minimum SNR. Ten log of 3 000 is 34.77, so the noise figure is minus 118 plus 174 minus 34.77 minus 17, about 4.2 dB. Because the SNR is read from a graph, a dB either way is acceptable, so write down the value you read and carry it through.

The dynamic range is the ratio between the largest and the smallest levels for which the system behaves as desired. The bottom is set by the noise floor, minus 174 plus the noise figure plus ten log of the bandwidth plus the gain, typically between minus 60 and minus 100 dBm over the system bandwidth. The top is set by compression, so the linear dynamic range at the output is the output 1 dB compression point minus the output noise floor. Referred to the input, it is the input 1 dB compression point minus the sensitivity. To move the compression point between input and output, remember that the gain there is already compressed by one dB, so the input point is the output point minus the gain plus one.

Two quick output dynamic ranges. In the example, the noise floor is minus 174 plus 7 plus ten log of ten to the 8, which is 80, plus 40, giving minus 47 dBm. The dynamic range is 25 minus minus 47, which is 72 dB; the slide also does it in linear units, 316 milliwatts over 2 times ten to the minus 5 milliwatts, 1.58 times ten to the 7, the same 72 dB. In the problem, the noise floor is minus 174 plus 6 plus 73.01 plus 30, which is minus 65.0 dBm. So the output dynamic range is 21 plus 65, 86 dB.

The last problem on the slides puts everything together. The gain is minus 1 plus 10 minus 4 plus 20, 25 dB, and Friis, with the passive filter and mixer at their losses, gives a noise factor of 2.66, 4.25 dB. The IF bandwidth is 500 megahertz, so the sensitivity is minus 174 plus 4.25 plus 87 plus the required 20 dB, minus 62.8 dBm. Reading the IP3 values as output intercepts, as in the previous budget problem, the input intercepts are 0 dBm for the RF amplifier, 13 plus 4, 17 dBm for the mixer, and 16 minus 20, minus 4 dBm for the IF amplifier. The cascade sum is 0.794 plus 7.94 over 50.1 plus 3.16 over 0.398, which is 8.89 per milliwatt, so the input IP3 is minus 9.5 dBm. The 1 dB compression point is about 9.6 dB lower, minus 19.1 dBm. And the input dynamic range is minus 19.1 minus minus 62.8, which is 43.7 dB.

The last problem of last year's exam was given as a graph of output against input power. At the 1 dB compression point the gain is 12 dB instead of 13, so the input point is 20 minus 12, 8 dBm; one solved copy wrote 7, forgetting that the gain is compressed, so the exact answer is 8. Then the IIP3 is about 9.6 dB higher, 17.6 dBm, and the OIP3 adds the gain, 30.6 dBm. For the noise floor, the bandwidth from 1 to 10 GHz is 9 GHz, ten log of which is 99.54, so the floor is minus 174 plus 3.5 plus 99.54 plus 13, minus 57.96 dBm. That gives an output dynamic range of 78 dB, a useful extra line if the question asks for it.

Before the summary, the traps that cost points last year. Never add levels in dBm: convert to milliwatts, add, and convert back. Both cascade formulas need linear values, noise factors, gains and intercepts in milliwatts. In the Friis formula the first stage dominates if its gain is high, while in the intercept cascade each stage weighs its preceding gain over its own IIP3, which often makes a late stage the bottleneck, but always compare the terms. A passive lossy stage has a gain of minus its loss and a noise figure equal to its loss. If a datasheet gives the output intercept, subtract the gain to get the input intercept before using the cascade formula. And remember that the input compression point is the output one minus the gain plus one, and that minus 174 dBm per hertz holds only at 290 kelvin; for another source temperature use k T B directly.

To summarise part two: convert between watts, dBW and dBm, between volts, dBV and dBµV, and use 10 log for powers and 20 log for voltages. Noise starts at minus 174 dBm per hertz, the equivalent temperature is T zero times F minus one, and cascades follow Friis, with the first stage dominating when its gain is high. Linearity is described by the compression point and the intercept point, with OIP3 equal to IIP3 plus the gain, the compression point about 9.6 dB below the IIP3, and the intercept cascade weighted by the gain in front of each stage. The sensitivity and the noise floor are the same decibel sum with the minimum SNR or the gain added, and the dynamic range is the distance from the floor to the compression point. In part three we move to transmission lines, reflection, the Smith chart and matching.