Part 2 ended with mixers producing far more than the wanted sum and difference. Mixer intermodulation products are the unwanted outputs created when two or more signals combine in a nonlinear device, and the third-order pair is the most damaging because it lands close to the signals that create it. An intercept point is a single number that predicts the level of these products at any input power. This part explains where the products come from, how to calculate them, and how intercept points cascade through a receiver.
Sum and Difference Products at the IF Port
A downconverting mixer delivers both fRF + fLO and fRF − fLO at its IF port, at about equal levels. Usually the difference is wanted. The sum is sometimes loosely called the image, but the two are different things. The sum product is an output at a high frequency, removed by filtering after the mixer. The image, from Part 2, is an input at another RF frequency, and it must be removed before the mixer.
What happens to the sum matters more than it first appears. A conventional IF filter reflects energy outside its passband, so the sum returns to the mixer. There it combines with the second harmonic of the LO that the mixer generates internally, and the result falls right back on the IF.
Take fRF = 1960 MHz and fLO = 1776 MHz. The IF is 184 MHz and the sum is 3736 MHz. The LO second harmonic is 3552 MHz, and the difference between that and the sum is |3552 − 3736| = 184 MHz, exactly the IF. This secondary IF arrives with a phase set by the path length to the filter, so it adds to or subtracts from the primary IF and makes the conversion loss ripple with frequency. It also raises intermodulation.
Terminating the IF port
The cure is to terminate the IF port in a constant impedance at all the frequencies the mixer generates, so that the sum is absorbed and not reflected. A diplexer or constant-impedance filter does this. Return loss at the sum frequency decides how much comes back: 10 dB of return loss reflects 10% of the power, and 20 dB reflects 1%. Diode rings are particularly sensitive to this, and a mixer that meets its data sheet on a 50 Ω test bench can fall well short when it drives a reactive filter directly.
How Intermodulation Products Are Formed
A weakly nonlinear device can be modelled by a power series, y = a1x + a2x² + a3x³. Applying two tones f1 and f2 gives terms at every combination mf1 ± nf2, and the order of each product is m + n. A mixer behaves the same way, with the LO and its harmonics as one of the inputs, so general spurious responses appear at mfRF ± nfLO.
The table lists the main two-tone products for tones at 1950 and 1955 MHz. Even-order products fall near DC or near twice the carrier, where a band filter usually removes them. Odd-order products of the form (k+1)f1 − kf2 stay next to the tones, and no filter ahead of the channel filter can separate them if the tones themselves fall in the passband.
| Order | Product | Frequency (MHz) | Near the tones? |
|---|---|---|---|
| 2 | f2 − f1 | 5 | No, near DC |
| 2 | f1 + f2 | 3905 | No, near 2f |
| 3 | 2f1 − f2 | 1945 | Yes |
| 3 | 2f2 − f1 | 1960 | Yes |
| 3 | 2f1 + f2 | 5855 | No, near 3f |
| 4 | 2f2 − 2f1 | 10 | No, near DC |
| 5 | 3f1 − 2f2 | 1940 | Yes |
| 5 | 3f2 − 2f1 | 1965 | Yes |
Figure 1 shows why this matters. A receiver tuned to 1960 MHz with strong signals at 1950 and 1955 MHz will reject both of them in its channel filter, and neither one is on the wanted channel. Yet both reach the first active stages, because the band filter ahead of those stages passes them. The product 2f2 − f1 is then created inside the LNA or mixer, at exactly 1960 MHz, and the channel filter passes it as if it were the wanted signal. Selectivity after the point of distortion cannot help.
Mixers add a further response. The 2×2 spur, 2fRF − 2fLO, responds to an input halfway between the wanted signal and the LO. With the plan above, an interferer at 1868 MHz gives 2 × 1868 − 2 × 1776 = 184 MHz, the IF. This half-IF spur sits only 92 MHz from the wanted signal. Frequency planning is the discipline of keeping these responses out of the way.
Intercept Points and the 3:1 Slope
Each third-order product depends on the cube of the input, so it rises 3 dB for every 1 dB of input, while the fundamental rises 1 dB. On a log plot the two straight lines would meet at a point called the third-order intercept. Expressed at the input this is IIP3, and at the output OIP3, and OIP3 = IIP3 + gain. The device would compress long before reaching it, so the point is a mathematical extrapolation and cannot be measured directly. Its use is that the whole low-level behaviour follows from it.
Two equations do almost all the work. With both tones at Pin dBm per tone, and all levels referred to the input:
- IIP3 = Pin + ΔdBc / 2, where ΔdBc is the measured gap between the tone and the product.
- PIM3 = 3Pin − 2 IIP3.
Worked example: measuring and using IIP3
The two-tone test is the standard measurement. Two signal generators are combined through an isolating network, so that they do not intermodulate with each other, and the output goes to a spectrum analyzer whose own distortion must be well below the device under test. Suppose tones at −30 dBm each produce products 70 dB below the tones. Then IIP3 = −30 + 70/2 = +5 dBm, the case drawn in Figure 2.
Now apply this to a whole receiver with a cascaded IIP3 of −5.0 dBm, the value derived in the next section, and two interferers at −40 dBm each. The product referred to the input is 3(−40) − 2(−5.04) = −109.9 dBm. For a 5 MHz channel, with 4.5 MHz of noise bandwidth and a 6 dB noise figure, the noise floor is −101.4 dBm, so the product is 8.5 dB below it and does little harm. Raise the tones until the product equals the noise floor and the level is −37.2 dBm per tone. Because of the 3:1 slope, the product climbs 3 dB for each decibel beyond that, and it soon dominates the channel.
Cascading Intercept Points
A receiver is a chain, and each stage adds distortion referred back to the input through the gain ahead of it. Assuming the products add in phase, which is the safe worst case, and working in linear power:
1 / IIP3total = 1 / IIP31 + G1 / IIP32 + G1G2 / IIP33 + …
| Stage | Gain (dB) | IIP3 (dBm) | Share of total distortion |
|---|---|---|---|
| LNA | +15 | 0 | 31.3% |
| Mixer | −7 | +12 | 62.5% |
| IF amplifier | +20 | +15 | 6.2% |
The result is a cascaded IIP3 of −5.0 dBm, and an output intercept of 23.0 dBm after 28 dB of total gain. Figure 3 shows how the stages share the total. The mixer, despite having the best IIP3 of the front two stages, contributes 62.5%, because the 15 dB of LNA gain ahead of it raises the level it sees. Its own +12 dBm referred to the input is only −3 dBm.
This is the opposite of noise, which Part 4 will show is decided by the first stages. Later stages dominate linearity, and earlier stages dominate noise. Adding LNA gain helps sensitivity and harms IIP3, so LNA gain is a compromise and not something to maximize. The cascaded IP3 page treats the calculation in more depth, including cases where products do not add in phase.
Second-Order Distortion and IIP2
Even-order products were dismissed above because they fall far from the tones, but that holds only when the signal stays at RF. In a direct-conversion receiver the channel sits at baseband, and the second-order product f2 − f1 lands right in it. A single amplitude-modulated blocker does the same thing, because its envelope is demodulated to baseband by the second-order term.
The second-order intercept point works like IIP3, with a 2:1 slope. Input-referred, PIM2 = 2Pin − IIP2. Two tones at −30 dBm into a mixer with an IIP2 of +40 dBm give a product at −100 dBm. That is comparable to the −101.4 dBm noise floor used above, so homodyne receivers need IIP2 figures of +40 dBm or better, and they rely on circuit balance and calibration to reach them. A superheterodyne receiver, by contrast, filters the product away at the IF, and for that architecture IIP2 usually receives far less attention.
Reducing intermodulation in practice
- Filter early. A signal removed ahead of the first active stage cannot intermodulate, so band filtering ahead of the LNA is the first defence.
- Keep gain before the mixer modest. As the cascade shows, every decibel of LNA gain raises the level at the mixer by a decibel.
- Choose a strong mixer. Double-balanced rings and passive FET mixers with high LO drive reach higher intercept points, at the cost of LO power.
- Terminate the mixer well. A broadband match at the IF port prevents the sum product from returning.
Compression and Spurious-Free Dynamic Range
A related measure of large-signal behaviour is the 1 dB compression point. For a memoryless third-order nonlinearity the input P1dB lies 9.64 dB below IIP3, so the receiver above compresses near −14.7 dBm. Real devices often differ from that figure, particularly mixers and amplifiers with bias that shifts under drive, so the relation is a guide and not a substitute for a measurement.
Intercept point and noise floor together give spurious-free dynamic range, the span over which a signal is detectable with its third-order products still below the noise:
SFDR = (2/3)(IIP3 − N)
Here N is the input noise floor in dBm. For this chain SFDR = (2/3)(−5.0 + 101.4) = 64.3 dB. The factor of 2/3 comes straight from the 3:1 slope.
In practice a receiver also has to survive a single strong signal, which raises a separate set of issues: gain compression and the transfer of LO noise covered in Part 2. Receiver blocking and desensitization looks at that case.
Next in the Series
Part 4, Receiver Sensitivity, Noise and Front-End Amplifiers, moves from the top of the dynamic range to the bottom: kTB, noise figure, the Friis formula and the sensitivity budget that sets the weakest signal a receiver can use.