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Why IIP2 Matters in a Zero-IF Phone Receiver

By John · Updated · 6 min read

The second-order product rises 2 dB per dB of input, so P_IM2 = 2 P_in − IIP2. Two −30 dBm tones into an IIP2 of +40 dBm give a −100 dBm product at f2 − f1, right where a zero-IF receiver's wanted signal sits.

For practitionersNew to this? Plain-language version

For practitioners

A zero-IF receiver converts the wanted channel straight down to DC. That removes the image problem and the IF filter. It also creates a weakness that a heterodyne receiver mostly avoids: any second-order distortion that demodulates a strong blocker drops its products right at DC, on top of the wanted signal. IIP2 is the number that tells you how bad this will be.

What a second-order term does

Model the receiver's weak nonlinearity with its first two terms:

y=a1x+a2x2y = a_1 x + a_2 x^2

Apply two tones, x = A cos ω1t + A cos ω2t. The x² term expands into a DC term, second harmonics at 2ω1 and 2ω2, and two cross products:

a2x2⊃a2A2cos⁡((ω2−ω1)t)+a2A2cos⁡((ω1+ω2)t)a_2 x^2 \supset a_2 A^2 \cos((\omega_2 - \omega_1)t) + a_2 A^2 \cos((\omega_1 + \omega_2)t)

The sum product at f1 + f2 lands far above the band and is easy to filter. The difference product at f2 − f1 is the problem. It sits at low frequency no matter how high the tones are, and its frequency depends only on how far apart they are.

From two tones to a modulated blocker

A real blocker is not two tones. It is a modulated signal that occupies some RF bandwidth BW. Write it as x(t) = A(t) cos(ωt + φ(t)). Squaring gives:

x2(t)=A(t)22+A(t)22cos⁡(2ωt+2φ(t))x^2(t) = \frac{A(t)^2}{2} + \frac{A(t)^2}{2}\cos(2\omega t + 2\varphi(t))

The second term sits near twice the carrier and is harmless. The first term is the squared envelope. The x² term has demodulated the blocker, much like a square-law detector does. You can see the same result in the frequency domain: x² mixes every pair of spectral components inside the blocker, and each pair produces a beat at their frequency difference. The largest possible difference is the full occupied bandwidth. So the low-frequency product is centered at DC and extends up to BW.

How much energy this hump holds depends on the modulation. A blocker with a constant envelope puts its x² output mostly into a DC term. A blocker whose amplitude varies a lot spreads energy across the whole hump. So the same IIP2 can give very different interference depending on what the blocker carries.

Why zero-IF makes this a direct hit

AntennaRF filterLNAMixer (x² term)LOLO at wanted channelBasebandSpectrum 1: at RF(antenna, RF filter, LNA)Power (dB)RF filter passbandwantedmodulatedblockerBWf_wf1f2Frequency (RF)Spectrum 2: after mixer(baseband)Power (dB)interference lands onthe wanted signaldemodulated envelope(x² of blocker)f1 + f2product0 (DC)BWf1 + f2Frequency (baseband)
How the x² term turns an in-band modulated blocker into a hump from DC to BW that lands on the wanted signal in a zero-IF receiver. Concept drawing.

The chain in the figure runs antenna, RF filter, LNA, mixer, baseband. The LO sits at the wanted channel, so the wanted signal comes out of the mixer at DC. The blocker of bandwidth BW sits beside the wanted signal, inside the RF filter's passband. The filter cannot remove it without also removing the wanted channel.

The figure lumps the x² term into the mixer. After the mixer, the blocker's demodulated envelope is a hump centered at DC that extends up to BW. The wanted signal also sits at DC, under that hump. The baseband channel filter passes both, because the interference is in-band. The f1 + f2 product shows up far above as a small hump. It does no harm.

In a heterodyne receiver the same x² term still produces the low-frequency hump, but the wanted signal is at an IF away from DC, so the hump can be filtered out. In zero-IF, the wanted signal and the interference share the same frequencies. The only defense is to keep the second-order term small, and that is what IIP2 specifies.

IIP2: definition and the 2:1 slope

Plot input-referred output power against input power per tone, both in dBm. The fundamental rises with slope 1 dB/dB. The second-order product rises with slope 2 dB/dB, because its amplitude goes as A². Extend both lines straight. The input power where they cross is IIP2. From the two-tone expansion, the amplitudes are equal when a1 A = a2 A², so A = a1/a2. That point is usually far above any power the circuit can actually handle, which is why it is called an extrapolated point.

For equal tones of power P_in each, everything input-referred in dBm:

PIM2=2Pin−IIP2P_{\text{IM2}} = 2P_{\text{in}} - \text{IIP2}

Solve for the IIP2 needed to keep the product at or below a chosen limit:

IIP2=2Pin−PIM2,max\text{IIP2} = 2P_{\text{in}} - P_{\text{IM2,max}}

Worked example

Take two equal tones at P_in = −30 dBm each into a receiver with IIP2 = +40 dBm:

  • P_IM2 = 2(−30) − 40 = −100 dBm, input-referred, at f2 − f1.

Raise each tone by 10 dB to −20 dBm:

  • P_IM2 = 2(−20) − 40 = −80 dBm.

The input went up 10 dB and the product went up 20 dB. That is the 2:1 slope at work, and it is why a blocker that gets only a little stronger can quickly swamp a weak wanted signal.

Now turn it around. Suppose you decide the product must stay at or below −110 dBm with −30 dBm tones. Then IIP2 = 2(−30) − (−110) = +50 dBm. Each 1 dB tightening of the limit costs 1 dB of IIP2. Each 1 dB increase in blocker level costs 2 dB of IIP2.

Keep all terms input-referred and use the power of one tone, not the combined power of both. Mixing input-referred and output-referred numbers, or using total two-tone power, shifts the answer by the gain or by 3 dB and makes the comparison meaningless.

How IIP2 is achieved and what limits it

Even-order terms cancel in a balanced circuit. In a differential LNA or mixer, the wanted signal appears with opposite sign on the two halves, while the x² product appears with the same sign on both. Taking the difference cancels it. The cancellation is only as good as the match between the halves. Device mismatch, layout asymmetry and LO duty-cycle errors leave a residual second-order term, and that residual sets the IIP2 you actually get. This is why IIP2 tends to vary from part to part more than IIP3 does: it depends on how imperfect the balance is, not only on the nominal design.

In plain words · for students and beginners

IIP2 in plain words

An everyday picture

An old AM radio has a detector that throws away the radio carrier and keeps only how loud the signal is from moment to moment. That loudness pattern is the "envelope", and it is the audio you hear. A receiver's x² term (the part of the circuit's response that grows with the square of the input) acts like an accidental AM detector. Feed it a strong unwanted signal, called a "blocker", and it pulls out that signal's loudness pattern as a low-frequency hum.

Where the picture breaks down: the AM radio does this on purpose, to one station. The receiver does it by accident, to whatever strong signal is nearby, and the result lands exactly where the phone is trying to listen.

Why a zero-IF phone suffers

A "zero-IF" receiver shifts the wanted signal straight down to 0 Hz (DC). The accidental hum also sits around 0 Hz. So the two overlap. A filter at the antenna cannot help, because the blocker is close to the wanted channel and passes through with it.

The same example in simple steps

IIP2 is a rating in dBm (power compared with 1 milliwatt). Higher is better.

−60−40−200+20+40+50+50+40+200−20−40−60−80−100−120Input power per tone, Pin (dBm)Input-referred power (dBm)IIP2 = +40 dBmFundamental: P = Pinslope 1 dB/dB2nd-order product:PIM2 = 2 Pin − 40slope 2 dB/dB+20 dB out+10 dB inPin = −30 dBmPIM2 = −100 dBmPin = −20 dBmPIM2 = −80 dBm
Fundamental and second-order product versus tone power for a receiver with IIP2 of +40 dBm, where the product line rises 2 dB for each dB of input. Computed from the formula.
  1. Each of two test tones is −30 dBm. The receiver's IIP2 is +40 dBm.
  2. Double the tone power in dBm: 2 × (−30) = −60.
  3. Subtract IIP2: −60 − 40 = −100 dBm. That is the unwanted product.
  4. Make each tone 10 dB stronger, −20 dBm: 2 × (−20) − 40 = −80 dBm. The product grew 20 dB.

In the plot, the wanted signal's line climbs 1 dB for each dB in, and the product's line climbs 2 dB. The two lines meet at +40 dBm. That meeting point is the IIP2.

Mistakes beginners make

  • Thinking a 10 dB stronger blocker gives 10 dB more interference. It gives 20 dB more.
  • Expecting the RF filter to remove the problem. The blocker sits inside the filter's passband, and the product is made after the filter.
  • Using the combined power of both tones instead of the power of one tone.
  • Confusing IIP2 with IIP3. IIP3 describes third-order distortion, which has a different slope and different products.

Remember

  • P_IM2 = 2 P_in − IIP2, all in dBm and all referred to the input.
  • The second-order product rises 2 dB for every 1 dB of blocker power.
  • Differential circuits cancel the x² term, and mismatch between the two halves sets the real IIP2.

Frequently asked questions

Why is IIP2 more critical in a zero-IF receiver than in a heterodyne receiver?

In a zero-IF receiver the wanted signal is converted to DC. Second-order distortion demodulates the envelope of a strong blocker into a component centered at DC, so the interference lands on the wanted signal, and no RF filter can remove it after the fact.

How do I calculate the IM2 product from IIP2?

For two equal tones of power P_in each, the input-referred product is P_IM2 = 2 P_in − IIP2, all in dBm. With −30 dBm tones and an IIP2 of +40 dBm, the product is −100 dBm.

How much does the IM2 product grow when the blocker gets stronger?

It rises 2 dB for every 1 dB of input power. A 10 dB stronger blocker gives 20 dB more second-order product.

What limits IIP2 in a differential mixer?

A differential design cancels the even-order terms, but only as well as the two halves match. Mismatch between the halves leaves some second-order term behind, and that residual sets the IIP2 you actually get.

References and further reading

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