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IQ Imbalance: Gain and Phase Errors vs Image Rejection

By John · Updated · 6 min read

For small gain error ε and phase error θ in radians, image rejection is approximately 4/(ε²+θ²). A 1% gain error with 1 degree of phase error limits image rejection to about 40 dB.

For practitionersNew to this? Plain-language version

For practitioners

A quadrature mixer separates positive and negative frequencies only if the I and Q paths have the same gain and sit exactly 90 degrees apart. Any departure from that leaves some of the image in the output. This page derives how much, gives the small-error rule that every IQ budget uses, and works one number through.

The model: two errors between I and Q

Take the LO driving the I mixer as the reference, cos(ωt). The Q mixer is driven at an amplitude (1 + ε) and with a phase that misses 90 degrees by θ:

LOI=cos⁡(ωt),LOQ=−(1+ε)sin⁡(ωt+θ)LO_I = \cos(\omega t), \quad LO_Q = -(1+\varepsilon)\sin(\omega t + \theta)

ε is the relative gain mismatch between the paths. It covers the mixers, the baseband filters, the amplifiers and the ADCs, not only the LO. θ is the phase error in radians, meaning how far the I/Q phase difference is from exactly 90 degrees. It does not matter physically where each error arises. Only the net mismatch between the two paths at the output counts.

Write the complex LO as LO_I + j·LO_Q. It splits into two terms. The wanted term converts the signal at +f. The unwanted term converts the mirror frequency −f into the same output. The gains of the two terms are:

K1=1+(1+ε)e−jθ2,K2=1−(1+ε)ejθ2K_1 = \frac{1 + (1+\varepsilon)e^{-j\theta}}{2}, \quad K_2 = \frac{1 - (1+\varepsilon)e^{j\theta}}{2}

With ε = 0 and θ = 0, K_2 is exactly zero. The image disappears completely, so image rejection is infinite in theory.

Image rejection ratio: exact and small-error forms

Image rejection ratio (IRR) is the power gain of the wanted path divided by the power gain of the image path, |K_1|² / |K_2|². Expanding the two magnitudes gives the exact form:

IRR=1+2(1+ε)cos⁡θ+(1+ε)21−2(1+ε)cos⁡θ+(1+ε)2\text{IRR} = \frac{1 + 2(1+\varepsilon)\cos\theta + (1+\varepsilon)^2}{1 - 2(1+\varepsilon)\cos\theta + (1+\varepsilon)^2}

For small errors the numerator is close to 4. In the denominator, use cos θ ≈ 1 − θ²/2. The constant terms cancel, as do the linear ε terms, and what remains is ε² + θ²(1 + ε) ≈ ε² + θ². So:

IRR≈4ε2+θ2\text{IRR} \approx \frac{4}{\varepsilon^2 + \theta^2}
IRRdB=10log⁡10(4ε2+θ2)\text{IRR}_{\text{dB}} = 10\log_{10}\left(\frac{4}{\varepsilon^2 + \theta^2}\right)

Three points follow directly from this form.

  • Gain and phase errors add in power, as squares. Each one sets a ceiling on its own, and the two together always do worse than either alone.
  • The exchange rate is one to one in natural units. A gain error of 0.01 (1%) costs exactly as much as a phase error of 0.01 rad, which is about 0.57 degrees.
  • Halving both errors quarters ε² + θ², which adds about 6 dB of IRR. Each further 6 dB needs the errors halved again.

Worked example: 1% gain error and 1 degree phase error

Convert both errors to natural units first:

  • ε = 0.01, so ε² = 0.0001
  • θ = 1 degree = 0.01745 rad, so θ² = 0.0003046
  • ε² + θ² = 0.0004046
  • IRR = 4 / 0.0004046 = 9886
  • IRR in dB = 10 log10(9886) ≈ 39.95 dB

So a 1% and 1 degree receiver gives about 40 dB of image rejection. Look at each error on its own. The 1 degree phase error alone gives 4 / 0.0003046 = 13,132, about 41.2 dB. The 1% gain error alone gives 4 / 0.0001 = 40,000, about 46.0 dB. Phase is the larger term here, so it is the error to attack first.

Now tighten the phase to 0.5 degree (0.008727 rad) with no gain error. θ² = 0.00007615 and IRR = 4 / 0.00007615 = 52,525, about 47.2 dB. Going in the other direction, a 3% gain error alone gives 4 / 0.0009 = 4444, about 36.5 dB, and adding any phase error only lowers it further.

20304050607000.511.522.53Phase error θ (degrees)IRR (dB)ε = 0ε = 0.01ε = 0.03ε = 0 curve risestoward ∞ as θ → 047.2 dB atε = 0, θ = 0.5°39.95 dB atε = 0.01, θ = 1°
Image rejection ratio IRR = 10 log10(4 / (ε² + θ²)) versus phase error for gain errors of 0, 1% and 3%. Computed from the formula.

The plot shows IRR from 0 to 3 degrees of phase error for ε = 0, 0.01 and 0.03. The ε = 0 curve rises without limit as θ goes to zero. The two marked points are the worked numbers: 39.95 dB at ε = 0.01 with θ = 1 degree, and 47.2 dB at ε = 0 with θ = 0.5 degree. Each curve with a gain error flattens at small θ toward its gain-only ceiling. When that happens, further work on the phase buys little.

What the image does in a zero-IF receiver

In a heterodyne or low-IF receiver, the image is a different channel. IRR then sets how much an adjacent or blocking signal at the mirror frequency leaks onto the wanted one. In a zero-IF receiver the LO sits at the channel centre. The mirror of the signal's upper half is its own lower half, so the image of the wanted signal falls on the wanted signal itself.

That self-image does not show up as a separate spur. It shows up as a distortion of the constellation. The leaked component is the complex conjugate of the wanted signal scaled by K_2 / K_1, so it behaves like an error vector whose power is 1/IRR relative to the signal. For the worked case, 1/IRR = 1/9886, which is −39.95 dBc. The corresponding error amplitude is √(1/9886) ≈ 0.0101, so IQ imbalance alone contributes roughly 1% to the EVM. That contribution adds in power to the EVM terms from phase noise, noise and nonlinearity.

Using the rule in a budget

  • Work in radians and in fractional gain, not in dB of gain mismatch or degrees, until the last step.
  • Allocate ε² + θ² from the required IRR: ε² + θ² = 4 / IRR. A 40 dB target means ε² + θ² ≤ 0.0004.
  • Split the allocation according to which error is cheaper to control in your chain, then check each path element against its share.
  • Remember that both errors can vary with frequency across the baseband bandwidth. A single ε and θ describe one frequency, or a narrow band.
The 4/(ε²+θ²) rule assumes ε and θ are well below 1. For errors of a few percent and a few degrees it is accurate to a small fraction of a dB. For large errors, use the exact form above.

In plain words · for students and beginners

IQ imbalance in plain words

An everyday analogy

Think of two people carrying a long table through a doorway. To keep it level, they must lift at the same height (equal gain) and walk exactly in step (a 90 degree spacing in time). If one lifts a little lower, or is a little early, the table tilts. In a receiver the "tilt" is a copy of the signal at the mirror frequency, called the image. The analogy breaks down in one way: a tilted table is a single error you can see. In a receiver, the gain error and the phase error each add their own amount of image, and the two amounts add up.

Zero-IF receiverRF inputI mixerQ mixerLO0°90° + θI1 + εgain error εQθ: phase error, ε: gain errorPerfect paths (equal gain, exactly90°): infinite image rejection in theorySpectrum sketchlevel−f0+ffrequencywantedsignalmirrorimageimage leakdegrades EVM
Zero-IF receiver with gain error ε and phase error θ between the I and Q paths, and the resulting image leak from −f onto the wanted signal at +f. Concept drawing.

The picture shows a zero-IF receiver. Zero-IF means the radio signal is shifted straight down to zero frequency. The RF input feeds two mixers. A mixer is a circuit that multiplies the signal by a local oscillator (LO) tone to shift its frequency. The I path gets the LO at 0 degrees and the Q path gets it at 90 degrees plus a small error θ. There is also a small gain difference ε between the two paths. In the spectrum sketch, the wanted signal sits at +f and its mirror image at −f. A small part of the image lands on the signal and spoils its EVM. EVM (error vector magnitude) measures how far received symbols are from where they should be. If both paths were perfect, the image would cancel completely.

The worked example in simple steps

  1. The gain error is 1%. Write it as a number: ε = 0.01. Square it: 0.0001.
  2. The phase error is 1 degree. Convert to radians (multiply by π/180): θ = 0.01745. Square it: 0.0003046.
  3. Add the squares: 0.0001 + 0.0003046 = 0.0004046.
  4. Divide 4 by that: 4 / 0.0004046 = 9886. This is the image rejection ratio. The wanted signal comes out 9886 times stronger in power than the image.
  5. Convert to dB: 10 log10(9886) ≈ 39.95 dB, so about 40 dB.

If you make the phase error 0.5 degree and the gain error zero, you get 52,525, or about 47.2 dB.

Mistakes beginners make

  • Putting degrees into the formula. θ must be in radians, or the answer is wrong by a large factor.
  • Adding the dB values of each error. The errors add as squares before you take the log, not as dB afterwards.
  • Thinking a zero-IF receiver has no image problem because there is no separate image channel. The image is the signal's own mirror, and it lands on the signal.
  • Fixing only one error. If the gain error stays at 3%, even a perfect phase gives only about 36.5 dB.

Remember

  • Image rejection ≈ 4 / (ε² + θ²), with θ in radians.
  • 1% gain error and 1 degree phase error give about 40 dB.
  • In zero-IF, the leaked image lands on the signal and lowers EVM.

Frequently asked questions

What image rejection does 1% gain error and 1 degree phase error give?

With ε = 0.01 and θ = 0.01745 rad, ε² + θ² = 0.0004046. The image rejection ratio is 4 / 0.0004046 = 9886, about 39.95 dB.

What is the formula for image rejection from IQ imbalance?

For small errors, IRR ≈ 4 / (ε² + θ²), where ε is the relative gain mismatch and θ is the phase error in radians. In dB, IRR = 10 log10(4 / (ε² + θ²)).

Why does IQ imbalance degrade EVM in a zero-IF receiver?

In a zero-IF receiver the mirror image of the wanted signal falls on the signal itself. IQ imbalance leaks part of that image into the demodulated signal, and the leak acts as an error vector that degrades EVM.

Is gain error or phase error worse for image rejection?

Neither by nature. Both enter as squares in ε² + θ², so 0.01 of gain error (1%) costs the same as 0.01 rad of phase error (about 0.57 degrees).

References and further reading

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