# IQ Imbalance and Compensation in RF Transmitter

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# IQ Imbalance and Compensation in RF Transmitter

A typical RF transmitter is shown as below.

Ideally in-phase and quadrature components of the desired local oscillator (LO) should be balanced to have equal amplitude and 90 degree phase shift as
$LO_I(t)=\cos(\omega_ct)$
$LO_Q(t)=\sin(\omega_ct)$

In practice, due to circuit imperfection, the phase and gain are mismatched which is referred to as IQ imbalance:
$LO_I(t)=(1+\epsilon)\cos(\omega_ct+\theta)$
$LO_Q(t)=\sin(\omega_ct)$
where $\epsilon,\theta$ are normally time invariant and is small.

Denote in-phase and quadrature components of baseband modulator output as I(t) and Q(t). With ideal LO, the final output s(t) is
$s(t)=I(t)LO_I(t)+Q(t)LO_Q(t)$

With the absence of IQ imbalance, s(t) becomes
$s(t)=I(t)\cos(\omega_ct)+Q(t)\sin(\omega_ct)$

Next we’ll see how to measure and quantize IQ imbalance and further how to compensate IQ imbalance.

Assume baseband modulator outputs a sine single tone as
$I(t)=\cos(\omega_{IF}t)$
$Q(t)=\sin(\omega_{IF}t)$

Without IQ imbalance, we have
$s(t)=\cos(\omega_{IF}t)\cos(\omega_ct)+\sin(\omega_{IF}t)\sin(\omega_ct)=\cos(\omega_c-\omega_{IF})t$
So the RF output only has $f_c-f_{IF}$ item.

However with IQ imbalance present, we have
$s(t)=\cos(\omega_{IF}t)(1+\epsilon)\cos(\omega_ct+\theta)+\sin(\omega_{IF}t)\sin(\omega_ct)$

After some derivation:
$s(t)=\frac{1}{2}((2+\epsilon)\cos(\omega_c-\omega_{IF})t-(1+\epsilon)\theta\sin(\omega_c-\omega_{IF})t)+\frac{1}{2}(\epsilon\cos(\omega_c+\omega_{IF})t-(1+\epsilon)\theta\sin(\omega_c+\omega_{IF})t)$

So RF output has both the desired item at $f_c-f_{IF}$ and the undesired image item at $f_c+f_{IF}$. The image item is

$s_{image}(t)=\frac{1}{2}(\epsilon\cos(\omega_c+\omega_{IF})t-(1+\epsilon)\theta\sin(\omega_c+\omega_{IF})t)$

and its power is
$P_{image}=\epsilon^2+(1+\epsilon)^2\theta^2\simeq\epsilon^2+\theta^2$

The desired item is
$s_{desired}(t)=\frac{1}{2}((2+\epsilon)\cos(\omega_c-\omega_{IF})t-(1+\epsilon)\theta\sin(\omega_c-\omega_{IF})t)$

and its power is
$P_{desired}=(2+\epsilon)^2+(1+\epsilon)^2\theta^2\simeq4(1+\epsilon)$

Then we have the image rejaction ratio, IRR, defined as
$IRR=10\log_{10}(\frac{P_{desired}}{P_{image}})=6+10\log_{10}(\frac{1+\epsilon}{\epsilon^2+\theta^2})dB$
IRR can be measured at RF ouput port to determine and quantize IQ imbalance.

Next we’ll see how to achieve and implement IQ imbalance compensation in baseband.

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## iq imbalancetransmitter

1. mutasem 12 months ago
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good one

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2. JLee 2 years ago
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Good tutorial

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3. DanDixon 2 years ago
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I m new to this. Nice introduction.

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4. jmong 2 years ago
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Good article Korolev!
I am thinking TSSI can be used in this topic. It triggers me to write a short blog to explain it, http://www.valpont.com/tssi-and-use-tssi-in-iq-imbalance-compensation/pst/.

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