# IQ Imbalance and Compensation in RF Transmitter

Make it to the Right and Larger Audience

Blog

# 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.

The following is site premium content.
Use points to gain access. You can either purchase points or contribute content and use contribution points to gain access.
Highlights: 1815 words, 37 images  Staff System Engineer I
Author brief is empty
Groups:

Tags:

## iq imbalancetransmitter

1. mazhar 3 months ago
0
-0

Great tutorial man! Thanks for sharing.

0
2. 0
-0

I m new to this. Nice introduction.

0
3. mutasem 4 years ago
0
-0

good one

0
4. JLee 4 years ago
0
-0

Good tutorial

4
5. DanDixon 5 years ago
0
-0

I m new to this. Nice introduction.

5
6. jmong 5 years ago
0
-0

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/.

5