# IQ Mismatch and Image

In this paper they compare the performance of direct conversion and heterodyne receiver with IQ mismatch.

The paper says that heterodyne architecture performs better with IQ mismatch because the image is separated by 2 times Intermediate frequency (IF). But for direct conversion receivers the image and the desired signal lie on top of each other (figure 6 in the document). The paper says that the image signal is created by the IQ mismatch. But I don't understand what is this image signal.

• How can IQ mismatch create an image signal? And:
• Why it is separated by IF in heterodyne systems?
• You are talking about the text around Figures 5 and 6? – TimWescott Apr 13 '20 at 0:00
• @TimWescott Yes. – sarthak Apr 13 '20 at 4:19

Consider an IQ signal with amplitude or quadrature mismatch such as these QAM waveforms depicted below:

These can be decomposed into the sum of a perfectly balanced IQ signal and another signal that is modulating 0 to 180° with any fixed phase rotation. The perfect balanced IQ signal will be translated from the positive RF spectrum to baseband (as I depicted in the other post here Can Carrier Offset cause Image Problems) as a complex signal and thus if there was a carrier offset that would be maintained as an offset in the complex baseband signal but easy to correct for with a subsequent complex frequency translation. The signal that is on one axis only is a (potentially phase rotated) real signal and this will have alias distortion.

Consider an example with:

$$A\cos(\omega t)- jB\sin(\omega t))$$

If $$A=B$$ then this is simply $$Ae^{-j\omega t}$$. But if $$B = (1+\Delta)A$$ for example representing an amplitude imbalance of $$\Delta$$ then we get:

$$A\cos(\omega t)- j(1+\Delta)A\sin(\omega t))$$

$$=Ae^{-j\omega t} - j\Delta\sin(\omega t)$$

So the down-conversion that would have been image free with perfect IQ balance for a real input as follows:

$$r(t) = 2\cos(\omega_c t)$$ $$LO = \hat r(t) = e^{-j\omega_{LO} t}$$

$$r(t)\hat r(t) = 2\cos(\omega_c t)e^{-j\omega_{LO} t} = (e^{j\omega_c t}+e^{-j\omega_c t})e^{-j\omega_{LO} t}= e^{(\omega_c-\omega_{LO})t}+ e^{(-\omega_c-\omega_{LO})t}$$

Where the first component $$e^{(\omega_c-\omega_{LO})t}$$ is the only one that would appear at baseband.

With the amplitude impbalance as depicted above this example now becomes:

$$(e^{j\omega_c t}+e^{-j\omega_c t})(e^{-j\omega_{LO} t}- j\Delta\sin(\omega t))$$

Consisting of the same result above with a new term given as:

$$(e^{j\omega_c t}+e^{-j\omega_c t})(-j\Delta\sin(\omega t))) = -(e^{j\omega_c t}+e^{-j\omega_c t})\frac{\Delta}{2}(e^{j\omega_{LO} t} - e^{-j\omega_{LO} t})$$

$$= -\frac{\Delta}{2}[e^{j(\omega_c+\omega_{LO})t}+ e^{j(\omega_c-\omega_{LO})t}+e^{j(-\omega_c+\omega_{LO})t}+e^{j(-\omega_c-\omega_{LO})t}]$$

Where we see we get two components at baseband, one $$-\frac{\Delta}{2}e^{j(\omega_c-\omega_{LO})t}$$ is constructive with our original signal (in this case decreasing the amplitude but no other distortion, while the other is an image $$-\frac{\Delta}{2}e^{j(-\omega_c+\omega_{LO})t}$$ which would cause serious distortion if the RF spectrums were not symmetric (as is typical for quadrature modulated signals) or if we had any frequency error between the carrier $$\omega_c$$ and the LO $$\omega_{LO}$$.

You can create a similar example with quadrature error, as well as the appearance of LO leakage with DC offset.

Consider this with the spectrums in a Zero-IF receiver as depicted below; the LO is depicted as a negative frequency only - with quadrature or amplitude error the positive frequency would start to appear starting with a very low level. You can follow the related spectrums when that occurs knowing the output is the convolution of the RF input spectrum with the LO spectrum to see the effect on the output.

With the super-heterodyne those overlapping images above would be separated appearing at their IF frequencies (and balanced since an IQ down-conversion would not be necessary from RF). But even a super-heterodyne receiver will have this same issue ultimately with quadrature and amplitude balance from the down-conversion at the IF frequency, since it does the down-conversion in two stages: first from RF to IF where we would then effectively have a Zero-IF down-converter from the IF frequency to baseband (either digitally or in the analog). However traditionally it has been a lot easier to achieve better amplitude and phase balance in the down-converter at the lower IF frequencies, as well as tighter filtering hence the superior performance of the super-heterodyne.

Thankfully with modern signal processing this is easy to detect and correct for, which has increased the popularity and use of Zero-IF receivers (and Direct RF transmitters). Below shows a typically correction approach for amplitude and phase imbalance:

How can IQ mismatch create an image signal?

IQ mismatch at receiver does not create image signal. Image signal is created by IQ Imbalance at transmitter. What IQ Imbalance at receiver does is, it is not able to nullify the effect of image signal if present. One big disadvantage of Direct Conversion receiver is the signal quality degradation due to mismatch of I/Q arms. Semiconductor companies spend millions of dollars (time+effort) just to come up with calibration algorithms to remove this imbalance. I digress, but here it is

Consider a single frequency baseband signal $$e^{j\omega_0t}$$, so baseband is $$\cos(\omega_0t) + j\sin(\omega_0t)$$. $$x_I = \cos(\omega_0t)$$, and $$x_Q = \sin(\omega_0t)$$. The pass band equivalent is $$x_P(t)=cos(\omega_0t)\cos(\omega_ct)-sin(\omega_0t)\sin(\omega_ct)$$. At the receiver, to get back In-phase component, we multiply by $$\cos(\omega_ct)$$ and Low Pass Filter to get back $$\cos(\omega_0t)$$. To get back quadrature component, we multiply by $$-\sin(\omega_ct)$$ and Low Pass Filter to get back $$\sin(\omega_0t)$$. In the following steps, I ignore the $$2\omega_c$$ terms for simplicity as they are rejected by LPF.

Consider an imbalanced quadrature LO signal $$(1+\delta)\sin(\omega_0t+\phi)$$, while the in-phase LO is still $$\cos(\omega_0t)$$.

For the in phase component we will obtain $$\cos(\omega_0t)$$. But at the quadrature arm $$\cos(\omega_0t+\omega_ct)(1+\delta)\sin(\omega_ct) \rightarrow (1+\delta)\sin(\omega_0t - \phi)$$ You can easily derive the above steps using basic trigonometric identities. So the baseband signals are $$x_I(t)=\cos(\omega_0t)$$ and $$x_Q(t) = (1+\delta)\sin(\omega_0t-\phi)$$. The final baseband signal is $$x_B(t) = x_I(t)+jx_Q(t) =$$

$$x_B(t) = \cos(\omega_0t)+j(1+\delta)\sin(\omega_0t-\phi)\\ = 0.5e^{j\omega_0t}+0.5e^{-j\omega_0t} + 0.5(1+\delta)e^{j\omega_0t-\phi}-0.5(1+\delta)e^{-j\omega_0t-\phi}$$ In the above expression, you can clearly contribution from $$-\omega_0$$ along side $$\omega_0$$. If $$\delta =0$$ and $$\phi=0$$, the term $$e^{-j\omega_0t}$$ would have got cancelled with 2nd and 4th terms in the above expression. Hence we prove the effect of image frequency interfering with desired frequency in Direct conversion receiver when there is I/Q Imbalance.

If your baseband signal is having bandwidth from $$-\omega_0$$ to $$+\omega_0$$, the baseband frequencies will interfere among each other along $$\omega =0$$. The interference is limited to within the desired channel as shown in Figure 6.

Why it is separated by IF in heterodyne systems?

In heterodyne system, you are not downconverting the passband signal using I/Q arm but using only $$\cos(\omega_ct-\omega_{IF}t)$$ which will result in $$\cos(\omega_0+\omega_{IF}t)$$ (after rejecting the component at $$2\omega_c$$). This signal has components at $$f_0+f_{IF}$$ as well as $$-f_0-f_{IF}$$. So the "images" are far apart at $$2 \times F_{IF}$$. Because next stage we have digital direct conversion and the images are apart by $$2\times f_{IF}$$, the Heterodyne receiver is more robust to IQ Imbalance.

• you said that image comes from IQ mismatchh at transmitter but explained it with IQ mismatch at receiver.... Also if the image is at -$\omega_0$ how is it in the signal band – sarthak Apr 13 '20 at 5:56
• To be clear, the final baseband signal will have $\omega_0$ component plus a fraction of whatever is present at $-\omega_0$. If there was noise at $-\omega_0$, it will get amplified by this fraction and added to signal at $\omega_0$. If there was some other signal at $-\omega_0$, it will also get amplified by this fraction and get added at $\omega_0$. Is it clear? $-\omega_0$ will be in signal band because your BPF is very wide in the Figure 6. $-\omega_0$ is only as far from $\omega=0$ as $\omega_0$. – jithin Apr 13 '20 at 6:05
• I considered a simple $e^{j\omega_0t}$ signal here. Actual signals will be having a composite of many such signals (example - OFDM) – jithin Apr 13 '20 at 6:06
• How IQ imbalance at Tx causes image - dsplog.com/2009/03/08/iq-imbalance-in-transmitter – jithin Apr 13 '20 at 6:07