In homodyne reciever, we read about LO Leakage causing DC Offset. Say my output at downconversion is in analog voltage range +5 and -2, and there is a DC offset of 1 unit So, it can be added right? To make it +6 and -1 But everywhere i see DC offset shown in frequency domain is there any practical interpretation of it which I'm missing?


1 Answer 1


A DC offset in the frequency domain will be an impulse at $f=0$. The interpretation of this is a single tone interference: "DC" is no different spectrally than any other frequency when viewed as the Fourier Transform of a complex signal with positive and negative frequencies as the baseband spectrum centered about $f=0$ is identical to that centered about any carrier frequency at $f=f_c$ (assuming linear processing is involved). The DC offset if indeed at $f=0$ is easily removed by subtracting the mean in post processing or using a DC nulling filter as described in this post and this post.

To see how LO leakage produces a DC offset, consider a simple demonstration with a BPSK signal, or more specifically a DSB-SC modulated sinusoid (which is similar in form to BPSK):

The baseband waveform as a simple cosine with frequency $\omega_m$ radians/sec is given as:

$$x_b(t)= \cos(\omega_m t)$$

And a higher frequency carrier at $\omega_c$ radians/sec is given as:

$$x_c(t)= \cos(\omega_c t)$$

We modulate this using DSB-SC (Double-Sideband Supressed Carrier) by simply multiplying the carrier with the baseband waveform (as we would or could do with an analog mixer, or digitally on the sampled signals with a time domain sample by sample product) to get:

$$x_{mod}(t)= x_b(t)x_c(t)= \cos(\omega_m t)\cos(\omega_c t)$$

Using the cosine product rule we get:

$$= \frac{1}{2}\cos((\omega_c-\omega_m) t)+ \frac{1}{2}\cos((\omega_c+\omega_m) t)$$

Confirming that the only frequencies present are the two sidebands at $\omega_c \pm \omega_m$ with the carrier at $\omega_c$ completely suppressed (DSB-SC).

To demodulate this modulated signal, we again multiply by the carrier and then low pass filter to remove the high frequency result, giving us the original baseband signal. The modulated signal is typically referred to as the "RF" (Radio Frequency signal) input to the product, while the carrier is typically referred to as the "LO" (Local Oscillator):

$$x_{demod}(t) = \text{LPF}\{x_{mod}(t)\cos(\omega_c t)\}$$

$$ = \text{LPF}\bigg\{\bigg(\frac{1}{2}\cos((\omega_c-\omega_m) t)+ \frac{1}{2}\cos((\omega_c+\omega_m) t)\bigg)\cos(\omega_c t)\bigg\}$$

Noting with the earlier result of the cosine product rule the resulting sum and difference of the frequencies, we see from above the sum frequency will be at $2\omega_c-\omega_m$ and the difference frequency will be at $\omega_m$, so after low pass filtering, which removes the sum term we get the scaled recovery of our baseband signal:

$$x_{demod}(t) = \frac{1}{4}\cos(\omega_m t)$$

With this we can now easily see the impact of "carrier leakage". If the carrier were to "leak" to the RF input such that instead of the product as given above, we get the following instead:

$$x_{dem2}(t) = \text{LPF}\{(x_{mod}(t)+\alpha\cos(\omega_c t))\cos(\omega_c t)\}$$

Where $\alpha$ represents an additional leakage term now present at the RF input and added to the modulated signal $x_{mod}$, with typical values $\alpha<<1$. If we desired we could also appropriately reduce the unity coefficient in front of the LO signal, but I will keep it simple as a unity coefficient which will still show mathematically the resulting DC offset:

$$x_{dem2}(t) = \text{LPF}\{(x_{mod}(t)\cos(\omega_c t)+\alpha\cos(\omega_c t)\cos(\omega_c t)\}$$

$$= \text{LPF}\{x_{mod}(t)\cos(\omega_c t)\}+\text{LPF}\{\alpha\cos(\omega_c t)\cos(\omega_c t)\}$$

$$=x_{demod}(t) +\text{LPF}\{\alpha\cos(\omega_c t)\cos(\omega_c t)\}$$

$$=x_{demod}(t) + x_{dc}(t)$$

Observe how the second term above, as another cosine product will have a sum and difference frequency, with the sum frequency removed by the low pass filter. The difference frequency is the resulting DC offset and given as:

$$x_{dc}(t) = \frac{\alpha}{2}\cos(0) = \frac{\alpha}{2}$$

Note that the above showed the effect of DC offset from carrier leakage using sinusoids which I generally try to avoid. The explanation is so much simpler using complex waveforms and phasor descriptions but that first requires an understanding of the equivalence between a phasor representation and the explanation give above, which may initially come across as more complicated. Rather than repeat another derivation I will simply demonstrate how a DC offset appears with the representation of a complex signal as phasors in the graphic below. Here we see the samples of a 16-QAM waveform at the correct sampling locations for demodulation however there is an additional DC offset (with arbitrary magnitude and phase) such that the entire waveform is shifted off of the origin. To properly demodulate the signal we need to remove this offset which will appropriately recenter the waveform on the origin and within each decision boundary. The DC offset itself is represented by the small blue vector, here at approximately +45°.

DC offset

  • $\begingroup$ In homodyne reciever, if we have RF Freq f and LO Freq f_LO where f_LO = f, so we will have RF SIGNAL * LO SIGNAL * LO SIGNAL( LEAKAGE ONE) and then we filter it, can you pls suggest any mathematical description of it to understand it even way better that how LO Leakage will produce DC OFFSET when this whole equation pass through LPF? It'll be a great help. $\endgroup$ Jan 1, 2022 at 16:26
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    $\begingroup$ Yes I can do that. I'll update my answer to show you this. $\endgroup$ Jan 1, 2022 at 16:27
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    $\begingroup$ @ReddiSuresh I think I see what your confusion was, in thinking of "LO Leakage" as leaking from the LO to the baseband output directly (instead of to the RF as I clarified). Note that this "leakage" often is dominantly due to analog imbalances in the mixer circuit in addition to actual radiated emission from one port to the other. We can compensate for such leakage by injecting a small and opposite DC current into the LO port directly (or in the case of IQ mixers as would be used for the QAM graphic I incuded, a complex IQ offset compensation) $\endgroup$ Jan 1, 2022 at 17:31
  • $\begingroup$ Thanks for the additional help for clarification. $\endgroup$ Jan 1, 2022 at 17:51

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