I recently designed the LPF of the IQ demodulator using the Butterworth LPF refering to https://dspillustrations.com/pages/posts/misc/baseband-up-and-downconversion-and-iq-modulation.html. But I have a question. If you look at the block in the picture below, you can see the baseband block after passing the A/D Conversion. I think I applied IIR digital filter, but the composition of the picture below shows that LPF is analog domain. So I'm suddenly very curious. Is this LPF analog filter or a digital filter?

enter image description here

I designed Lowpass filter corresponding to LP block using Butterworth IIR filter design method in the IQ demodulator block diagram above. I first calculated analog butterworth filter coefficient, and then I calculated IIR filter coefficient b,a of z-domain using the bilinear transform. And with this, we calculated the waveform through the lpf using the filter(b,a,x) function of the matlab. So, we got the spectrum for Inphase and Quadrature phase as below. But in this process, I have a question. Obviously, I designed a digital filter of z-domain, and I wonder why digital filter exists in analog domain because it is before ADC in block diagram. enter image description here enter image description here

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    $\begingroup$ The filters given by "LP" and $g(t)$ are in the analog domain. The article from that link is just showing you how to simulate it. $\endgroup$
    – Envidia
    Apr 15 '21 at 16:44
  • $\begingroup$ I designed LPF using the method of Butterworth IIR lowpass filter synthesis. so I consider LPF is digital filter. by the way why digital lowpass filter is in a analog domain? This is my question. $\endgroup$
    – agile
    Apr 16 '21 at 6:57

The A/D can be placed as a single real A/D before the multipliers, OR as shown in the diagram as two A/Ds one after each multiplier to sample the I and Q channels. In either case, an analog filter is required before any A/D conversion as an anti-alias filter. This can be a bandpass filter or a low-pass filter, depending on which image in the analog domain's frequency spectrum is desired as the signal of interest. Even if there is no actual additional signal in the image bands, with no filter (and gain in the receiver) the noise figure will be significantly degraded due to the accumulated noise floor from each of the images. An analog filter prior to any sampling is important.

So in this case given the placement of the A/D converter, the filters as shown would be the analog anti-alias filter that I describe.

If the reason for image rejection is confusing, please refer to this posts that details how aliasing occurs in the digital sampling process: Higher order harmonics during sampling

So in summary, for the implementation of an IQ demodulator, there could be two A/D converters if I and Q are to be sampled separately: one for the I channel and one for the Q channel resulting in a complex representation of the baseband signal. Alternatively there can be a single A/D converter prior to the multipliers and then the multiplier and filter structures as shown can be implemented all digitally (referred to as a Digital Down-Converter or DDC). In this case, a single analog anti-alias filter (band-pass or low pass as desired for the implementation) is still needed in front of the A/D converter, in addition to the digital low pass filters after the multipliers to select the low frequency component after the multipliers.

The block diagram as shown in the question does not make that clear that $v(t)$, $y(t)$ and $\hat d[k]$ are complex signals (although clearer if you read through the linked reference).

  • $\begingroup$ Thank you. Please read the contents below. I couldn't put the picture in the comments, so I wrote it in the answer box. $\endgroup$
    – agile
    Apr 18 '21 at 13:29
  • $\begingroup$ @JunghakAhn is my updated answer now clearer to you? $\endgroup$ Apr 19 '21 at 18:22

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