# Digital Downconversion of Signal

DSP-Newbie here... Why are we using the ratio of the center frequency of the signal to the sampling frequency of the signal spectrum when we downconvert the signal?

I have an Spectrum with Carriers which reaches from $-f_{s}/2 ... f_{s}/2$ ($f_{s}$ is the Sampling-frequency) (see picture)

The downconversion/shifting of the spectrum is

$$e^{-j2\pi(LO/f_s)}$$ • Perhaps adding some context such as where you are using the ratio of center frequency to sampling frequency (write out a little as to what is being multiplied by what else and what happens next (e.g. filtering or integration or whatever) will help the reader understand what you are talking about. – Dilip Sarwate Jul 31 '18 at 17:13
• Do you mean downsampling or downconversion? Downconversion, to me, would be mixing two signals together, like what’s done when you’re lowering an RF signal into some IF which is over lower frequency; the opposite of this process is upconversion. Those operations do not change the number of samples, but they do change where they frequencies are centered. Conversely, upsampling and downsampling will change how many samples you have through an effective sample rate conversion. Which do you mean? – matthewjpollard Jul 31 '18 at 20:49
• Digital down conversion involves both center frequency shifting and sample rate reduction. Which stage are you referring to in this question ? – Fat32 Jul 31 '18 at 20:58
• @Fat32 I am wondering why we can shift the spectrum with $exp(-j2\pi f{c}/f{s})$ – John75 Aug 5 '18 at 12:04

For a centre frequency of $F_C$ for a modulated signal $v(t)$. $$r(t) = v_I(t) \sqrt{2} \cos 2\pi F_C t - v_Q(t) \sqrt{2}\sin 2\pi F_C t$$

Assuming a direct RF sampling at a rate of $F_S=1/T_S$ before downconversion, we plug $t=nT_S$ above. $$r(nT_S) = v_I(nT_S) \sqrt{2} \cos 2\pi F_C nT_S - v_Q(nT_S) \sqrt{2}\sin 2\pi F_C nT_S$$

The expression $F_C nT_S$ above becomes $\frac{F_C}{F_S}n$ to yield $$r(nT_S) = v_I(nT_S) \sqrt{2} \cos 2\pi \frac{F_C }{F_S} n - v_Q(nT_S) \sqrt{2}\sin 2\pi \frac{F_C }{F_S} n$$

Perhaps this is the ratio of the centre frequency to the sample rate you are asking about.

EDIT: Downconversion can be embedded in a downsampling operation as follows. In a conventional operation, we will have an LO downconverting the signal to baseband, filter it and then downsample it according to the symbol rate, as shown below. Here, the filter is operating at a higher rate which is unnecessary when we have to throw $M-1$ out of every $M$ samples anyway. This is a hint for what we need to do.

First, we interchange the operations of downconvesion and filtering so our filter needs to be implemented in passband now. If the lowpass filter impulse response is $h[n]$, the passband filter response is $$g[n] = h[n] e^{j2\pi \frac{F_C}{F_S} n}$$

The interchange of downconversion and filtering is shown below. Next, we slide the downsampling operation past the LO whose frequency gets multiplied by $M$, i.e., $F_C/F_S \cdot M$. We assume that $$F_C = \frac{F_S}{M}$$ then our LO frequency after sliding it past the downsampler becomes $$\frac{F_C}{F_S}M = \frac{F_S}{M}\frac{M}{F_S}=1$$ which implies that no downconversion is actually required! However, remember that filter is still operating it at a higher rate. So we slide the downsampler past the filter now and implement it as a polyphase filter. The original spectrum is shown below. This operation changes the spectrum such that all spectral replicas from $-F_S/2$ to $F_S/2$ end up at baseband overlapped on each other. The interesting part is that at the input of the polyphase arms, they possess a unique phase profile across the whole $360^{\circ}$ around the frequency IQ spectrum. A rough sketch is below. This is only true in an exact manner for the sinusoidal signals, so the rest of the task for a signal with bandwidth is performed by the filter. When implemented as a polyphase filter, this filter operates with the delays matching those of the spectral replicas shown above. A final summation at the end cancels everything out leaving the desired spectrum.

This also implies that if the desired spectrum is at $F_S/M$ and we want to employ the initial lowpass filter $h[n]$ instead of the bandpass filter $g[n]$, a corresponding rotation is required after each polyphase arm to line everything up before the final cancelation.

• I think that your answer has cleared my question @QasimChaudhari ! :) Why can we sample with a lower sampling-frequency than the highest signal frequency (Sampling at IF or something...)? And $n$ would be here in my case =1? – John75 Aug 5 '18 at 11:40
• So you are indeed asking about downconversion embedded in downsampling of the signal, as pointed out by @Fat32 earlier. This is a rather lengthy topic but I try to summarize it above. – QMC Aug 7 '18 at 2:05