I have two conflicting concepts when looking at the spectrum of a analog to digital signal that has been sampled.

  1. The nyquist zones. So any sampling frequency fs will make the spectrum replicate at baseband and integer multiples of fs causing the following:

enter image description here

  1. When a signal is sampled, the frequency is reduced by factor of fs. So the carrier fc will get divided by fs in this manner: enter image description here

Im confused as to how these two concepts interact. Why is new central frequency after sampling not showing this fc/fs in the first image? The signal in nyquist zone 1 centered at fs/4 or any other replica, why don't they fall at fc/fs?

This becomes even harder to understand when studying the digital down converter with this diagram: enter image description here

Which shows the mixer following the AD stage using the frequency of fo/fs implying the AD stage placed the sampled signal at fo/fs which is why it will end up in base band after going through the mixer. (*note i am considering fo and fc refer to the same frequency)


1 Answer 1


Sampling does not change the frequency of the signal. Normalized frequency ($f/f_s$) is just converting the frequency axis from units of Hz to units of cycles/sample. Once done all frequencies that extend from DC to the sampling rate in Hz would be converted to normalized units going from 0 to 1 cycle/sample. (So the normalized frequency 0.5 cycles/sample corresponds to half the sampling rate). Either is correct, it is just a matter of what units you want to give to the frequency axis. Another form of normalized frequency commonly used is radians/sample in which case DC to the sampling rate is given as 0 to $2\pi$ radians/sample.

Please see this other post where I go into deeper detail on what normalized frequency is and why we use it:

What is normalized frequency

That post also introduces the concept and use of complex signals, and the bottom diagram the OP provides above is a great reason to for clearly understanding this. Below shows the same drawing drawn with complex datapaths each having an in-phase and quadrature signal as $I + j Q$ :

complex diagram

The diagram shows the use of a quadrature splitter to convert a real signal into a complex signal (real in and $I + jQ$ out), which is done here with the "local oscillator" as $2\cos(\omega_{LO}t)$ and in the OP's case provided as an input at a normalized frequency of $f_o/f_s$. This is then multiplied by the incoming real signal using a complex multiplier. Multiplying a real signal times a complex signal requires two real multipliers as shown in the OP's diagram, and as we can see with the math below given a real signal as $I_R$ and the complex signal as $I_{LO}+jQ_{LO}$:

$$ I_R (I_{LO}+jQ_{LO}) = I_R I_{LO} + j I_R Q_{LO} $$

Reviewing the OP's diagram and the above formula and we see that is exactly what is occurring.

Also included in my graphic above are the spectrums with the top one being the input spectrum given as $2\cos(\omega_R t)$ which we see has a Fourier Transform with a positive and negative frequency component, consistent with Euler's formula with a $e^{+j\omega_R t}$ and $e^{-j\omega_R t}$. (Every impulse on the frequency domain is a "spinning phasor" in the time domain given as $e^{j\omega t}$). With the exponential complex form, the processing is quite simple to visualize since the exponents add when we multiply the signals:

$$(e^{j\omega_Rt}+e^{-j\omega_Rt})e^{-j\omega_{LO}t} = e^{j(\omega_R-\omega_{LO})t}+e^{-j(\omega_R +\omega_{LO})t}$$

After low pass filtering we get the spectrum given by $e^{j(\omega_R-\omega_{LO})t}$.

  • 1
    $\begingroup$ Hello Dan, thank you for this detailed response. It's very clear to me now where I went wrong. Also, really appreciate all the content you put out on DSP, it's been helping me so much and got very excited to see that you answered. If I had enough reputation I would have marked this as the perfect answer! $\endgroup$ Apr 15, 2022 at 21:58

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