# Doppler frequency when down-converting a receiving signal

Suppose we have a monostatic radar system, with only one target crossing at a speed $$v$$ at the main lobe of the antenna. The Doppler frequency for this case will be given by

$$f_d = \frac{2v}{\lambda} = \frac{2v \cdot f}{c}$$

With $$f$$ being the carrier frequency of the received signal. However, if we downconvert, $$f = 0$$, which would give a $$f_d = 0$$, which does not sound right to me. So, how does downconversion affect the Doppler signature?

• Um, what are you doing to end up with $f=0$? that's wrong. Commented Apr 29, 2020 at 20:37

When there is relative motion between the transmitter and receiver the frequency seen by the receiver is not $$F_c$$, so let's say you incur a Doppler shift of $$\Delta_f$$, then the shifted signal frqeuency is $$F_c - \Delta _f$$ now when you downconvert with $$F_c$$ the frequency offset of $$\Delta_f$$ still remains in the signal. That means the baseband spectrum "centred around freqeuency" of $$0$$ will be shifted by this value

In the receivers, this offset is estimated by the baseband signal processing (by performing correlation of time domain samples, linear shift in freqeuency domain means phase shift in time domain, so these correlations of samples simply leave the phase corresponding to the frequency offset due to doppler), there are other methods as well. Once these are estimated the clock at the receiver is further corrected with the estimation to tune the receiver to the "seen" signal freqeuency. So that the baseband spectrum is again centred around $$0$$ frequency

The simple answer is down-conversion is just a frequency translation. Consider the downcoversion explained by the cosine product rule:

$$\cos(\alpha)\cos(\beta) = \frac{1}{2}\cos(\alpha+\beta) + \frac{1}{2}cos(\alpha-\beta)$$

If we are down-converting a we are interested in the difference, which we get by low pass filtering. So consider the down-conversion of our carrier at $$f_1$$ shifted in frequency due to Doppler by $$f_\Delta$$.

$$y(t) = \cos(2\pi (f_1+f_\Delta) t) cos(2\pi f_{LO} t)$$

The difference term after low pass filtering is:

$$y_{LPF}(t) = \cos\big(2\pi (f_1-f_{LO}+f\Delta)t\big)$$

If we want to down-convert to "0" our local oscillator would also be at $$f_1$$:

$$y(t) = \cos(2\pi (f_1+f_\Delta) t) cos(2\pi f_1 t)$$

After low pass filtering the difference term would be:

$$y_{LPF}(t) = \frac{1}{2}\cos(2\pi (f_\Delta))$$

Notice that the down-conversion process does nothing to change the offset frequency from Doppler, regardless of what frequency we translate to.

When you downconvert your radar signal, it is based on the center frequency of your transmitted signal, $$f_c$$ (assuming that is the frequency you choose to downconvert with). However, if your received signal ($$f_r$$) has a doppler shift on it, the signal moves away from your center frequency. So the f on the right side of your equation becomes the difference between your original transmitted signal and the frequency of your received signal that is doppler shifted, $$f = f_c - f_r$$.

If no doppler shift occurs on the signal, then your $$f$$ becomes 0 when you downconvert.

Answer : Down-converting from $$f_c$$ to DC involves multiplying with locally generated $$f_c$$. Which would create 2 copies of the original rx signal at $$f_c$$. One at DC and one at $$2f_c$$. Then you pass this though a Lowpass filter to suppress the copy at $$2f_c$$. What you are left with it the copy at DC ($$f=0$$).

When there is a Doppler shift in the rx signal the frequency of the rx signal becomes $$f_c + f_d$$. Now if you perform the above processing to downconvert the rx signal, you will get one copy at $$2f_c + f_d$$ and one at $$f_d$$. Lowpass filtering removes the copy at $$2f_c+f_d$$ and what remains is the copy at $$f_d$$.