0
\$\begingroup\$

I was reading about Frequency modulated Continuous wave radar but still struggling to understand it. enter image description here

I have a few questions in the slide above. The blue line in the graph represents the transmitted signal and red the received signal. If the receiver output is pure sinusoid as is shown in lower graph why is its frequency vs time graph linear (red line). Shouldn't it contain a single frequency?
I also do not understand why the FFT of the received output gives us the range of the obstacle. And again, how its FFT gives us the velocity. Could anyone please explain this to me.
The snippet above is from slides from TU-Delft, which can be found here: http://www.slideshare.net/tobiasotto/principle-of-fmcw-radars

\$\endgroup\$
1
\$\begingroup\$
  1. The receiver's output is not the red line. Instead, it is the beat frequency between red line and blue line. f(beat) = f(blue)-f(red) = fb;

  2. The receiver calculates fb, which is the center frequency shown in FFT diagram. Knowing fb, you can calculate R from the formula $$R=c*Ts*Fb/2Bsweep $$

  3. Check out slide page 9. The diagram you posted indicates the condition when the detected object does not move. What may happen if the object moves? If the detected moves relative to receiver, doplar effect will start to work. Therefore you have a different frequency, which is doplar frequency.

Check the bottom diagram of slide page 9, it shows the condition when the object move: combination of doplar effect and previous frequency shift.

  1. I guess you may need to know one point. The transmitter is keeping sweeping the frequency, which is indicated by the blue line. In the diagram in page 9, the frequency increases and then decreases and keep doing that in cycles. As a result, beat frequency has the pattern shown. You can get both distance and speed from the beat frequency.
| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Thanks it makes clear how we calculate the distance....But in these slides I read that FFT of the range (distance) will give us the speed, could you explain this also. \$\endgroup\$ – sarthak Nov 7 '15 at 9:45
  • \$\begingroup\$ Just edited the answer. \$\endgroup\$ – richieqianle Nov 7 '15 at 9:59
  • \$\begingroup\$ I understood the doplar effect but how does the fft of the range helps to calculate this frequency. \$\endgroup\$ – sarthak Nov 7 '15 at 10:41
  • \$\begingroup\$ it is not only fft of the range. It is the fft of the receiver. The frequency profile includes effects from both range and dopler. \$\endgroup\$ – richieqianle Nov 7 '15 at 10:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.