Excellent presentation Sandeep. Small technical details on Fourier transform computation and resolution are valuable and needed in successful implementation.
Answer at 14.40: Chirp A has better frequency resolution than Chirp B, of course. On the other hand, two objects with the same Δd will result in a larger frequency separation in Chirp B because of the higher slope rate, S (see S*τ in 7.20). So, Chirp B has lower frequency resolution than Chirp A but the IF frequencies of the two objects are farther apart between each other in the case of Chirp B than in Chirp A. The net result is that the range resolution is a function only of bandwidth and the two Chirps will have the same range resolution.
Thank you very much for the tutorial! I have a question on page 13. Shouldn't the 3 f of the IF signal (the horizontal lines) start at the time each of them is received, and end at the same time (where Tx ends)? Why are they all lined up at the beginning? In my understanding it should be like a upstair staircase. Bottom line should be the longest as it's the closet object Rx and receives the earliest. And such for the top two lines... Or am I understanding it wrong? Thank you!
Thanks for the amazing lecture! But I have a question on 14:31, page 19. Though chirpA has a longer duration which is good for differentiating two tones, this is only when the distance between two tones stays the same. With chirpB, even though it has a shorter duration, its two tones are now farther and easier to distinguish. p.s. I found other comments also point this out. Besides that, I wonder about the relation between capture duration Vs. frequency difference. For example, does double capture duration increase double frequency resolution?
Thanks a lot for this video. I just have some comments: Minute 13:50. In the plot it is written S = BTc , but since it is the slope it should be S = B/Tc. Minute 20:35 d_res = c/2B. This means that d_res is conversely proportional to Bandwidth. This implies that, by increasing the Bandwidth reduces de d_res. However in the blue rectangle it is written that Larger Chirp Bandwidth leads to better range resolution.
in 20:35 Note that the goal is to detect objects as close as possible. d_res expresses the distance between the objects and you want it to be as small as possible and that means B as large as possible
Better the range resolution mean smaller should be its value. That's why if you increase the Bandwidth the range resolution will go smaller in value which mean a much better range resolution.
Very good and interesting video, well explained. I was kind of confused by the derivation in slide 17 because of the mentioned formula S=BTc, which is incorrect. It should be S=B/Tc.
Thanks so much for the great sharing! Just one slight confusion on slide 11: where does this delta_f > 1/T come from? For instance, is the FT of finite long continuous sine function equivalent to the FT of product of sine function and rectangular window function of T?
it is due to actually performing FFT rather than FT, since FT is applicable only continuous systems (i.e., real life). Using a digital system (my guess is that these figures are produced with MATLAB, but it could be any other tool), the horizontal axis resolution in Hz of the right figure (frequency) is equal to 1/T, where T is the total length in sec of the left figure (time). Furthermore, the total length in Hz of the right figure is equal to 1/ts, where ts is the sampling period or equivalently the horizontal axis resolution in sec of the left figure. I hope this helps and feel free to correct or improve.
Excellent stuff! But, at 9:35, you have slight error in the diagram of "Multiple tones in the IF signal"; Lower tone should be longer than one above, and middle one longer than last upper one. Not a big deal, but people who experimenting with some modules may find it confusing. I am waiting my "cheap" $6US 24 GHz module to arrive... then fun can begin. Thanks.
yes, i totally agree, sir. That's great. Can you tell me which one you bought, cause it seems really cheap and i would be interested to have a look. Regards
Hi! Thanks so much for posting this! Little side note, I think there is a small mistake at slide 21. The sampling frequency should be more than double the f_IF_max in order to adhere to Shanon's sampling theorem. This means that : Fs >= 4(s*d_max/c). Feel free to point out if I made a mistake in my thought process! :) Cheers
@@sandeeprao3753 Wow, I didn't know there was a difference between a complex broadcast and a real broadcast regarding the sampling frequency! I'm about to finish my engineering degree and I'm only now realizing that it's true lol Thank you!
Shannon's sampling theorem states that a signal can be accurately reconstructed from its samples if the sampling frequency is at least twice the maximum frequency component present in the signal. In the context of a radio receiver, the intermediate frequency (IF) is the frequency of the signal after it has been down-converted from the original frequency to a lower frequency for further processing. The maximum frequency component of the signal at the IF is known as the maximum IF frequency (f_IF_max). The formula you provided, Fs >= 4(s*d_max/c), relates the sampling frequency (Fs) to the maximum IF frequency (f_IF_max) and the distance between the transmitter and receiver (d_max). By rearranging the formula, we can derive an expression for the maximum IF frequency: f_IF_max = c/(2sd_max) Substituting this expression into Shannon's sampling theorem, we get: Fs >= 2f_IF_max = c/(sd_max) Multiplying both sides by 2, we get: Fs >= 4*(s*d_max/c) Therefore, the formula you provided is correct.
On slide 21 (Digitizing IF signal), shouldn't the ADC sampling frequency be always greater than 2B? At 16:24 it is mentioned that complex baseband signal is assumed (hence half the nyquist rate of real signal), what is meant by real signal here?
Can someone please explain? ADC sampling frequency must be double the maximum signal frequency. On slide 21 it can be seen that Fs and Fif_max are equal.
For real-valued signals, the ADC sampling should be > 2B. However, for a complex valued signal, an ADC sampling of greater B suffices. See dsp.stackexchange.com/questions/672/complex-sampling-can-break-nyquist
Thanks for the great video! A question: at 4:25, the output of the mixer should also have a frequency part (w1+w2) right? This part is just filtered by the followed low-pass filter.
Yes, that is right. The w1+w2 component is filtered out, The w1-w2 component is a few 10's of MHz while the w1+w2 component is (for a 77GHz Radar) ~144GHz
I am confused between B and IF, where B is the bandwidth as subtraction between the frequencies of x1 and x2, while Intermediate frequency is a new signal out of the mixer that is having a frequency equal to the subtraction of the frequencies and phases of x1 and x2. I see them as the same while it seems they are not because when you gave the distance formula you said d = Fs c / (2S), while S = B / Tc and Fs = 2 x IF, which shows that IF is not the same as B.
B refers to the bandwidth of the transmitted chirp(see for e.g ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-8cHACNNDWD8.html ) . The signals x1 and x2 refer to the transmitted chirp and the return chirp (i.e., a delayed version of the transmitted chirp) respectively. The frequency of the difference between the two (x1-x2) is IF and is different from B.
Thanks alot! But there are a lot of things that I don't understand or that don't work out for me even after watching the video about 8 times **First of all, I need to understand what the answer is to the 2 questions asked in the video. **What is the effect of Tc vs a larger slope? **What is the difference between frequency separation vs frequency resolution? **What defines the IF bandwidth **Which of the parameters (B,Tc,IF,d,S,τ,Fs) is responsible for: - Maximum distance - Ability to distinguish between nearby objects
Well, after some thought, if I understand correctly: The IF resolution, meaning the amount of bandwidth to sample the IF frequency which means higher\lower Fs and sample components If B is constant, the larger Tc is, the less the requirement for sampling frequency is, and this indicates a longer sampling time, which means something (??) If B is constant, the larger S is, the less the requirement for a high sampling time, which means a high sampling frequency, and this indicates a higher IF resolution, which means a greater distance discrimination ability (??) The concept responsible for the maximum distance is the IF resolution, because there the Fs is set, which is the sampling frequency that dmax is limited to, and the parameter responsible for the IF resolution are the parameters S, Tc by determining the necessary sampling components The parameter responsible for the ability to distinguish between nearby objects is only B that expressed in the formula they developed in the video (speed of light divided by 2B) Regarding the differences between frequency separation and frequency resolution I still haven't understood, even ChatGPT doesn't give a logical answer.
For small chances in delta_d, the change in frequency is too small to be measured accurately. Change in phase is much larger . See for e.g. module 2 at: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-bB-SGw9uRgQ.html
Yes, you can use an electrical spectrum analyzer (ESA) to measure the FFT signal, which is a frequency-power graph of a signal. An ESA is a device that measures the spectrum of an electrical signal, i.e., it provides a frequency-domain representation of the signal. The ESA works by downconverting the signal to a lower frequency range, filtering out unwanted frequencies, and then amplifying and measuring the resulting signal. The output of an ESA is typically a frequency-power graph, which shows the power of the signal at each frequency. To measure the FFT signal using an ESA, you would need to first convert the time-domain signal to a frequency-domain signal using an FFT algorithm. Once you have the FFT signal, you can then feed it into the ESA and measure its frequency spectrum. This would give you a frequency-power graph of the FFT signal. Note that the accuracy of the measurement would depend on the frequency range and resolution of the ESA, as well as the sampling rate and length of the FFT used to compute the FFT signal.
Sorry for the late reply. This is because if two tones have a very less frequency difference they can't be resolved as two different tones in the frequency domain. Longer the observation period the better the range resolution. In general, an observation of window of T can separate frequency components that are separated by more than 1/T Hz. Watch the same video at 08 mins 22sec
As shown in page 11, in one observation window T, it must contain at least 1 cycle difference. cycles = freq * observation Window T. so | f1 * T - f2*T| > 1 then delta Freq > 1/T
I think they are the same resolution. the one with longer T has smaller delta freq compare to the steeper slope one, which has a larger delta freq, with the same delta time. With delta freq > 1/Tc, Tc has to be longer for the smaller delta freq to achieve the separating peaks.
