When you separate the collected signals, each car’s channel must attenuate the other cars’ signals at least by 20 dB.

A speaker will be able to generate a sound which contains 3 different frequency values of sine waves. And also the amplitudes of the sine waves will be tunable, so that at each instant;

A(f1) = 0 (meaning does not exist at that time)

A(f2) = above_threshold ( higher amplitude w.r.t f3 )

A(f3) = below_threshold

Note: The f1,f2,f3 will be change their states for different scenarios. A means amplitude, f_i correspondes to the sine wave at that frequency.

Then a microphone(seperated by 10 cm from the speaker) should be able to collect this sound, distinguish different frequency components by also indicating their relative amplitudes, because "which of frequency components has higher amplitude which one has lower amplitude and which one is non-existing" is an important information that should be obtained.

This information will be used in a next step of decison mechanism, so I will need DC converted signal levels. To achieve that I plan to amplify signals and then full wave rectify them.

So, relating this part, there is a requirement which is quoted above. Please notice that car's signal means one of the three different frequencies.

  • \$\begingroup\$ You'll have to provide more context (more of the text you got that quote from.) \$\endgroup\$ – JRE Nov 30 '19 at 10:27
  • \$\begingroup\$ Better yet, more context and a link to the original text if it is online somewhere. \$\endgroup\$ – JRE Nov 30 '19 at 10:32
  • \$\begingroup\$ I don't think I can explain more, I have edited and added that the fact that incoming signals to the microphone first will be amplified. \$\endgroup\$ – muyustan Nov 30 '19 at 10:47
  • \$\begingroup\$ Still have no idea what you are working on. \$\endgroup\$ – JRE Nov 30 '19 at 10:51
  • \$\begingroup\$ Sure you can explain more. What does "each car's channel" mean? What are the sirens? Is their amplitude constant or is it going to vary with distance and reflections from buildings, etc.? Where is the microphone? What is this circuit supposed to do? Are you trying to measure the relative amplitude of each signal or what? Where did the quote come from? "I have edited and added that the fact that incoming signals to the microphone first will be amplified." Your question implies that they will first be filtered, then amplified but that probably doesn't make any difference. \$\endgroup\$ – Transistor Nov 30 '19 at 11:10


simulate this circuit – Schematic created using CircuitLab

Figure 1. Block diagram of the system. Because you have only three channels BP (band-pass 0.7 kHz could be replaced with an LP (low-pass) 0.7 kHz filter and PB 1.3 kHz could be replaced by a HP 1.3 kHz filter.

When you separate the collected signals, each car’s channel must attenuate the other cars’ signals at least by 20 dB.

This is telling you that the band-pass filters must attenuate the adjacent frequencies by at least 20 dB or 1/10. So if each band-pass filter gave an output of X Vp-p for its tuned frequency then it would give < X/10 Vp-p for either of the other two frequencies.

To achieve this with such close channel spacing you will have to calculate the attenuation rate required. For example, an attenuation rate of 20 dB/decade on a 1 kHz filter would attenuate by 20 dB at 100 Hz or 10 kHz. Because your channels are so close to each other you will need much steeper attenuation.

Since you are working with sine waves there is no need to full-wave rectify in the peak detection circuits. Half-wave rectification will give you the same accuracy.

  • \$\begingroup\$ if this is not an answer, I don't know what is! Thanks a lot, I will examine this answer and maybe comment here about it later also. 300 Hz is the maximum seperation between two adjacent freq. signals allowed. I also don't understand how fair is demanding this narrow spacing from students... \$\endgroup\$ – muyustan Nov 30 '19 at 12:54
  • \$\begingroup\$ if a 10x freq. displacement means 1 decade, then ,from 1kHz to 1.3kHz, will I need 20 dB / (0.13 decade) which means 153.8 dB/decade ? Am I doing right by behaving decade in a linear way? If it is right, then 153.8 dB/decade is, I guess, harder to achieve than 20dB/decade right? \$\endgroup\$ – muyustan Nov 30 '19 at 13:01
  • \$\begingroup\$ I don't work with filters and my knowledge is hobby related from many decades ago. I recommend you un-accept my answer (thanks) for a day or two to encourage other answers which will give different perspectives and insights. The frequency scale is going to be logarithmic so we'll have to adjust that 0.13 decade calculation. Let me think ... \$\endgroup\$ – Transistor Nov 30 '19 at 13:44
  • \$\begingroup\$ ok, I un-accepted your answer for now. \$\endgroup\$ – muyustan Nov 30 '19 at 13:56

It means that the filter for each channel must attenuate the other frequencies from the other channels by a factor of 10 of the their amplitudes.

So e.g. the 1kHz channel must attenuate the signals with frequencies of 0.7kHz and 1.3kHz by a factor of 10. (Not an easy filter task)


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