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Suppose you transmit at a single tone $\omega\textrm{ rad/sec}$ and you have two receiving microphones $i=1,2$ which sample at $r$ samples per second. Assume you have a direct path with coefficients $h_{d,i}$ and a delay of $\tau_{d,i}$ and one reflected component with coefficients $h_{r,i}$ and a delay of $\tau_{r,i}$ at each microphone.

  • How does the received signal look in analog and digital time domain and also how does the FFT look like?

  • Do we still perceive only one spike in a particular bin in FFT? Or should you sample appropriately to achieve this?

  • How does the FFT look if we have two tones?

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  • $\begingroup$ This feels like you're asking us about a very basic introduction on what the spectrum of a signal and the spectral properties of a tapped delay line model are. Are you sure reading a good book isn't the preferred path here? $\endgroup$ Jul 13, 2016 at 19:31
  • $\begingroup$ @MarcusMüller It is not very explicit in books $\endgroup$
    – Turbo
    Jul 13, 2016 at 20:30
  • $\begingroup$ @Turbo: Any chance of marking answers to your questions as accepted? $\endgroup$
    – jojek
    Nov 21, 2016 at 16:38

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OK, so suppose your transmitted signal is $s(t)$ and the received signal is $r_i(t)$ in continuous (analog) time .

Then the received signal at the $i^\textrm{th}$ microphone is $$ r_i(t) = h_{d,i} s(t-\tau_{d,i}) + h_{r,i} s(t-\tau_{r,i}) $$

How does the received signal look in analog and digital time domain and also how does the FFT look like?

Well if $s(t) \stackrel{\textrm{FT}}{\longleftrightarrow} S(\omega)$ then $$ r(t) \stackrel{\textrm{FT}}{\longleftrightarrow} R(\omega) = \left [ h_{d,i} e^{-j\omega\tau_{d,i}} + h_{r,i} e^{-j\omega\tau_{r,i}} \right ] S(\omega) $$ which is simple application of the time shift property.

Do we still perceive only one spike in a particular bin in FFT? Or should you sample appropriately to achieve this?

Once you start talking about 'bins' they we move to discrete-time. In general, unless you are very careful with sampling, it's almost impossible to sample a single sinusoid and achieve a spike in just two bins (positive and negative frequencies). Achieving a spike in just one bin requires a complex exponential which is not really possible with real-world sampling unless you use a specific modulation scheme (which you don't mention).

How does the FFT look if we have two tones?

Again FFT implies discrete time. If you send two tones and sample them, then unless you choose your sampling scheme very cleverly, you will end up with energy in many bins. See this discussion for more details.

he math of this is what I am seeking. Assume baseband audio such as speech.

@Turbo If the signal is baseband audio, then there will never be just a single tone. Speech is generally modelled as an autoregressive (LPC) model driven by either white noise or a periodic pulse train. See figure 2.40 of this reference (screenshot included below for clarity).

enter image description here

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  • $\begingroup$ Thank you for the great post. "Achieving a spike in just one bin requires a complex exponential which is not really possible with real-world sampling unless you use a specific modulation scheme (which you don't mention)" The math of this is what I am seeking. Assume baseband audio such as speech. $\endgroup$
    – Turbo
    Jul 13, 2016 at 21:24
  • $\begingroup$ @Turbo See my updated answer. $\endgroup$
    – Peter K.
    Jul 13, 2016 at 21:32
  • $\begingroup$ I understand that but assume we have a tone say at $347$ and $401$ Hz. $\endgroup$
    – Turbo
    Jul 13, 2016 at 21:34
  • $\begingroup$ @Turbo So you'll get four spread-out peaks: $\pm 347$Hz and $\pm 401$Hz. They'll be spread out depending on how many samples you take and whether your sampling is synchronous with the sinusoids' generation. $\endgroup$
    – Peter K.
    Jul 13, 2016 at 22:12
  • $\begingroup$ "whether your sampling is synchronous with the sinusoids' generation" this is the math I am looking for (what happens at different sampling rates and when you are off synchronization and correctly synchronized) $\endgroup$
    – Turbo
    Jul 13, 2016 at 23:07

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