In digital communications systems we say that the received signal can be expressed as

$$y(t)= x(t)+ n(t)\tag1$$

if x(t) is the transmitted signal and $n(t)$ is the noise.

I am wondering whether a better model would be

$$y(t)= x(t-t_0)+ n(t)\tag2$$

To account for delay caused by channel. Is my logic correct?

If so then, what are the assumptions made when using (1) and what are we accounting for if we use (2)? ie what type of errors are we accounting for (delay spread, or time acquistion error...)

I am looking forward to hearing your responses and discussion on this. thanks


1 Answer 1


Equation (1) in your question is a pure additive noise channel without any further distortion of the signal. Equation (2) takes a delay into account, but this is usually not relevant. A pure delay does not affect the quality of the communication link in terms of bit error rate. (I'm not talking about the annoying effect of delay in voice communication).

So the model in Equation (2) is often not relevant and can be replaced by Equation (1). A more general and useful model, which includes Eq. (2) as a special case, is


where $h(t)$ is the impulse response of the channel, and $*$ denotes convolution. This model represents a linear time-invariant channel with additive noise. Eq. (2) is obtained as a special case with $h(t)=\delta(t-t_0)$.

  • $\begingroup$ Thanks, actually I have come across (2) in LTE and IEEE 802.11 and in particular when discussing time synchornization problems. That is the start of FFT part of the signal. As I understand this is a major problem for OFDM. Do you have a good understanding of what time synchornization if OFDM systems is? Is it similar to timing offset. $\endgroup$
    – Tyrone
    Commented May 27, 2015 at 20:52
  • $\begingroup$ The reason I ask, is because 2 is used to model this problem the time synchronization problem. $\endgroup$
    – Tyrone
    Commented May 27, 2015 at 20:59
  • $\begingroup$ @Tyrone: I'm afraid I can't help with OFDM time synchronization. $\endgroup$
    – Matt L.
    Commented May 28, 2015 at 6:58
  • $\begingroup$ @Tyrone this has been discussed here: dsp.stackexchange.com/questions/7724/… $\endgroup$
    – Deve
    Commented May 28, 2015 at 7:00

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