I have been studying the two ray model in communication systems.

The text I'm studying has two signals, $u(t)$, which is Line of Sight and $u(t-t')$, which is the reflected signal.

It quotes,

If the transmitted signal is narrowband relative to the delay spread ($B \ll \frac{1}{t'}$) then $u(t)$ is approximately equal to $u(t-t')$.

  • What is the meaning of this statement?
  • How is the delay related to the bandwidth?

Sorry if the question wasn't clear the first time. For the source, one can refer to https://en.wikipedia.org/wiki/Two-ray_ground-reflection_model or Andrea Goldsmith's Wireless Communication (mine is page 34, chapter 2) which quotes the above.

Inverse signal bandwidth is ($\frac{1}{B}$) and inverse delay spread is ($\frac{1}{t'}$). So either ($B \ll \frac{1}{t'}$) or equivalently ($t' \ll \frac{1}{B}$).

So let me rephrase. How can I show that $u(t)$ is approximately equal to $u(t-t')$ given either ($B \ll \frac{1}{t'}$) or equivalently ($t' \ll \frac{1}{B}$). In other words, what is the proof of the sentence in quotes (or highlights)?

  • $\begingroup$ Regarding your first question: the meaning of the statement is... pretty much exactly written there. Can you elaborate on what you not understand? I think we could help you better then! Regarding your second question: Imagine your channel as a FIR filter. A larger delay spread means a longer filter, which "allows" you finer "selectivity" in frequency domain. $\endgroup$ Commented Jan 24, 2019 at 13:00
  • $\begingroup$ In addition to what @MarcusMüller said, your question title is about the meaning of "inverse signal bandwidth," but that phrase isn't mentioned in the question anywhere. $\endgroup$
    – Jason R
    Commented Jan 24, 2019 at 13:39
  • $\begingroup$ The first question isn't clear. I think we need more context. Can you provide the source with the page number? $\endgroup$
    – BlackMath
    Commented Jan 24, 2019 at 17:19
  • $\begingroup$ I have made edits to the questions based on your comments. $\endgroup$
    – Kartik
    Commented Jan 24, 2019 at 19:27

1 Answer 1


OK, now it's more clear. In a two ray model, you have tow versions of the signal; the line of sight $u(t)$, and the reflected version $u(t-\tau)$, where $\tau$ is the delay spread of the channel.

If $\tau<<\frac{1}{B_s} \simeq T_s$, where $B_s$ and $T_s$ are the bandwidth and duration of the signal, then the shift in the reflected version would be very small compared to the symbol duration. So, the LOS version would be received at time $t_0$ up to $t_0+T_s$, and the reflected version would be received at time $t_0+\tau$ up to $t_0+T_s+\tau$. But $\tau<<T_s$, and thus the two versions are received very close to each others. This answers your first question.

As for the second question, the maximum delay spread of the channel is related to something called the coherence bandwidth of the channel. Roughly speaking, the coherence bandwidth $B_c$ is inversely proportional to the maximum delay spread, in this case $\tau$. If the signal bandwidth $B_s\simeq \frac{1}{T_s}<B_c\simeq \frac{1}{\tau}$, then the channel is said to be frequency-flat fading, which means that the channel frequency response remains contact withing the signal bandwidth. In the time domain, this is translated to interference-free channel, since different symbols don't interfere with each others. On the other hand, if $B_s > B_c$, then the channel is called frequency-selective channel, as different frequency components of the signal are affected differently by the channel. In the time domain, this causes inter-symbol interference (ISI), since the maximum delay spread is larger that the symbol duration, and different symbols interfere with each others at the receiver.


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