I am following the book: Blind equalization and Identification by Zhi Ding and Ye Li. In Chapter 2, the concept of T spaced equalizers is presented. It is mentioned that the output of the channel is $x[k] = \sum_{i=-\infty}^\infty h[i]s[k-i] + w[k]$ where $h$ is the impulse response, $s$ are the symbols / source input and $w[k]$ is iid noise. Then it is defined that *

"When the channel is non-deal, its impulse response $h[k]$ has more than one nonzero component. Consequently, undesirable signal distrotion as the channel output $x[k]$ depends on multiple symbols in $\{s[k]\}$. This phenomenon is known as inter-symbol interference (ISI)."


I do not understand why the impulse response has to be zero when it is ideal channel - no ISI. If $h$ is zero then $h[i].s[k-i] = 0$ where $dot$ is the multiplication operator. Or am I mistaken and in the equation $h$ is convolved with $s$ and not multiplied and essentially impulse response will be zero for ideal channel. What is the concept and reason? I do know that convolution in time domain is equivalent to multiplication in frequency domain. But the above representation is in time domain. A brief explanation will work and then I can follow on from the key points mentioned in the answer. Thank you.


The book doesn't say that the impulse response must be zero for an ideal channel. It says that an ideal channel has exactly one, and not more than one, non-zero component, i.e. the ideal channel's impulse response is an impulse, which means that the signal is only delayed but not distorted.

| improve this answer | |
  • $\begingroup$ This is correct, except that the signal can also be attenuated or amplified, in addition to delayed. $\endgroup$ – MBaz Oct 30 '15 at 21:20
  • $\begingroup$ @MBaz: That could be, it all depends on the definition of 'distortionless'. I believe to remember that some sources define it as a pure delay, with a gain of unity. Anyway, good you pointed it out. $\endgroup$ – Matt L. Oct 30 '15 at 21:24
  • $\begingroup$ Yes, the strict definition of 'distortionless' is what you say. Since all physical channels attenuate the input anyway, in communications 'distortion' is usually reserved to describe a modification to the signal's shape beyond a constant gain. $\endgroup$ – MBaz Oct 30 '15 at 21:50
  • $\begingroup$ ISI is caused by smearing out of the data symbols over time by a dispersive channel, i.e. a channel with an impulse response that is not a single impulse. The longer the channel's impulse response, the more severe the ISI will be on average. In the time/sample domain, the signal is always convolved with the impulse response, that's how linear time-invariant systems work. You get multiplication in the frequency domain. $\endgroup$ – Matt L. Nov 2 '15 at 8:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.