Intersymbol Interference

Module by: Don Johnson

One drawback of the general strategy employed here is that the bit interval has been lengthened to accommodate channel dispersion. In many cases, we cannot accept such limitations, or the transmitter cannot do anything about the dispersion. Consequently, there is "spillover" of what occurred in one bit interval into another.

For example, let the transmitted signal set, defined over the interval 0≤t<T 0 t T , be s 0 t=0 s 0 t 0 s 1 t=ET s 1 t E T and the channel be characterized by h CH tτ=ⅇ-at-τut-τ h CH t τ a t τ u t τ . A typical received signal would then resemble that shown in Figure 1. The effect of what happens in other bit intervals affecting the present interval is termed intersymbol interference (ISI). Many techniques have been developed to combat ISI; we explore only one here.

Let the signal set be an antipodal signal set. The entire transmitted signal is given by: st=∑k=-∞∞ B k s 0 t-kT s t k B k s 0 t k T where B k =±1 B k ± 1 according to the value of the k th k th bit. The signal s 0 t s 0 t is defined over 0T 0 T only. The received waveform is given by rt=∫-∞t h CH tτsτdτ+nt r t τ t h CH t τ s τ n t where nt n t is additive, white Gaussian noise. The received signal can be rewritten as rt=∑k=-∞∞ B k s 0 * t-kT+nt r t k B k s 0 * t k T n t where s 0 * t=∫0T h CH tτ s 0 τdτ s 0 * t τ 0 T h CH t τ s 0 τ . The presumption now is that the duration of s 0 * t s 0 * t is longer than TT. Let us further assume that the receiver has a front end containing a matched filter having impulse response h f tτ h f t τ , the output of which is sampled every t=lt t l t to produce the sequence r ~ l T r ~ l T .

r ~ l T =∑k=-∞l-1 B k s ~ lT-kT+ n ~ l T r ~ l T k l 1 B k s ~ l T k T n ~ l T If there were no intersymbol interference, s ~ lT-kT=0 s ~ l T k T 0 , k≠l k l , and we would compare this sequence with zero to determine optimally what was transmitted. When intersymbol interference is present, we can usually do better than that. Let s ~ l = s ~ lT s ~ l s ~ l T and n ~ l = n ~ lT n ~ l n ~ l T be the sample values of the signal and noise, respectively. We have r ~ l =∑k=-∞∞ B k s ~ l - k + n ~ l r ~ l k B k s ~ l - k n ~ l . As s ~ l =0 s ~ l 0 , l<0 l 0 , we have r ~ l =∑k=-∞l B k s ~ l - k + n ~ l r ~ l k l B k s ~ l - k n ~ l . The sum is a convolution sum; thus the transmitted sequence B l B l is filtered by a discrete-time filter having unit sample response s ~ l s ~ l .

One method of "recovering" B l B l would be to find the best inverse filter to remove the ISI. Another method is to use previous determinations of B l B l to remove the interference. Let B ^ l B ^ l denote the receiver's version of the transmitted sequence B l B l . Consider subtracting from the filtered and sampled received signal the terms corresponding to previous bit determinations. In essence, we are feeding-back previous decisions to help determine what the present bit might be. r ~ l -∑k=-∞l-1 B ^ k s ~ l - k = B l +∑k=-∞l-1 B k - B ^ k s ~ l - k + n ~ l r ~ l k l 1 B ^ k s ~ l - k B l k l 1 B k B ^ k s ~ l - k n ~ l (Note that we have normalized s ~ l s ~ l so that s ~ 0 =1 s ~ 0 1 ). Obviously, if the receiver does not make an error ( B ^ l = B l B ^ l B l ), this quantity equals B l + n ~ l B l n ~ l , the simple signal-in-noise situation, and we are left with the digital transmission scheme without ISI but with the unavoidable additive noise. One usually only "feedbacks" those B ~ l B ~ l for which s ~ l s ~ l is significant. This method is termed decision-feedback equalization.

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This work is licensed by Don Johnson under a Creative Commons Attribution License (CC-BY 1.0).

Last edited by Liqun Wang on Aug 29, 2003 2:30 pm GMT-5.