# Will signal get distorted if channel bandwidth is lesser than the signal bandwidth?

This statement is given in the book "digital communicatinon by Taub schilling " Now one point I am unable to understand that if the channel bandwidth is lesser than the signal bandwidth,how it is possible that the signal will not get distorted? can someone explain this to me in simple word Please?

• It doesn't say that the signal won't be distorted, only that you can still receive it with arbitrarily small probability of error. This is an advantage of digital communications; even if the signal is distorted in the channel, there are conditions under which you can theoretically receive it perfectly with no bit errors. – Jason R Oct 6 '17 at 12:50

In line with DilipSarwate let me put what I understand from your question and the excerpt you posted.

First of all the excerpt does not say that one can transmit a signal without distortion through a channel whose bandwidth is less than the signal's bandwidth. Rather it says that one can, in principle, perfectly reverse the distortions that occured during transmission and hence recover the original signal by a suitable processing provided that there is no noise in the channel. Of course this is effectly a distortionless transmission whatever.

The author gives this (idealized) example probably to support the idea that in digital communication systems one can transmit message signals over noisy channels without errors (or very small error rates). The example in the second paragraph however is not what you would normally put forward to support the main idea as the aternate answer describes.

Furthermore, the idea in the second paragraph actually is an example of signal recovery by inverse filtering; a signal $x(t)$ is distorted by a filter $H(\omega)$ while passing through a channel as $y(t) = h(t) \star x(t)$ and $Y(\omega) = H(\omega) X(\omega)$. Then you can apply an inverse filtering on $y(t)$ which would recover the original signal $x(t)$ as: $$X_i(\omega) = Y(\omega) H_{i}(\omega) \longleftrightarrow x_i(t) = x(t) \star h_i(t)$$

Which suggests that $$H_i(\omega) = \frac{1}{H(\omega)}$$ for all $\omega$ such that $H(\omega) \neq 0$.

Now it's obvious that even in principle you cannot invert an arbitrary filter $H(\omega)$ into $1/H(\omega)$ because of the possibility of zeros of the forward filter; i.e. if the forward filter has nulls, then the inverse filter will approach $\infty$ at those frequencies and hence make the processed output $x_i(t)$ be unbounded (useless)

That last observation is further enhanced by the introduction of noise into the system. In any practical system there is always noise. So a more correct inversion relation is:

$$y(t) = h(t)\star x(t) + n(t) \leftrightarrow x_i(t) = y(t) \star h_i(t) = x(t)\star h(t) \star h_i(t) + n(t) \star h_i(t)$$

As you can see the noise term at the output is what contributes to the imperfectness of the inversion process in addition to any imperfectness of the inverse system impulse response $h_i(t)$.

In the example, the author therefore stresses the fact that there is no noise through the channel $n(t) = 0$ and furthermore he implies (but not clearly states) that the forward system (the channel) frequency response has no null frequencies; i.e. there is no $\omega$ for which $H(\omega) = 0$ at least through the signal bandwidth. Under these two conditions, then, it's possible that you can perfectly recover the original transmitted signal $x(t)$ from the received signal $y(t)$ whatever the frequency response of the transmission channel is.

One (pretty unseccesful) example of such reversible channel could be obtained by a simple electronic RC filter whose impulse response is $$h_{RC} = \frac{1}{RC} e^{-t/{RC}} u(t)$$ and the corresponding frequency response $H_{RC}(\omega)$ does not have any zeros, hence invertable as $H_i(\omega) = 1/H(\omega)$

I leave the MATLAB simulation of this to you.

The key here is the title of the book you are reading: "digital communicatinon by Taub schilling " as you say

What the book is talking about is digital communications and not analog communications in which the fidelity of the transmitted signal is of great importance; we want the output signal to be as close a replica of the input signal as is possible. In digital communications, it does not matter diddly-squat if the transmitted waveform is distorted as long as from the received waveform one can determine which of the two (more generally, few) possible transmitted waveforms $s_0(t)$ and $s_1(t)$ caused this received signal to occur. If the receiver can make this determination with high accuracy, then we have managed to transmit one bit (or more generally $\log_2$(few bits)) with high accuracy by choosing $s_0(t)$ or $s_1(t)$ according as we wish to send a $0$ or a $1$. Even if the received signal $r(t)$ is badly distorted from what was sent, the receiver can still determine whether the transmitter sent $s_0(t)$ or $s_1(t)$: fidelity of transmission is unimportant.

With the above in mind, note that the Taub and Schilling description is dreadfully ambiguous, and in many spots downright incorrect. An equalizer (a concept from digital communications) cannot perfectly undo the distortion caused by forcing a 1 kHz analog signal through a low-pass filter with 3 dB bandwidth $1$ Hz. An inverse (analog) filter is trivial to write down --- it is a filter whose transfer function is $\frac{1}{H(f)}$ where $H(f)$ is the channel response --- but in most cases that inverse filter is unrealizable. The truth of the matter is that any filtering by a channel causes the transmitted signal to be spread out in time and reversing this time spreading and bringing the signal back to its pristine shape is in almost all cases impossible: once the genie is out of the lamp, he cannot be stuffed back in, ditto Pandora's box for those preferring a more Western analogy. That last sentence "The signal is then recoverable precisely as transmitted" is sheer nonsense.

• If the S/N is good enough that you can recover the original digital data without error (possibly after ECC etc.), then you can feed that data through an identical modulator to the one that was earlier used for transmission, and thus recover the original transmitter output signal without channel distortion. – hotpaw2 Oct 7 '17 at 2:27
• @hotpaw2 But what you say is not what the Taub and Schilling text is saying, viz. the transmitted signal can be recovered by amplification (to get rid of the attenuation) and equalization (inverse filtering?). No mention of digital data, detection followed by signal reconstruction, and the claim about equalization/inverse-filtering is nonsense. – Dilip Sarwate Oct 9 '17 at 19:21

My interpretation of that text (the full book is not available to me) is that he is mixing somehow digital and analog signal concepts (probably, on purpose). He is actually saying three things:

1. (Digital-domain) The trade-off $BW$ vs $S/N$ is not unconditionally bounded by BW, this is clear when we check the Shannon–Hartley theorem: increasing $S/N$, we can decrease $BW$.
2. (Digital-domain) The Shannon's theorem is true: since the signal bit-rate is lower than the channel capacity, arbitrarily low error transmission is possible.
3. (Analog-domain) A RC filter is invertible: ideally, this is clear, we just have to multiply by the inverse channel response (Fat32's answer already states it nicely).