Design Of Pulse Shapes for no ISI


So far we have considered an ideal communication channel which only adds additive white gaussian noise. Another way to look at this is that the channel has an ideal frequency response of 1 for all frequencies and in the absence of noise the received signal would be the same as the transmitted signal. However, many practical channels will not have ideal frequency response and hence will distort the transmitted signals and then noise will get added at the receiver front end. Understanding communication through these channels is our current topic of interest.

General Baseband communication system

Consider a general baseband communication system shown in the below figure,


g(t) is the baseband shaping pulse. Let us consider a linear modulation format so that,

\begin{align} s_l(t) = v(t) = \sum I_n g(t-nT) = \left( \sum I_n \delta(t-nT) \right) * g(t) \end{align}

sl(t) is the Low-pass (Baseband) waveform. The baseband received waveform is,

\begin{eqnarray} r_l(t) &=& v(t)*c(t) + z(t) \\ &=& \left( \sum I_n \delta(t-nT) \right) * g(t)*c(t) + z(t) \end{eqnarray}

let $g(t)*c(t)=h(t)$, therfore

\begin{align} r_l(t) = \sum I_n h(t-nT) + z(t) \end{align}

We pass rl(t) through a receiver front end filter and we will sample the output once every T sec with a fixed offset of $\tau_0$,

\begin{align} y(t) = \left( \sum T_n \delta(t-nT) \right) * g(t) * c(t) * g_R(t) + z(t)+ g_R(t) \end{align}

$g(t)* c(t)*g_R(t) = x(t)$, x(t) is the equivalent Impulse response and $z(t)*g_R(t) = \nu(t)$ being the equivalent noise.

\begin{align} y(t) = \sum I_n x(t-nT) + \nu(t) \end{align}

we sample y(t) at kT + $\tau_0$ to get yk ,

\begin{eqnarray} y_k &=& y(kT+\tau_0) \\ &=& \sum_{n=o}^\infty I_n x(kT-nT+\tau_0) + \nu(kT+ \tau_0) \end{eqnarray}

Let $x_k = x(kT+\tau_0)$, $\nu_k=\nu(kT+\tau_0)$, then

\begin{align} y_k=\sum_{n=0}^\infty I_n x_{n-k} + \nu_{n-k} \hspace{5mm} k=0,1,2,\ldots \end{align}
\begin{align} y_k=x_0 I_k + \sum_{n\neq0} I_n x_{n-k} + \nu_k \end{align}

The first term in the right hand side of the equation is the desired symbol, the middle term being the ISI part and the last term is the noise component. We seek to design a transmitter and a receive filters such that

  • Inter Symbol interference is eliminated.
  • Noise is minimized.

Let us look at few examples to reinforce our intuition.

Ideal Channel

C(t) = $\delta(t)$, then $x(t)=g(t)*g_R(t)$. Suppose we want to transmit a symbol rate of 1/T symbols per sec. We choose

\begin{align} g(t)= sinc \left( \frac{t}{T} \right) , g_R(t)= sinc \left( \frac{t}{T} \right), x(t)= g(t)*g_R(t)=c * sinc \left( \frac{t}{T} \right) \end{align}

At the receiver if we choose $\tau_0=0$, we can see that,

\begin{align} x_0 = c, x_n= 0 , \hspace{5mm} \forall n\neq0 \end{align}

Thus we can recover I_n.Notice that we can choose an arbitrarily small T, i.e since the channel is ideal the achievable data rate is unbounded. Also note that $\nu_k$ is sequence of of uncorrelated random variables.

Ideal Channel with delay

Suppose $c(t)= \delta(t-t_0)$. If we choose $g(t)= sinc \left( \frac{t}{T} \right)$ and $g_R(t)=sinc \left( \frac{t}{T} \right)$ then $x(t)= c*sinc \left( \frac{t- \tau_0 }{T} \right)$
Now sampling at kT+$\tau_0$ we can recover In. We have ,

\begin{align} x_0= x\left( \tau_0 \right)=c, x_k=x \left( kT+\tau_0 \right) \hspace{4mm}k\neq0 \end{align}

Non-ideal channels

For Non-Ideal channels if we choose g(t) and gR(t) to be $sinc \left( \frac{t}{T} \right)$, we may not have xk to be zero for $k\neq0$

Ideally Band-limited channel


Design of pulses for no ISI- The Nyquist Criterion


\begin{align} y_k=x_0 I_k + \sum_{n\neq0} I_n x_{n-k} + \nu_k \end{align}

In-order to have no ISI, we must satisfy the following condition.

\begin{align} x_k= \left\{ \begin{array}{cc} 1 \mbox{ or any c} & k=0 \\ 0 & k\neq0 \end{array} \end{align}

Recall, $x(t)= g(t)c(t)g_R(t)$. Let us choose gR(t) to be the pulse matched to g(t).Thus we observe that,

\begin{eqnarray} x(f) &=& G(f)C(f)G^*(f) \\ &=& \left\{ \begin{array}{cc} |G(f)|^2 & |f|\leq W \\ 0 & |f|>W \end{array} \end{eqnarray}

Nyquist theorem: The necessary and sufficient condition for no ISI is that,

\begin{align} B(f)=\sum_{m=-\infty}^{\infty} X \left( f+\frac{m}{T} \right) =constant= cT \end{align}

See the text book for detailed proof. B(f) is called the folded spectrum.We will discuss a simple intuitive proof below.

The discrete time sequence {xk} or the equivalent countinous time signal $\sum x_m \delta (t-mT)$ can be obtained by multiplying x(t) with $\delta(t-mT)$

Let $\tilde{x}(t) = x(t) \left( \sum_{m=-\infty}^{\infty} \delta(t-mT) \right)$

\begin{eqnarray} \tilde{X}(f) &=& X(f)* \left( \frac {1}{T} \sum_{m=-\infty}^{\infty} \delta(f- \frac{m}{T}) \right) \\ &=& \frac{1}{T} \sum_{m=-\infty}^{\infty} X \left( f-\frac{m}{T} \right) = \frac{1}{T} \sum_{m=-\infty}^{\infty} X \left( f+\frac{m}{T} \right) \end{eqnarray}

We want $\tilde{X}(f) =1 \hspace{3mm} \forall f$ or $\tilde{x}(t) = \delta (t)$. Therefore,

\begin{align} \frac{1}{T} \sum_{m=-\infty}^{\infty} X \left( f+\frac{m}{T} \right) = T \end{align}

This shows the necessary part.Think about the sufficient part of the proof


\begin{align} C(f)= \left\{ \begin{array}{cl} 1 & |f| \leq W \\ 0 & otherwise \end{array} \end{align}

Relationship between Symbol duration and Bandwidth

Let us look at the cases for the relationship between T and W, Rs=1/T is the symbol rate.

\begin{align} C(f)= \left\{ \begin{array}{ccl} 1 & for &|f| \leq W \\ 0 & for & otherwise \end{array} \end{align}

Case 1: Rs>2W ( signaling rate > Bandwidth(2W)), then 1/Ts>2W.


The folded spectrum will always be zero for $W<|f|<\frac{1}{T}-W$.Thus it is possible to obtain ISI-free communication.

Case 2: Rs=2W or 1/T=2W then X(f) and X(f-1/T) will barely touch.


In-order to make the spectrum flat we need to choose

\begin{align} X(f)= \left\{ \begin{array}{cl} 1 & |f| < W \\ 0 & outside \end{array} \end{align}

Case 3: Rs < 2W or Ts>1/2W. Here we are under utilizing bandwidth or we are using excess bandwidth over the Nyquist bandwidth. there are many choices for X(f) that make the folded spectrum to be flat


In practice a popularly used shape for X(f) and hence for $|G(f)|^2$ is a raised cosing pulse given by,

\begin{align} X_{RC}(f) = \left\{ \begin{array}{lc} c & |f|< \frac{1-\beta}{2T}\\ \frac{c}{2} \left[ 1+ \cos \left[ \frac{\pi T_s}{\beta} \left( |f| - \frac{1-\beta}{2T} \right) \right] \right] & \frac{1-\beta}{2T}< |f|< \frac{1+\beta}{2T} \\ 0 & |f|>\frac{1+\beta}{2T} \end{align}
\begin{align} x_{RC}(t)= sinc \left( \frac{t}{T} \right) \frac{\cos \left( \pi \beta \frac{t}{T_s} \right)}{ 1- \left( \frac {2\beta t}{T_s}\right) ^2} \end{align}

Practical Considerations and why excess bandwidth is commonly used

From the previous section we can see that one can communicate through a ideal band-limited channel of bandwidth W without ISI at the same rate of 2W samples /sec if we used a sinc pulse as the pulse shape.


However there are problems in using this pulse in practice. A small error in timing (the time at which the samples are taken) can lead to large (unbounded actually) error due to ISI. Similarly truncation of the pulse can also cause large sidelobes in the spectrum.


Practical solution: Use square root Raised cosine pulses at transmitter and receiver so that x(t) is a Raised cosine pulse with excess band-width $\beta$. The Raised cosine pulse has a much faster decay of the tail with time. For $\beta > 0$ the decay is $\frac{1}{t^3}$. Consequently mistiming error does not cause a lot of ISI.

EYE diagram:

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License