change function from continuous to discrete

Question

I have function that I want to sample with sampleing rate 20 Msamples/sec.

The funciton is the one below

$w_T(t)= \left\{ \begin{array}{ll} \sin^2(\frac{\pi}{2}(0.5+\frac{t}{T_{TR}})) & \mbox{if $-T_{TR}/2 < t<T_{RT}/2$};\\ 1 & \mbox{if $T_{TR}/2 < t<T-T_{TR}/2$}\\ \sin^2(\frac{\pi}{2}(0.5-\frac{t-T}{T_{TR}})) & \mbox{if $T-T_{TR}/2 < t<T+T_{TR}/2$};\\ \end{array} \right. $

I saw the solution is

$w_T(n)=\left\{ \begin{array}{ll} 1 & \mbox{if $1<n<79$};\\ 0.5 & \mbox{if $0,80$}.\end{array} \right. $

I tried to see why it so so I put $t=n T_s= n \frac{1}{20\times10^{-6}}$. It is also given that $T_{TR}= 100 ns$ and $T=4 \mu sec$. But i fail to reach the discrete funciton.

Is there any way to reach solution?

Answer 1

if you take sampling frequency as 20Msamples/sec then sampling time Ts is $$T_s=\frac{1}{f_s}\\T_s=\frac{1}{20\times10^6}\\T_s=50ns$$since the solution is till n=79 we can see that t=79*50=3950
From the continues time function this Trt/2=50ns and T-Trt/2=3950ns hence we get 1 n=1 to 79 when n=0 we get the first case which is equal to 0.5 similarly of n=80.This can be see as$$ w_T(nT_s)= \begin{cases} \sin^2(\frac{\pi}{2}(0.5 +\frac{n50ns}{100ns})),& \text{if } -50\times10^{-6}<n50\times10^{-6}<50\times10^{-6}\\ 1, & \text{if } 50\times10^{-6}\leq n50\times10^{-6}\leq 3950\times 10^{-6}\\ \sin^2(\frac{\pi}{2}(0.5 -\frac{n50ns-4000ns}{100ns})),&\text{if } other case \end{cases}$$ which comes down to
$$ w(n)= \begin{cases} \sin^2(\frac{\pi}{2}(0.5 +\frac{n}{2})),& \text{if } -1<n<1\\ 1, & \text{if } 1\leq n\leq79\\ \sin^2(\frac{\pi}{2}(0.5 -\frac{n-79}{2})),&\text{if } 79<n<81 & \text{otherwise} \end{cases} $$ In the above equation we can remove the $$-1<n<1$$as n=0 when n=0 w(0)=0.5 for n=1to78 w(n)=1.When n=80 the third one become zero.Hence you get the finalized expression.I hope this answers your question