David83
Advanced Member level 1
Hello all,
I am trying to understand sampling from the following:
Suppose the channel impulse response is given by:
\[h(\tau)=h\delta(\tau)\]
and the transmitted signal is:
\[s(t)=\sum_ks_kg(t-kT_s)\]
where {s_k} are the transmitted symbols and g(t) is a rectangular pulse of duratuion Ts the sample time. The received signal can by written as:
\[v(t)=h\sum_ks_kg(t-kT_s)+w(t)\]
where w(t) is the additive noise. We can sample the above received signal at t=nTs resulting in:
\[v_n=hs_n+w_n\] .... (1)
which is obvious. Now suppose I want to take another approach, say to find the optimum receiver from the received signal v(t) as:
\[v_n=h^*\int_{-\infty}^{\infty}v(t)g(t-nT_s)\,dt=h^*\int_{nT_s}^{(n+1)T_s}v(t)\,dt\] .... (2)
which must be the same as multiplying (1) by conj(h). Does this mean that the integration in (2) is effectively sampling, which means that:
\[\int_{nT_s}^{(n+1)T_s}v(t)\,dt=v(nT_s)\]??
Thanks in advance
I am trying to understand sampling from the following:
Suppose the channel impulse response is given by:
\[h(\tau)=h\delta(\tau)\]
and the transmitted signal is:
\[s(t)=\sum_ks_kg(t-kT_s)\]
where {s_k} are the transmitted symbols and g(t) is a rectangular pulse of duratuion Ts the sample time. The received signal can by written as:
\[v(t)=h\sum_ks_kg(t-kT_s)+w(t)\]
where w(t) is the additive noise. We can sample the above received signal at t=nTs resulting in:
\[v_n=hs_n+w_n\] .... (1)
which is obvious. Now suppose I want to take another approach, say to find the optimum receiver from the received signal v(t) as:
\[v_n=h^*\int_{-\infty}^{\infty}v(t)g(t-nT_s)\,dt=h^*\int_{nT_s}^{(n+1)T_s}v(t)\,dt\] .... (2)
which must be the same as multiplying (1) by conj(h). Does this mean that the integration in (2) is effectively sampling, which means that:
\[\int_{nT_s}^{(n+1)T_s}v(t)\,dt=v(nT_s)\]??
Thanks in advance