X内色X(e“)儿w=袋= r[nle-j2xkn/N For convenience in notation,we define Wwei祭 then 十00 X内=∑cmw. n=-00 However,we still can not compute X[k]since xn]can be infinitely long and we are summing for all n.Even if x[n]is finite,we can not always recover x[n]perfectly from X[k]because X[k]s are essentially the samples of the DTFT.So there are some necessary conditions for which we will be able to recover x[n]from X[k]. In order to find these conditions,first speculate the sampling theorem in the continuous-time domain: X.(j2) 0 X,(j) △T () 2m 2 △T △T Sampling in time corresponds to replication in the frequency domain.In order to recover Xe(jn) perfectly by low-pass filtering Xs(jn),frequency aliasing should be avoided.Therefore,x(t) should be bandlimited to o,and sampling interval should be short enough so that △T< 2n We can apply the time-frequency duality to illustrate the sampling process in the frequency domain.If the sampling interval in frequency is not short enough,we get time aliasing.If the 2+∞ X[k] � X(ejω)| ω= 2πk = � x[n]e−j2πkn/N N n=−∞ For convenience in notation, we define −j 2π WN � e N then +∞ X nk [k] = � x[n]WN . n=−∞ However, we still can not compute X[k] since x[n] can be infinitely long and we are summing for all n. Even if x[n] is finite, we can not always recover x[n] perfectly from X[k] because X[k]s are essentially the samples of the DTFT. So there are some necessary conditions for which we will be able to recover x[n] from X[k]. In order to find these conditions, first speculate the sampling theorem in the continuous-time domain: Sampling in time corresponds to replication in the frequency domain. In order to recover Xc(jΩ) perfectly by low-pass filtering Xs(jΩ), frequency aliasing should be avoided. Therefore, x(t) should be bandlimited to Ω0, and sampling interval should be short enough so that 2π ΔT < . Ω0 We can apply the time-frequency duality to illustrate the sampling process in the frequency domain. If the sampling interval in frequency is not short enough, we get time aliasing. If the 2