The lowpass filter H(u) has a cutoff frequency wc=205T rad/ sec. Thus, c(t)is r(t) where all terms with frequency above we are removed by the lowpass filter. The terms which are kept have kwol 205T rad /sec k|< 10.25, so the output, ac(t),is r(t)= To obtain n, we sample c(t) every T=5 10-3 seconds with an impulse train The sampling frequency is 400T=2 x maximum frequency in c(t). Therefore
Problem set SolutioN Exercise for home study o&W8.35 (a) From the system diagram, we see that a(t)=a(t)cos(wet) Using the multiplication property ZGu)= o(X(u)* FT(cos wct)) FT of cos wt is two impulses with area T at +wc. Therefore, Z(w)is the spectrum
Introduction X Focuses on the structure of programs that use RPC, and shows how program can be divided along procedure boundaries Introduces the stub procedure concept and a program generator tool that automates much of the code generation associated with ONC RPC
What needed? to combine signals and/or to split them multiple ways up to 1000X1000 for WDM LANS three important characteristics Return Loss the amount of power that is reflected and thus lost Insertion LoSs the amount of signal lost in the total transit through the device Excess Loss additional loss of a device
