Question

2.51 (a) The root mean-square (rms) bandwidth of a low-pass signal g(t) of finite energy is defined by 0O 1/2 rms where G()2 is the energy spectral density of the signal. Correspondingly, the root mean- square (rms) duration of the signal is defined by rms- &(t) dt Using these definitions, show that Assume that Ig(t)| → 0 faster than 1 / Vlt! as lt-oo (b) Consider a Gaussian pulse defined by g(t) exp(-I2) Show that, for this signal, the equality rms w rms can be reached Hint: Use Schwarz's inequality (see Appendix 5) in which we set g1(t) tg(t) and 82(t) de( dt

Answer 1

An saer ne:1 The given voot mean-S9μα e(xms) bandwidth of a low pas siynad go) os defined by rute ene spectral densi of the Signal Smiauy Reot an suan (oms) duzation of the Signal is /2 ob definad b mS uing the above bandtotdth ond fine duation definitions no ω (oe- show that Tims Wns 4 T Fom ven data ooe can diefine thot the innex product oo ncins okich is complex valued |cP-3>}2s(//> < g-3> schudorta ineguality

So, it is then checked as P,3, > Bu using the above depinition,we can axite. niude of above definition く,3,2(t) dt Ipat)dt let us expand the above expresston, t is eal numben e nteipants ae haue the inteqsation

e ne 3 as dt use tnau that by considering tho given stote mant the Aisst component wil be Zero. So the remaini component. Nou , Substitute the Yesult in the previous Part which w let ie the pars vels velation 3(t) dt The PoureansPosmfCEs j2χ pGC) and also apry paxsvels relation to da Ct) dt -04 Naw, 月ivide. bhe yight part of equality in previows paxt y the left part and e the de fimition

2 not o Gaussian pulse, it be checkes The shape of the pulse s qpare d is similar to the and also it needs so Sorne Gaussian distribution ai th nomalisation Tms on 15 YmS ←to ims uring Schwon3 ne od 9,(t)d dt LETT LfT