Consider the set of complex exponentials {ejk omega 0t : k = 0,1,2,3....}. Note
ID: 1809143 • Letter: C
Question
Consider the set of complex exponentials {ejk omega 0t : k = 0,1,2,3....}. Note they are all periodic over any length of time T0 = 2 pi / omega 0. Show that if you pick any two of the exponentials in this set. say with k = k1 and with k = k2, that they are orthogonal over any time interval of length T0 = 2 pi / omega 0, that is. show they are orthogonal for any time window [t0, t0 + T0] where to can be arbitrarily chosen. Let x(t) = for some constant T. Find the coefficients ak in the Fourier series expansion of x(t).Explanation / Answer
(a)
(integral) e^(j k1 wo t).(e^(j k2 wo t))* dt from t = t0 to t0 + T0
= (integral) e^(j k1 wo t).e^(-j k2 wo t) dt from t = t0 to t0 + T0
= (integral) e^(j wo t (k1 - k2)) dt from t = t0 to t0 + T0
= [ e^(j wo t (k1 - k2)) / (wo (k1 - k2)) ] from t = t0 to t0 + T0
= (1/(wo (k1 - k2))) x ( e^(j wo (t0+T0) (k1 - k2)) - e^(j wo t0 (k1 - k2)) )
= (1/(wo (k1 - k2))) x ( e^(j ( wot0(k1 - k2) + woT0(k1-k2) )) - e^(j wo t0 (k1 - k2)) )
= (1/(wo (k1 - k2))) x ( e^(j ( wot0(k1 - k2) + 2pi(k1-k2) )) - e^(j wo t0 (k1 - k2)) )
= (1/(wo (k1 - k2))) x ( e^j(wot0(k1 - k2)).e^j(2pi(k1-k2)) - e^(j wo t0 (k1 - k2)) )
= (1/(wo (k1 - k2))) x ( e^j(wot0(k1 - k2)).1 - e^(j wo t0 (k1 - k2)) )
= (1/(wo (k1 - k2))) x ( 0 )
= 0
(using e^j (2pi x integer) = 1 )