To evaluate the integral in the second term, integration by parts is again used and
will result in another expression of a similar form containing a higher-order derivative
and another integral. This result can be expressed as the following infinite sum.
/ RlooptRload I RlooptRload
dBnorm (1 loop, (RlOPodp
S(_)n+l loop d(n)Bnorm) (3-105)
S1 ) \Rloop +R load ) dl(n) (3 )
The term d(n)Bnorm(l)/dl(n) denotes the nth derivative of Bnorm () with respect to
1. Substituting the above expression into Equation 3-100 and reintroducing the limits of
integration yields
( RAoop+Rload .
t(t) RloadAloop e loo p lp (3-106)
Lloop
e ooRload n dn) Borm1
loo \RRloop +Rloadj dl(n)B
If d(n)Bnorm ()/dl(n) < for all n, then the term resulting from the lower limit of
integration will be zero. Therefore, the expression for Vout(t) becomes
R oadAloop R+RlooRoad t Rloop+Rload t
VOW(t) p e 1 oo e )
Lloop
I-)n-+1 loop d(n)Bnorm(t) (3-107)
n= / \Rloop +Rload dt(n) (3-107)
The multiplication of the two exponential terms results in unity, therefore
Vot (t)= RloadAloop ( l 1 Lloop n d(n) Bnorm (t) (3-108)
Lloop Rloop +Rload dt (n)
Hence, the time-domain voltage output of a loop antenna is the weighted sum of
all of the time-derivatives of Bnorm(t). If the first term in the summation is expanded,