Appl. Sci. 2023, 13, 1872 
150f 17 
Substituting Equations (A4) and (A5) in Equation (A3) and by knowing the trigono- 
metric addition formula, we obtain 
s3 = Az3[cos(wt) cos(93) — sin(wt) sin(93)] = Az cos(wt + 03). (A6) 
= 
To derive the formula which shows the dependency of Az on amplitude and phase of 
and s,, we square and sum Equations (A4) and (A5). After some passages, we obtain 
Az = A/ A3 + A3 + 22147 cos(81 — 9) 
To tind 93, we divide Equation (A5) by (A4), obtaining 
A1sin(81) + A, sin(8,) 
t. 8 ALL 
an) A1cos(91) + A» cos(8,) 
By inverting Equation (A8), we can write 93, as follows 
1A sin(9;) +A2 sin(9,) 
O3 = tan! (| — DAL 
3 7 Ta * cos(81) + A, cos(8,) 
(A7) 
(AZ 
(A9) 
We are now interested in the maximum phase error caused by the signal s» when it 
is summed to s;. For simplicity, we assume 91 = 0. Therefore, the phase error is directly 
represented by 03. To find the maximum, we compute the first derivative of Equation (A9) 
and find the zeros. We first define the ratio between A and A2 as p and then, remembering 
‘hat 1 = 0, we write Equation (A9), as follows 
(A10) 
We also assume p > 1, which means A1 > A. Then, we compute the derivative with 
‚espect to 8, 
do3 _ 1 cos 8, (p + cos 8 + sin? 8) 
de 1 ( sin 82 ) (0 + cos 02)? 
. cos 85 
(0 + cos 82)? cos 8 (0 + cos 8 + sin? 9) 
(0 + cos 8,)2 + sin? 8, (0 + cos 8,)? 
_ pCc0os%2 + cos? 8, + sin? 6, 
p2 +20 cos 92 + cos? 8, + sin? 6, 
a 1 +20 cos 8, 
p2 +20cos92 +1 
rom which we find 
cos > = -1/0 — 9, = arccos(-1/p) 
Substituting Equation (A12) in (A10), we obtain 
(A111) 
442) 
Y1- 1 
a sin[arccos(—1/p)] ) _ 4an-l p ) — tan! (=>): (A13) 
03 = tan a ta tan ( 0-1 an /0? — 1 
By observing Equation (A13), it can be noted that the maximum error depends purely 
on the ratio p. This formula cannot be applied if 0 < op < 1. In such a case, we consider the 
four-quadrant arctangent instead of the normal arctangent operator. Thus, Equation (A10) 
becomes 
02 — atan2(sin6>, 0 + cos 9») 
(A14)