直角位相振幅変調
r i ( t ) = s ( t ) cos  ( 2 π f 0 t ) = I ( t ) cos  ( 2 π f 0 t ) cos  ( 2 π f 0 t ) + Q ( t ) sin  ( 2 π f 0 t ) cos  ( 2 π f 0 t ) {\displaystyle {\begin{aligned}r_{i}(t)=&s(t)\cos(2\pi f_{0}t)\\=&I(t)\cos(2\pi f_{0}t)\cos(2\pi f_{0}t)+Q(t)\sin(2\pi f_{0}t)\cos(2\pi f_{0}t)\end{aligned}}}
標準的な三角恒等式を使用して、以下のように書き直せる:
r i ( t ) = 1 2 I ( t ) [ 1 + cos  ( 4 π f 0 t ) ] + 1 2 Q ( t ) sin  ( 4 π f 0 t ) = 1 2 I ( t ) + 1 2 [ I ( t ) cos  ( 4 π f 0 t ) + Q ( t ) sin  ( 4 π f 0 t ) ] {\displaystyle {\begin{aligned}r_{i}(t)=&{\frac {1}{2}}I(t)\left[1+\cos(4\pi f_{0}t)\right]+{\frac {1}{2}}Q(t)\sin(4\pi f_{0}t)\\=&{\frac {1}{2}}I(t)+{\frac {1}{2}}[I(t)\cos(4\pi f_{0}t)+Q(t)\sin(4\pi f_{0}t)]\end{aligned}}}