مكونات متوافقة الطور ومتعامدة

في الهندسة الكهربائية ، يمكن تفكيك أو توليف الدالة الجيبية [ ${\displaystyle \sin(x)}$] ذات تشكيل زاوي (angle modulation) من دالتين جيبية مشكّلة بالسعة (amplitude-modulated) التي يتم تعويضها في الطور بربع دورة (π / 2 راديان). جميع الدوال الثلاث لها نفس التردد. تعرف الدوال الجيبية المشكلة بالسعة بالموجات العامودية متوافقة الطور. في بعض السياقات يكون من الملائم أكثر الإشارة إلى تشكيل السعة (القاعدي) فقط بهذه الشروط.^[1]

في تطبيقات تشكيل الزاوية، مع تردد الموجة الحاملة f, φ هي أيضاً دالة متغير الزمن،^[2] :

*يحتاج تعديل هذا القسم*

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$^[3]

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mo stretchy="false">

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mn>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mn><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mo>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mo stretchy="false">

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mo stretchy="false">

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mo stretchy="false">

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mo>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mo>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mo stretchy="false">

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mn>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mn><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

mi><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mo stretchy="false">

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mo>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mo>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mo stretchy="false">

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mo stretchy="false">

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mo stretchy="false">

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mo stretchy="false">

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo></mrow><mo>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo></munder></mrow><mrow class="MJX-TeXAtom-ORD"><mtext>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mtext></mrow></munder><mo>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mo>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mrow><mo>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mrow><mn>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mn><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mo>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mrow class="MJX-TeXAtom-ORD"><mfrac><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mn>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mn></mfrac></mrow></mrow><mo>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo></mrow></mrow><mo>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo></mover></mrow><mrow class="MJX-TeXAtom-ORD"><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mo>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mo stretchy="false">

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mn>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mn><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mo stretchy="false">

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo></mrow></mover><mo>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mo>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mo stretchy="false">

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mo stretchy="false">

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mi>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mi><mo stretchy="false">

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo><mo stretchy="false">

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo></mrow><mo>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo></munder></mrow><mrow class="MJX-TeXAtom-ORD"><mtext>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mtext></mrow></munder><mo>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

</mo></mrow>

${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$

${\displaystyle }$${\displaystyle \sin[2\pi ft+\phi (t)]\ =\ \underbrace {\sin(2\pi ft)\cdot \cos[\phi (t)]} _{\text{in-phase}}\,+\,\underbrace {\overbrace {\sin \left(2\pi ft+{\tfrac {\pi }{2}}\right)} ^{\cos(2\pi ft)}\cdot \sin[\phi (t)]} _{\text{quadrature}}.}$^[4]

</img>

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