[1] [2] [1]
以下是數學公式的測試。
R e q = 50 Ω {\displaystyle R_{eq}=50\,\Omega }
R e q / [ R e q + X s ( n ω 0 ) ] {\displaystyle R_{eq}/[R_{eq}+X_{s}(n\omega _{0})]}
X p ( n ω 0 ) / [ X p ( n ω 0 ) + X s ( n ω 0 ) + R e q ] {\displaystyle X_{p}(n\omega _{0})/[X_{p}(n\omega _{0})+X_{s}(n\omega _{0})+R_{eq}]}
X p ( ω 0 ) / [ X p ( ω 0 ) + X s ( ω 0 ) + R e q ] {\displaystyle X_{p}(\omega _{0})/[X_{p}(\omega _{0})+X_{s}(\omega _{0})+R_{eq}]}
1 I o u t ( t ) = I a cos ( ω 0 t + ϕ ) {\displaystyle I_{out}(t)=I_{a}\cos {(\omega _{0}t+\phi )}}
2 V 1 ( t 1 ) = 0 {\displaystyle V_{1}(t_{1})=0}
3 d V 1 ( t 1 ) d t = 0 {\displaystyle {\frac {dV_{1}(t_{1})}{dt}}=0}
4 C e q = C 0 + C 1 / / C 2 = C 0 + C 1 C 2 C 1 + C 2 {\displaystyle C_{eq}=C_{0}+C_{1}//C_{2}=C_{0}+{\frac {C_{1}C_{2}}{C_{1}+C_{2}}}}
5 α = C 1 C 1 + C 2 {\displaystyle \alpha ={\frac {C_{1}}{C_{1}+C_{2}}}}
6 β = C 0 C e q {\displaystyle \beta ={\frac {C_{0}}{C_{eq}}}}
7 q = 1 ω 0 L 0 C e q {\displaystyle q={\frac {1}{\omega _{0}{\sqrt {L_{0}C_{eq}}}}}}
8 V D D − V 1 ( t ) = L 0 d i L ( t ) d t {\displaystyle V_{DD}-V_{1}(t)=L_{0}{\frac {di_{L}(t)}{dt}}}
9 i L ( t ) = C 0 d V 1 ( t ) d t + C 1 ( d V 1 ( t ) d t − d V 2 ( t ) d t ) {\displaystyle i_{L}(t)=C_{0}{\frac {dV_{1}(t)}{dt}}+C_{1}\left({\frac {dV_{1}(t)}{dt}}-{\frac {dV_{2}(t)}{dt}}\right)}
10 C 1 ( d V 1 ( t ) d t − d V 2 ( t ) d t ) = C 2 d V 2 ( t ) d t + I a cos ( ω 0 t + ϕ ) {\displaystyle C_{1}\left({\frac {dV_{1}(t)}{dt}}-{\frac {dV_{2}(t)}{dt}}\right)=C_{2}{\frac {dV_{2}(t)}{dt}}+I_{a}\cos {(\omega _{0}t+\phi )}}
11 V D D = L 0 d i L ( t ) d t {\displaystyle V_{DD}=L_{0}{\frac {di_{L}(t)}{dt}}}
12 ( C 1 + C 2 ) d V 2 ( t ) d t = − I a cos ( ω 0 t + ϕ ) {\displaystyle (C_{1}+C_{2}){\frac {dV_{2}(t)}{dt}}=-I_{a}\cos {(\omega _{0}t+\phi )}}
13 V 1 ( t ) = A 1 V D D cos ( ω 1 t ) + A 2 V D D sin ( ω 1 t ) + κ V D D 1 q 2 − 1 sin ( ω 0 t + ϕ ) + V D D {\displaystyle V_{1}(t)=A_{1}V_{DD}\cos {(\omega _{1}t)}+A_{2}V_{DD}\sin {(\omega _{1}t)}+\kappa V_{DD}{\frac {1}{q^{2}-1}}\sin {(\omega _{0}t+\phi )}+V_{DD}}
14 V 2 ( t ) = C 1 C 1 + C 2 V 1 ( t ) − I a ( C 1 + C 2 ) ω 0 sin ( ω 0 t + ϕ ) {\displaystyle V_{2}(t)={\frac {C_{1}}{C_{1}+C_{2}}}V_{1}(t)-{\frac {I_{a}}{(C_{1}+C_{2})\omega _{0}}}\sin {(\omega _{0}t+\phi )}}
15 i L ( t ) = C e q q ω 0 [ A 2 V D D cos ( q ω 0 t ) − A 1 V D D sin ( q ω 0 t ) ] + κ V D D C e q ω 0 q 2 q 2 − 1 cos ( ω 0 t + ϕ ) {\displaystyle i_{L}(t)=C_{eq}q\omega _{0}[A_{2}V_{DD}\cos {(q\omega _{0}t)}-A_{1}V_{DD}\sin {(q\omega _{0}t)}]+\kappa V_{DD}C_{eq}\omega _{0}{\frac {q^{2}}{q^{2}-1}}\cos {(\omega _{0}t+\phi )}}
16 κ = α I a C e q ω 0 V D D {\displaystyle \kappa ={\frac {\alpha I_{a}}{C_{eq}\omega _{0}V_{DD}}}}
17 V 2 = − I a ( C 1 + C 2 ) ω 0 sin ( ω 0 t + ϕ ) + A 3 V D D {\displaystyle V_{2}=-{\frac {I_{a}}{(C_{1}+C_{2})\omega _{0}}}\sin {(\omega _{0}t+\phi )}+A_{3}V_{DD}}
18 i L ( t ) = V D D L 0 t + i L ( t 1 ) {\displaystyle i_{L}(t)={\frac {V_{DD}}{L_{0}}}t+i_{L}(t_{1})}
19 A 1 = q ( sin a 2 − sin a 1 + sin a 3 − sin a 4 ) + sin a 1 + sin a 2 q sin a 1 − q sin a 2 − 2 q sin ϕ − sin a 2 − sin a 1 {\displaystyle A_{1}={\frac {q(\sin {a_{2}}-\sin {a_{1}}+\sin {a_{3}}-\sin {a_{4}})+\sin {a_{1}}+\sin {a_{2}}}{q\sin {a_{1}}-q\sin {a_{2}}-2q\sin {\phi }-\sin {a_{2}}-\sin {a_{1}}}}}
20 A 2 = q ( cos a 1 − cos a 2 + cos a 4 − cos a 3 ) − cos a 1 − cos a 2 + 2 cos a 5 q sin a 1 − q sin a 2 − 2 q sin ϕ − sin a 2 − sin a 1 {\displaystyle A_{2}={\frac {q(\cos {a_{1}}-\cos {a_{2}}+\cos {a_{4}}-\cos {a_{3}})-\cos {a_{1}}-\cos {a_{2}}+2\cos {a_{5}}}{q\sin {a_{1}}-q\sin {a_{2}}-2q\sin {\phi }-\sin {a_{2}}-\sin {a_{1}}}}}
21 κ = − q ( cos ( 2 q π ( 1 − D ) ) q 2 − cos ( 2 q π ( 1 − D ) ) − q 2 + 1 ) k 1 + k 2 + k 3 + k 4 + k 5 {\displaystyle \kappa ={\frac {-q(\cos {(2q\pi (1-D))}q^{2}-\cos {(2q\pi (1-D))-q^{2}+1})}{k_{1}+k_{2}+k_{3}+k_{4}+k_{5}}}}
22 ϕ = π D + π + arctan ( g n 1 + g n 2 + g n 3 + g n 4 g d 1 + g d 2 + g d 3 + g d 4 + g d 5 ) {\displaystyle \phi =\pi D+\pi +\arctan {\left({\frac {g_{n1}+g_{n2}+g_{n3}+g_{n4}}{g_{d1}+g_{d2}+g_{d3}+g_{d4}+g_{d5}}}\right)}}
23 V 2 ( t ) = V 2 _ 0 + V 2 _ 1 cos ( ω 0 + ϕ 1 ) + V 2 _ 2 cos ( 2 ω 0 + ϕ 2 ) + V 2 _ 3 cos ( 3 ω 0 + ϕ 3 ) + ⋯ {\displaystyle V_{2}(t)=V_{2\_0}+V_{2\_1}\cos {(\omega _{0}+\phi _{1})}+V_{2\_2}\cos {(2\omega _{0}+\phi _{2})}+V_{2\_3}\cos {(3\omega _{0}+\phi _{3})}+\cdots }
24 Z o u t = V 2 _ 1 I a exp [ ( ϕ 1 − ϕ ) j ] = Z o u t _ r e a l + j ⋅ Z o u t _ i m a g {\displaystyle Z_{out}={\frac {V_{2\_1}}{I_{a}}}\exp {[(\phi _{1}-\phi )j]}=Z_{out\_real}+j\cdot Z_{out\_imag}}
25 Z o u t = ( R 0 / / 1 C 3 ω 0 j ) + L 1 ω 0 j = R 0 1 + ( R 0 C 3 ω 0 ) 2 + ( L 1 ω 0 − R 0 2 C 3 ω 0 1 + ( R 0 C 3 ω 0 ) 2 ) j {\displaystyle Z_{out}=\left(R_{0}//{\frac {1}{C_{3}\omega _{0}j}}\right)+L_{1}\omega _{0}j={\frac {R_{0}}{1+(R_{0}C_{3}\omega _{0})^{2}}}+\left(L_{1}\omega _{0}-{\frac {R_{0}^{2}C_{3}\omega _{0}}{1+(R_{0}C_{3}\omega _{0})^{2}}}\right)j}
26 L 1 ω 0 = Z o u t _ i m a g + R 0 2 C 3 ω 0 1 + ( R 0 C 3 ω 0 ) 2 {\displaystyle L_{1}\omega _{0}=Z_{out\_imag}+{\frac {R_{0}^{2}C_{3}\omega _{0}}{1+(R_{0}C_{3}\omega _{0})^{2}}}}
27 C 3 ω 0 = ( 1 Z o u t _ r e a l R 0 − 1 R 0 2 ) {\displaystyle C_{3}\omega _{0}={\sqrt {({\frac {1}{Z_{out\_real}R_{0}}}-{\frac {1}{R_{0}^{2}}})}}}
28 L x n = L x ⋅ ω 0 {\displaystyle L_{xn}=L_{x}\cdot \omega _{0}}
29 C x n = C x ⋅ ω 0 {\displaystyle C_{xn}=C_{x}\cdot \omega _{0}}
30 P o u t = V D D ⋅ 1 T ∫ 0 T i L 0 ( t ) d t = V D D 2 C e q n h ( q , D ) {\displaystyle P_{out}=V_{DD}\cdot {\frac {1}{T}}\int _{0}^{T}i_{L_{0}}(t)dt=V_{DD}^{2}C_{eqn}h(q,D)}
31 P o u t ≈ ( p O 3 q 3 + p O 2 q 2 + p O 1 q + p O 0 ) C e q n V D D 2 {\displaystyle P_{out}\approx (p_{O_{3}}q^{3}+p_{O_{2}}q^{2}+p_{O_{1}}q+p_{O_{0}})C_{eqn}V_{DD}^{2}}
32 L i n ≈ p L 3 q 3 + p L 2 q 2 + p L 1 q + p L 0 {\displaystyle L_{in}\approx p_{L_{3}}q^{3}+p_{L_{2}}q^{2}+p_{L_{1}}q+p_{L_{0}}}
33 C 3 n ≈ p C 3 q 3 + p C 2 q 2 + p C 1 q + p C 0 {\displaystyle C_{3n}\approx p_{C_{3}}q^{3}+p_{C_{2}}q^{2}+p_{C_{1}}q+p_{C_{0}}}
34 P L o s s _ M 1 _ n o r m ( t ) = P L o s s _ M 1 ( t ) P D C , P A {\displaystyle P_{Loss\_M1\_norm}(t)={\frac {P_{Loss\_M1}(t)}{P_{DC,PA}}}}
35 P A C C _ M 1 _ n o r m ( t ) = 1 T ∫ 0 t P L o s s _ M 1 _ n o r m ( t ) d t {\displaystyle P_{ACC\_M1\_norm}(t)={\frac {1}{T}}\int _{0}^{t}P_{Loss\_M1\_norm}(t)dt}