是否有可能像这样对齐方程:
我尝试过这个,但是:
\begin{align*}
& u_a(t) &= u_e(t) - u_{R1}(t) \\
& &= u_e(t) - R_1 \cdot i_1(t) = u_e(t) - R_1(i_2(t) + i_3(t)) \\
& &= u_e(t) - R_1 \cdot i_2(t) - R_1 \cdot i_3(t) \\
& &= u_e(t) - R_1 \cdot \frac{u_a(t)}{R_2} - R_1 \cdot C \cdot \ddt[t]{u_a(t)}\\
<=> & u_a(t) + u_a(t) \cdot \frac{R_1}{R_2} + C \cdot R_1 \ddt[t]{u_a(t)} = u_e(t) \\
\end{align*}
答案1
对于多重比对,您可以使用aligned
内部align
。
\begin{align*}
& \begin{aligned}
u_a(t) &= u_e(t) - u_{R1}(t) \\
&= u_e(t) - R_1 \cdot i_1(t) = u_e(t) - R_1(i_2(t) + i_3(t)) \\
&= u_e(t) - R_1 \cdot i_2(t) - R_1 \cdot i_3(t) \\
&= u_e(t) - R_1 \cdot \frac{u_a(t)}{R_2} - R_1 \cdot C \cdot
ddt[t]{u_a(t)}\\
\end{aligned}\\
<=> & u_a(t) + u_a(t) \cdot \frac{R_1}{R_2} + C \cdot R_1 ddt[t]{u_a(t)} = u_e(t) \\
\end{align*}