Filtering inductor for power electronic converter design formula

I have following power electronic converter

and I need to determine inductance of the filtering chokes. I have deduced following formula $$L=\frac{V_{dc}\cdot(1-d)}{f_s\cdot\Delta i_L}$$ (based on $v=L\frac{di}{dt}$), where $V_{dc}$ is dc link voltage, $f_s$ is switching frequency (=12 kHz), $\Delta i_L$ is desired current ripple and $d$ is duty ratio ($0\leq d\leq1$). My problem is that I would like to use Sinewave Pulse Width Modulation so the duty ratio is changing sinusoidally. I don't know what value to substitute. The solution could be to choose the worst case but I don't know what value of $d$ coresponds to the worst case. Can anybody give me an advice how to solve that? Thanks in advance.

• Possible duplicate of Filtering inductance calculation – JRE Oct 18 '16 at 10:07
• Why did you delete the original question? It had some comments that may have been of use in answering your question. – JRE Oct 18 '16 at 10:16
• I used the comments in the new question. – Steve Oct 18 '16 at 10:18
• Still, though, it wasn't necessary to create the new question at all. – JRE Oct 18 '16 at 10:20
• That's what editing is for. – JRE Oct 18 '16 at 10:35

• I have started with this assumption $v_L=+V_{dc}$ for $0\leq t \leq d\cdot T_s$ and $v_L=-V_{dc}$ for $d\cdot T_s\leq t \leq T_s$. The inductor current at the end of the first interval is $i_L(d\cdot T_s=\frac{1}{L}V_{dc}\cdot d\cdot T_s$. This value of inductor current is initial condition for integration the voltage over the other interval. The inductor current at the of the other interval is $i_L(T_s)=-\frac{1}{L}V_{dc}\cdot T_s(1-d)+\frac{1}{L}V_{dc}\cdot d\cdot T_s$. – Steve Oct 18 '16 at 11:24