Chosing a suitable switch mode power supply
\$ 550W \times 300 \mu s = 165mJ \$ (energy per pulse)

\$ 165mJ \times 50/s = 8.25W \$ (average power, 20ms pulse period so 50 pulses per second)

35–48Vdc range needed so use a 48V 10W PSU and a bank of capacitors for storing energy. Caps will slowly charge up to \$ V_{max} = 48V \$ and during the pulse, they will be discharged down to \$ V_{min} = 35V \$. Energy needed in caps \$ \Delta Ε = 165mJ\$. The caps will give this energy while discharged from \$V_{max}\$ to \$V_{min}\$:

\$ \DeltaΕ = C\dfrac{(V_{max})^2}{2} − C\dfrac{(V_{min})^2}{2} \$

solve for C

\$C = 2 \Delta Ε((V_{max})^2−(V_{min})^2) = 300 \mu F \$

(330μF closest cap value)

If you use a bigger cap, the voltage will be more stable, plus a cap this size costs almost nothing. Remember to get one with at least 63V rating.