Solution for adding around 60V dc-offset on digital signal(0 and 5v) of 10MHz frequency
You did not give us a lot of details, so I don't know whether my solution is really applicable in your case, but the simplest is to AC-couple your 10MHz digital signal using a capacitor:







simulate this circuit &ndash; Schematic created using CircuitLab

On the output, you end up with a signal that moves between about 59.5 and 64.5V (due to the diode drop). The diode is here to maintain the capacitor charged to 60V (when the input is at low level). It charges very quickly, and, once charged, the output will move in the same way as the input, because the voltage across the capacitor changes very slowly compared to the input voltage (so when the input changes, the output changes also). You could put a resistor instead (a few kOhm), but this would lead to the output voltage average being centered around 60V instead, (so the voltage levels would actually depend on the duty cycle). With the diode, we have the low level fixed at 60V (minus the diode drop).

Note that the input signal must keep changing all the time. If it stays high for too long, you'll end up with a low level on the output after some time (a few ms, with the circuit shown). Because the capacitor will have discharged too much.

Also, be careful about the voltage rating of the capacitor. It must be at least 60V.

Edit: I've added (see comments below) a decoupling cap at the diode's anode. This ensures that the short current pulses required to recharge the capacitor when the input goes low can be provided fast enough, and avoid some additional voltage drop due to the resistance of the wire between the supply and the circuit (in case the supply is located far away).