Because the DC voltage from a fullwave rectifier is equal to the peak voltage of the AC waveform minus the diode drops.
DCVout = (1.414 x VAC) - DiodeDrops
49.5 = (1.414 x 36) - 1.4
Use a transformer with a lower voltage or add a voltage regulator after the filter cap. What is the voltage rating of the stepper motor?
and what is the diffrence to use one filter or 2 filter
i am using now one cap 10.000 uf
so if i put other one in parallel will be the same or the out will change
Well, if you have a 12V stepper and you don't intend to PWM it, then you should be running it on 12V! Since the l298 drops nominal 3.7V under full load (2A), you need to feed the l298 with around 15.7Vdc to get 12V out for the stepper. A 12VAC transformer with a bridge rectifier and cap would then give you a nominal unregulated 15.6Vdc.
Are you going the PWM the stepper for more torque?
Why would a unregulated supply be used over a regulated one? Cost? Efficiency? If it's efficiency can you bring the input voltage as close to the regulated output voltage as possible to lower the loss?