Implement this equation: x(n) = x(n-1) + a, where a = const and x(n-1) is a start point.

Your signal seems to have the same slope at every step. One of the simplest method will be to generate a ramp signal with the desired slope and to substract from it, at the desired pace a stair signal with a desired step voltage.

If you want to build it as a circuit, you can generate a ramp with an opamp integrator.

The stair signal can be generated with a counter and a multiplexer (16 stairs) or with a DAC if you need more stairs. You could eventually generate a stairs by feeding very short pulses ( "dirac function") to an integrator (which could be the same). The energy of the pulse will give the amplitude of the stair.

So, the simplest method to obtain a signal like that is to make an adder-integrator in inverter configuration and to feed it trough one resistor with negative voltage (the slope of the ramp) and trough another resistor with short positive functions (the downclimbing stairs).

Short positive pulses, not functions, sorry

Thanks to Yuriy Shmaliy and Teodor Petrita for their valuable suggestion.

Hi,

You can generate it using transition filter in Verilog-AMS.

electrical o_ramp;

real real_ramp;

analog initial begin

real_ramp = 0;

#5ns real_ramp = 1.8;

end

analog_begin

V(o_ramp)