Add two signals without op-amp

I'd measure its response to a step input.

On a related note: How would such a device ever lead an input 180° (or any other angle with its output?

Phil, I take it you an an amateur?

--- Try this:

Version 4 SHEET 1 880 680 WIRE 144 32 96 32 WIRE 288 32 288 -64 WIRE 288 32 224 32 WIRE 304 32 288 32 WIRE 352 32 304 32 WIRE 496 32 432 32 WIRE 96 80 96 32 WIRE 496 80 496 32 WIRE 304 96 304 32 WIRE 96 192 96 160 WIRE 496 192 496 160 WIRE 96 304 96 272 WIRE 304 304 304 176 WIRE 304 304 96 304 WIRE 496 304 496 272 WIRE 496 304 304 304 WIRE 96 368 96 304 FLAG 96 368 0 FLAG 288 -64 OUT SYMBOL voltage 96 176 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value PULSE(10 20 0 1u 1u 1ms 2ms) SYMBOL voltage 496 288 R180 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V2 SYMATTR Value 10 SYMBOL res 240 16 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R1 SYMATTR Value 100 SYMBOL res 448 16 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R2 SYMATTR Value 100 SYMBOL res 80 64 R0 SYMATTR InstName R3 SYMATTR Value 100 SYMBOL res 480 64 R0 SYMATTR InstName R4 SYMATTR Value 100 SYMBOL res 288 80 R0 SYMATTR InstName R5 SYMATTR Value 10K TEXT 104 336 Left 0 !.tran 10ms uic

R3 and R4 are the source impedances of the square wave source (V1) and the offset supply (V2), and R1 and R2 are used to sum the currents from V1 and V2.

In order to work properly, assuming equal and opposite offsets from V1 and V2, R1+R3 must be equal to R2+R4 for the offsets to sum to 0V.

With a very high load impedance, (R5) the amplitude of the output will be about 1/2 the amplitude of the square wave.

-- JF