Correct Option: D

Redrawing the first stage,

Assuming ideal OPAMP’s

V- = V+ =	V2
	2

V+ =	V2 R	=	V2
	R + R		2

Using KCL at inverting terminal, we get

	V1 - V-	=	V- - V01
	R		R

⇒ V₀₁ = 2V_– – V₁ = V₂ – V₁
= – V_i [as V₁ – V₂ = V_i]
For the second-stage schmitt-trigger.

For V₀₁ < V₊ or V_i > V₊, V₀ = 12 Volts

v+ =	R	VR +	R	V0 ≡ V1
	R + R		R + R

As V_R = 0 [V_R is grounded]

v+ =	V0	≡ V1 .......(A)
	2

For v₀₁ > v₊
or v_i < v₊ , v₀ = – 12 volts
The voltage at the non-inverting terminal is,

v+ =	R VR	-	R	V0 ≡ V2
	R + R		R + R

v+ = -	V0	≡ V2
	2