Figure 17 which is discontinuous; hence, dq/dp is discontinuous, by recalling that g also equals C(v, — v). Now, since the charge sensitivity is the integral of the current sensitivity, there must be an impulse in the current sensitivity at the time when u(t) = 3. What this example demonstrates, is that when there are loops of capacitors and voltage sources, and one of the capacitor charge models is of the type assumed here, then the current can possibly have dis- continuities and the current sensitivity can possibly have impulses. which is continuous. Computing the time derivatives it can be seen that 0(4) and i(t) are discontinuous. The voltage sensitivity response can be computed as