Abstract
The Nyquist diagram technique is examined under very general assumptions; in particular, the linear subsystem is represented by a convolution operator, thus, the case of any linear time-invariant distributed circuit is included. It is shown that if there are no encirclements of the critical point, then the impulse response of the closed-loop system is bounded and absolutely integrable on[0, \infty); it also tends to zero ast \rightarrow \infty. For any initial state, the zero-input response of the closed-loop system is also bounded and goes to zero. If, on the other hand, there are one or more encirclements of the critical point, then the closed-loop impulse response tends asymptotically to a growing exponential.

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