0
\$\begingroup\$

enter image description here

Say e(t) is sine wave and y(t) is sine output.

If y(t) has the same frequency as e(t) with phase shift of 180 degrees, we can only get stability of the system, if the gain of open loop at this frequency is less than 1.

This makes sense to me, in order for y(t) to gradually descend toward 0, the gain obviously has to be less than 1.

In order for that to happen, the polar plot of open loop G(iw) at this frequency (where it crosses the (-x)) has to lie between points (-1,0) and (0,0).

Why? What does the location of that exact point tell me about gain of this system? enter image description here

\$\endgroup\$
  • \$\begingroup\$ It tells you that the loop gain is less than unity when the phase lag is 180 degrees. Therefore the closed loop system will not have sufficient gain to maintain a steady state oscillation. \$\endgroup\$ – Chu Apr 5 '16 at 21:35
1
\$\begingroup\$

The Nyquist plot of a transfer function is just a plot of the imaginary part versus the real part of the transfer function. If you imagine a vector that starts at the origin and traces the curve starting at omega=0, then the angle of the vector is the phase shift and its length is the magnitude.

In the plot given as an example you can see that the gain at DC (the starting point of the curve) is close to 4 and the phase shift is zero. For higher frequencies the gain goes towards zero (because the curve terminates at the origin) and for a phase shift of -180 the distance from the origin to the curve is smaller than 1 because it is left to the -1 point and therefore the distance to the origin is smaller than 1.

\$\endgroup\$
0
\$\begingroup\$

In your example, you have plotted the Nyquist curve for the gain function G(s). In this case, the gain of the active device is identical to the so-called LOOP GAIN (gain of the complete loop if it is open) because of 100% feedback.

Pleae note, that for such stability analyses based on Nyquist plots you ALWAYS must use the loop gain function (which only in this case is identical to the gain function of the active block).

For a circuit with feedback to be stable,we require that the loop gain for rising frequencies NEVER assumes a phase shift of 360deg at a loop gain magnitude of equal or larger than unity.

Hence, the Nyquist curve of the product G(s)*feedback factor (excluding the minus sign at the summing node !!) must not cross the negative-real axis (-180 deg) at a point where the magnitude (length of the vector from the origin to the curve) is equal or larger than "1".

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.