# How does a time delay affect the Gain and Phase Margin?

How does a time delay affect the gain and phase margin. Answers using nyquist plots would be helpful

I know that a time delay of t corresponds to gain multiplying by 1 and phase subtracting omega*T . So as omega goes to infinity it will rotate around the origin getting closer and closer.

• Can you show what you have tried so I can see if I can add to that? Thanks
– rrz0
Jan 14, 2019 at 17:26
• @Rrz0 hope this helps a bit apologies if not Jan 14, 2019 at 17:28
• When you are driving a car with eyes giving you error feedback, can you visualize what happens if the feedback is a second late on say a web based camera with windshield covered when driving on a curved road? This delay can induce instability if the compensation is too fast or too high gain so the effects of this improved compensation shifts the poles and by anticipation you can add some zeros to optimize your driving so you can predict changes ahead and not over-react to errors. THis prediction method is often called lead-lag phase compensation. Jan 14, 2019 at 20:28
• In Nyquist plots often simulated with pole-zero cancellation filter transforms as long as you do not make a circle around the origin. dsp.stackexchange.com/questions/38480/… Jan 14, 2019 at 20:31
• Please try investigating this yourself before just memorizing answers from us. I suggest you make a really simple system with a loop gain of $100/s$ -- so in closed loop it has a loop closure frequency of 100 radians/sec, a phase margin of 90 degrees, and no gain margin because there's no phase crossing points. Now add a delay of $\tau = 1 \mathrm{ms}$, and do it over again (the transfer function for a pure delay in the Laplace domain is $e^{- \tau s}$). Now try it again with delays of $\tau = 2 \mathrm{ms}, 5 \mathrm{ms}, 10 \mathrm{ms}, 20 \mathrm{ms}, 50 \mathrm{ms}$, etc. Jan 14, 2019 at 21:11