# limits of performance

I am trying to understand a topic on limits of performance.

So there is a delay of tau between input and output so the transfer function becomes G(s)*e^-(tau *s).

I am unable to understand the statement "The phase contribution of delay is negative and at crossover frequency is -(tau * omega_c) ,where omega_c is crossover frequency" I don't understand how the phase contribution due to this delay is -(tau*omega_c)?

## 2 Answers

$e^{-\tau s} \rightarrow e^{-j\omega \tau}=cos(\omega \tau)-jsin(\omega \tau) = 1\angle( -\omega \tau)$

This means unity gain, and a phase lag proportional to angular frequency. At the cut-off frequency of the other the component of the TF, the phase angle contributed by the delay will be found by setting $\omega =\omega_c$, thus: $\phi _{delay}=-\omega_c \tau$ rad.

The phase shift due to a time delay, measured as a fraction of a cycle of the signal frequency, is equal to the time delay divided by the period of the signal frequency. Since the period is the reciprocal of the frequency, the phase shift can be expressed as the product of the time delay and the signal frequency. This is the way your book is presenting it. To calculate the phase shift in degrees, you would multiply this value by 360 since one cycle of the signal represents 360 degrees of phase.