# Frequency response based on a transfer function

The transform function: $$T(s) = \frac{1-sRC}{1+sRC}$$ Polynomial form: $$-\frac{s-\frac{1}{RC}}{s + \frac{1}{RC}}$$ Since magnitudes of the zero Sn = 1/RC and pole Sp = -1/RC are equal,amplitude gain is 0. What about the phase? How does '-' sign affect the phase?

Without the minus: Sn i positive and real and has a $\Pi$ phase while the negative Sp has a $0$, at $\omega=0$.

As $\omega\rightarrow\infty$, $\pi \rightarrow \pi/2$ and $0\rightarrow\pi/2$. After subtracting the phases from zero and poles we have that phase changes from $\pi\rightarrow0$.

What does the minus affect?

• For w=0 the phase is 0 and and approaches -180deg for rising frequencies. It is simply a first-order allpass.
– LvW
Oct 26, 2016 at 12:35
• If $z = x + j y$, what is the difference in phase compared to $-z = -x - j y$? Oct 26, 2016 at 12:36
• @LvW Why is it -180deg, shouldn't it be +, since phase of -1 is +180? Oct 26, 2016 at 12:53
• @Arnfinn + $\pi$ ? Oct 26, 2016 at 14:07
• For stable systems the phase always goes to negative values (phase lag,falling chareacteristic).
– LvW
Oct 26, 2016 at 14:09