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?

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

1 Answer 1



simulate this circuit – Schematic created using CircuitLab

Here is a schematic of a first-order allpass. As one can see, the phase starts at 0 deg (cap is open circuit) and goes to negative values (low pass response). For very large frequencies, the non-inv. input is grounded and we have negative unity gain (inverter) - equivalent to -180deg phase shift.

The transfer function is

H(s)=(1+R/R)/(1+sRC) -R/R; H(s)=(1-sRC)/(1+sRC)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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