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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?

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  • \$\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

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schematic

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)

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