Conjugates and Symmetry Explained

Key Rules: Transfer Function & Conjugation

1. Magnitude Squared Definition

∣G(s)∣2=G(s)G∗(s)|G(s)|^2 = G(s)G^*(s)

  • This is a general rule that applies to ALL functions.

  • This gives the power response of the filter.

  • The question now is, how do we define the conjugate of a function?

  • Using the next rule, we can find the conjugate of a real valued filter, and then use this rule to find the magnitude squared.

2. Conjugation on the Imaginary Axis

If the filter only has real coefficients and we are evaluating on the imaginary axis, we can use this rule to define what a conjugate is.

G∗(jω)=G(−jω)G^*(j\omega) = G(-j\omega)

  • For filters with real coefficients

  • This implies (subbing back into the top rule):

    ∣G(jω)∣2=G(jω)G(−jω)|G(j\omega)|^2 = G(j\omega)G(-j\omega)

Example:
Let

G(s)=s+2s2+2s+5G(s) = \frac{s + 2}{s^2 + 2s + 5}

At s=js = j:

G(j)=j+2(j)2+2j+5=j+2−1+2j+5=j+24+2jG(j) = \frac{j + 2}{(j)^2 + 2j + 5} = \frac{j + 2}{-1 + 2j + 5} = \frac{j + 2}{4 + 2j}

Now:

G∗(j)=−j+24−2j=G(−j)G^*(j) = \frac{-j + 2}{4 - 2j} = G(-j)

4. Symmetric Poles and Zeros (Quadrantal Symmetry)

  • Poles and zeros of G∗(s)G^*(s) are the reflections of those in G(s)G(s):

    • If pp is a pole of G(s)G(s), then −p-p is a pole of G∗(s)G^*(s)

  • This gives symmetry in the s-plane about both the real and imaginary axes

Example:
Let

G(s)=1(s+1)(s+2)G(s) = \frac{1}{(s + 1)(s + 2)}

Poles of G(s)G(s): s=−1,−2s = -1, -2
Poles of G∗(s)=G(−s)G^*(s) = G(-s): s=1,2s = 1, 2

So G(s)G(s) and G∗(s)G^*(s) are mirror images in the s-plane.

It is obvious these are symmetrical across the imaginary axis. But they are ALSO symmetrical across the real axis. They just reflect on to themselves.

This example also only valid when we evaluate s to be imaginary, because of the definition of the conjugate we used.

5. Even Powers in ∣G(s)∣2|G(s)|^2

∣G(s)∣2=G(s)G(−s)|G(s)|^2 = G(s)G(-s)

  • Result: all odd powers of ss cancel out

  • Final expression contains only even powers of ss

  • This is only valid for real coefficient filters

Example:
Let

G(s)=s+1s+2G(s) = \frac{s + 1}{s + 2}

Then:

G(−s)=−s+1−s+2G(-s) = \frac{-s + 1}{-s + 2}

Now:

∣G(s)∣2=s+1s+2⋅−s+1−s+2=(s+1)(−s+1)(s+2)(−s+2)|G(s)|^2 = \frac{s + 1}{s + 2} \cdot \frac{-s + 1}{-s + 2} = \frac{(s + 1)(-s + 1)}{(s + 2)(-s + 2)}=−(s2−1)−(s2−4)=s2−1s2−4= \frac{-(s^2 - 1)}{-(s^2 - 4)} = \frac{s^2 - 1}{s^2 - 4}

→ Only even powers of ss appear.