Simple question here, How do I rearrange this:
$$T(s) = \frac{1}{(s + 1)(s^2 + s + 1)}$$
to this?
$$\left |T(j\omega)\right |= \frac{1}{\sqrt{1 + \omega^6}}$$
Wolfram does it and says it works when omega is positive but I'm still lost, doing the calculation for squares I get a huge string of exponents of omega such as the denominator after squaring:
$$-\omega^6 + 4j\omega^5 +8\omega^4 -10j\omega^3 -8\omega^2 +4j\omega + 1$$
The denominator first becomes this and I don't like to proceed further because all directions I have followed I've ended up with a large series such as that above.
$$\left|(1+2 j w-2 w^2-j w^3)\right|$$
Best Answer
First off, we transform the s domain to frequency domain with \$s=j\omega \$ which gives us:
$$T(j\omega) = \frac{1}{(j\omega+1)((j\omega)^2+j\omega+1)} = \frac{1}{(j\omega+1)(j\omega+1-\omega^2)} = \frac{1}{1-2\omega^2+j(-\omega^3+2\omega)}$$
Notice i wrote the denominator in the form \$a+bj\$. Taking the magnitude of this function gives:
$$\left |T(j\omega)\right |= \frac{1}{\sqrt{a^2+b^2}} = \frac{1}{\sqrt{(1-2\omega^2)^2+(-\omega^3+2\omega)^2}} = \frac{1}{\sqrt{(4\omega^4-4\omega^2+1)+(\omega^6-4\omega^4+4\omega^2)}} =\frac{1}{\sqrt{1+\omega^6}} $$