Magnitude of complex function

3.1k Views Asked by At

I was going through an example in a book and and it says to take the magnitude of the function. What it shows is

$$X(\omega)=\frac 1{\alpha+j\omega} \implies |X(\omega)|=\frac 1{\alpha^2+\omega^2}$$

I thought if you were to take the magnitude using $|X(\omega)| = \sqrt{a^2+b^2}$

the result would come out to be something like this.

$$|X(\omega)|=\frac 1{\sqrt{\alpha^2+\omega^2}}$$

I think I am not understanding how to calculate the magnitude of a complex function.

--The entire question was

Calculate the energy of $X(t) = e^(-at)u[t]$ Via FT

$$X(\omega) =\frac 1{a+j\omega} \implies |X(\omega)|=\frac 1{\alpha^2+\omega^2}$$

$$E = \frac 1{2\pi}\int\frac 1{\alpha^2+\omega^2}$$ Some integral math $= \frac 1{2a}$

I understand how the integral is solved and all, I just don't understand the part where they took the magnitude.

1

There are 1 best solutions below

0
On

The magnitude of the function is correct here is why: $$X(w) = \frac{1}{a+jw} = \frac{1}{a+jw}*\frac{a-jw}{a-jw} = \frac{a-jw}{a^2+w^2}$$ $$X(w) = \frac{a}{a^2+w^2}+ j*\frac{-w}{a^2+w^2}$$ $$|X(w)| = \sqrt{\frac{a^2}{(a^2+w^2)^2} + \frac{w^2}{(a^2+w^2)^2}} = \sqrt{\frac{1}{a^2+w^2}}$$ $$|X(w)| = \frac{1}{\sqrt{a^2+w^2}}$$