# Polar form of eigenvalue

I greatly appreciate the opportunity afforded by this forum to submit a query.

It has been suggested that for a sinusoidal input, represented as 2 phasors, into a Linear Shift-invariant system, the output...

...can also be written as follows:

where $$H(\Omega) = \int_{-\infty}^{\infty} h(\lambda) e^{-j\Omega\lambda} d\lambda$$

Accompanying brief comment suggests the above to be on account of "polar form representation of H".

Although I am cognizant of polar form representation of complex numbers i.e.

$$z = |z|(\cos(\theta)+i \sin(\theta)) = |z|e^{i \theta}$$

...I am completely unable to fathom the alternative above.

All advice would be greatly appreciated.

Best regards,

wirefree


## 1 Answer

I think all it's saying is that we can write $H(\Omega)$ as:

$$H(\Omega) = \left| H(\Omega) \right | e^{j \angle H(\Omega) }$$

It also seems to be assuming that $h(t)$ is real-valued so that $H(-\Omega) = H^*(\Omega)$.

The function $H(\Omega)$ is just a map from $\mathbf{R} \mapsto \mathbf{C}$ (from the reals to the complex numbers). For any specific value of $\Omega$ this generates a single complex number $H(\Omega)$. And, as you say, any complex number resolves to its exponential notation.

• Appreciate your response, @peter-k I refer to the exponential notation again: any complex number resolves to its exponential notation due to Euler's formula; but how does H(Omega) resolve to it? Would be grateful if you'd drop in a word. Is there a concept I should read up on that's clearly ostensible to you? Best regards, wirefree – Gaurav Sobti Sep 28 '15 at 9:56