0
$\begingroup$

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...

1

...can also be written as follows:

2

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
$\endgroup$
1
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Gaurav Sobti Sep 28 '15 at 9:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.