2
$\begingroup$

Define moving average process $y_t := 0.5 x_t + 0.5 x_{t-1}$ where $x_t := e^{i2 \pi t}$. Its frequency response is then:

$$H(f) = 0.5 + 0.5 e^{-i2\pi f}$$

Recall that the frequency response in polar notation is:

$$H(f) = G(f)e^{i \theta}$$

where $G(f)$ is the gain function. We then have that:

$$ H(f) = 0.5 \big( e^{-i \pi f} + e^{i \pi f} \big)e^{-i \pi f}$$.

QUESTION: Can someone please explain why the last step is true? An explanation with some detail is much appreciated (I am new to signal analysis).

$\endgroup$
2
  • $\begingroup$ Is this homework? $\endgroup$
    – datageist
    May 25, 2013 at 5:29
  • $\begingroup$ @datageist Not homework. It is from the body of text in Chapter 2 in "An Introduction to Wavelets and Other Filtering Methods in Finance and Economics." $\endgroup$
    – Jase
    May 25, 2013 at 5:34

1 Answer 1

3
$\begingroup$

One way to see the equivalence is to first factor out the $0.5$ in your initial transfer function, i.e.

$$ H(f) = 0.5(1 + e^{-i2\pi f}). $$

Then you can rewrite the $1$ and $e^{-i2\pi f}$ as follows:

$$ H(f) = 0.5(e^{i\pi f}e^{-i\pi f} + e^{-i\pi f}e^{-i\pi f}) = 0.5(e^{i\pi f} + e^{-i\pi f})e^{-i\pi f}. $$

Your result then follows as a trivial transform.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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