# Gain function calculation (frequency response)

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).

• Is this homework? – datageist May 25 '13 at 5:29
• @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." – Jase May 25 '13 at 5:34

## 1 Answer

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.