# Insertion loss equality - proof

From this paper (slide 11), the insertion loss (or attenuation) is defined as

$$L(\omega^2)=\frac{\lvert V_i(j\omega)\rvert^2}{\lvert V_o(j\omega)\rvert^2}=\frac{1}{\lvert H(j\omega)\rvert^2}=10 \log\left(\frac{1}{H(j\omega)H(-j\omega)}\right)$$

I'm interested specifically in, where does the last inequality follow?

• The last equality does not follow. The first three terms are in linear scale, the last one is logarithmic. – Peter K. Jun 27 '16 at 11:17
• @PeterK. This is given here d-filter.ece.uvic.ca/SupMaterials/Slides/DSP-Ch10-S1-7.pdf p. 11. – mavavilj Jun 27 '16 at 11:39
• Well then it's complete rubbish. – Peter K. Jun 27 '16 at 12:02
• Looks like a typo. The 10 log() crept in one line to early – Hilmar Jun 28 '16 at 20:02

As pointed out in PeterK.'s comment, the way it is written this does not make sense. What is meant here is that the insertion loss can be measured on a linear scale (i.e., as a factor) as

$$L(\omega^2)=\frac{1}{|H(j\omega)|^2}$$

or it can be given in decibels:

$$L_{dB}(\omega^2))=10\log_{10}(L(\omega^2))$$

The remaining question could be why $|H(j\omega)|^2$ can be replaced by $H(j\omega)H(-j\omega)$. The reason is the symmetry of the frequency response $H(j\omega)$ for real-valued filters:

$$H(j\omega)=H^*(-j\omega)$$

From this it follows that

$$|H(j\omega)|^2=H(j\omega)H^*(j\omega)=H(j\omega)H(-j\omega)$$