How to approximate $\ln \left( {{e^{{x_1}}} - {e^{{x_2}}}} \right)$?

64 Views Asked by At

In the Max-Log-Map algorithm for channel decoding, the approximation $\ln \left( {{e^{{x_1}}} + {e^{{x_2}}}} \right) \approx \max \left( {{x_1},{x_2}} \right)$ can be considered because $\ln \left( {{e^{{x_1}}} + {e^{{x_2}}}} \right) = \max \left( {{x_1},{x_2}} \right) + \ln \left( {1 + {e^{ - \left| {{x_1} - {x_2}} \right|}}} \right)$. In fact, this approximation is ussually use in practice because it can be done easily using simple circuit.

Since my application change this algorithm a little bit, what should I do if I need to approximate $\ln \left( {{e^{{x_1}}} - {e^{{x_2}}}} \right)$ ?

Would you kindly help me with this ?

Thank you for your enthusiasm !

1

There are 1 best solutions below

0
On

If $x_2 - x_1 = t > 0$, $$ \ln\left(e^{x_2} - e^{x_1}\right) = \ln\left(e^{x_2} \left(1 - e^{-t}\right) \right) = x_2 + \ln(1-e^{-t}) $$ How you might want to approximate $\ln(1-e^{-t})$ depends on what you know about $t$. For $t$ near $0$, $1 - e^{-t} \approx t$, so $\ln(1-e^{-t}) \approx \ln(t)$. For large $t$, $1-e^{-t} \approx 1$ so $\ln(1-e^{-t}) \approx 0$. If you approximate $\ln(e^{x_2} - e^{x_1})$ as $x_2 + \ln(x_2 - x_1)$ for $0 < x_2 - x_1 < 1$ and $x_2$ for $x_2 - x_1 \ge 1$, the maximum absolute error is approximately $0.46$.