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Performance

What kind of results can we expect from a delay-and-sum beamformer? For now we'll assume that the source angle is known and we can delay the signals without any error. First some definitions:
$\displaystyle s_i(t)$ $\textstyle \doteq$ $\displaystyle \mbox{real valued signal}$  
$\displaystyle n_i(t)$ $\textstyle \doteq$ $\displaystyle \mbox{white gaussian noise}$  
$\displaystyle \sigma_x^2 \mbox{(average power)}$ $\textstyle \doteq$ $\displaystyle \frac{1}{N}\sum_{k=0}^{N-1}{x(t)^2}$  
$\displaystyle SNR_{s+n}$ $\textstyle \doteq$ $\displaystyle \frac{\sigma_s^2}{\sigma_n^2}$  

Now put $s_1(t)+s_2(t)$ into the average power definition. Then,
\begin{displaymath} \sigma_{s_1+s_2}^2 & = & \frac{1}{N}\sum_{k=0}^{N-1}{(s_1(t)+s_2(t))^2}\nonumber \end{displaymath}  

If we delay $s_1(n)$ so that $s_1(n) & = & s_2(n)$, then we get $\sigma_{s_1+s_1}^2 & = & 4\sigma_{s_1}^2\nonumber$.

Repeat the above procedure with the noise and expand.

\begin{displaymath} \sigma_{n_1+n_2}^2 & = & \frac{1}{N}\sum_{k=0}^{N-1}{(n_1^2(... ...sum_{k=0}^{N-1}{(n_1^2(t)+n_2^2(t))}+2<n_1(t),n_2(t)>\nonumber \end{displaymath}  

The time-averaged cross-correlation, $<n_1(t),n_2(t)>$, is zero for noise. Then, $\sigma_{n_1+n_2}^2 & = & \sigma_{n_1}+\sigma_{n_2} & = & 2\sigma_{n_1}^2\nonumber$. And
\begin{displaymath} SNR_{s_1+n_1+s_1+n_2} & = & 2SNR_{s_1+n_1}\nonumber \end{displaymath}  

And in general, for M microphones, we get M times the SNR [MIK]. For the Signal to Interference Ratio (SIR), the cross-correlation term will take a value between zero and the interference power. So the array SIR will be better than the one element microphone SIR and worse than M times the SIR for one microphone.
next up previous contents
Next: Finding the Delay Up: Delay-and-Sum Beamforming Previous: Basic Idea   Contents
Todd A Goldfinger 2004-11-22