Question 6.4: Consider, as in Example 6.3, a generator with internal resis......
Consider, as in Example 6.3, a generator with internal resistance R_G>0 connected to a load R_L. Investigate the behavior of the product \left|\Gamma_G \Gamma_L\right| (1) when R_L assumes positive and negative values; (2) for different values of the normalization resistance R_0.
We have:
\left|\Gamma_G \Gamma_L\right|=\left|\frac{R_G-R_0}{R_G+R_0} \frac{R_L-R_0}{R_L+R_0}\right|=\left|\frac{r_G-1}{r_G+1} \frac{r_L-1}{r_L+1}\right| .
Investigation of the values assumed by the product \Gamma_G \Gamma_L (real in this case) allows us to clarify that when \Gamma_G \Gamma_L=1, then, correctly, r_G+r_L=0 corresponding to the oscillation condition. However r_G+r_L>0 does not imply necessarily \left|\Gamma_G \Gamma_L\right|<1, nor r_G+r_L<0 implies \left|\Gamma_G \Gamma_L\right|>1. This is clearly seen from the result shown in Fig. 6.7. In fact, imagine that R_G=R_0 and therefore \Gamma_G=0; in such case \left|\Gamma_G \Gamma_L\right|=0<1 independent from the value of R_L. In Example 6.3 we had the condition r_L=-r_G / 2 which always implies of course r_G+r_L>0 but may correspond to \left|\Gamma_G \Gamma_L\right|>1 or to \left|\Gamma_G \Gamma_L\right|<1 according to the choice of the normalization resistance. In conclusion, while \Gamma_G \Gamma_L=1 implies r_G+r_L=0 and vice versa the condition \left|\Gamma_G \Gamma_L\right|<1 does not necessarily imply r_G+r_L>0.
