Conjugate Impedance Matching

Consider the AC system shown above. There is a source with impedance RS+jXSR_S + jX_S and a load with impedance RL+jXLR_L + jX_L. The resistive (R)(R) components must be positive (per standard physics), and the reactive (X)(X) components can be either positive or negative (corresponding to inductive / capacitive components respectively). The source voltage angle is shown as zero degrees, but this angle is actually irrelevant.

Suppose we want to maximize the active power (Watts) dissipated by the load. What should the load impedance parameters be?

Derive an expression for the magnitude of the load current:

I=V(RS+RL)2+(XS+XL)2I2=V2(RS+RL)2+(XS+XL)2|I| = \frac{V}{\sqrt{(R_S + R_L)^2 + (X_S + X_L)^2}} \\ |I|^2 = \frac{V^2}{(R_S + R_L)^2 + (X_S + X_L)^2}

Load power:

PL=I2RL=V2RL(RS+RL)2+(XS+XL)2P_L = |I|^2 R_L = \frac{V^2 R_L}{(R_S + R_L)^2 + (X_S + X_L)^2}

Since VV is a constant, we therefore want to maximize the following quantity:

α=RL(RS+RL)2+(XS+XL)2\alpha = \frac{R_L}{(R_S + R_L)^2 + (X_S + X_L)^2}

Assume that this function is both continuous and continuously differentiable in both variables (RL,XL)(R_L, X_L). Take partial derivatives with respect to both variables and set them equal to zero. This is a multi-variate version of the standard calculus-based optimization strategy. I will also neglect to perform the equivalent of a "second derivative test" here. The mathematically inclined reader is free to try these as side problems. I don't want to get bogged down in mathematical details.

αRL=(RS+RL)2+(XS+XL)2RL(2(RS+RL))β=0    (RS+RL)2+(XS+XL)2=2RL(RS+RL)\frac{\partial{\alpha}}{\partial{R_L}} = \frac{(R_S + R_L)^2 + (X_S + X_L)^2 - R_L (2(R_S + R_L))}{\beta} = 0 \\ \implies (R_S + R_L)^2 + (X_S + X_L)^2 = 2 R_L (R_S + R_L)

αXL=2RL(XS+XL)β=0    RL(XS+XL)=0\frac{\partial{\alpha}}{\partial{X_L}} = \frac{-2 R_L (X_S + X_L)}{\beta} = 0 \\ \implies R_L (X_S + X_L) = 0

Since we are maximizing the load active power, we must obviously neglect the case in which RL=0R_L = 0.

    XS+XL=0XL=XS\implies X_S + X_L = 0 \\ \boxed{X_L = -X_S}

The resulting expression from the RLR_L partial derivative thus reduces to:

(RS+RL)2=2RL(RS+RL)(R_S + R_L)^2 = 2 R_L (R_S + R_L)

Since the resistive terms cannot be negative, and RLR_L cannot be zero, the following results:

RS+RL=2RLRL=RSR_S + R_L = 2 R_L \\ \boxed{R_L = R_S}

We see, therefore, that in order to maximize the active power dissipated by the load, the load resistance must match the source resistance, and the load reactance must be the negative of the source reactance. In more technical language, the load impedance must be the complex conjugate of the source impedance.

Note by Steven Chase
3 years, 10 months ago

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