Consider an AC voltage source is placed in series with a complex impedance (z) and a load impedance. The maximum power transfer to the load impedance occurs when the load impedance equals the complex conjugate of the series impedance (z).

What I want to know is what is the (mathematically rigorous) derivation of this?

I've looked through my electrical engineering textbooks and can't find a good derivation. If you could provide one, I would be very appreciative.

Thanks in advance for your help.

  • 1
    \$\begingroup\$ Actually, to transfer the maximum power into a fixed load impedance, you make the source impedance as small as possible (so as not to waste power in the source). The equal/conjugate impedances comes into play when you have a fixed source impedance and can only vary load impedance such that it sucks the most power out of the source. \$\endgroup\$
    – Kaz
    May 17 '12 at 4:56
  • \$\begingroup\$ I.e. this problem assumes that you cannot lower the total impedance. There is an implied constraint that the total load + source impedance must stay fixed, and within this constraint, you solve for the maximum power transfer. Similar to any other maximization problem, like: given a string of so many feet, what is the largest area you can enclose which looks like a rectangle with a semi-circular cap. This is the key; I think you can work it out now. \$\endgroup\$
    – Kaz
    May 17 '12 at 5:07
  • \$\begingroup\$ @Kaz of course if we can make the sum of the load and series equivalent impedances lower, the power transfer will be higher. But the question is regarding the derivation for a constant series resistance, z. \$\endgroup\$
    – JonaGik
    May 18 '12 at 1:37

The reference is Desoer & Kuh, Basic Circuit Theory.

First, notation and an expression for average power Pav. For a sinusoidal voltage v and current i at the same frequency:

$$v(t)=V_m cos(\omega t + \measuredangle{V})= Re(Ve^{j \omega t}) \text{ where } V \triangleq V_m e^{j \measuredangle{V}}$$

$$i(t)=I_m cos(\omega t + \measuredangle{I})= Re(Ie^{j \omega t}) \text{ where } I \triangleq I_m e^{j \measuredangle{I}}$$

$$p(t)=v(t)i(t)=\frac{1}{2}V_mI_m cos(\measuredangle{V}-\measuredangle{I})+\frac{1}{2}V_mI_m cos(2\omega t +\measuredangle{V}+\measuredangle{I})$$

Averaging over a period, the average power Pav is:

$$P_{av}=\frac{1}{2}V_mI_m cos(\measuredangle{V}-\measuredangle{I})=\text{Re}(\frac{1}{2}V \overline{I})$$

If V is related to I by a complex impedance Z, V=IZ, then:

$$P_{av}=\frac{1}{2}\text{Re}(I \overline{I} Z) = \frac{1}{2}|I|^2 \text{Re}(Z)$$

With that out of the way, on to the maximization. With source voltage vs, load voltage vl, and current i as above, fixed source impedance Zs=Rs+jXs, and to-be-determined load impedance Zl=Rl+jXl, the average power delivered to the load Pav is:

$$P_{av}=\frac{1}{2} |I|^2 R_l$$



it follows that

$$P_{av}=\frac{1}{2} |V_s|^2 \frac{R_l}{|Z_s+Z_l|^2}=\frac{1}{2} |V_s|^2 \frac{R_l}{(R_s+R_l)^2+(X_s+X_l)^2}$$

You can now maximize this expression by separately differentiating with respect to the imaginary and real parts of Zl:

  1. With respect to Xl, which only appears in one location, the maximum is achieved at Xl=-Xs.
  2. With respect to Rl, which appears in two locations, the maximum is achieved at Rl=Rs.

So, for maximum power delivery, set Zl to:

$$Z_{l,opt}= R_s-jX_s = \overline{Z_s}$$ The maximum average power delivered to that load is:


  • \$\begingroup\$ You are most welcome. \$\endgroup\$
    – Art Brown
    May 18 '12 at 2:25

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