Simple Nature - Light and Matter

a purely inductive or capacitive load, P av equals zero, so the problemisn’t very interesting!)Since we’re assuming both the load and the transmission line areresistive, their impedances add in series, and the amplitude of thecurrent is given bysoI o =V oZ o + Z T,P av = I o V o /2= I 2 oZ o /2=V 2o Z o(Z o + Z T ) 2 /2 .The maximum of this expression occurs where the derivative is zero,0 = 1 [d V 2 ]o Z o2 dZ o (Z o + Z T ) 20 = 1 []d Z o2 dZ o (Z o + Z T ) 20 = (Z o + Z T ) −2 − 2Z o (Z o + Z T ) −30 = (Z o + Z T ) − 2Z oZ o = Z TIn other words, to maximize the power delivered to the load, weshould make the load’s impedance the same as the transmissionline’s. This result may seem surprising at first, but it makes senseif you think about it. If the load’s impedance is too high, it’s likeopening a switch and breaking the circuit; no power is delivered.On the other hand, it doesn’t pay to make the load’s impedance toosmall. Making it smaller does give more current, but no matter howsmall we make it, the current will still be limited by the transmissionline’s impedance. As the load’s impedance approaches zero, thecurrent approaches this fixed value, and the the power delivered,IoZ 2 o , decreases in proportion to Z o .Maximizing the power transmission by matching Z T to Z o iscalled impedance matching. For example, an 8-ohm home stereospeaker will be correctly matched to a home stereo amplifier withan internal impedance of 8 ohms, and 4-ohm car speakers will becorrectly matched to a car stereo with a 4-ohm internal impedance.You might think impedance matching would be unimportant becauseeven if, for example, we used a car stereo to drive 8-ohmspeakers, we could compensate for the mismatch simply by turningthe volume knob higher. This is indeed one way to compensatefor any impedance mismatch, but there is always a price to pay.614 Chapter 10 Fields