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In case the two measurement ports use the same reference impedance, the insertion loss is defined as: [1] [2]= | |. Here is one of the scattering parameters.Insertion loss is the extra loss produced by the introduction of the DUT between the 2 reference planes of the measurement.
Typical transformer insertion loss measurements are taken at 1,000 Hz to optimize the transformer's specifications. [4] Using this method, typical insertion losses are about 1 dB, a 20% power loss. Most of the power in voice-application audio systems is below 400 Hz, meaning that insertion loss at lower frequencies would be greater.
The equations above find the impedance and loss for an attenuator with given resistor values. The usual requirement in a design is the other way around – the resistor values for a given impedance and loss are needed. These can be found by transposing and substituting the last two equations above; If = =
The extra loss may be due to intrinsic loss in the DUT and/or mismatch. In case of extra loss the insertion loss is defined to be positive. The negative of insertion loss expressed in decibels is defined as insertion gain and is equal to the scalar logarithmic gain (see: definition above).
The L-pad computation assumes that port 1 has the highest impedance. If the highest impedance happens to be the output port, then use this figure. Unique resistor designations for Tee, Pi and L pads. The attenuator two-port is generally bidirectional. However, in this section it will be treated as though it were one way.
The insertion loss is not such a problem for an unequal split of power: for instance -40 dB at port 3 has an insertion loss less than 0.2 dB at port 2. Isolation can be improved at the expense of insertion loss at both output ports by replacing the output resistors with T pads. The isolation improvement is greater than the insertion loss added ...
One simply needs to know the input impedance R p and to choose the output impedance R s. Or conversely know R s and choose R p. Keep in mind that R p must be larger than R s. Because reactance is frequency dependent the L network will only transform the impedances at one frequency. Inclusion of two L networks back to back creates what is known ...
When Z 1 Z 2 = R 0 2 it becomes a constant resistance network, which has an insertion loss given by () = + When normalized to 1ohm, the source, load and R 0 are all unity, so Z 1.Z 2 = 1, and the insertion loss becomes = + ()