EMC Question of the Week: August 10, 2026
A signal traveling on a 50-Ω coaxial cable passes through an unshielded connector whose conductors are two identical blades with a 50-Ω characteristic impedance before continuing along another 50-Ω coaxial cable. At frequencies where the connector is short relative to a wavelength, the connector can be accurately modeled as a
- series inductor
- shunt capacitor
- common-mode voltage source
- common-mode current source
Answer
The best answer is "c". When the electrical balance of a transmission line changes, a common-mode voltage is generated that is equal to the differential-mode voltage at that point times the change in the imbalance factor, h. In this case, there are two changes in the electrical balance, one at each end of the connector. These changes are equal in magnitude with opposite polarity. The equivalent common-mode sources nearly cancel each other except for the phase delay in the differential-mode voltage at the two points.
At frequencies where the connector is short relative to a wavelength, the two common-mode sources can be combined into a single common-mode source with amplitude,
where is the length of the connector, and λ is the wavelength at the signal frequency.
The connector cannot be modeled as a series inductor or a shunt capacitance, because the characteristic impedance is unchanged. The differential-mode signal is unaffected by the discontinuity unless the power lost to the common-mode becomes significant.
To illustrate the importance of maintaining a constant level of electrical balance, a signal in the coaxial cable with an amplitude of 0.2 V at 80 MHz passing through a 1.2-cm balanced connector would generate 2 mV of common-mode voltage. This is often more than enough to exceed an FCC or CISPR 32 radiated emissions specification.
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