I have been struggling with this issue: https://electronics.stackexchange.c...onance-frequency-go-above-the-input-frequency

Does anyone have an idea why that is the case

 

Because at resonance, the parallel tank has an infinite impedance (not counting losses). If the signal generator sees an infinite impedance (like a 1MΩ resistor), what would that do to the voltage? Absolutely nothing! Now, if you were to connect a load across a source and their impedances were a conjugate match, you would see half the sig gen open circuit voltage.

Edit: I misread the answer offered on that site, he was somewhat correct, but it didnt answer the question If the source is 50Ω and there was a 50Ω resistor in parallel with that tank, then at resonance, the voltage would be half the source voltage. A mismatch between the source and load will give you anything from no voltage to twice the expected voltage because the source and load impedance form a voltage divider.

 

Signal generators will always produce twice the voltage they claim when not terminated into their designed impedance, so at resonance, you are seeing the open circuit voltage of the generator.

 

Signal generators will always produce twice the voltage they claim when not terminated into their designed impedance, so at resonance, you are seeing the open circuit voltage of the generator.

But everything was terminated 50 R impedance coaxial cable as stated at the end of the question. That is what I do not understand because everything seems to be done correctly.

 

But everything was terminated 50 R impedance coaxial cable

Thats not what his schematic shows, and coax cables on their own are not terminations. There would have to be 50Ω at the other end of the coax for the input to be 50Ω. If the impedance at the load end is not matched to the coax impedance, the coax acts like a transformer changing the complex impedance at the input. You cannot just pass a signal through a 50Ω coax and assume the input to that coax is 50Ω.

If the coax was between the source and the tank circuit, at resonance, the tank disappears putting an open circuit on the coax. When there is an open circuit on a coax, the signal will hit the end, reflect back, and add to the source voltage. A TDR demonstrates this with a DC pulse nicely. Lets say you have a coax that is a quarter wavelength at the sig gen frequency and that coax has an open circuit on the end. To the sig gen, it looks like a dead short. If the other end of the (1/4 wavelength) coax is shorted, the input appears as an open at that frequancy. Coax cables are impedance transformers (when not matched to their characteristic impedance), they are not terminations.

 

RE:""Does anyone have an idea why that is the case>""
WHAT is the problem????
Why there happens resonance? Because X(L) fully or at least particularly neutralize the -X(C).
Why the resonance is not "eternal"? Because You have R.
Why the resonance is lower or higher than signal? Because You switched signal generator on another frequency than L and C sets frequency to tank.
Why the resonance multiplication of V (or i for parallel tank) particularly happens? Because the resonance tail catches the signal tail with some 0.00X percent of energy.