Measuring the ESR of capacitor with Feeltech FY3200S

[Pygmalion]

You have probably heard of the method of measuring the ESR of a capacitor using pulses (https://www.electronicdesign.com/technologies/power/article/21199428/a-simple-method-for-a-capacitors-esr-measurement). The method assumes a constant internal impedance of the function generator.

On paper, the FY3200S should actually be a good choice. If you set a square wave of amplitude 10Vpp with an offset of 5V and a duty cycle of 1%, this will give perfect pulses. Although the impedance of the FY3200S should be about 50Ω, with a 10Ω resistor at the output the voltage drops from 10V to only about 1V. This means that the impedance is not constant and can be as high as 100Ω with large loads.

Is such behaviour normal for signal generators or is it just the result of using a cheap signal generator? If so, does this mean that the FY3200S cannot be used to measure the ESR of capacitors?

[Aldo22]

First, I have to say that I am learning, or in other words, I have no idea what I am talking about. 
But if no one else says anything....

I have a FY3224S and a 8.2 Ω resistor and I wanted to check your question.
Also, I found a calculator for internal resistance.
https://electronicbase.net/de/innenwiderstand-berechnen/#innenwiderstand-online-rechner

So, I measure a voltage under load (8.2 Ω) of 0.448 Vrms at 3.536 Vrms source voltage (10Vpp).
That gives 56.52 Ω internal resistance according to the above calculator.

I don't know if it's correct to calculate it this way, but the result seems halfway reasonable.
However, the waveform suffers a bit at 8.2Ω.
I think that a limit is soon reached anyway.
The question is what you expect or what can be expected.

Please correct me if I am wrong.

[Pygmalion]

First, I have to say that I am learning, or in other words, I have no idea what I am talking about. 
But if no one else says anything....

I have a FY3224S and a 8.2 Ω resistor and I wanted to check your question.
Also, I found a calculator for internal resistance.
https://electronicbase.net/de/innenwiderstand-berechnen/#innenwiderstand-online-rechner

So, I measure a voltage under load (8.2 Ω) of 0.448 Vrms at 3.536 Vrms source voltage (10Vpp).
That gives 56.52 Ω internal resistance according to the above calculator.

I don't know if it's correct to calculate it this way, but the result seems halfway reasonable.
However, the waveform suffers a bit at 8.2Ω.
I think that a limit is soon reached anyway.
The question is what you expect or what can be expected.

Please correct me if I am wrong.

Your calculation is correct. If I use approximately the same conditions 10Vpp, 10Ω, I get an internal resistance of 58Ω.

However, if I use 20Vpp, the internal resistance increases significantly to 92Ω. Can you repeat the process with 20Vpp?

[Aldo22]

However, if I use 20Vpp, the internal resistance increases significantly to 92Ω. Can you repeat the process with 20Vpp?

Yes, I measure that too.
But I don't have enough experience to draw conclusions from it.
I also don't know if "better" signal generators behave differently in this point.