I have this circuit :

The prob is stick to 1mH Inductor.

Now, I found that the current flowing in inductor makes a low pass filter.
However I cannot show how does the inductor makes a low-pass filtering.

All I know is that an inductor is familiar with low frequency and a conductor is familiar with high frequency. That's all I got.
Can anybody add some information for me? Thank you.

 

What does it mean to be "familiar" with a frequency?

Why would a conductor be more "familiar" with high frequencies than low frequencies?

What is it that is being filtered? Voltage or current? Which voltage or current?

 

You need to give us some context as to what course you are studying and where this problem comes from.

 

I guess that the "gain" is 0.1A/V ?? That means: It is an OTA with a current output, correct?
In such a case, it is always helpful (better: necessary) to find the transfer function (output/input) of the circuit.
However, you need to define the OUTPUT node!!

 

I'm studying with a book "Fundamentals of Electric Circuit" by Alexander.

The output node is the 1mH inductor.

 

All I know is that an inductor is familiar with low frequency and a conductor is familiar with high frequency.

The word "familiar" is not a normal way to describe the action of inductors and capacitors (at least not in idomatic English).

More correctly, the reactive impedance of an inductor increases with frequency and that of a capacitor decreases with frequency.
Thus if you put a resistor in series with an inductor to ground as you show, the attenuation of the voltage divider formed by the resistor and inductor will be reduced as the frequency increases, giving the high-pass function.
At DC the attenuation is infinite (zero output) and at a high frequency, the attenuation goes to essentially zero (the same as if the inductor were removed).

The attenuation is -3dB when the frequency is such that the inductive reactance equals the series equivalent resistance, which is called the corner frequency of the filter.
Above that frequency the attenuation will increase by 6dB/octave of frequency increase.

 

More correctly, the reactive impedance of an inductor increases with frequency and that of a capacitor decreases with frequency.
Thus if you put a resistor in series with an inductor to ground as you show, the attenuation of the voltage divider formed by the resistor and inductor will be reduced as the frequency increases, giving the high-pass function.
At DC the attenuation is infinite (zero output) and at a high frequency, the attenuation goes to essentially zero (the same as if the inductor were removed).

Thank you for your sincere answer. I am not in a English-speaking country thus I'm not very fluent in using these terms. I'll investigate on it.

I think your answer is about the Voltage output. I throughly understood your words and wondered if, when it comes to frequency response in 1mH inductor from the view of current, it concludes that the circuit is a low-pass filter. Am I thinking it right way?

 

Thank you for your sincere answer. I am not in a English-speaking country thus I'm not very fluent in using these terms. I'll investigate on it.

I think your answer is about the Voltage output. I throughly understood your words and wondered if, when it comes to frequency response in 1mH inductor from the view of current, it concludes that the circuit is a low-pass filter. Am I thinking it right way?

Yes. At DC an inductor passes all of the current (it looks like a short). As the frequency increases less current will flow.

Do you see why it is critical to state whether your signal is a voltage signal or a current signal (in addition to where that signal is being measured)?