Filter function judgment

1nnocent

Filter function judgment [Copy link]

Determine what function the signal output in the figure implements (high pass, low pass, band pass, etc.)

Uo1 is low pass, Uo2 is band pass, Uo3 is band pass; how do you tell? It's a bit hard to tell

jojoee

In your picture, uo1 should be high-pass, uo2 should be band-pass, and uo3 should be low-pass, right?

1nnocent

jojoee posted on 2021-7-13 16:21 In your picture, uo1 should be high-pass, uo2 should be band-pass, and uo3 should be low-pass, right?

How is this determined? Exercises in the analog electronics book, the answers are not available here

jojoee

1nnocent posted on 2021-7-13 19:00 How is this specifically judged? The exercises in the analog electronics book, the answers are not available here now

Consider separately

1) At low frequencies (frequency approaches DC, 0 Hz), the capacitor impedance is infinite, so the two capacitors are open-circuited;

2) At high frequencies (frequency approaches infinity Hz), the impedance of the capacitor approaches 0, so the two capacitors are short-circuited.

Analyze the circuit outputs corresponding to these two situations. Note: The three op amps in your diagram have closed negative feedback loops in both high-frequency and low-frequency situations, so you can use virtual short and continuous interrupt conditions.

For example: 1) When the capacitor is open-circuited at low frequency, A2 forms a negative feedback loop through R5 and A1. The virtual open condition of A2 indicates that the current of R6 must be 0; and the virtual short condition of A2 indicates that the potential on the right side of R6 must be 0; so you can see that the potential at the left end of R6 = the potential at the right end, that is, uo1 is 0 at low frequency.

2) When the capacitor is short-circuited at high frequency, the virtual short conditions of A2 and A3 immediately show that uo2 and uo3 are both equal to 0 voltage.

At this time, the right end of R5 is equivalent to grounding, the right end of R6 is grounded, and the lower end of R3 is also grounded, so if you only look at A1 and the connected resistors, it is a common-mode amplifier.

In summary, the output of uo1 is 0 at low frequency, and at high frequency, the output is u1 first divided and then amplified in phase to produce uo1 signal, so it is a high-pass filter response.

Just like other methods, you can analyze it yourself by following their example.

1nnocent

jojoee posted on 2021-7-13 20:36 Consider 1) at low frequency (frequency tends to DC, 0hz) the capacitor impedance is infinite, so open the two capacitors; 2) at high frequency (frequency...

For example: 1) When the capacitor is open-circuited at low frequency, A2 forms a negative feedback loop through R5 and A1. The virtual open condition of A2 indicates that the current of R6 must be 0; and the virtual short condition of A2 indicates that the potential on the right side of R6 must be 0; so you can see that the potential at the left end of R6 = the potential at the right end, that is, uo1 is 0 at low frequency.

"So you can see that the potential at the left end of R6 = the potential at the right end" Is this because the current through R6 is zero?

jojoee

1nnocent published on 2021-7-15 10:05 For example: 1) When the capacitor is open at low frequency, A2 has a loop through R5 and A1 to form a negative feedback. The virtual disconnection condition of A2 means that the current of R6 must be 0; and A2 ...

Yes, because the low-frequency capacitor C is open, the current flowing through the right end of R6 can only pass through A2, and the virtual open condition of A2 means that this current must be 0.

According to Ohm's law, V6/R6=I6=0, which means that the potential difference V6 across R6 must be 0, and the potential on the left can only be equal to the potential on the right.

These are all basic concepts. If you can’t see them quickly, it’s probably because you’re not familiar with the basics. You need more practice.