# Homework Help: Diagramming an Active Filter

1. Dec 10, 2013

### BenBa

1. The problem statement, all variables and given/known data

2. Relevant equations

See picture

3. The attempt at a solution

I am unsure how this filter works (i am in an intro to analog circuits class and we barely covered active filters).

Here is my attempt at drawing the input voltage:

you can see I used the 20KHz wave as an envelope for the 100Khz signal, drawing 5 full waves of the 100KHz within one half wave of the 20KHz, but i am not fully sure if this is the correct thing to draw as i have never worked with overlapping signals...

As for the voltage at the point specified, i believe that all that does is cut off the bottom portion of what I drew as well as decrease the voltage by 0.7 volts (The standard voltage drop of a diode). But i am not sure if this is correct.

For part B I have no clue how to approach this. I didn't know this filter had gain, i thought active filters just filtered...

EDIT: For part a i just realized that there is a capacitor acting as ripple control, so the output should not just be the top half of the input, but how do i calculate this ripple control?

Still super confused about part b.

EDIT 2: I just realized my scaling is off, i should have 2.5 waves of the 100KHz within on half wave of the 20KHz, so 5 waves total over the full length of the 20KHz.

Last edited: Dec 10, 2013
2. Dec 11, 2013

### rude man

First, the problem is badly stated. "Add" 20 KHz to 100 Khz would look something like what you drew for the input voltage, not the cited node. But then there's the word "carrier" which suggests the 100 KHz is modulated somehow (how is not stated) by the 20 KHz.

Looking at the time constant R1 C1 ~ 1 KHz it would seem that the intent is to provide a modulated signal with several sidebands so that the LSB is pushed down to near dc. Then the + part of the LSB would get through which is a half-wave rectification, sort of.

Thanks to D1 I can see no way a Bode plot of this circuit could be conceived, let alone generated. Bode plots assume linear systems and this is highly non-linear ...

3. Dec 11, 2013

### BenBa

I asked this question on another forum, and i seem to have got a correct answer. But i may have made a mistake in my assumptions, do you see anything wrong with my logic?

Also i think the bode plot is referring to the second stage that acts as a high pass active filter with f0 = 1/2piRC?

4. Dec 11, 2013

### rude man

I looked at the link and it's incorrect. Any frequency in the neighborhood of 20 KHz and up will be squashed at the cathode of D1. That's because the R1-C1 time constant is so large (0.15 ms.) that after a few cycles of input C1 is charged to near the positive peak of the input voltage and then sits there with only a small ripple on it, since R1 does not have time to discharge it significantly between input cycles.

I repeat, unless this is a modulation process with some very much lower frequencies generated by that process, in the neighborhood of 1 KHz or less, then this circuit does effectively nothing.
That is much more believable! So go ahead and have a stab at it? What's the transfer function from U1 output to U2 output? (I am labeling the two op amps U1 and U2 from left to right).

5. Dec 11, 2013

### BenBa

I'm sorry, can you explain what you mean about the "squashing" of the 20Hz signals and up due to the RC time constant? The capacitor charges up some and then slowly charges down but according to the ripple control formula the change in V is only equal to 0.02 volts, right? What is wrong with that analysis?

As for the second part, i believe the corner frequency is at 1/(2*pi*R1*C2) = 106.1 Hz, so any signals lower than this will be cut off and any higher will be passed, correct?

6. Dec 11, 2013

### rude man

20 Hz? You said 20 KHz! 20 Hz is an entirely different matter. I pointed out that only low frequencies could get past the R1-C1 time constant and 20 Hz certainly qualifies!
That is entirely correct.

7. Dec 11, 2013

### BenBa

Sorry i meant to say 20KHz, i was refferng to where you said:

Can you explain this?

I am trying to figure out why my diagram on the other forum is incorrect, i know the capacitor discharges slowly, but how does that effect the overall V of the ripple control?

8. Dec 11, 2013

### rude man

I already explained. The capacitor charges up quickly to the peak of the input voltage, then because it has almost no time to even partially discharge all you get at D1 cathode is a small ripple at the peaks' frequency, plus the rectified dc voltage. The dc voltage ~ the peak of the input voltage. But the dc does not pass through the second stage hi-pass filter.