Identifying the time domain behavior of the circuit is easy from the impedance analysis.
Because if you recall, for our RLC circuit, Vc over Vi from the previous video was given by omega naught squared divided by s squared plus 2 alpha s plus omega naught squared. The s was equal to j omega. Now, you can get the time domain behavior completely from this equation here. OK, if you recall, at a given point in our analysis, we had substituted e to the st for a response, and when we
differentiated e to the st with respect to time, we got s e raised to st. And the e raised to st things canceled out. So each time you saw a d by dt, we ended up with an s in our equation.
So what we can do is a reverse mapping. So wherever we see an s, we can go and stick in a d by dt
there and derive the differential equation in a backward step. Wherever you see an s squared, stick in a second derivative.
However, just for kicks, let's say we care about a step input. So for example, let's try a step input.
So for a step input, the input signal will simply be some dc value starting at time t equal to 0.
There you go. So I've very quickly written my differential equation for a step input for the same circuit completely by using frequency domain analysis.
As before, even if I did the time domain analysis, my Q is the same, omega naught by 2 alpha, and
for the series RLC circuit, 2 alpha is R divided by L. So I ended up with omega naught divided by R over L, which is Q. We will look at using Q at both time domain and frequency domain behavior.
For the frequency domain, we would apply a sinusoid given by Vi cosine omega t. For the time domain, just for fun, we are going to use a unit step. And as we have seen in the past, we can do the analysis in three cases.
So one case is for overdamped behavior. So for this overdamped behavior, alpha is greater than omega naught. And in terms of Q, Q-- which is equal to omega naught by 2 alpha-- is less than 1/2.
Then for alpha less than omega naught, or Q--which is still omega naught by 2 alpha-- is greater than 1/2. this is high Q. This is underdamped.
And then, finally, a look at alpha equals omega naught. And in this case, Q equals 1/2. This is critically damped.
With the over-damped case, alpha is greater than omega naught. This is the over-damped case. In one case, a sinusoid is applied at the input and the frequency response is measured. We are going to measure the magnitude of Vc/Vi, the magnitude of the transfer function for the frequency response. For the time domain part, we are going to apply a unit step and observe for what v_C looks like.
For the frequency domain behavior, it looks like a low-pass filter. It's low-Q, no peakiness.
And at the time domain, there is a pretty sluggish circuit, which just goes up lazily and saturates at unity.
For the underdamped case, alpha is less than omega-naught. And so therefore, Q will be greater than half. Again, this is the case where I'm going to get a high Q circuit. So in this case, what's going to happen is that at omega-naught, because I have a high Q filter, I am going to get a peaky response.
The bandwidth is defined as the interval between the points in the curve where the response falls to 1 by square root 2 of the peak value. And 1 by square root 2 is 0.707. And sometimes it's also called the 0.707 frequency at which you reach the 1 by square root 2 value. Either way, it's a narrow band, and it's a very high Q filter. What happens in the time domain? Underdamped circuits gave rise to a lot of ringing. They were not sluggish. So for this capacitor's circuit, it would eventually settle down at 1.
So for instance, if I have a Q of 50, then the circuit is going to sit around ringing for about 50 rings.
And of course, notice that if I make R close to 0, if R goes close to 0, then this just becomes LC.
And recall, for an LC circuit, it will ring forever. Provide a unit step, they will simply ring forever
and ever and ever. Because Q for that circuit would be infinity when R goes to 0. So the key here is that in the time domain, Q relates to how ringy the circuit is. Q relates to ringiness. And in the frequency domain, Q relates to peakiness.
Finally, let me look at the critically damped case where alpha equals omega naught or, equivalently, Q equals half. Same circuits, input sinusoid for the frequency domain analysis and unit step for the time domain analysis. So in the case of my frequency domain analysis, let me first draw out clearly, at the very low and very high frequencies, the circuit behaves similarly. And it will look something like . . . 'bloop'. Next, let's look at the time domain behavior of the same circuit and for the step input. I just get a little, little blip. It may be seen that correlated to a flipped version of the time domain
step response but don't get confused by that.
