# Analyzing The Circuit By Laplace Domain Analysis

#### Yunus Emre SERT

Joined Jul 11, 2019
13
Hello,

I need to analyze and find v(t) of the circuit below. I did it by determining the type of response and using differential equations, however, I couldn't get the same result by using Laplace domain analysis. I attached the circuit diagram Figure (1), simulation graph Figure (2), my solution with Laplace domain method Figure (3) and MATLAB draw of Vc(t) that I've found in Figure (3) . While v(t) must approach 1 V as time goes infinity, v(t) approaches 1.5 V in my solution as seen in Figure (4). Could you check my solution steps in Figure (3) and inform me about what I'm doing wrong?

Thanks.

Figure 1. Circuit Diagram

Figure 2. v(t) versus time, simulation output

Figure 3. My Solution with Laplace Domain Method

Figure 4. MATLAB Output of Vc(t) Found in Figure (3)​

#### ZCochran98

Joined Jul 24, 2018
53
I think I see where you've made a small mistake. At t = 0, you have i(0) = 1 A. That is correct. However, the inductor voltage is not 0.5V at t = 0, but 0V at t = 0, and that is because at t = 0 you can assume it's been in steady state DC (so, at this point, the inductor is acting as a wire and current has been constant for some time). In fact, the way you determined i(0) = 1A you had already (correctly) assumed this condition. The only reason I assume it's been at a steady state is because your voltage source is v(t) = 1V - not 1*u(t) V.

So, with that in mind, getting rid of the extra 0.5V from your math brings the DC level back down to 1V (like in your simulation) instead of the 1.5V.

As far as I can tell, everything else looks right.

Hope that helps (and was a clear-enough explanation)!

#### Yunus Emre SERT

Joined Jul 11, 2019
13
I think I see where you've made a small mistake. At t = 0, you have i(0) = 1 A. That is correct. However, the inductor voltage is not 0.5V at t = 0, but 0V at t = 0, and that is because at t = 0 you can assume it's been in steady state DC (so, at this point, the inductor is acting as a wire and current has been constant for some time). In fact, the way you determined i(0) = 1A you had already (correctly) assumed this condition. The only reason I assume it's been at a steady state is because your voltage source is v(t) = 1V - not 1*u(t) V.

So, with that in mind, getting rid of the extra 0.5V from your math brings the DC level back down to 1V (like in your simulation) instead of the 1.5V.

As far as I can tell, everything else looks right.

Hope that helps (and was a clear-enough explanation)!
Thanks for your help, sir. I got the point and I reached the accurate equation for v(t). In fact, I'm confused about the s-domain equivalent of the inductor. When I set it as in Figure (5) below, I couldn't achieve the correct result and I could by using the equivalent in Figure(6). However, I couldn't understand the logic behind that why we use the s-domain equivalent in Figure(6) while the inductor has an initial current i(0) = 1A. How can it be explained?
Regards.

Figure 5. s-Domain Equivalent of Inductor

Figure 6. s-Domain Equivalent of Inductor​

#### ZCochran98

Joined Jul 24, 2018
53
I figured it out here. Technically the method I said was more time-domain-based (and partially wrong for the s-domain comments I made, which is why you couldn't follow the logic), which I only just realized after re-reading what I wrote (sorry about that!).

So, I sat down and worked out the problem myself, and this is what I got, including that -Li(0) term:

$-\frac{1}{s} + I(s) + \frac{1}{2}sI(s) - \frac{1}{2} + \frac{5}{s}I(s)=0$

The -1/s is from the voltage source, the I(s) is from the resistor, the 1/2s*I(s) is from the inductor, the -1/2 is from -Li(0), and 5/s I(s) is from the capacitor. Upon checking what you did, I saw that you put an extra 1/s term in there: for your -Li(0), you wrote down -Li(0)/s, which is incorrect, and is why I was discussing initial conditions initially (thus my more time-domain-based explanation I gave). What -Li(0)/s would correspond to is as though you had an initial constant voltage across the inductor, which is not the case. What -Li(0) would correspond to is the initial s-domain response at t = 0 when the switch is flipped - the -Li(0) term is only an s-domain parameter, so it's not treated like a proper voltage source (like the 1V), but is used "as-is" in the s-domain calculations (it is an s-domain voltage source, for lack of a better term, corresponding to its initial response). Using what I wrote down (which is only different by that single "/s" term), we get, for the capacitor voltage:

$V(s) = \frac{5}{s}I(s) = \frac{5}{s}\frac{s+2}{(s+1)^2+3^2}$

Check the initial value theorem and final value theorems, and you will see that this gives the appropriate values (0 and 1, respectively).

Basically, in short, the mistake you made was just as small as the one I had originally said, but less due to initial-value calculations and more due to just writing down the wrong equation.

Hope this helps, and, again, I apologize for the confusion!

#### Yunus Emre SERT

Joined Jul 11, 2019
13
I figured it out here. Technically the method I said was more time-domain-based (and partially wrong for the s-domain comments I made, which is why you couldn't follow the logic), which I only just realized after re-reading what I wrote (sorry about that!).

So, I sat down and worked out the problem myself, and this is what I got, including that -Li(0) term:

$-\frac{1}{s} + I(s) + \frac{1}{2}sI(s) - \frac{1}{2} + \frac{5}{s}I(s)=0$

The -1/s is from the voltage source, the I(s) is from the resistor, the 1/2s*I(s) is from the inductor, the -1/2 is from -Li(0), and 5/s I(s) is from the capacitor. Upon checking what you did, I saw that you put an extra 1/s term in there: for your -Li(0), you wrote down -Li(0)/s, which is incorrect, and is why I was discussing initial conditions initially (thus my more time-domain-based explanation I gave). What -Li(0)/s would correspond to is as though you had an initial constant voltage across the inductor, which is not the case. What -Li(0) would correspond to is the initial s-domain response at t = 0 when the switch is flipped - the -Li(0) term is only an s-domain parameter, so it's not treated like a proper voltage source (like the 1V), but is used "as-is" in the s-domain calculations (it is an s-domain voltage source, for lack of a better term, corresponding to its initial response). Using what I wrote down (which is only different by that single "/s" term), we get, for the capacitor voltage:

$V(s) = \frac{5}{s}I(s) = \frac{5}{s}\frac{s+2}{(s+1)^2+3^2}$

Check the initial value theorem and final value theorems, and you will see that this gives the appropriate values (0 and 1, respectively).

Basically, in short, the mistake you made was just as small as the one I had originally said, but less due to initial-value calculations and more due to just writing down the wrong equation.

Hope this helps, and, again, I apologize for the confusion!
This is what I exactly missed... I couldn't believe myself that I wrote down the equation incorrectly. Although checking the formulas in sources again and again, I was unsuccessful to figure my mistake out and I kept doing the same mistake. At the same time, I was sure that something was wrong because initial value and final value theorems never satisfied the condition. That's why I couldn't understand the logic, because the results and the way I got them didn't seem to be correct besides. Furthermore, you explained very well why s-domain voltage source is not treated like a proper voltage source and why -Li(0)/s must be -Li(0) accordingly.

I'm grateful to you! Your explanations helped me figuring out the problem and refreshing my background.

Thanks again & Regards

#### ZCochran98

Joined Jul 24, 2018
53
Glad to have been of assistance! The most annoying and hardest-to-catch mistakes are the little ones that, upon looking back at them, make you go "what was I doing?" They're great learning experiences, too.

Good luck with future work!