Laplace Step Response Circuit Analysis

In summary: That would be my guess.When I was a high school student, I built a 500 microhenry inductor that was wound on a wooden spool that was 4 inches in diameter. It had a DC resistance of 16 ohms. Your intuition is good, but you're off by about a factor of 10. It's not unusual for an inductor to have a DC resistance of a few tens of ohms, and it can be even higher if it's a high Q inductor. In an inductor as high in value as 150 mH, the wire gauge needs to be pretty hefty to keep the number of turns down to a reasonable number. The more turns you have, the
  • #1
Meadman23
44
0

Homework Statement



Shown in attachment

Homework Equations


The Attempt at a Solution



I'm trying to analyze the circuit in the attached picture. This is a step response with a 3V input or 3u(t).

What I've done so far is:

1. convert all of the components to the s-domain.

R = R, L = sL, C = 1/sC

2. Combine the L and C into one impedence

[sL*(1/sC)]/[sL + (1/sC)] = Z

\[\frac{7.4999999999999985\,{10}^{7}}{0.15\,s+\frac{4.9999999999999994\,{10}^{8}}{s}}\]

3. Using voltage divider formula, solve for the voltage across R

Vout = (R*Vin)/(R+Z)

Vout = [R*(3/s)]/(R+Z)4. Simplified all calculations

Vout = (3R/s)/R+Z

\[\frac{9000.0}{\left( \frac{7.4999999999999985\,{10}^{7}}{0.15\,s+\frac{4.9999999999999994\,{10}^{8}}{s}}+3000.0\right) \,s}\]

5. Evaluated inverse laplace transform of Vout using Maxima

\[3-\frac{30\,{e}^{-\frac{250000\,t}{3}}\,\mathrm{sinh}\left( \frac{50000\,\sqrt{13}\,t}{3}\right) }{\sqrt{13}}\]

After plugging in certain times and comparing them to actual measured values this circuit provides at the same times, I get step responses with percent differences ranging from 4-30%.

I feel like this is wrong since most formulas I've derived for earlier types of circuits resemble the real life results quite closely.
 

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  • #2
Meadman23 said:

Homework Statement



Shown in attachment

Homework Equations





The Attempt at a Solution



I'm trying to analyze the circuit in the attached picture. This is a step response with a 3V input or 3u(t).

What I've done so far is:

1. convert all of the components to the s-domain.

R = R, L = sL, C = 1/sC

2. Combine the L and C into one impedence

[sL*(1/sC)]/[sL + (1/sC)] = Z

\[\frac{7.4999999999999985\,{10}^{7}}{0.15\,s+\frac{4.9999999999999994\,{10}^{8}}{s}}\]

3. Using voltage divider formula, solve for the voltage across R

Vout = (R*Vin)/(R+Z)

Vout = [R*(3/s)]/(R+Z)


4. Simplified all calculations

Vout = (3R/s)/R+Z

\[\frac{9000.0}{\left( \frac{7.4999999999999985\,{10}^{7}}{0.15\,s+\frac{4.9999999999999994\,{10}^{8}}{s}}+3000.0\right) \,s}\]

5. Evaluated inverse laplace transform of Vout using Maxima

\[3-\frac{30\,{e}^{-\frac{250000\,t}{3}}\,\mathrm{sinh}\left( \frac{50000\,\sqrt{13}\,t}{3}\right) }{\sqrt{13}}\]

After plugging in certain times and comparing them to actual measured values this circuit provides at the same times, I get step responses with percent differences ranging from 4-30%.

I feel like this is wrong since most formulas I've derived for earlier types of circuits resemble the real life results quite closely.

Can you fix your LaTex? It's hard to follow what you've done with the equations so battered looking :smile:

Even so, It doesn't appear to me that your resulting expression for the parallel impedance of the cap and coil looks right. C1 is 0.002 μF, right?
 
  • #3
Your basic math looks good.
As gneill says, hard if not impossible to check the details.

As far as corroborating your empirical data with your analysis I would point out that, while the capacitor probably acts very close to an ideal capacitor, such is not true of the inductor, which has very appreciable equivalent resistance in series with it. Actually, that resistance is a function of frequency, and since your step input spans a large frequency range (|1/jw|) it is wishful thinking for you to hope that you will get the exact results you expect.

As an example: set your input sine generator to about 9.2 KHz. You should see a dip in the output around that frequency. The fact that you can't attain zero output is ascribable almost entirely to the finite "Q" of the inductor.

P.S. also vary the input voltage to make sure you're not saturating the inductor which at that value most probably has a high-permeability core.
 
  • #4
Also following along the lines of what Rude Man has stated about the inductor, capacitors can have a rather large tolerance variations in their values. Some inexpensive capacitors can have values that stray -20% to +80% of their marked values. You may be able to locate a Tolerance Code on your capacitor to give you an idea of what range the actual capacitance will lie. Failing that, find a capacitance meter and measure it.
 
  • #5
gneill said:
Also following along the lines of what Rude Man has stated about the inductor, capacitors can have a rather large tolerance variations in their values. Some inexpensive capacitors can have values that stray -20% to +80% of their marked values. You may be able to locate a Tolerance Code on your capacitor to give you an idea of what range the actual capacitance will lie. Failing that, find a capacitance meter and measure it.

Sure, but a 0.002uF capacitor is 2000 pF which will probably be a good-quality ceramic or something equally free from those problems. Totally ignorable compared to the inductor IMO.
 
  • #6
rude man said:
Sure, but a 0.002uF capacitor is 2000 pF which will probably be a good-quality ceramic or something equally free from those problems. Totally ignorable compared to the inductor IMO.

True, it depends upon the type of capacitor. Still, there will be some variance from the stated value. For a lab where one should do error analysis, it would make sense to check.

I would think that an inductor as large as 150 mH would have a DC resistance on the order of 200 to 300 Ohms.
 

Related to Laplace Step Response Circuit Analysis

What is a Laplace step response circuit?

A Laplace step response circuit is a mathematical model used to analyze the behavior of an electrical circuit when a sudden change (step) is applied to the input. It takes into account the circuit's components, their relationships, and their time-dependent behavior.

Why is Laplace step response analysis important?

Laplace step response analysis is important because it allows us to predict how a circuit will respond to a sudden input change. This can help us design and troubleshoot circuits, as well as understand their behavior over time.

What is the Laplace transform and how is it used in step response analysis?

The Laplace transform is a mathematical tool used to convert a function of time into a function of complex frequency. In step response analysis, it is used to transform the time-domain circuit equations into the frequency domain, making it easier to solve and analyze the circuit's behavior.

How do you calculate the step response of a circuit using Laplace analysis?

To calculate the step response of a circuit using Laplace analysis, you first need to transform the circuit's differential equations into the frequency domain using the Laplace transform. Then, you can use circuit analysis techniques such as Kirchhoff's laws and Ohm's law to solve for the output in the frequency domain. Finally, you can use the inverse Laplace transform to convert the frequency domain solution back into the time domain, giving you the step response of the circuit.

What are some practical applications of Laplace step response analysis?

Laplace step response analysis is widely used in the design and analysis of electronic circuits, control systems, and signal processing systems. It is also used in various engineering fields such as electrical, mechanical, and chemical engineering to study the behavior of systems over time and make predictions about their performance.

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