# Impulse Response of an RLC Circuit

• Engineering
• RoyalFlush100
In summary: I am still unsure how to calculate the Impulse Response without delta involved in the equation. I believe the circuit is under-damped because: sqrt(16^2 - 4(2)(50)) is not a real solution. However, I don't know how that would relate to the Impulse Response.Why don't you look up what an impulse response is in the frequency domain. Hint, its very simple. The table below should help
RoyalFlush100
I bolded the portions I need help with.

1. Homework Statement

A series circuit consists of a resistor with a resistance of 16 ohms, an inductor with inductance of 2 H, and a capacitor with a capacitance of 0.02 F.
At time t = 0 there is no charge on the capacitor and no current in the circuit. If a 9 V battery is connected at at t=0 and disconnected at t=2
(a) The charge, q(t), on the capacitor at all t
(b) Compute the impulse response of the circuit as a function of time and classify the response as under-damped, over-damped, or critically-damped, explaining your rationale.
(c) Suppose the resistor were changed to make the circuit response critically-damped. What would be the value of R? Compute the new impulse response with this value B.

## Homework Equations

The equation:
2q'' + 16q + 50 = 9(1 - U(t-2)). Where U is the unit step function.

## The Attempt at a Solution

So, I found the q(t) by applying the Laplace Transform on the equation and then the inverse Laplace Transform Q(s). My work for the first part is in the image attached.

However, I am unsure how to calculate the Impulse Response without delta involved in the equation. I believe the circuit is under-damped because: sqrt(16^2 - 4(2)(50)) is not a real solution. However, I don't know how that would relate to the Impulse Response.

#### Attachments

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donpacino said:
Why don't you look up what an impulse response is in the frequency domain. Hint, its very simple. The table below should help

https://image.slidesharecdn.com/lap...pp02/95/laplace-table-1-638.jpg?cb=1395841667
I still don't get it. There's no delta function, which means there's never an instantaneous impulse. Or is the answer simply the transfer function (e.g. Laplace Transform of the above equation)?

You're right for the wrong reasons!

asking for the impulse response is asking how your system will react when an impulse is put in, just like asking for the "step response" would be asking what would happen if there is a step function at the input.

that being said, the laplace transform of an impulse at zero is... 1. So in the frequency domain you multiply your function by 1 to get the impulse response, therefore the transfer function is equal to the impulse response. If you wanted to get the step response, in the frequency domain you would multiple your function by 1/s. Does that make sense?

As far as the damping, my answer will depend on your level of knowledge in electrical engineering.

If you are well versed in the frequency domain, you will know that the poles of your system (the denominator in your transfer function) will determine your damping.

You can also determine damping by looking at a plot of the data. Do you know the difference between over/under/critical damped?

donpacino said:
You're right for the wrong reasons!

asking for the impulse response is asking how your system will react when an impulse is put in, just like asking for the "step response" would be asking what would happen if there is a step function at the input.

that being said, the laplace transform of an impulse at zero is... 1. So in the frequency domain you multiply your function by 1 to get the impulse response, therefore the transfer function is equal to the impulse response. If you wanted to get the step response, in the frequency domain you would multiple your function by 1/s. Does that make sense?

Okay, so the reason the Laplace Transform of the impulse is 1 at zero is because it occurs at t=0, right? If it occurred at t=1, it'd be e^-s, correct? Then, the Impulse will be a function of s (dependent on what the frequency of the signal is, assumed to be at t=0, because that's when the impulse occurs)?

RoyalFlush100 said:
Okay, so the reason the Laplace Transform of the impulse is 1 at zero is because it occurs at t=0, right? If it occurred at t=1, it'd be e^-s, correct? Then, the Impulse will be a function of s (dependent on what the frequency of the signal is, assumed to be at t=0, because that's when the impulse occurs)?
Yes that is correct! in general impulse response just means transfer function unless they specify.

RoyalFlush100
How are you doing on parts B and C?

donpacino said:
As far as the damping, my answer will depend on your level of knowledge in electrical engineering.

If you are well versed in the frequency domain, you will know that the poles of your system (the denominator in your transfer function) will determine your damping.

You can also determine damping by looking at a plot of the data. Do you know the difference between over/under/critical damped?

I know in the time domain, over-damped means a real, non-repeating root, under means imaginary root, and critical means real, repeating root. Underdamped will oscillate, the rest decay (same way as the exponential curve), but they will usually converge to some value.

However, I'm not really aware of how this translates to the frequency domain.

RoyalFlush100 said:
I know in the time domain, over-damped means a real, non-repeating root, under means imaginary root, and critical means real, repeating root. Underdamped will oscillate, the rest decay (same way as the exponential curve), but they will usually converge to some value.

However, I'm not really aware of how this translates to the frequency domain.
You know everything you need to know! the roots (poles) of a system are the best way to tell

donpacino said:
You know everything you need to know! the roots (poles) of a system are the best way to tell
Based off my formula for Q(s) poles exist at s=0 and two complex roots (-4+3i & -4-3i). The complex roots suggest underdamped. Does the simple pole at 0 impact this?

oooo I just found a mistake in your work. unfortunately you are making life difficult on yourself Your actual system olny has poles 2 poles at -4 +-3. The pole at zero is caused by you incorperating the step function (which in an input) into your transfer function. Remember tf = output / input. the 9*(1-u(t-2)) is your input and should not be incorperated into your tf. And yes, it is underdamped. If there was a pole at zero in your system, the system would be marginally stable, ie at steady state it would oscillate.

donpacino said:
oooo I just found a mistake in your work. unfortunately you are making life difficult on yourself Your actual system olny has poles 2 poles at -4 +-3. The pole at zero is caused by you incorperating the step function (which in an input) into your transfer function. Remember tf = output / input. the 9*(1-u(t-2)) is your input and should not be incorperated into your tf. And yes, it is underdamped. If there was a pole at zero in your system, the system would be marginally stable, ie at steady state it would oscillate.

2q'' + 16q' + 50q = 9(1 - U(t-2))
So, we just apply the Laplace Transform to 2q'' + 16q' + 50q = 9? How would we then solve for q(t), to account for U? Or do we only omit the unit step function when determining whether it's damped (i.e. only take the Laplace Transform of the left of the above equation)?

RoyalFlush100 said:
Okay, so the reason the Laplace Transform of the impulse is 1 at zero is because it occurs at t=0, right? If it occurred at t=1, it'd be e^-s, correct? Then, the Impulse will be a function of s (dependent on what the frequency of the signal is, assumed to be at t=0, because that's when the impulse occurs)?
Bit confusing.
The L. transform of an impulse input delayed by time T is e-sT. Your problem does not deal with delayed inputs.

What "frequency of the signal"? It's the same impulse at T as at 0, just time-delayed. The spectrum of an impulse anywhere is constant with respect to frequency, all the way to infinity. But I don't think that is what you meant.

The input pulse described is an approximation to an impulse input δ(t). It approaches an impulse as the pulse width approaches zero while keeping the time-voltage product constant (so it gets taller and thinner). It would be significantly more cumbersome to have to determine the output with the given pulse than with a Dirac delta input δ(t).

rude man said:
Bit confusing.
The L. transform of an impulse input delayed by time T is e-sT. Your problem does not deal with delayed inputs.

What "frequency of the signal"? It's the same impulse at T as at 0, just time-delayed. The spectrum of an impulse anywhere is constant with respect to frequency, all the way to infinity. But I don't think that is what you meant.

The input pulse described is an approximation to an impulse input δ(t). It approaches an impulse as the pulse width approaches zero while keeping the time-voltage product constant (so it gets taller and thinner). It would be significantly more cumbersome to have to determine the output with the given pulse than with a Dirac delta input δ(t).

So, the Laplace Transform is just e^0 = 1. If you meant when I asked if it at t=1, I was more making sure I understood the more general case.

By "frequency of the signal," I meant that s would be the variable, so the magnitude of the impulse response would be dependent on s. Did I misunderstand something?

RoyalFlush100 said:
So, the Laplace Transform is just e^0 = 1. If you meant when I asked if it at t=1, I was more making sure I understood the more general case.
.OK. Just that e-s is dimensionally incorrect. You set T=1, can't argue with that!
By "frequency of the signal," I meant that s would be the variable, so the magnitude of the impulse response would be dependent on s.
Right, but just to be sure, the magnitude of the impulse response would be dependent on s even if the input is δ(t). Except the whole output would be time-delayed by 1 sec.

rude man said:
.Right, but just to be sure, the magnitude of the impulse response would be dependent on s even if the input is δ(t). Except the whole output would be time-delayed by 1 sec.
Wait, so why would δ(t) delay one second? Wouldn't δ(t-1), delay a second, because that's when the impulse would occur?

RoyalFlush100 said:
Wait, so why would δ(t) delay one second? Wouldn't δ(t-1), delay a second, because that's when the impulse would occur?
I didn't mean that. As you say, δ(t) is an impulse that occurs at t=0 and δ(t-1) occurs at t=1 sec.
What I did mean is that the output transform is a function of s whether your input is δ(t) or δ(t-1).

If L-1[Y(s)] is the time response for an input of δ(t) then L-1[Y(s)e-sT] is the response for an input of δ(t-T).

## 1. What is an RLC circuit?

An RLC circuit is an electrical circuit that contains a resistor (R), an inductor (L), and a capacitor (C). These three components are connected in series or in parallel and can be used to create various types of filters and oscillators in electronic devices.

## 2. What is impulse response in an RLC circuit?

The impulse response of an RLC circuit refers to the output response of the circuit when an impulse signal is applied to it. It is a measure of how the circuit responds to a sudden change in input. It can be calculated by applying the Laplace transform to the circuit's transfer function.

## 3. How does the value of each component affect the impulse response in an RLC circuit?

The value of each component (resistor, inductor, and capacitor) affects the impulse response in different ways. The resistor affects the damping of the response, the inductor affects the frequency of resonance, and the capacitor affects the amplitude of the response. Changing the values of these components can alter the shape and characteristics of the impulse response.

## 4. What is the significance of the impulse response in an RLC circuit?

The impulse response is important in understanding the behavior and performance of an RLC circuit. It can help predict the output response to different input signals and can be used to design and optimize the circuit for specific applications. It is also used in signal processing and control systems to analyze and improve system performance.

## 5. How is the impulse response of an RLC circuit different from other types of circuits?

The impulse response of an RLC circuit is unique due to the presence of both energy storing (inductor and capacitor) and energy dissipating (resistor) elements. This results in a complex response with characteristics such as resonance, damping, and oscillation. In contrast, other types of circuits may only have one type of element, resulting in simpler impulse responses.

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