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The product LC controls the bandpass frequency while RC controls how narrow the passing band is. The Bode plot is a convenient tool for investigating the bandpass characteristics of the RLC network. Use tf to specify the circuit's transfer function for the values. However, the attenuation is only dB half a decade away from this frequency. To get a narrower passing band, try increasing values of R as follows:. The waves at 0. The long transient results from the poorly damped poles of the filters, which unfortunately are required for a narrow passing band:.

To analyze other standard circuit configurations such as low-pass and high-pass RLC networks, click on the link below to launch an interactive GUI. A modified version of this example exists on your system. Do you want to open this version instead?

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Off-Canvas Navigation Menu Toggle. No, overwrite the modified version Yes. Select a Web Site Choose a web site to get translated content where available and see local events and offers. Select web site.Today I am going to make a brief description of the step response of a RLC series circuit. This is the schematic made with LTspice. As you can see the components used are a resistor, an inductor and a capacitor connected in series.

By applying Kirchhoff voltage law we obtain the following equation. This equation may look a little bit hard to solve, thankfully the Laplace transform comes to rescue. If I apply the Laplace integral transform to both sides this is what I get. This equation is much easier to work with. We can then solve by X and obtain the transfer function for our system.

As expected we obtained a transfer function for a second order system with resonance frequency at. The resonance frequency is about Remember that at this frequency we expect the current to have a maximum, that is, the current amplitude should be at its highest if we apply a sinusoidal voltage u whose frequency is the same as the resonance frequency.

Be careful not to forget that we are analysing an AC circuit so all the physical quantities will be sine or cosine waves since this is the input signal. The bode plot confirms this behaviour. Note that the horizontal axis in Matlab is in radiant per second. We can make the same simulation with LTspice using the. As expected, there is a peak in the absolute value of the response at the resonance frequency while on the more extreme ends of the frequency spectrum the response tends to zero.

As far as the phase shift is concerned, we start from a 90 degree phase shift because the transfer function has a zero in the origin of the s plane, then there is degree phase shift as the frequency increase, however near the resonance frequency we can expect a small phase shift of the response signal.

This is the response to a sine wave, however, if we suddenly turn on the voltage to a constant value say 1 Volt what would our system do? Well, in order to know this we need to simulate this behaviour.

The circuit needed is the following. As you can see the voltage source is set to provide a 1 Volt signal for 0. By running the simulation we obtain the following response signal.Using the Laplace transform as part of your circuit analysis provides you with a prediction of circuit response.

Analyze the poles of the Laplace transform to get a general idea of output behavior. Real poles, for instance, indicate exponential output behavior. Apply the Laplace transformation of the differential equation to put the equation in the s -domain. Apply the inverse Laplace transformation to produce the solution to the original differential equation described in the time-domain. To get comfortable with this process, you simply need to practice applying it to different types of circuits such as an RC resistor-capacitor circuit, an RL resistor-inductor circuit, and an RLC resistor-inductor-capacitor circuit.

Here you can see an RLC circuit in which the switch has been open for a long time. Next, formulate the element equation or i-v characteristic for each device. Substituting the element equations, v R tv C tand v L tinto the KVL equation gives you the following equation with a fancy name: the integro-differential equation :.

The next step is to apply the Laplace transform to the preceding equation to find an I s that satisfies the integro-differential equation for a given set of initial conditions:.

The preceding equation uses the linearity property allowing you to take the Laplace transform of each term. For the first term on the left side of the equation, you use the differentiation property to get the following transform:. Because the switch is open for a long time, the initial condition I 0 is equal to zero. To get the time-domain solution i tuse the following table, and notice that the preceding equation has the form of a damping sinusoid. For this RLC circuit, you have a damping sinusoid.

The oscillations will die out after a long period of time. John M.

### Analyzing the Response of an RLC Circuit

Santiago Jr. During that time, he held a variety of leadership positions in technical program management, acquisition development, and operation research support. About the Book Author John M.Sign in to comment. Sign in to answer this question. Unable to complete the action because of changes made to the page.

### Analyze an RLC Circuit Using Laplace Methods

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You may receive emails, depending on your notification preferences. RLC circuit impulse response. Nicolas on 2 Apr Vote 0.

Hello, i'm really new to Matlab and i don't understand how it works. I need to plot the impulse of a series rlc circuit, using discrete analysis. Here is my code:.By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service. Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It only takes a minute to sign up. I am not too sure of doing it the way using differential equations.

I am more familar iwth laplace transforms. The 5 that you use in square 5, 50 is actually interpreted as a single item time vector and simply resolves to the integer -1 when evaluated.

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You might be able to trick Matlab in to working with the square function by performing some symbolic toolbox trickery, but I'd rather suggest you rewrite your problem and use a numeric solver like ode45 to plot the response for a given time period. Additionally, to plot an analytical function named solution you should run ezplot solution without the quotes. Sign up to join this community. The best answers are voted up and rise to the top.

Home Questions Tags Users Unanswered. Asked 8 years ago. Active 1 year, 3 months ago. Viewed 24k times. I've read a bit around it, but I can't get it to work. Help would be much appreciated.

Null 5, 10 10 gold badges 23 23 silver badges 34 34 bronze badges. No output? Wrong output? Active Oldest Votes. Jugurtha Hadjar Jugurtha Hadjar 3 3 bronze badges. Do you know of a way to generate the response to a sine wave? So step is for step function. What would be for sine?

Do I have to use lsim?Goals To build RLC circuits and to observe the transient response to a step input.

**LCA 8.5 Step Response Series RLC Circuit (In English)**

You will study and measure the overdamped, critically damped, and underdamped circuit responses. Background RLC circuits are widely used in a variety of applications such as filters in communications systems, ignition systems in automobiles, and defibrillator circuits in biomedical applications.

The analysis of RLC circuits is more complex than that of the RC circuits we have seen in previous labs. RLC circuits have a much richer and interesting response than the previously studied RC circuits. A summary of the response is given below. Let's take a series RLC circuit as shown in Figure 1.

The discussion is also applicable to other RLC circuits such as a parallel circuit. Let's focus on the complementary solution. The case of a critically damped response to a unit input step function is shown in Figure 2.

Case 2 : Overdamped response: two real and unequal roots: s 1 and s 2 4 Figure 2 shows an overdamped response to a unit input step function. Figure 2: Critically and overdamped response to a unit input step function. Case 3 : Underdamped response: two complex roots 5 Figure 3 shows an underdamped response to a unit input step function.

Figure 3: Underdamped response to a unit input step function. Pre-Lab Reading and Questions.

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