For a series RLC circuit, derive an expression for the reactance slope at resonance.

Name: Problem 11, Chapter 9: Small Antenna Design (Communications Engineering Series)
Uploaded: 2020-08-14T15:54:30-08:00
Duration: 14 min 21 s
Channel: Donald Albin
Description: For a series RLC circuit, derive an expression for the reactance slope at resonance.

step 1 I'm going to sketch a graph of the phase angle versus frequency. So my phase angle, I'm going to

For a series RLC circuit, derive an expression for the...

Key Concepts

Slope of the Reactance (Differentiation)

The slope of the reactance refers to the rate of change of the net reactance with respect to angular frequency. It is determined by differentiating the net reactance expression (?L - 1/?C) with respect to ?. Evaluating this derivative at the resonant frequency provides insight into how rapidly the reactance changes around resonance, which is important in understanding the circuit’s sensitivity and dynamic behavior near resonance.

Resonance

Resonance in a series RLC circuit occurs when the inductive and capacitive reactances cancel out, leading to a condition where the total reactive impedance is zero. This happens at the resonant frequency, ?? = 1/?(LC), where the circuit's behavior is dominated by the resistor, resulting in maximum current flow and minimal impedance.

Series RLC Circuit

A series RLC circuit is composed of a resistor (R), an inductor (L), and a capacitor (C) connected in series. Understanding this circuit involves analyzing how the resistive, inductive, and capacitive components interact to determine the overall impedance of the circuit and its response to AC signals.

Reactance

Reactance is the component of impedance that arises from the inductors and capacitors in an AC circuit. The inductive reactance (?L) increases with frequency, whereas the capacitive reactance (1/?C) decreases with frequency. In a series RLC circuit, the net reactance is the difference between these two quantities, which is crucial for assessing the circuit's frequency response.

Transcript

00:01 I'm going to sketch a graph of the phase angle versus frequency.

00:10 So my phase angle, i'm going to go from negative 90 degrees to 90 degrees.

00:23 And so this is going to be phase angle.

00:27 And then this is going to be frequency.

00:33 Frequency.

00:35 Now, to get an idea of what this would look like, i actually used values of l equal 0 .1, henry, c equals 10 to the negative 5th ferrets.

00:59 So i went to desmos .com and i this and i see that it never reaches negative 90 or 90.

01:15 But with these exact values, let me see here, i should put in omega.

01:31 So what i'm going to, what i'm graphing is the inverse tangent of omega.

01:41 Which would be 2 pi f l minus 1 over omega c 2 pi f c and then i actually used a resistance of 100 oms when i put this into desmos i didn't put in the 2 pi but i'm going to do that now and so i see that there is an x intercept at 150 and i think that a thousand divided by 2 pi is 159.

02:40 So you have an x intercept at r over 2 pi.

02:57 Now, i'm going to make sure that i have this from negative 90 to 90, which i do.

03:04 And so what i see in the graph is something like this, where at asymptotes, to 90 degrees.

03:30 Okay, so that's the graph.

03:32 Now, in the second part, the value of phi at resonance frequency.

03:47 Well, at resonant frequency, xl equals xc.

03:57 So at resonant frequency, the phase angle would be the inverse tangent of zero, because that numerator becomes zero.

04:21 And the inverse tangent of zero is zero, and so zero would be the phase angle, at the resonant frequency.

04:33 Now, the slope of the graph has to be 2q over omega -0.

04:45 Prove that the slope is 2 -q over omega -0.

04:53 The slope of this graph, and this is a graph of this.

05:04 Well, first of all, i know that q is omega 0 l over r because that's how you calculate quality factor and so according to this equation q over omega 0 is going to be l over r so this would be the equivalent of saying 2 l over r is the slope okay but this is a graph of the inverse tangent of omega -l minus 1 over omega -c over r.

06:20 So i'm just going to change this to tangent of phi equals omega -l minus 1 over omega -c.

06:36 And you know what? i'm going to go ahead and multiply by r on both sides.

06:45 So if i take the derivative of this, that would give me the slope.

07:04 Derivative of the tangent is the secant squared.

07:15 Oops, i meant to write phi.

07:19 I don't see how this is going to get us to the answer, but i'm doing what seems obvious.

07:26 And then it's the slope with respect to the frequency.

07:34 So i guess i really don't want omega -in.

07:38 Here.

07:40 Let me look at the question again.

07:43 Prove the slope.

07:44 Oh, versus omega.