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Lecture 28 Review: Frequency domain circuit analysis Superposition Frequency domain system characterization Frequency Response Related educational modules: –Section 2.7.4, 2.7.5, 2.8.0

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Note during summary that we will be changing our mindset, rather than doing anything fundamentally new – I will keep re-introducing the same example circuit, but with minor modifications and extensions

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Frequency domain analysis – review The analysis techniques we used for time domain analysis of resistive networks are applicable to frequency domain analysis of general circuits E.g. KVL, KCL, circuit reduction, nodal analysis, mesh analysis, Thevenin’s and Norton’s Theorems… In general, we simply need to: Subsitute impedances for resistances Use phasor voltages & currents in place of time domain voltages & currents

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Superposition Superposition in frequency domain: If multiple signals exist at different frequencies, superposition is the only valid frequency domain approach Effects of individual sources can be analyzed independently in the frequency domain Summation of individual contributions must be done in the time domain (unless all contributions have same frequency)

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Example 1 – Superposition Determine the steady-state voltage across the inductor in the circuit below

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In previous slide, show time domain superposition. V(t) = v1(t) +v2(t) – Note that we can determine v1(t) and v2(t) individually using frequency domain analysis – We can’t however, superimpose these individually, for reasons that will be made clear on the next slide

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Example 1 – continued The associated frequency domain circuits are shown below:

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Derive the phasors V1 and V2 on the previous slide Point out that they can’t be added directly – they don’t even apply to the same frequency domain circuit – We will superimpose the results in the time domain

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Example 1 – continued again Superimpose results in the time domain:

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On previous slide, V1 has frequency of 9 rad/sec and V2 has frequency of 3 rad/sec

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Notes on superposition Superposition of sinusoidal signals is extremely important in circuit analysis! In later courses, we will see that (nearly any) signal can be represented as a superposition of sinusoidal signals, using Fourier Series and Fourier Transforms If we determine the circuit’s response to each of these sinusoids and superimpose them, we can determine the circuit’s response to (nearly any) input function We will spend the next couple of lectures looking at circuit responses to multiple sinusoidal inputs

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Example 2 Determine the capacitor voltage, v(t), for the circuit below if v S1 (t) = 3cos(2t+20 ) and v S2 (t) = 5cos(2t-45 )

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On the previous chart, draw single-source circuit and point out that VS = vs1+vs2 and v(t)=v1+v2 – Draw two frequency-domain circuits ; note that they both have the same appearance

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Try a trick to do example 2 easily Let’s look at an “arbitrary” input with the same frequency as our individual inputs:

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Determine input-output relationship (gain, phase) => point out that this is only good for 2 rad/sec frequency sinusoids

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Example 2 – continued Now we can use our trick to easily determine the circuit’s response to both inputs (since they are at the same frequency)

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Important result The steady-state sinusoidal response of a circuit, at a frequency 0, can be characterized by two values: Gain: output-to-input amplitude ratio Phase difference: difference between the output and input phases

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On previous slide, annotate output to show on block diagram: – Y = H*U, where H is a complex number that depends only on frequency – Circuit “looks like” H, at the frequency of interest

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Example 3 – Multiple input frequencies Determine the capacitor voltage, v(t), for the circuit below if v S1 (t) = 3cos(2t+30 ) and v S2 (t) = 5cos(4t+60 )

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Annotate slide to show two frequency-domain circuits; note that they have different frequency-domain characteristics – I don’t want to analyze two circuits, so let’s leave frequency in the circuit model!

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Try a similar trick with example 3 Let’s characterize the circuit’s gain and phase as a function of frequency:

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Example 3 – continued Now we can use our trick to determine the circuit’s response to both inputs and superimpose the results v S1 (t) = 3cos(2t+30 ) v S2 (t) = 5cos(4t+60 )

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On the previous slide, write H, U, and Y as functions of frequency: – H(j2), U(j2), Y(j2) – H(j4), U(j4), Y(j4) Note that we analyzed the circuit only once!

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Super-important result The steady-state sinusoidal response of a circuit is characterized by the frequency response, H(j ) Magnitude response: output-to-input amplitude ratio vs. frequency Phase response: difference between the output and input phases as a function of frequency

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Frequency Response In the time domain, we characterize systems by the differential equation relating the input and output In the frequency domain, we characterize systems by their frequency response Magnitude response and phase response give the gain and phase difference relating the input and output The frequency of the signal (rather than time) is now our independent variable

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