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Lecture 10a’ Types of Circuit Excitation Why Sinusoidal Excitation? Phasors

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Types of Circuit Excitation Linear Time- Invariant Circuit Steady-State Excitation Linear Time- Invariant Circuit OR Linear Time- Invariant Circuit Digital Pulse Source Transient Excitation Linear Time- Invariant Circuit Sinusoidal (Single- Frequency) Excitation

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Why is Sinusoidal Single-Frequency Excitation Important? 1.Some circuits are driven by a single-frequency sinusoidal source. Example: The electric power system at frequency of 60+/-0.1 Hz in U. S. Voltage is a sinusoidal function of time because it is produced by huge rotating generators powered by mechanical energy source such as steam (produced by heat from natural gas, fuel oil, coal or nuclear fission) or by falling water from a dam (hydroelectric). 2.Some circuits are driven by sinusoidal sources whose frequency changes slowly over time. Example: Music reproduction system (different notes).

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Why … (continued) 3.You can express any periodic electrical signal as a sum of single-frequency sinusoids – so you can analyze the response of the (linear, time-invariant) circuit to each individual frequency component and then sum the responses to get the total response.

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Representing a Square Wave as a Sum of Sinusoids (a)Square wave with 1-second period. (b) Fundamental compo- nent (dotted) with 1-second period, third-harmonic (solid black) with1/3-second period, and their sum (blue). (c) Sum of first ten components. (d) Spectrum with 20 terms.

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PHASORS You can solve AC circuit analysis problems that involve Circuits with linear elements (R, C, L) plus independent and dependent voltage and/or current sources operating at a single angular frequency = 2 f (radians/s) such as v(t) = V 0 cos( t) or i(t) = I 0 cos( t) By using any of Ohm’s Law, KVL and KCL equations, doing superposition, nodal or mesh analysis, and Using instead of the terms below on the left, the terms below on the right:

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Resistor I-V relationship v R = i R R ………….V R = I R R where R is the resistance in ohms, V R = phasor voltage, I R = phasor current (boldface indicates complex quantity) Capacitor I-V relationship i C = Cdv C /dt...............Phasor current I C = phasor voltage V C / capacitive impedance Z C: I C = V C /Z C where Z C = 1/j C, j = (-1) 1/2 and boldface indicates complex quantity Inductor I-V relationship v L = Ldi L /dt...............Phasor voltage V L = phasor current I L / inductive impedance Z L V L = I L Z L where Z L = j L, j = (-1) 1/2 and boldface indicates complex quantity

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SAME RULE: “Sinusoid in”-- “Same-frequency sinusoid out” is true for linear time-invariant circuits. (The term “sinusoid” is intended to include both sine and cosine functions of time.) Intuition: Think of sinusoidal excitation (vibration) of a linear mechanical system – every part vibrates at the same frequency, even though perhaps at different phases. Circuit of linear elements (R, L, C) Excitation: Output: v S (t) = V S cos( t + ) I out (t) = I 0 cost( t + ) Given ??

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Example 1 We’ll explain what phasor currents and voltages are shortly, but first let’s look at an example of using them: Here’s a circuit containing an ac voltage source with angular frequency , and a capacitor C. We represent the voltage source and the current that flows (in boldface print) as phasors V S and I -- whatever they are! V S I + - C We can obtain a formal solution for the unknown current in this circuit by writing KVL: -V S + Z C I = 0 We can solve symbolically for I: I = V S /Z C = j CV S

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Note that so far we haven’t had to include the variable of time in our equations -- no sin( t), no cos( t), etc. -- so our algebraic task has been almost trivial. This is the reason for introducing phasors! In order to “reconstitute” our phasor currents and voltages to see what functions of time they represent, we use the rules below. Note that often (for example, when dealing with the gain of amplifiers or the frequency characteristics of filters), we may not even need to go back from the phasor domain to the time domain – just finding how the magnitudes of voltages and currents vary with frequency may be the only information we want.

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Rules for “reconstituting” phasors (returning to the time domain) Rule 1: Use the Euler relation for complex numbers: e jx = cos(x) + jsin(x), where j = (-1) 1/2 Rule 2: To obtain the actual current or voltage i(t) or v(t) as a function of time 1.Multiply the phasor I or V by e j t, and 2.Take the real part of the product For example, if I = 3 amps, a real quantity, then i(t) = Re[Ie j t] = Re[3e j t ] = 3cos( t) amps where Re means “take the real part of” Rule 3: If a phasor current or voltage I or V is not purely real but is complex, then multiply it by e j t and take the real part of the product. For example, if V = V 0 e j , then v(t) = Re[Ve j t ] = Re[V 0 e j e j t ] = Re[V 0 e j( t + ) ] = V 0 cos( t + )

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Apply this approach to the capacitor circuit above, where the voltage source has the value v S (t) = 4 cos( t) volts. The phasor voltage V S is then purely real: V S = 4. The phasor current is I = V S /Z C = j CV S = ( C)V S e j /2, where we use the fact that j = (-1) 1/2 = e j /2 ; thus, the current in a capacitor leads the capacitor voltage by /2 radians (90 o ). The actual current that flows as a function of time, i(t), is obtained by substituting V S = 4 into the equation for I above, multiplying by e j t, and taking the real part of the product. i(t) = Re[j ( C) x 4e j t ] = Re[4( C)e j( t + /2) ] i(t) = 4( C)cos( t + /2) amperes + - C i(t) v S (t) = 4 cos( t) Finishing Example 1

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Analysis of an RC Filter + - V out V in + - R C Consider the circuit shown below. We want to use phasors and complex impedances to find how the ratio |V out /V in | varies as the frequency of the input sinusoidal source changes. This circuit is a filter; how does it treat the low frequencies and the high frequencies? Assume the input voltage is v in (t) = V in cos( t) and represent It by the phasor V in. A phasor current I flows clockwise in the circuit.

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Write KVL: -V in + IR +IZ C = 0 = -V in + I(R + Z C ) The phasor current is thus I = V in /(R + Z C ) The phasor output voltage is V out = I Z C. Thus V out = V in [Z C /(R + Z C )] If we are only interested in the dependence upon frequency of the magnitude of (V out / V in ) we can write | V out / V in | = |Z C /(R + Z C )| = 1/|1 + R/ Z C | Substituting for Z C, we have 1 + R/ Z C = 1 + j RC, whose magnitude is the square root of ( RC) 2 + 1. Thus,

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Explore the Result If RC << 1 (low frequency) then | V out / V in | = 1 If RC >> 1 (high frequency) then | V out / V in | ~ 1/ RC If we plot | V out / V in | vs. RC we obtain roughly the plot below, which was plotted on a log-log plot: The plot shows that this is a low-pass filter. Its cutoff frequency is at the frequency for which RC = 1.

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Why Does the Phasor Approach Work? 1.Phasors are discussed at length in your text (Hambley 3 rd Ed., pp. 195-201) with an interpretation that sinusoids can be visualized as the real axis projection of vectors rotating in the complex plane, as in Fig. 5.4. This is the most basic connection between sinusoids and phasors. 2.We present phasors as a convenient tool for analysis of linear time-invariant circuits with a sinusoidal excitation. The basic reason for using them is that they eliminate the time dependence in such circuits, greatly simplifying the analysis. 3.Your text discusses complex impedances in Sec. 5.3, and circuit analysis with phasors and complex impedances in Sec. 5.4.

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Motivations for Including Phasors in EECS 40 1.It enables us to include a lab where you measure the behavior of RC filters as a function of frequency, and use LabVIEW to automate that measurement. 2.It enables us to (probably) include a nice operational amplifier lab project near the end of the course to make an “active” filter (the RC filter is passive). 3.It enables you to find out what impedances are and use them as real EEs do. 4.The subject was also supposedly included (in a way) in EECS 20.

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