Lecture 10a’. Types of Circuit Excitation Why Sinusoidal Excitation? Phasors. Linear Time- Invariant Circuit. Linear Time- Invariant Circuit. Linear Time- Invariant Circuit. Linear Time- Invariant Circuit. Digital Pulse Source.
Types of Circuit Excitation
Why Sinusoidal Excitation?
Types of Circuit Excitation
Why is Sinusoidal Single-Frequency Excitation
Why … (continued)
Representing a Square Wave as a Sum of Sinusoids
PHASORSYou can solve AC circuit analysis problems that involveCircuits with linear elements (R, C, L) plus independent and dependent voltage and/or currentsourcesoperating at asingle angular frequency w = 2pf (radians/s)such asv(t) = V0cos(wt) or i(t) = I0cos(wt)By using any of Ohm’s Law, KVL and KCL equations, doingsuperposition, nodal or mesh analysis, andUsing instead of the terms below on the left, the terms below on the right:
Resistor I-V relationship
vR = iRR ………….VR = IRR where R is the resistance in ohms,
VR = phasor voltage, IR = phasor current
(boldface indicates complex quantity)
Capacitor I-V relationship
iC = CdvC/dt ...............Phasor current IC = phasor voltage VC /capacitive impedance ZC: IC = VC/ZC where ZC = 1/jwC , j = (-1)1/2 and boldface indicates complex quantity
Inductor I-V relationship
vL = LdiL/dt ...............Phasor voltage VL = phasor current IL/inductive impedance ZL VL = ILZL where ZL = jwL, j = (-1)1/2 and boldface
indicates complex quantity
Circuit of linear elements
(R, L, C)
vS(t) = VScos(wt + f)
Iout(t) = I0cost(wt + a)
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 w, and a capacitor C. We represent the voltage source and the current that flows (in boldface print) as phasors VS and I -- whatever they are!
We can obtain a formal solution for the unknown current in this circuit by writing KVL: -VS + ZCI = 0
We can solve symbolically for I: I = VS/ZC = jwCVS
Note that so far we haven’t had to include the variable of time in our equations -- no sin(wt), no cos(wt), 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 w may be the only information we want.
Finishing Example 1
Apply this approach to the capacitor circuit above, where the voltage source has the value vS(t) = 4 cos(wt) volts.The phasor voltage VS is then purely real: VS = 4. The phasor current is I = VS/ZC = jwCVS = (wC)VSejp/2, wherewe use the fact that j = (-1)1/2 = ejp/2; thus, the current in a capacitor leads the capacitor voltage by p/2 radians (90o). The actual current that flows as a function of time, i(t), is obtained by substituting VS = 4 into the equation for I above, multiplying by ejwt, and taking the real part of the product. i(t) = Re[j (wC) x 4ejwt] = Re[4(wC)ej(wt + p/2)] i(t) = 4(wC)cos(wt + p/2) amperes
vS(t) = 4 cos(wt)
Analysis of an RC Filter
Consider the circuit shown below. We want to use phasors and complex impedances to find how the ratio |Vout/Vin| 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 vin(t) = Vincos(wt) and represent
It by the phasor Vin. A phasor current I flows clockwise in the
Write KVL: -Vin + IR +IZC = 0 = -Vin + I(R + ZC)
The phasor current is thus I = Vin/(R + ZC)
The phasor output voltage is Vout = IZC.
Thus Vout = Vin[ZC /(R + ZC)]
If we are only interested in the dependence upon frequency
of the magnitude of (Vout / Vin) we can write
| Vout / Vin | = |ZC/(R + ZC)| = 1/|1 + R/ ZC |
Substituting for ZC, we have 1 + R/ ZC = 1 + jwRC,
whose magnitude is the square root of (wRC)2 + 1. Thus,
Explore the Result
If wRC << 1 (low frequency) then | Vout / Vin | = 1
If wRC >> 1 (high frequency) then | Vout / Vin | ~ 1/wRC
If we plot | Vout / Vin | vs. wRC 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 w for which wRC = 1.
Why Does the Phasor Approach Work?
Motivations for Including Phasors in EECS 40