Response_to_AC - EE201 Response of the basic elements to AC...

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Unformatted text preview: EE201 Response of the basic elements to AC / Phasor Notation Let’s go back and look at the voltage w.r.t. time in a R-L-C series circuit: 47Ω Take, for now, as a given: i(t) = 2.48sin(377t)A i=2.48sin(377t+00)A + The voltage is also sinusoidal, of the same frequency, and a magnitude of e=170sin(377t+?)V 170Vpeak, but unknown phase. First, find R, XL and XC: R = 47Ω (given) XL = ωL = (377)(.1H) = 37.7Ω XC = 1/ωC = 1/{(377)(100X10-6F)} = 26.5Ω Next, find the maximum voltages across the elements: VR = IMR = (2.48A)(47Ω) = 116.6V VL = IMXL = (2.48A)(37.7Ω) = 93.5V VC = IMXC = (2.48A)(26.5Ω) = 65.7V If we simply add these up we get 275.8V!! Clearly this would violate Kirchhoff’s voltage law as we know it! To help understand why, let’s plot these voltages: 100 µF 0.1H We can now see, as we noted earlier, that the various voltages all have differing phases with respect to each other. The problem we face is how to add sine waves of differing phases and frequencies? Here are three options to consider: 1) Graphically: Use the plotted waveforms and graphically add them to get the resultant waveform. Looking at the plot above, we can easily check the validity of this method by testing a few points. For example, if we look at values of the component waves at t=0: vR = 0V, vL = max = 93.5V and vC = min = -65.7V. If we add these we see the result in the ‘sum’ waveform: 27.8V. Also, if we look at the point in time when vR = vC = 0V, the ‘sum’ = vR and we can see that the curves cross at that point. We could continue this type of Denard Lynch Page 1 of 4 Oct 2009 EE201 Response of the basic elements to AC / Phasor Notation process until we plotted the entire result, but this would obviously be a very laborious and time consuming endeavour, although it would work for any phase and any frequency. 2) Mathematically: Consider two waveforms: V1sin(ω1t+φ1), and V2sin(ω2t+φ2). We can use the identities: sin(α+β) = sinαcosβ + cosαsinβ and sin(α-β) = sinαcosβ - cosαsinβ, where we can let ω = α and φ = β, and if we impose the restriction that ω1 = ω2, we can, with some manipulation, reduce the sum to the following expression: (V1cosφ1 + V2cosφ2)sinωt + (V1sinφ1 + V2sinφ2)sin(ωt+900), and recalling that sin(ωt+900) = cosωt: (V1cosφ1 + V2cosφ2)sinωt + (V1sinφ1 + V2sinφ2)cosωt. Note that we have 2 terms dependent only on V and φ (both not time varying!), that are 900 to each other. This method could reduce the work considerable, although it is simplest if we restrict its use to sine waves of the same frequency but differing phase. However, a minor extension of this math leads to a very popular and simple method for dealing with AC sinusoids of the same frequency; phasors! 3) Phasor notation: Charles Proteus Steinmetz is credited with developing this “phasor notation” method of representing AC sinusoidal waveforms, and it has reduced the complexity of working with sinusoids of differing phases to almost that of dealing with DC values! “Phasor notation” , is a transformation of a sinusoidal, time-varying signal into a complex, two-dimensional domain where combination is relatively trivial. (Works for any sinusoidally varying quantities of the same frequency.) It can be visualized as a counter-clockwise rotating vector in a two-dimensional, complex plane. How it works… General transformation of a sinusoidal signal in the time domain to the phasor domain: Time: v(t) = Asin(ωt+φ)V A A A Phasor: V = ∠ϕ V (polar form), = cos( ϕ ) + j sin( ϕ ) (rectangular form) 2 2 2 Note: phasors are conventionally shown as RMS values, but the concept works equally well for maximum (i.e. peak) values. Reverse transformation from the phasor domain back into the time domain: Phasor: V = B∠θV (polar form) or V = (x +jy)V (rectangular form) Time: v( t ) = 2 B sin( ω t + θ )V , or v( t ) = (x 2 ⎛ ⎛ y⎞⎞ + y 2 sin ⎜ ω t + tan−1 ⎜ ⎟ ⎟ V ⎝ x⎠⎠ ⎝ Oct 2009 ) Denard Lynch Page 2 of 4 EE201 Response of the basic elements to AC / Phasor Notation Of course in the reverse case, there is know way of determining ω without additional information! The Phasor Diagram: Phasors are just vectors in a 2-D complex plane. A Phasor Diagram can be used to illustrate their relationship and addition. An example: Phasor Diagram j V2sinφ2 V1 VTot V1sinφ1 φ1 φ2 V2 V1cosφ1 V2cosφ2 A couple more things to complete our “toolbox” for AC circuits… Define one more new term: Impedance, Z, which represents the “opposition to flow” in the phasor domain. It is also measured in Ohms (Ω), and is defined as: phasor _ voltage V , or Z = phasor _ current I (Note: since Z (or X) is just a ratio, it can be calculated from either RMS or Max values.) Consider the impedance, Z of the basic (ideal) elements (again using current as a reference): Resistors, V is in phase with I, therefore: V ∠0 ZR = = R∠0 0 Ω I ∠0 Inductors, V leads I (eLi): V ∠90 ZL = = X L ∠90 0 Ω I ∠0 Capacitors, V lags I (iCe): V ∠ − 90 ZC = = X C ∠ − 90 0 Ω I ∠0 Denard Lynch Page 3 of 4 Oct 2009 EE201 Response of the basic elements to AC / Phasor Notation Note, that reactance(X) and resistance(R) are scalars and not complex. While impedance is a complex vector which can be represented on a 2-D plane, Z does not represent a timevarying quantity (like a phasor does). Impedance, Z, can also be represented and added on a 2-D plane called an Impedance Diagram (similar to, but not a Phasor Diagram!): Impedance Diagram j ZL ZTot ZR ZC Finally our favourite laws. These are virtually identical to those for the DC world, except we use phasors: Ohms Law for AC Circuits: V = IZ , and Z = V I Kirchhoff’s Laws for AC circuits: KVL: The Σ phasor voltages around a loop = 0 KCL: The Σ phasor currents into a node = 0 Denard Lynch Page 4 of 4 Oct 2009 ...
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