WEBVTT NOTE Phasors, Rectangular and Polar Coordinates 1 00:00:00.000 --> 00:00:12.772 Two ways to write the same impedance. An impedance has two independent parts, so it needs two numbers. There are two conventional ways to supply them. 2 00:00:12.772 --> 00:00:18.733 Rectangular: Z equals R plus jX, a resistive part and a reactive part. 3 00:00:18.733 --> 00:00:27.773 Question. When using rectangular coordinates to graph the impedance of a circuit, what do the axes represent? 4 00:00:28.673 --> 00:00:37.944 Answer. The X axis represents the resistive component, and the Y axis represents the reactive component. 5 00:00:38.544 --> 00:00:46.184 Question. Where is the impedance of a pure resistance plotted on rectangular coordinates? 6 00:00:47.084 --> 00:00:50.804 Answer. On the horizontal axis. 7 00:00:51.404 --> 00:01:03.103 Zero reactance means zero on the vertical axis, so a pure resistance sits on the horizontal axis. Polar: magnitude and angle. 8 00:01:03.103 --> 00:01:08.463 Question. How are impedances described in polar coordinates? 9 00:01:09.363 --> 00:01:13.523 Answer. By magnitude and phase angle. 10 00:01:14.123 --> 00:01:28.222 Question. What coordinate system is often used to display the phase angle of a circuit containing resistance, inductive, and/or capacitive reactance? 11 00:01:29.122 --> 00:01:32.562 Answer. Polar coordinates. 12 00:01:33.162 --> 00:01:47.562 The j operator, and its sign. j marks the reactive part, and its sign says which kind: +j is inductive. −j is capacitive. 13 00:01:47.562 --> 00:01:57.031 Question. Which of the following represents pure capacitive reactance of 100 ohms in rectangular notation? 14 00:01:57.931 --> 00:02:03.211 Answer. 0 - j100. 15 00:02:03.811 --> 00:02:11.517 No resistance, so the real part is zero; capacitive, so the sign is negative. 16 00:02:11.517 --> 00:02:19.077 Question. What does the impedance 50 - j25 ohms represent? 17 00:02:19.977 --> 00:02:28.057 Answer. 50 ohms resistance in series with 25 ohms capacitive reactance. 18 00:02:28.657 --> 00:02:34.259 Read it straight off: 50 resistive, 25 capacitive. 19 00:02:34.259 --> 00:02:41.708 Question. Which of the following represents a pure inductive reactance in polar coordinates? 20 00:02:42.608 --> 00:02:46.969 Answer. A positive 90 degree phase angle. 21 00:02:47.569 --> 00:03:01.115 Pure reactance means no resistive part, so the impedance lies on the vertical axis, at +90° for inductive and −90° for capacitive. Which form for which job. 22 00:03:01.115 --> 00:03:20.045 The two forms are equivalent, and each makes a different operation easy. Knowing which to use is the practical skill. Operation, Adding impedances in series; Easier in, rectangular, add real parts, add imaginary parts. 23 00:03:20.045 --> 00:03:37.151 Operation, Multiplying or dividing; Easier in, polar, multiply magnitudes, add angles. Operation, Taking a reciprocal (impedance → admittance); Easier in, polar, reciprocal magnitude, negate angle. 24 00:03:37.151 --> 00:03:47.311 Operation, Reading off R and X; Easier in, rectangular. Operation, Reading off magnitude and phase; Easier in, polar. 25 00:03:47.311 --> 00:04:00.249 That is why the admittance conversion in the previous lesson is stated in polar form: it is one step there and messy in rectangular. Phasor diagrams. 26 00:04:00.249 --> 00:04:09.929 Question. What kind of diagram is used to show the phase relationship between impedances at a given frequency? 27 00:04:10.829 --> 00:04:13.689 Answer. Phasor diagram. 28 00:04:14.289 --> 00:04:27.005 A phasor diagram plots each impedance as a vector, length for magnitude, angle for phase, so a series combination becomes vector addition you can see. 29 00:04:27.005 --> 00:04:40.484 It is the picture that makes "the reactances cancel at resonance" obvious: two vectors of equal length pointing exactly opposite. At a given frequency matters. 30 00:04:40.484 --> 00:04:46.249 Reactance changes with frequency, so a phasor diagram is a snapshot. 31 00:04:46.249 --> 00:05:02.811 Reading figure E5-1. Three pool questions ask you to locate an impedance on this figure. This question refers to one of the exam's circuit diagrams, which you will need to look at in the written lesson. 32 00:05:02.811 --> 00:05:14.945 The method is mechanical: Compute the reactance at the stated frequency, X_L = 2πfL or X_C = 1/(2πfC). The resistance is the horizontal coordinate.. 33 00:05:14.945 --> 00:05:29.621 The reactance is the vertical coordinate, positive up for inductive and negative down for capacitive. Find the point with roughly those coordinates. Nothing more subtle than that. 34 00:05:29.621 --> 00:05:40.689 The arithmetic is the reactance formula from the General track; the figure just asks you to plot the answer. Frequency response graphs. 35 00:05:40.689 --> 00:05:48.996 Question. What type of Y-axis scale is most often used for graphs of circuit frequency response? 36 00:05:49.896 --> 00:05:52.816 Answer. Logarithmic. 37 00:05:53.416 --> 00:06:00.381 Because response spans orders of magnitude and because decibels are logarithmic. 38 00:06:00.381 --> 00:06:14.399 A filter with 60 dB of ultimate rejection has a stopband a millionth of its passband in power, invisible on a linear axis, clearly readable on a logarithmic one. 39 00:06:14.399 --> 00:06:32.857 The mode-bandwidth figure in the Technician signals module makes the same point about the frequency axis. Check yourself. Write "30 ohms resistance in series with 40 ohms inductive reactance" in rectangular form. 40 00:06:32.857 --> 00:06:43.740 What is its magnitude? Where does 0 + j75 plot, and what is it? You need to add two series impedances. Which coordinate form? 41 00:06:49.740 --> 00:07:09.280 30 + j40. Magnitude = √(30 to the power of 2 + 40 to the power of 2) = 50 ohms. On the vertical axis, above the origin, a pure inductive reactance of 75 ohms, at +90°. 42 00:07:09.280 --> 00:07:16.300 Rectangular, add the real parts and add the imaginary parts.