Laboratory Experiment 1 on Waveform Measurements | ECE 225, Lab Reports of Electrical Circuit Analysis

Download Laboratory Experiment 1 on Waveform Measurements | ECE 225 and more Lab Reports Electrical Circuit Analysis in PDF only on Docsity! Boise State University Department of Electrical and Computer Engineering ECE 225L – Circuit Analysis and Design Lab Experiment #1: Waveform Measurements 1 Objectives The objectives of this laboratory experiment are: • to review and apply the definitions of performance measures of periodic waveforms. 2 Theory minV maxV ot +Tot t2TT 0 v(t) Figure 1: Example of a Periodic Waveform Consider the above periodic waveform with period T in seconds. Recall that the period is equal to the smallest time interval over which the waveform repeats itself. The period also includes any dead time when the waveform is equal to zero. Mathematically, f(t + T ) = f(t) for any time t (1) The frequency f of this waveform is equal to the number of cycles over one second. Mathematically, the frequency of a periodic waveform is equal to the reciprocal of the period and it has units of cycles per second (cps) or Hertz (Hz). Thus, f = 1 T (2) The angular frequency of the waveform is equal to ω = 2πf = 2π T (3) and has units of radians per second (rad/s). Note that θ = ωT = 2π = 360o (4) 1 over a period T. The zero-to-peak or peak value Vp of this waveform is defined as the largest absolute value (peak absolute value) of this waveform. Mathematically, Vp = max {|Vmax|, |Vmin|} (5) The peak-to-peak value Vpp of this waveform is equal to the range of this waveform between its maximum and minimum values. Mathematically, Vpp = Vmax − Vmin (6) The average value of a periodic waveform is defined as Vavg = 1 T ∫ T 0 v(t) dt ⇐⇒ TVavg = ∫ T 0 v(t) dt (7) Using the relationship on the right, this average value can be interpreted as the value of a constant (DC) waveform that covers the same area between itself and the time axis over a period. Note that the area under the original waveform is counted positively if the waveform is positive and negatively if the waveform is negative. Thus, a sine waveform has zero average value over a period as the positive area cancels the negative area under the sine wave. The effective value Veff of a periodic waveform is defined as the root of the mean (average) of the squared waveform. For this reason, it is often referred to as the rms value of the waveform or Vrms. Mathematically, Veff = Vrms = √ 1 T ∫ T 0 v2(t) dt ⇐⇒ V 2rms = 1 T ∫ T 0 v2(t) dt (8) Using the above relationships, the rms value squared can be interpreted as the average value of the waveform squared over a period T. Note that the rms value of a DC (constant) waveform is equal to the absolute value of the DC waveform. A periodic waveform v(t) having a nonzero average voltage can be considered as the superposition of a DC (constant) waveform and an AC waveform having zero average. Thus, v(t) = Vdc + vac(t) (9) where Vavg = 1 T ∫ T 0 v(t) dt = 1 T ∫ T 0 Vdc dt + 1 T ∫ T 0 vac(t) dt = Vdc + 0 = Vdc (10) The rms value of this mixed waveform is then computed as Vrms = √ 1 T ∫ T 0 v2(t) dt = √ 1 T ∫ T 0 [Vdc + vac(t)]2 dt (11) = √ 1 T ∫ T 0 [ V 2dc + v2ac(t) + 2vac(t)Vdc ] dt = √ 1 T ∫ T 0 [ V 2dc + v2ac(t) ] dt (12) = √ V 2dc + V 2ac (13) where Vac = √ 1 T ∫ T 0 v2ac(t) dt (14) 2 Start Point Value End Point Value 00000 1.0000 00001 1.0000 00002 0.7500 00003 0.7500 00004 0.5000 00005 0.5000 00006 0.2500 00007 0.2500 00008 0.0000 00009 0.0000 Hook up the output of the waveform generator directly to Channel 1 of your scope. Observe and record the same performance measures for this periodic waveform using the scope and the benchtop and handheld multimeters. Oscilloscope Benchtop Handheld Multimeter Multimeter Waveform Vavg DCVrms ACVrms DC V AC V V̄ Ṽ Arbitrary #1 5 Report Questions Q1: Compare the theoretical rms values of the four built-in waveforms (sine, square, triangular, sawtooth) with measurements obtained using the oscilloscope and the two multimeters. Assuming that the function generator is 100% accurate, which of the measuring instruments was the most accurate? Q2: Verify analytically that the waveform in Figure 2 has the following performance measures: Vavg = 2 (V) Vrms = √ 6 ∼= 2.45 (V) Compare these values with the measurements obtained using the oscilloscope and the two multi- meters. Q3: Compute analytically the form factor (FF) and the crest factor (CF) for the arbitrary waveform in Figure 2. 5 Boise State University Department of Electrical and Computer Engineering ECE 225L – Circuit Analysis and Design Lab Experiment #1: Waveform Measurements Date: Data Sheet Recorded by: Equipment List Equipment Description BSU Tag Number or Serial Number HP/Agilent 54810A Infinium Oscilloscope HP/Agilent 33120A Function/Arbitrary Waveform Generator HP/Agilent 34401A Benchtop Multimeter Fluke 111 True RMS Multimeter Oscilloscope Benchtop Handheld Multimeter Multimeter Waveform Vavg DCVrms ACVrms DC V AC V V̄ Ṽ Sine Square Triangular Sawtooth Arbitrary #1 Note: Also, save a screenshot of the arbitrary waveform on a diskette together with a menu showing V avg, DCV rms and ACV rms. Appendix: Average and Effective Values of Common Periodic Waveforms Vavg = 0 Vrms = Vm√ 2 Vavg = 0 Vrms = Vm Vavg = 0 Vrms = Vm√ 3 Vavg = 0 Vrms = Vm√ 3 Vavg = 2Vm π Vrms = Vm√ 2 Vavg = Vm π Vrms = Vm 2