Question

Problems 1 through 8 refer to the sinusoidal waveforms on the right, which are displayed on an oscilloscope having the following settings: Channel 1 Vertical Axis: 1 V/Division Channel 2 Vertical Axis: 50 mV/Division Horizontal Axis: 1 ms/Division The zero-volt level is at the center of the screen for both channels. Consider the left edge of the screen to be the t=0 axis. Consider the waveform displayed on Channel 1 as the output voltage of a function generator. In ze order to use this voltage waveform as a reference for phase angle measurements, the oscilloscope could be adjusted so that the positive-slope zero-crossing of the sinewave is at the t=0 position as shown above. Described mathematically, vi(t) = Vpsin(t+0v), where Vp is the amplitude or peak value of the voltage, is the frequency in radians/s, and Oy is the phase angle in degrees. Since the positive-slope zero-crossing of the sinewave is at the t=0 axis, Ov = 0º and the voltage would be described mathematically as v.(t) = 4sin(ot). Answer the following questions regarding the waveform observed on Channel 2, vz(t) = Vpsint+Ove). Important: Since both waveforms have the same period and frequency, the two waveforms differ only in amplitude (VP) and phase angle (Ov). 1. What is the period of the voltage waveform? 2. Calculate the frequency in Hz and radians/s. Hz (1) rad/s. (1) (1) 3. What is the amplitude? (Notice: The positive peak is 3.2 divisions above the v=0 axis.) 4. Calculate the horizontal scale factor in degrees per division. -*/div (1) - (1) 5. Find Ov2 in degrees. Indicate whether the angle is positive or negative. Ov2 Check the values for Vp and Oy, after solving problems 6 and 7. 6. Substitute the values of Vp, , and @y into the general form: v(t) = Vpsin(ot+0v). v:(t) = _V (1) 7. Use this equation to calculate the value of v2 to three significant figures at t = 0. Does this value agree with the point "v2 @t=0" circled on the diagram? If not, reevaluate your answers to problems 3 and 5. _V (1) V2 =

Answer 1

Given you slowl from the given diagram Total No of horizantal 10 divisom horizantal aais, Imal divisom T = 10x1ms 10 ms. T = loma © Tomi 10 100 la w = 27 T = 27 Tomi. - 628.3 radlec 7 VO) - Vp Sin(wł + 0) divisons Vertical from zero to Peak = 3.2 divisons . Vp =3.2 x 50m v = 160mV = 0.16 V [ channel 2: verticel a sis: Somul divison] 10 dúvisons 360° I divison > 36' V is lagging v, by I divison. = OV = -36 - 7

V (1) = 0.16 sin(wl -36) at to V, (+) = 0.86 sin (0-36) =-0.16 sin 36 = -0.1528 v V CH = 0.16 Sin (wt - 36') v w = 628.3 radi sec - at t=0 V, (o) = 0.16 sin (0-36) = -0.16 sin360 = -0.094 N -0.1 =-100 mv Both are similar