Friday, July 8, 2011

Solutions to Part 24 (Oscilloscope Problems)

We now look at the solutions to the problems at the end of Part 24, starting with giving the problems again:

1) In a test of an oscilloscope the Y2 plate is connected to a positive (d.c.)potential, sketch what would be observed.

2) The time base is connected to a positive potential (X-plates only involved). Sketch what would be observed.

3) When the time variation of a quantity is to be studied, an alternating p.d. representing the quantity is applied to the Y-plates via the Y-amplifier to the time -base. This generates a "saw-tooth" pattern. Illustrate such a pattern, and label its "sweep" and "flyback" aspects.

(Hints: The sweep must be linear so the deflection of the beam is proportional to the time. The 'saw-tooth' results from a combination of the alternating input p.d. and the steady speed of the image spot going left to right)

4) Say a 2 cm long line (peak-to-peak, from above to below reference horizontal) is generated on the screen in a lab oscilloscope by an a.c. input of 60V. What would the peak voltage be? What will the rms voltage be?

5) A signal generator producing 1,000 Hz is connected to the vertical Y-input of an oscilloscope while a 2nd signal generator producing 500 Hz is connected to the horizontal. Sketch the Lissajous figure that would be produced. Explain why you sketched it as you did.

6) The frequency ratio analyzed from a Lissajous figure is found to be:

f(v)/f(h) = 1

Sketch how this pattern would appear on the oscilloscope screen.


(1) The CRO spot appears exactly as shown in Fig. 4 of Part 24.

(2) The spot is displaced to the top of the CRO screen, as in Fig. 4, but moves to the right side, e.g.

0- - - - - - - - - - - -- - ----------

(3)The key step is to note from Fig. 3(b) how the sweep characteristic appears as a time-voltage vs. deflection. Thus, given the hint, the result is shown in the diagram for problem 3.

(4) For this CRO, 60V generates a 2 cm line, so peak voltage = 60V/2 = 30 V

The rms voltage is: V(peak)/[2]^½ = 30V/[2]^½ = 21.2V

(5) We need to obtain the numbers of tangency points in order to sketch the figure. We know:

f(v)= 1000 Hz, and f(h) = 500 Hz. Then:

f(v)/ f(h) = 1000 Hz/ 500 Hz = 2/1

But, the ratio of the tangency points is the inverse ratio of the input frequencies.

So T-points = 500/ 1000 = 1/2

In other words, there is ONE vertical tangency point, and TWO horizontal tangency points.

The result is sketched in the diagram for Problem 5.

(6) We have:

f(v)/f(h) = 1

The inverse of this also yields 1. Therefore there is one horizontal and one vertical tangency point.

The pattern is sketched in the diagram for Problem 6. It's a circle.

No comments: