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[ E X A M P L E ]
In the example s h o w n in Fig. 4 7 , the V O L T S / D I V set-
ting is 1 V/div and the reference signal is 2 V r m s . A d -
just the V A R I A B L E so that the amplitude
reference signal is 4 div.
Variable coefficient =
Then, measure the unknown signal. Supposing the ver-
tical amplitude is 3 divisions,
Effective value of the unknown signal
= 3 (div) x 0 . 5 x 1 (V/div) = 1.5 V r m s
* Period
Setting of relative s w e e p coefficient using reference fre-
quency
1)
Apply the reference signal to the input connector. A d -
just each control for normal s w e e p display. T h e n , ad-
just the V O L T S / D I V and V A R I A B L E to obtain proper
w a v e f o r m observation.
Adjust the S W E E P T I M E / D I V and V A R I A B L E controls
accurately so that one cycle width coincides with a f e w
divisions of the graduation line.
Do not change V A R I A B L E setting after the above ad-
justment.
2)
Divide the period of the reference signal frequency by
the product of the horizontal distance (in divisions) and
S W E E P T I M E / D I V setting given in item 1) above.
Horizontal coefficient
Period of reference signal (sec)
Horizontal distance (div) x S W E E P TIME/DIV setting
3)
Discontinue applying the reference signal and apply the
unknown signal to the input connector. Adjust the
S W E E P T I M E / D I V for easy observation. Measure the
horizontal distance of one cycle and calculate the peri-
od of the unknown signal using the following equation:
Period of unknown signal =
Horizontal distance (div) x horizontal coefficient
x S W E E P T I M E / D I V setting
[ E X A M P L E ]
In the example s h o w n in Fig. 4 8 , S W E E P TIME/DIV set-
ting is 0.1 ms/div and a reference signal of 1.75 kHz
is applied. Adjust the V A R I A B L E so that the distance
of one cycle is 5 divisions.
Horizontal coefficient =
The unknown signal is then measured. If the horizon-
tal distance is 7 divisions,
Pulse width =
7 (div) x 1 . 1 4 3 x 0.1 (ms/div) = 0.8 ms
of the
2 V r m s
= 0 5
4 ( d i v ) x 1 (V(div)
1
- - -
- - - - -
[
—
z
-
5 (div) x 0.1 (ms/div)
=
1 . 1 4 3
R e f e r e n c e
A d j u s t e d r e f e r e n c e s i g n a l
s i g n a l
U n k n o w n s i g n a l
Fig.
4 8
35
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