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[ E X A M P L E ]
A period of 4 0
is observed and measured. (See Fig. 13)
Substituting the given value:
Freq = 1 / [ 4 0 x 1 0 "
While the above method relies on the measurement directly of the period of one c y c l e ,
the frequency may also be measured by counting the number of c y c l e s present in a given
time period.
1 . Apply the signal to the INPUT jack. Set the V or Y MODE to the channel to be used
and adjusting the various controls for a normal display. Set the V A R I A B L E to C A L .
2. Count the number of cycles of waveform between a chosen set of vertical graduation
lines.
Using the horizontal distance between the vertical lines used above and the S W E E P
T I M E / D I V , the time span may be calculated. Multiply the reciprocal of this value by
the number of cycles present in the given time span. If " x 5 M A G " is used multiply
this further by 5.
Note that errors will occur for displays having only a f e w c y c l e s .
Using the formula:
# of cycles x " x 5 M A G " value
Freq =
Horizontal distance (div) x S W E E P T I M E / D I V setting
[ E X A M P L E ]
For the example, within 7 divisions there are 10 c y c l e s .
The S W E E P T I M E / D I V is 5 jis/div. (See Fig. 14)
Substituting the given value:
1 cycle = 4 0 /*s (5 /ts/div. x 8 div.)
6
] = 2 . 5 x 1 0
4
= 2 5 kHz
F r e q =
Count cycles between this portion
10
=
7
)
5 ( „
/ d i v )
( d i v
x
s
Fig. 13
2
8
5
7
k
H
z
Fig. 1 4
25
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