CRO and Electronic Measurements

1.\;\;10\frac{1}{2} digital time counter:
Frequency mode of operation: f=\frac{n}{t}

Let f = frequency of input signal
n = number of cycles of repetitive signal
\Rightarrow \;\;100 \times 10^6
Let t \Rightarrow Gate time \Rightarrow \;t=1 sec.
f=\frac{100 \times 10^6}{1 sec}=10^8 Hz
\;\;=10^8 cycles/sec
On 10\frac{1}{2} digit display
100000000.00Hz

2.\;\; \text{Period mode of operation:}
P=\frac{1}{f}=\frac{t}{n}=\frac{1sec}{100 \times 10^6}
Let P=Period of input signal
P=0.01 \times 10^{-6}
\;\;=10 \times 10^{-9}
\;\;=1 \times 10n-sec
\;\;=10.000000000 n-sec

\begin{aligned} \frac{f_y}{f_x} &= \frac{\text{Horizontal Tangencies}}{\text{VerticalTangencies}}\\ \Rightarrow \; \frac{f_y}{3} &=\frac{3 }{2}\\ \Rightarrow \; f_y&= 4.5kHz \end{aligned}

A delay line of 150ns has to be inserted into the y-channel to get synchronous between sweep signal applied to x-plate and test signal applied to y-plate.

The waveform pattern appearing on the screen of CRO is shown below. Here, S_1 is applied to vertical deflecting plate and S_2 to horizontal deflecting plate.

A CRO uses a horizontal input voltage which is an internally generated ramp voltage (i.e. linear sweep voltage) called "Time base of CRO". An oscilloscope is used to disply waveform which varies as a function of time. To reproduce the waveform accurately, the beam must have a constant horizontal velocity i.e. the deflecting voltage must increase linearly with time. This voltage is called a ramp voltage. Thus , a sawtooth waveform also called linear sweep voltage is applied to the horizontal deflection system ( or horizontal axis) which provides a time base to CRO.

4\frac{1}{2} digit

No. of full digits in case of 4\frac{1}{2} digit display=4
So, maximum count with 4\frac{1}{2} digits display=20,000
Full scale reading =200V
1 count =\frac{200V}{20,000}
E_1= Error corrsponding to 10 counts =10\times \frac{200}{20,000}=0.1V
Reading = 100V
Error corresponding to 0.2% of reading
E_2=0.2 \times \frac{100}{100}=0.2V
\text{Total error}=E_1+E_2
\;\;=0.1+0.2=0.3V i.e. 0.3%

For triangular wave,
\begin{aligned} \text{Average value} &=\frac{V_m}{2}\\ \text{RMS value } &= \frac{V_m}{\sqrt{3}} \\ \therefore \;\; \frac{V_m}{2}&=10V\; \Rightarrow \; V_m=20\\ \text{RMS value } &= \frac{V_m}{\sqrt{3}}=\frac{20}{\sqrt{3}}V \end{aligned}

Because phase difference only patterns changes from ellipse to circle and back to ellipse.

Here, P(\omega _1 t)=A \sin \omega _1 t
y_1-line cut '4' times the Lissajous pattern
x_1-line cut '2' times the Lisaajous pattern

\frac{\omega _y}{\omega _x}=(maximum number of intersection of a horizontal line with Lissajous pattern)/(maximum number of intersection of a vertical line with Lissajous pattern)
\begin{aligned} \therefore \;\; \frac{\omega _y}{\omega _x}&= \frac{2}{4}=\frac{1}{2} \\ \therefore \;\; \omega _x &= 2 \omega _y\\ \omega _1 &= 2\omega _2 \end{aligned}
and q(\omega _2,t) will lead q(\omega _1,t) by 90^{\circ} as trace is a circle
\begin{aligned} q(\omega _2,t) &=A \sin (\omega _2 t +90^{\circ}) \\ q(\omega _2,t) &= A \cos \omega _2 t \end{aligned}