Cro And Electronic Measurements Practice Questions

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.

$1.\;\;10\frac{1}{2}$ 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 $P=0.01 \times 10^{-6}$ $\;\;=10 \times 10^{-9}$ $\;\;=1 \times 10n-sec$ $\;\;=10.000000000 n-sec$

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.

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$4\frac{1}{2}$ 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 $E_2=0.2 \times \frac{100}{100}=0.2V$ $\text{Total error}=E_1+E_2$ $\;\;=0.1+0.2=0.3V i.e. 0.3$ %

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

For triangular wave,

Single slope integrating type ADC utilize digital counter techniques to measure time required for a voltage ramp to rise from zero to the input voltage. If conversion time for input voltage $5V=T_A$ So, conversion time for input voltage $2.5V=T_A/2$ Conversion time in successive type ADC does not depend on input voltage. So, conversion time for input voltage 2.5V is also $T_B$ .

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)

and $q(\omega _2,t)$ will lead $q(\omega _1,t)$ by $90^{\circ}$ as trace is a circle

Square wave is of low frequency. So, it can be assumed that time during which the wave forms are displayed on the screen, the voltage across R and L is $V_{\in}$

So, $S_1$ is connected to point A. and, $G_1$ is connected to point B. Voltage across inductor $=V_{BC}=L\frac{di}{dt}$ $\;\;\;=L\frac{d}{dt}\left ( \frac{V_{\in}}{R} (1-e^{-Rt/L})\right )$ $V_{BC}=V_{\in}e^{-Rt/L}$ So, $S_2$ is connected to point C. and $G_2$ is connected to point B.

Peak-peak (p-p) division of upper trace voltage= 2 and value of (p-p) voltage =5V $\therefore \;\;\frac{\text{Voltage}}{Division}=2.5V$ Now it will be same for unknown voltage (p-p) division of unknown voltage=3 Therefore, p-p voltage $3 \times 2.5 =7.5 V$ Frequency of upper trace=1 kHz $\therefore \;\; \text{Time period}=\frac{1}{10^3}=1ms$ Division of x-axis(upper)=4 Division of x-axis(lower)=8 Therefore, Period of unknown signal=2ms

Resolution of digital to analog coverter $R=\frac{V_0}{2^N}=14mV$ Where, $V_0=$ Full scale output voltage=3.5V N = bit-size

$f_y=$ frequency of signal applied to y plates $f_x=$ frequency of signal applied to x plates $x(t)=P \sin (4t+30)$ So, $f_x=\frac{4}{2 \pi}Hz$ $\frac{f_y}{f_x}=\frac{\text{No. of \times tangent touches top or bottom}}{\text{No. of \times tangent touches either side}}=\frac{1}{2}$ $f_y=\frac{f_x}{2}=\left ( \frac{4}{2\pi} \right )\times \frac{1}{2}=\frac{2}{2\pi}$ $\omega _y=2 \pi f_y=2 \pi\left ( \frac{2}{2 \pi} \right )=2 rad/sec.$ $Q \sin (2t+15)$ has angular frequency. $\omega =2 rad/sec$

$Q=\frac{\omega _0 L}{R}$ where $\omega _0$ is the resonant angular frequency, L is the inductance and R is the effective resistance of the coil. The priciple of working of Q-meter is based upon series resonance of R, L, C circuit.

Taking supply voltage as the reference $V_s=10\angle 0 V$ Using KCL, $\frac{V_p-10}{100 \times 10^3}+\frac{V_p}{100 \times 10^3}$ + $\frac{V_p}{500 \times 10^3}+\frac{V_p}{-j159.15 \times 10^3}=0$ $V_p=4.37\angle -15.94V$

when two voltages are in phase i.e. $[\phi =0^{\circ}\; \text{or} \;180^{\circ}]$ , the trace is straight line. But for $\phi =0^{\circ}$ , the trace has positive slope and $\phi =180^{\circ}$ , the trace has negative slope. When the two voltage has phase displacement of $\phi =90^{\circ}$ , the trace is a circle. If the direction of the trace is in the clockwise direction then the phase difference is $\phi =90^{\circ}$ . If the direction of the trace is anti-clockwise direction the phase difference is $(2 \pi-\phi )=\left ( 2\pi-\frac{\pi}{2} \right )=3\frac{\pi}{2}$ When the difference is not equal to $0^{\circ}$ or $90^{\circ}$ , the trace is an ellipse. For $\pi \leq \phi \leq \frac{3\pi}{2}$ , the trace is an ellipse in anti-clockwise direction.

The wave form of dual slope integrater is shown. The maximum frequency can be attained when $T_2-T_1=0$ and as $T_1=2^NT_C\;\;\;(T_C=\text{clock period})$ $\therefore \;\;f_{max}=\frac{1}{2^N\cdot T_C}$ $\;\;\;=\frac{f_C}{2^N}=\frac{10^6}{1024}=1kHz$