# CRO and Electronic Measurements

 Question 1
A $10\frac{1}{2}$ digit timer counter possesses a base clock of frequency 100 MHz. When measuring a particular input, the reading obtained is the same in: (i) Frequency mode of operation with a gating time of one second and (ii) Period mode of operation (in the x 10 ns scale). The frequency of the unknown input (reading obtained) in Hz is _______.
 A 100 B 1000 C 10000 D 100000
GATE EE 2017-SET-2   Electrical and Electronic Measurements
Question 1 Explanation:
$1.\;\;10\frac{1}{2}$ 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
$P=0.01 \times 10^{-6}$
$\;\;=10 \times 10^{-9}$
$\;\;=1 \times 10n-sec$
$\;\;=10.000000000 n-sec$
 Question 2
A stationary closed Lissajous pattern on an oscilloscope has 3 horizontal tangencies and 2 vertical tangencies for a horizontal input with frequency 3 kHZ. The frequency of the vertical input is
 A 1.5kHz B 2kHz C 3kHz D 4.5kHz
GATE EE 2017-SET-2   Electrical and Electronic Measurements
Question 2 Explanation:
\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}
 Question 3
The slope and level detector circuit in a CRO has a delay of 100 ns. The start-stop sweep generator has a response time of 50 ns. In order to display correctly, a delay line of
 A 150 ns has to be inserted into the y-channel B 150 ns has to be inserted into the x-channel C 150 ns has to be inserted into both x and y channels D 100 ns has to be inserted into both x and y channels
GATE EE 2017-SET-1   Electrical and Electronic Measurements
Question 3 Explanation:
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.
 Question 4
The two signals S1 and S2, shown in figure, are applied to Y and X deflection plates of an oscilloscope. The waveform displayed on the screen is A A B B C C D D
GATE EE 2014-SET-3   Electrical and Electronic Measurements
Question 4 Explanation:
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. Question 5
In an oscilloscope screen, linear sweep is applied at the
 A vertical axis B horizontal axis C origin D both horizontal and vertical axis
GATE EE 2014-SET-1   Electrical and Electronic Measurements
Question 5 Explanation:
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.
 Question 6
A $4\frac{1}{2}$ digit DMM has the error specification as: 0.2% of reading + 10 counts. If a dc voltage of 100 V is read on its 200 V full scale, the maximum error that can be expected in the reading is
 A $\pm 0.1$% B $\pm 0.2$% C $\pm 0.3$% D $\pm 0.4$%
GATE EE 2011   Electrical and Electronic Measurements
Question 6 Explanation:
$4\frac{1}{2}$ 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
1 count $=\frac{200V}{20,000}$
$E_1=$ Error corrsponding to 10 counts $=10\times \frac{200}{20,000}=0.1V$
Error corresponding to 0.2% of reading
$E_2=0.2 \times \frac{100}{100}=0.2V$
$\text{Total error}=E_1+E_2$
$\;\;=0.1+0.2=0.3V i.e. 0.3$%
 Question 7
A dual trace oscilloscope is set to operate in the ALTernate mode. The control input of the multiplexer used in the y-circuit is fed with a signal having a frequency equal to
 A the highest frequency that the multiplexer can operate properly B twice the frequency of the time base (sweep) oscillator C the frequency of the time base (sweep) oscillator D half the frequency of the time base (sweep) oscillator
GATE EE 2011   Electrical and Electronic Measurements
 Question 8
 A $\frac{20}{\sqrt{3}}$ B $\frac{10}{\sqrt{3}}$ C $20\sqrt{3}$ D $10\sqrt{3}$
GATE EE 2009   Electrical and Electronic Measurements
Question 8 Explanation:
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}
 Question 9
The two inputs of a CRO are fed with two stationary periodic signals. In the X-Y mode, the screen shows a figure which changes from ellipse to circle and back to ellipse with its major axis changing orientation slowly and repeatedly. The following inference can be made from this.
 A The signals are not sinusoidal B The amplitudes of the signals are very close but not equal C The signals are sinusoidal with their frequencies very close but not equal D There is a constant but small phase difference between the signals
GATE EE 2009   Electrical and Electronic Measurements
Question 9 Explanation: Because phase difference only patterns changes from ellipse to circle and back to ellipse.
 Question 10
Two sinusoidal signals $p(\omega_1 t)=A\sin \omega _1t$ and $q(\omega_2 t)$ are applied to X and Y inputs of a dual channel CRO. The Lissajous figure displayed on the screen shown below : The signal $q(\omega_2 t)$ will be represented as A $q(\omega _{2}t)=Asin\omega _{2}t,\omega_{2}= 2 \omega _{1}$ B $q(\omega _{2}t)=Asin\omega _{2}t,\omega_{2}= \omega _{1}/2$ C $q(\omega _{2}t)=Acos\omega _{2}t,\omega_{2}= 2 \omega _{1}$ D $q(\omega _{2}t)=Acos\omega _{2}t,\omega_{2}= \omega _{1}/2$
GATE EE 2008   Electrical and Electronic Measurements
Question 10 Explanation:
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}
There are 10 questions to complete.