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
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%
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
An average-reading digital multi-meter reads 10 V when fed with a triangular wave, symmetric about the time-axis. For the same input an rms-reading meter will read
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.
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