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 _______.

100 | |

1000 | |

10000 | |

100000 |

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

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

1.5kHz | |

2kHz | |

3kHz | |

4.5kHz |

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

150 ns has to be inserted into the y-channel | |

150 ns has to be inserted into the x-channel | |

150 ns has to be inserted into both x and y channels | |

100 ns has to be inserted into both x and y channels |

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

The waveform displayed on the screen is

A | |

B | |

C | |

D |

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

vertical axis | |

horizontal axis | |

origin | |

both horizontal and vertical axis |

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

\pm 0.1% | |

\pm 0.2% | |

\pm 0.3% | |

\pm 0.4% |

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%

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

the highest frequency that the multiplexer can operate properly | |

twice the frequency of the time base (sweep) oscillator | |

the frequency of the time base (sweep) oscillator | |

half the frequency of the time base (sweep) oscillator |

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

\frac{20}{\sqrt{3}} | |

\frac{10}{\sqrt{3}} | |

20\sqrt{3} | |

10\sqrt{3} |

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}

\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.

The signals are not sinusoidal | |

The amplitudes of the signals are very close but not equal | |

The signals are sinusoidal with their frequencies very close but not equal | |

There is a constant but small phase difference between the signals |

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

q(\omega _{2}t)=Asin\omega _{2}t,\omega_{2}= 2 \omega _{1} | |

q(\omega _{2}t)=Asin\omega _{2}t,\omega_{2}= \omega _{1}/2 | |

q(\omega _{2}t)=Acos\omega _{2}t,\omega_{2}= 2 \omega _{1} | |

q(\omega _{2}t)=Acos\omega _{2}t,\omega_{2}= \omega _{1}/2 |

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}

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.