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.

There are 5 questions to complete.