period of a signal is 100 ms. Then the frequency The of this signal in kilohertz is ______.
10 -2
The question asks us to find the frequency of a signal when its period is given. The period (T) and frequency (f) of a signal are inversely related. This means if you know one, you can easily calculate the other using a simple formula.
The fundamental relationship between the period (T) of a signal and its frequency (f) is given by:
$$\text{f} = \frac{1}{\text{T}}$$
Here, T is the time taken for one complete cycle of the signal, and f is the number of cycles the signal completes in one second.
We are given the period of the signal as 100 milliseconds (ms). To use the formula $f = \frac{1}{T}$, the period must be in seconds (s). We also need the final frequency in kilohertz (kHz).
Let's perform the calculation step-by-step:
The given period is $T = 100 \text{ ms}$.
We know that $1 \text{ second (s)} = 1000 \text{ milliseconds (ms)}$.
So, $1 \text{ ms} = \frac{1}{1000} \text{ s} = 10^{-3} \text{ s}$.
Therefore, $T = 100 \text{ ms} = 100 \times 10^{-3} \text{ s} = 0.1 \text{ s}$.
Now that the period is in seconds, we can calculate the frequency in Hertz (Hz) using the formula $f = \frac{1}{T}$.
$$f = \frac{1}{0.1 \text{ s}} = \frac{1}{\frac{1}{10} \text{ s}} = 10 \text{ Hz}$$
So, the frequency of the signal is 10 Hz.
The question asks for the frequency in kilohertz (kHz).
We know that $1 \text{ kilohertz (kHz)} = 1000 \text{ Hertz (Hz)}$.
So, $1 \text{ Hz} = \frac{1}{1000} \text{ kHz} = 10^{-3} \text{ kHz}$.
To convert 10 Hz to kHz, we multiply by $10^{-3}$:
$$f = 10 \text{ Hz} = 10 \times 10^{-3} \text{ kHz} = 10^{-2} \text{ kHz}$$
The frequency of the signal is $10^{-2}$ kHz.
| Quantity | Unit Conversion |
|---|---|
| Period (Time) | $1 \text{ ms} = 10^{-3} \text{ s}$ |
| Frequency | $1 \text{ Hz} = 10^{-3} \text{ kHz}$ |
Let's quickly verify the calculation steps one more time:
The final result, $10^{-2}$ kHz, matches one of the given options.
| Concept | Formula | Key Conversion |
|---|---|---|
| Period & Frequency | $f = \frac{1}{T}$ | Units must be consistent (e.g., seconds for T, Hertz for f) |
| Milliseconds to Seconds | $1 \text{ ms} = 10^{-3} \text{ s}$ | Divide milliseconds by 1000 to get seconds |
| Hertz to Kilohertz | $1 \text{ Hz} = 10^{-3} \text{ kHz}$ | Divide Hertz by 1000 to get kilohertz |
Understanding signal characteristics like period and frequency is fundamental in many areas of physics and engineering, especially in fields like electronics, telecommunications, and signal processing.
The inverse relationship $f = \frac{1}{T}$ highlights that a shorter period means a higher frequency (more cycles per second), and a longer period means a lower frequency (fewer cycles per second).
In the standard Ethernet with transmission rate of 10 Mbps, asssume that the length of the medium is 2500 m and size of a frame is 512 bytes. The propagation speed of a signal in a cable is normally 2 × 108 m/s. The transmission delay and propogation delay are
Which of the following statement(s) is/are correct?
I. Metropolitan area network is a collection of Local area network.
II. Metropolitan area network support both data and voices transmission.
Data transmission using multiple pathways simultaneously is known as:
Consider the following statements about RS-232 ports and connectors.
S1: DB9 connector is associated with RS-232
S2: RS-232 uses RJ45 connector for standard serial communication
S3: RS-232 communication is point-to-point.
Which statements are TRUE?