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).
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.
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
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?
Which of the following statements is TRUE about data transmission modes?
Consider the following statements regarding modems used in computer network:
S1: Modems are required when communication medium carries analog signals
S2: Modulation converts digital data into analog form
S3: Demodulation converts analog data into digital form
S4: Modems assign MAC addresses to computers.
Which statements are TRUE?