Minimum Sampling Rate Calculation for Signal m(t)

To truthfully represent a continuous-time signal by its samples, we must determine the minimum sampling rate required. This is based on the Nyquist-Shannon sampling theorem, which states that the sampling frequency must be at least twice the highest frequency component present in the signal.

Signal Analysis and Frequency Identification

The given signal is:

\(m(t) = 2\cos(6000 \pi t) + 4\cos(8000 \pi t) + 6\cos(10000 \pi t)\)

A general sinusoidal signal can be expressed as \(A\cos(2\pi ft)\), where \(f\) is the frequency in Hertz (Hz). By comparing each term in \(m(t)\) with this standard form, we can identify the individual frequencies:

• For the first term, \(2\cos(6000 \pi t)\):
  • The angular frequency \(\omega_1 = 6000 \pi\) radians/second.
  • We know that \(\omega = 2\pi f\). So, \(2\pi f_1 = 6000 \pi\).
  • Therefore, the frequency \(f_1 = \frac{6000 \pi}{2\pi} = 3000 \text{ Hz}\).
• For the second term, \(4\cos(8000 \pi t)\):
  • The angular frequency \(\omega_2 = 8000 \pi\) radians/second.
  • So, \(2\pi f_2 = 8000 \pi\).
  • Therefore, the frequency \(f_2 = \frac{8000 \pi}{2\pi} = 4000 \text{ Hz}\).
• For the third term, \(6\cos(10000 \pi t)\):
  • The angular frequency \(\omega_3 = 10000 \pi\) radians/second.
  • So, \(2\pi f_3 = 10000 \pi\).
  • Therefore, the frequency \(f_3 = \frac{10000 \pi}{2\pi} = 5000 \text{ Hz}\).

We can summarize these frequencies in a table:

Maximum Frequency Identification

From the frequencies identified above, the highest frequency component present in the signal \(m(t)\) is \(f_{max} = 5000 \text{ Hz}\).

Nyquist-Shannon Sampling Theorem Application

According to the Nyquist-Shannon sampling theorem, to avoid aliasing and ensure that the original signal can be perfectly reconstructed from its samples, the sampling rate (\(f_s\)) must be at least twice the maximum frequency (\(f_{max}\)) present in the signal. This minimum sampling rate is also known as the Nyquist rate.

The formula for the minimum sampling rate is:

\(f_s \ge 2f_{max}\)

Calculation of Minimum Sampling Rate

Using the identified maximum frequency \(f_{max} = 5000 \text{ Hz}\):

\(f_s \ge 2 \times 5000 \text{ Hz}\)

\(f_s \ge 10000 \text{ Hz}\)

Converting Hertz to kilohertz (kHz):

\(10000 \text{ Hz} = 10 \text{ kHz}\)

Therefore, the minimum sampling rate required is \(10 \text{ kHz}\).

Conclusion

The minimum sampling rate for the given signal \(m(t)\) to be truthfully represented by its samples is 10 kHz.