Math / Algebra

QuestionA binary PAM wave is to be transmitted over a low-pass channel with bandwider of 75 kHz . The bit duration is $10 \mu \mathrm{~s}$ . Find a raised-cosine pulse spectrum that satisfies these requirements.

Studdy Solution

STEP 1

1. The channel is a low-pass channel with a bandwidth of 75 kHz.
2. The bit duration is $10 \, \mu \mathrm{s}$ .
3. We need to find a raised-cosine pulse spectrum that fits these parameters.

STEP 2

1. Understand the relationship between bit duration and bandwidth.
2. Determine the symbol rate.
3. Calculate the Nyquist bandwidth.
4. Determine the roll-off factor.
5. Define the raised-cosine pulse spectrum.

STEP 3

The bit duration $T_b$ is given as $10 \, \mu \mathrm{s}$ . The bit rate $R_b$ is the reciprocal of the bit duration:
$R_b = \frac{1}{T_b} = \frac{1}{10 \times 10^{-6}} = 100,000 \, \text{bits per second (bps)}$

STEP 4

The symbol rate $R_s$ for binary PAM is the same as the bit rate $R_b$ because each symbol represents one bit. Therefore:
$R_s = 100,000 \, \text{symbols per second (sps)}$

STEP 5

The Nyquist bandwidth $B_N$ is half the symbol rate for a raised-cosine filter:
$B_N = \frac{R_s}{2} = \frac{100,000}{2} = 50 \, \text{kHz}$

STEP 6

The channel bandwidth is given as 75 kHz. The excess bandwidth (roll-off factor $\beta$ ) can be calculated as:
$\beta = \frac{\text{Channel Bandwidth} - B_N}{B_N} = \frac{75,000 - 50,000}{50,000} = 0.5$

STEP 7

The raised-cosine pulse spectrum $H(f)$ is defined by the roll-off factor $\beta$ and the Nyquist bandwidth $B_N$ . The spectrum is:
$H(f) = \begin{cases} 1, & |f| \leq B_N \\ \frac{1}{2} \left[ 1 + \cos\left(\frac{\pi (|f| - B_N)}{\beta B_N}\right) \right], & B_N < |f| \leq (1 + \beta)B_N \\ 0, & |f| > (1 + \beta)B_N \end{cases}$
For this problem, $B_N = 50 \, \text{kHz}$ and $\beta = 0.5$ .
The raised-cosine pulse spectrum that satisfies these requirements is defined by the above equations with the given parameters.

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QuestionA binary PAM wave is to be transmitted over a low-pass channel with bandwider of 75 kHz . The bit duration is $10 \mu \mathrm{~s}$ . Find a raised-cosine pulse spectrum that satisfies these requirements.

Studdy Solution

STEP 1

1. The channel is a low-pass channel with a bandwidth of 75 kHz.
2. The bit duration is $10 \, \mu \mathrm{s}$ .
3. We need to find a raised-cosine pulse spectrum that fits these parameters.

STEP 2

1. Understand the relationship between bit duration and bandwidth.
2. Determine the symbol rate.
3. Calculate the Nyquist bandwidth.
4. Determine the roll-off factor.
5. Define the raised-cosine pulse spectrum.

STEP 3

The bit duration $T_b$ is given as $10 \, \mu \mathrm{s}$ . The bit rate $R_b$ is the reciprocal of the bit duration:
$R_b = \frac{1}{T_b} = \frac{1}{10 \times 10^{-6}} = 100,000 \, \text{bits per second (bps)}$

STEP 4

The symbol rate $R_s$ for binary PAM is the same as the bit rate $R_b$ because each symbol represents one bit. Therefore:
$R_s = 100,000 \, \text{symbols per second (sps)}$

STEP 5

The Nyquist bandwidth $B_N$ is half the symbol rate for a raised-cosine filter:
$B_N = \frac{R_s}{2} = \frac{100,000}{2} = 50 \, \text{kHz}$

STEP 6

The channel bandwidth is given as 75 kHz. The excess bandwidth (roll-off factor $\beta$ ) can be calculated as:
$\beta = \frac{\text{Channel Bandwidth} - B_N}{B_N} = \frac{75,000 - 50,000}{50,000} = 0.5$

STEP 7

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QuestionA binary PAM wave is to be transmitted over a low-pass channel with bandwider of 75 kHz . The bit duration is 10μ s10 \mu \mathrm{~s}10μ s. Find a raised-cosine pulse spectrum that satisfies these requirements.

Studdy Solution

STEP 1

1. The channel is a low-pass channel with a bandwidth of 75 kHz.2. The bit duration is 10 μs10 \, \mu \mathrm{s}10μs.3. We need to find a raised-cosine pulse spectrum that fits these parameters.

STEP 2

1. Understand the relationship between bit duration and bandwidth.2. Determine the symbol rate.3. Calculate the Nyquist bandwidth.4. Determine the roll-off factor.5. Define the raised-cosine pulse spectrum.

STEP 3

STEP 4

The symbol rate Rs R_s Rs​ for binary PAM is the same as the bit rate Rb R_b Rb​ because each symbol represents one bit. Therefore:Rs=100,000 symbols per second (sps) R_s = 100,000 \, \text{symbols per second (sps)} Rs​=100,000symbols per second (sps)

STEP 5

The Nyquist bandwidth BN B_N BN​ is half the symbol rate for a raised-cosine filter:BN=Rs2=100,0002=50 kHz B_N = \frac{R_s}{2} = \frac{100,000}{2} = 50 \, \text{kHz} BN​=2Rs​​=2100,000​=50kHz

STEP 6

STEP 7

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QuestionA binary PAM wave is to be transmitted over a low-pass channel with bandwider of 75 kHz . The bit duration is $10 \mu \mathrm{~s}$ . Find a raised-cosine pulse spectrum that satisfies these requirements.

1. The channel is a low-pass channel with a bandwidth of 75 kHz.
2. The bit duration is $10 \, \mu \mathrm{s}$ .
3. We need to find a raised-cosine pulse spectrum that fits these parameters.

1. Understand the relationship between bit duration and bandwidth.
2. Determine the symbol rate.
3. Calculate the Nyquist bandwidth.
4. Determine the roll-off factor.
5. Define the raised-cosine pulse spectrum.

The symbol rate $R_s$ for binary PAM is the same as the bit rate $R_b$ because each symbol represents one bit. Therefore:
$R_s = 100,000 \, \text{symbols per second (sps)}$

The Nyquist bandwidth $B_N$ is half the symbol rate for a raised-cosine filter:
$B_N = \frac{R_s}{2} = \frac{100,000}{2} = 50 \, \text{kHz}$