
A diode detector is used to detect an amplitude modulated wave of $60\%$ modulation by using a condenser of $250pF$ in parallel with the load resistance $100k\Omega$. Find the maximum modulated frequency, which could be detected by it.
A) $10.62KHz$
B) $5.31Mz$
C) $5.31Kz$
D) $10.62Mz$
Answer
135.9k+ views
Hint: From the question given, we consider the combined value for resistance and capacitance by multiplying the resistance with the capacitance. The modulation factor is also given. Therefore using the values in the equation of frequency, we get the required solution.
Formula Used:
$f = \dfrac{1}{{2\pi m\tau }}$
$f$ is the modulated frequency
$m$ is the modulation factor
$\tau$ is the relaxation time
Complete step by step solution:
We know, Modulation refers to the superimposition of a low-frequency signal on a high-frequency signal. This is done at the transmitter end to minimize losses.
Now, we know, the modulated frequency can be obtained by using the formula:
$f = \dfrac{1}{{2\pi m\tau }}$
The relaxation time $\tau$, is defined as the amount of time taken by the capacitance to charge and discharge itself.
In the given question, we can write:
$\tau = RC$
Where,
$R$ is the resistance
$C$ is the capacitance
From the question, we obtain the following values:
$R = 100 \times {10^3}\Omega$
$C = 250 \times {10^{ - 12}}F$
Thus, we get the relaxation time,
$\Rightarrow \tau = 100 \times {10^3} \times 250 \times {10^{ - 12}}$
$\Rightarrow \tau = 2.5 \times {10^{ - 5}}\Omega F$
And we have, $m = 0.6$
Therefore, on putting the values in the equation, we obtain:
$\Rightarrow f = \dfrac{1}{{2\pi \times 0.6 \times 2.5 \times {{10}^{ - 5}}}}$
On solving the above equation, we obtain:
$\Rightarrow f = 10.62KHz$
This is the maximum value of modulated frequency that can be detected.
Thus, option (A) is correct.
Additional Information:
Frequency modulation is done at the transmitter end that modifies the properties of a periodic waveform.
Similarly, a demodulator is present at the receiver end to obtain the actual wave as the output after transmission.
Note: There are various types of modulation, like analog modulation, the purpose of which is to transfer the analog signal, digital modulation is performed to transfer digital signals, and pulse modulation is used to transfer narrowband analog signals.
Formula Used:
$f = \dfrac{1}{{2\pi m\tau }}$
$f$ is the modulated frequency
$m$ is the modulation factor
$\tau$ is the relaxation time
Complete step by step solution:
We know, Modulation refers to the superimposition of a low-frequency signal on a high-frequency signal. This is done at the transmitter end to minimize losses.
Now, we know, the modulated frequency can be obtained by using the formula:
$f = \dfrac{1}{{2\pi m\tau }}$
The relaxation time $\tau$, is defined as the amount of time taken by the capacitance to charge and discharge itself.
In the given question, we can write:
$\tau = RC$
Where,
$R$ is the resistance
$C$ is the capacitance
From the question, we obtain the following values:
$R = 100 \times {10^3}\Omega$
$C = 250 \times {10^{ - 12}}F$
Thus, we get the relaxation time,
$\Rightarrow \tau = 100 \times {10^3} \times 250 \times {10^{ - 12}}$
$\Rightarrow \tau = 2.5 \times {10^{ - 5}}\Omega F$
And we have, $m = 0.6$
Therefore, on putting the values in the equation, we obtain:
$\Rightarrow f = \dfrac{1}{{2\pi \times 0.6 \times 2.5 \times {{10}^{ - 5}}}}$
On solving the above equation, we obtain:
$\Rightarrow f = 10.62KHz$
This is the maximum value of modulated frequency that can be detected.
Thus, option (A) is correct.
Additional Information:
Frequency modulation is done at the transmitter end that modifies the properties of a periodic waveform.
Similarly, a demodulator is present at the receiver end to obtain the actual wave as the output after transmission.
Note: There are various types of modulation, like analog modulation, the purpose of which is to transfer the analog signal, digital modulation is performed to transfer digital signals, and pulse modulation is used to transfer narrowband analog signals.
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