What should be the value of ω for a low pass filter to have magnitude of transfer function as 0.707? 

Explanation:

To determine the value of ω (omega) for a low pass filter to have a magnitude of the transfer function as 0.707, let's first understand the transfer function of a low pass RC filter and the conditions given in the problem statement.

Low Pass RC Filter:

A low pass RC (Resistor-Capacitor) filter is an electronic circuit that allows signals with a frequency lower than a certain cutoff frequency to pass through and attenuates frequencies higher than the cutoff frequency. The transfer function H(jω) of a low pass RC filter is given by:

H(jω) = \(\dfrac{1}{1 + jωRC}\)

where:

• j is the imaginary unit (j = \(\sqrt{-1}\))
• ω is the angular frequency (in radians per second)
• R is the resistance (in ohms)
• C is the capacitance (in farads)

The magnitude of the transfer function |H(jω)| is given by:

|H(jω)| = \(\dfrac{1}{\sqrt{1 + (ωRC)^2}}\)

According to the problem statement, the magnitude of the transfer function is 0.707. Therefore, we have:

\(\dfrac{1}{\sqrt{1 + (ωRC)^2}} = 0.707\)

To solve for ω, we need to square both sides of the equation to eliminate the square root:

\(\dfrac{1}{1 + (ωRC)^2} = (0.707)^2\)

We know that (0.707)^2 is approximately 0.5. So the equation becomes:

\(\dfrac{1}{1 + (ωRC)^2} = 0.5\)

Next, we take the reciprocal of both sides:

1 + (ωRC)^2 = 2

Subtract 1 from both sides to isolate (ωRC)^2:

(ωRC)^2 = 1

Take the square root of both sides:

ωRC = 1

Therefore, the angular frequency ω is given by:

ω = \(\dfrac{1}{RC}\)

Correct Option Analysis:

The correct option is:

Option 4: \(\dfrac{1}{RC}\)

This option correctly represents the value of ω for the low pass filter to have a magnitude of the transfer function as 0.707.

Important Information

To further understand the analysis, let’s evaluate the other options:

Option 1: 0

This option is incorrect because if ω = 0, the magnitude of the transfer function |H(jω)| would be 1, not 0.707. This represents the DC gain of the filter, where no attenuation occurs.

Option 2: infinity

This option is incorrect because if ω approaches infinity, the magnitude of the transfer function |H(jω)| would approach 0. This is due to the fact that high frequencies are highly attenuated by the low pass filter.

Option 3: CR

This option is incorrect because it does not conform to the derived formula. The correct relation involves the reciprocal of the product of resistance and capacitance, not the direct product.

Conclusion:

Understanding the transfer function and the behavior of low pass filters is essential for accurately determining the conditions for specific magnitudes of the transfer function. The correct value of ω for a low pass filter to have a magnitude of the transfer function as 0.707 is \(\dfrac{1}{RC}\), which is derived from the standard transfer function of a low pass RC filter. This analysis highlights the importance of correctly interpreting the filter's transfer function and its frequency response characteristics.