Low Pass Filter MCQ Quiz - Objective Question with Answer for Low Pass Filter - Download Free PDF

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.

Classification of filters:

1.) Low pass filter:

A low-pass filter is a filter that passes signals with a frequency lower than a selected cutoff frequency and attenuates signals with frequencies higher than the cutoff frequency.

2.) High pass filter:

A high pass filter is a filter that passes signals with a frequency higher than a selected cutoff frequency and attenuates signals with frequencies lower than the cutoff frequency.

3.) Band pass filter:

A band-pass filter is a filter that passes frequencies within a certain range ( f_L and f_H ) and rejects (attenuates) frequencies outside that range.

4.) Band reject filter:

A band-pass filter is a filter that attenuates frequencies within a certain range ( fL and fH ) and passes frequencies outside that range.

Concept:

• Active Low Pass Filter uses an op-amp for amplification and gain control.
• The simplest form of a low pass active filter is to connect an inverting or non-inverting amplifier.
• Applications of Active Low Pass Filters are in audio amplifiers, equalizers or speaker systems to direct the lower frequency bass signals to the larger bass speakers or to reduce any high frequency noise or “hiss” type distortion. 
• When used like this in audio applications the active low pass filter is sometimes called a “Bass Boost” filter.

Calculation:
Given;

\(f_c = \frac{1}{2\pi R_2C}\) ; where \(f_c\) = cut-off corner frequency

Therefore; \(R_2 = \frac{1}{2\pi f_c C}\)

Putting on the respective values, we get:

\(R_2 = \frac{1}{2\pi \times5\times10^3\times10\times10^-9}\)

R₂ = 3.10 kΩ

Note: It is important to understand the formula for cut-off frequency and remember it for once. It

saves a lot of time without the analysis to conclude that:

\(\omega_c = \frac{1}{R_2C}\)

Low pass filter:

This filter passes the low frequency and blocks or attenuates the high frequency.

It has only one cut off frequency above which it starts to attenuate the signals.

With the increase in frequency, we can see that the amplitude response of the filter changes and attenuates to 0 at high frequencies. 

The gain to frequency plot of a low pass filter is shown below:

With reference to frequency response of first order low frequency response:

• At f = 0, gain = A which is the maximum gain. Thus it has maximum gain at frequency of 0 Hz.
• The higher cutoff frequency occurs at 0.707 (1/\( \sqrt{2}\)) times the maximum gain.
• For frequency greater than higher cutoff frequency, the gain decreases at a constant rate of -20 dB/decade. 

Classification of filters:

1.) Low pass filter:

A low-pass filter is a filter that passes signals with a frequency lower than a selected cutoff frequency and attenuates signals with frequencies higher than the cutoff frequency.

2.) High pass filter:

A high pass filter is a filter that passes signals with a frequency higher than a selected cutoff frequency and attenuates signals with frequencies lower than the cutoff frequency.

3.) Band pass filter:

A band-pass filter is a filter that passes frequencies within a certain range ( f_L and f_H ) and rejects (attenuates) frequencies outside that range.

4.) Band reject filter:

A band-pass filter is a filter that attenuates frequencies within a certain range ( fL and fH ) and passes frequencies outside that range.

Concept:

The name of the filter is decided by the type of frequency it passes:

Low pass filter:

This filter passes the low frequency and blocks or attenuates the high frequency.

Analysis:

\({V_0} = \left( {\frac{R}{{R + jwL}}} \right){V_1}\)

ω = 0,

V₀ = V₁

ω = ∞

V₀ = 0

As, low frequencies are passed and high frequencies are blocked therefore it is a low pass filter. 

High Pass Filter:

This filter passes the high frequency and blocks the low frequency.

Band Pass Filter:

This filter passes a certain band of frequencies and blocks low and high frequencies.

Band Stop Filter:

This filter blocks a certain band of frequencies and passes low and high frequencies.

Concept:

The transfer function of a low pass filter is,

\(\Rightarrow G\left( {j\omega } \right) = \frac{1}{{1 + j\omega RC}}\)

\(\Rightarrow \left| {G\left( {j\omega } \right)} \right| = \frac{1}{{\sqrt {1 + {\omega ^2}{R^2}{C^2}} }}\)

And

∠G (jω) = - tan^-1 (ωRC)

With Bandwidth, \(B = \frac{1}{{RC}}\) 

at ω = 0, |G (jω)| = 1 and ∠G (jω) = 0°

From the phase response, at positive values of ω, the phase angle decreases linearly because of the negative sign.

Concept:

\({Z_1} = {R_1}\)

\({Z_2} = {R_2}||C\)

\({Z_2} = \frac{{{R_2}\left( {\frac{1}{{j\omega c}}} \right)}}{{{R_2} + \frac{1}{{j\omega c}}}}\;\)

\({Z_2} = \frac{{{R_2}}}{{1 + j\omega {R_2}C}}\)

For the inverting amplifier:

\(\frac{{{V_o}}}{{{V_i}}} = - \frac{{{Z_2}}}{{{Z_1}}}\)

\(\frac{{{V_o}}}{{{V_i}}} = \frac{{ - \frac{{{R_2}}}{{1\_j\omega {R_2}C}}}}{{{R_1}}} = \frac{{\frac{{{R_2}}}{{{R_1}}}}}{{1 + j\omega {R_2}C}}\)

The magnitude of the voltage gain will be:

\(\left| {\frac{{{V_o}}}{{{V_i}}}} \right| = \frac{{\frac{{{R_2}}}{{{R_1}}}}}{{\sqrt {1 + {{\left( {\omega {R_2}C} \right)}^2}} }}\)

At ω = 0:

\(\left| {\frac{{{V_o}}}{{{V_i}}}} \right| = \frac{{{R_2}}}{{{R_1}}}\)

Let it be A_max

\(\therefore \frac{{{v_o}}}{{{V_i}}} = \frac{{{A_{msx}}}}{{1 + j\omega {R_2}C}}\)

\(\left| {\frac{{{V_o}}}{{{V_i}}}} \right| = \frac{{{A_{max}}}}{{\sqrt {1 + {{\left( {\omega {R_2}C} \right)}^2}} }}\)

At ω = ω_c (Cut-off frequency)

\(\left| {\frac{{{V_o}}}{{{V_i}}}} \right| = \frac{{Amax}}{{\sqrt 2 }}\)

\(\frac{{Amax}}{{\sqrt 2 }} = \frac{{Amax}}{{\sqrt {1 + {{\left( {{\omega _c}{R_2}C} \right)}^2}} }}\)

2 = 1 + (ω_cR₂C)²

1 = ω_cR_2­C

\({\omega _c} = \frac{1}{{{R_2}C}}\)

The frequency response is plotted as:

Calculation:

\({f_c} = \frac{1}{{2\pi {R_2}C}}\)

\({R_2} = \frac{1}{{2\pi {f_c}C}}\)

Putting on the respective values, we get: 

\({R_2} = \frac{1}{{2\pi \; \times \;5\; \times\; {{10}^3}\; \times \;10\; \times\; {{10}^{ - 9}}}}\)

R₂ = 3.10 kΩ

Note: It is important to understand the formula for cut-off frequency and remember it for once. It saves a lot of time without the analysis to conclude that:

\({\omega _c} = \frac{1}{{{R_2}C}}\)

Low pass filter:

This filter passes the low frequency and blocks or attenuates the high frequency.

It has only one cut off frequency above which it starts to attenuate the signals.

With the increase in frequency, we can see that the amplitude response of the filter changes and attenuates to 0 at high frequencies. 

The gain to frequency plot of a low pass filter is shown below:

With reference to frequency response of first order low frequency response:

• At f = 0, gain = A which is the maximum gain. Thus it has maximum gain at frequency of 0 Hz.
• The higher cutoff frequency occurs at 0.707 (1/\( \sqrt{2}\)) times the maximum gain.
• For frequency greater than higher cutoff frequency, the gain decreases at a constant rate of -20 dB/decade. 

Concept:

1) PSD and Autocorrelation function form a Fourier pair

2) \(PS{D_{output}} = {\left| {H\left( \omega \right)} \right|^2}PS{D_{input}}\)

Calculation:

\(H\left( {j\omega } \right) = \frac{{\frac{1}{{j\omega C}}}}{{R + \frac{1}{{j\omega C}}}}\)

\(= \frac{1}{{1 + j\omega RC}}\)

output PSD

\({S_y}\left( \omega \right) = {\left| {H\left( \omega \right)} \right|^2}{S_x}\left( \omega \right)\)

\(= \frac{1}{{{R^2}{C^2}{\omega ^2} + 1}}{S_x}\left( \omega \right)\)

\({S_x}\left( \omega \right) = \frac{{{N_o}}}{2}\) (input PSD)

\(= \frac{1}{{{R^2}{C^2}{\omega ^2} + 1}}\frac{{{N_o}}}{2}\)

The Inverse Fourier Transform will give ACF.

Autocorrelation Function:

\({R_y}\left( \tau \right) = \frac{{{N_o}}}{{4RC}}\;{e^{ - \;\frac{\tau }{{RC}}}}\)

Average output power: is obtained from ACF by subsisting τ = 0

R = 1 kΩ and C = 0.1 μF

\({R_y}\left( {\tau = 0} \right) = \frac{{{N_o}}}{{4\left( {RC} \right)}} = \frac{{{{0.001\times10}^{ - 6}}}}{{\begin{array}{*{20}{c}} {4\times10^3\times0.1 \times {{10}^{ - 6}}} \end{array}}}\)

⇒ 2.5 μW

option 3 is correct. 

• A LPF is used in circuits that only allow low frequencies to pass through. It is often used to block high frequencies and AC current in a circuit.
• In other words, low-pass filters help in removing short-term fluctuations and provide a smoother form of signal.
•  An ideal integrator acts as a low pass filter.

The low-pass filter is as shown:-

\( \frac{{{V_o}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{1}{{1 + sCR}} \)

\(= \frac{1}{{1 + s\tau }} = \frac{1}{{1 + j\omega \tau }}\)

\({V_o}\left( t \right) = {V_o}{e^{ - \frac{t}{\tau }}}\) .

For the capacitor to act as an integrator the capacitor should discharge slowly. And this discharging time depends upon the time constant (τ).

Less the time constant → faster will be the discharge

More the time constant → Slower will be the discharge

So, for the capacitor to act as an integrator,

ωτ ≫ 1

Concept:

Normalized transition width

The region which is in between the passband and stopband is known as the "Transition region".

The transition width of a filter is defined as:

Δω = ω_s - ω_p 

Where,

ω_s : Stopband frequency

ω_p : Passband frequency

The normalized transition width is defined as:

\(Normalized\; transition\; width = \frac{Δ ω}{F_s}\)

F_s: Sampling frequency

Calculation:

The passband frequency is: 0 to 50 Hz and the stopband frequency is 60 to 280 Hz.

ωs : 60 Hz

ωp : 50 Hz

From the formula the Transition width is:

Δω = 2π(60 - 50)

Δ2πf_n = 2π (60 - 50)

The normalized transition width is:

\(f_n = \frac{10}{400}\)

f_n = 0.025 Hz

Important points:

Amplitude Parameters

δp = Passband Ripple, Rp = -20log[(1- δp )/(1+ δp )] in dB

Peak ripple, ap = -20log[(1- δp )] in dB

δs = Stopband Ripple/Attenuation,

As = -20log[(δs / 1+ δp )] in dB

Minimum stopband attenuation, as = -20log[(δs )] in dB

Frequency Parameters

Sampling frequency = F_s

Passband frequency, ω_p = 2πf_p , Normalized Passband frequency = ω_p /F_s

Stop-band frequency, ω_s = 2πf_s , Normalized Passband frequency = ω_s /F_s

Transition width, Δω = ω_s - ω_p 

Cut-off frequency, ω_c = (ωs - ωp ) / 2

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.