Filters MCQ Quiz - Objective Question with Answer for Filters - 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.

The correct option is c.

Detailed Solution:

Concept:

• High-pass filter: Those filter which passes high frequency and block low frequency i.e. known as high-pass filter.
• Low-pass filter:  Those filter which passes low frequency and blocks high frequency i.e. known as high-pass filter.
• Band-pass filter: Those filter which passes high frequency as well as low frequency i.e. known as band-pass filter.
• Band-rejects filter: Those filter which blocks high frequency as well as low frequency i.e. known as band-pass filter.​

Calculation:

At ω = 0, circuit is behave like:

V_P = V_N = \({1k\over(1k + 1k)}\cdot V_{in}\)

VP = VN = \(V_{in}\over 2\)

Apply KCL at node V_N :

\({V_{N} -V_{in}\over 1k} + {V_{N} - V_{out}\over 1k} = 0\)

2V_N - V_in = V_out

V_out  = 0

At ω = ∞, 

Circuit behave like :

V_P = \(V_{in}\over2\)

V_P = V_N = V_out

V_out = \(V_{in}\over2\)

On the above circuit, it's block at low frequency and passes at high frequency, so the filter is High-pass filter.

Critical Frequency:

At 3 dB power of the amplifier becomes half of its total power

It is often known as the “Cut-off frequency”, and it is also known as the “Corner frequency” or “break frequency”

It is defined as the frequency at which the ratio of output to input has a magnitude of 0.707 of the maximum amplitude.

When converted in decibels it is equal to – 3 dB.

\(Magnitude = 20\log \left( {\frac{{output}}{{input}}} \right)\)

The cut-off frequency is a characteristic of the filtering devices, such as RC circuits.

After this cut-off frequency point, the amount of attenuation due to the filter begins to increase rapidly.

Frequency Response of Band Reject filter is given by:

Now above magnitude response can be obtained by the following scheme:

The combination of the frequency response of LPF with frequency f1 with HPF with frequency f2, for  f2 > f1 will be:

Concept

The Q factor for a series RLC circuit is given by:

\(QF={1\over R_o}\sqrt{L_o\over C_o}\)

Calculation

Given, R0 = 50 Ω, L0 = 1 mH, and C0 = 10 nF

\(QF={1\over 50}\sqrt{1\times 10^{-3}\over 10\times 10^{-9}}\)

QF = 6.32

1) LC filter Ripple factor:

For FWR with LC filter, Ripple is given by:

\(r = \frac{{\sqrt 3 }}{3}\frac{{\left| {{X_C}} \right|}}{{\left| {{X_L}} \right|\;}}\)

So, the Ripple factor is independent of load resistance R_L and load current I_L.

Important Point:

1) Ripple factor for L filter:

\(r = \frac{{{R_L}}}{{3\sqrt 2 {W_o}L}} \to r \propto {R_L}\)

R_L = load Resistance

2) Ripple factor for C filter:

\( \Rightarrow r = \frac{1}{{2\sqrt 3 {f_o}c{R_L}}} \to HWR\)

f₀ = supply frequency

\( \Rightarrow r = \frac{1}{{4\sqrt {3} {f_o}c{R_L}}} \to FWR\)

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.

Choke input filter circuit:

• The choke filter consists of an inductor connected in series with the rectifier output circuit and a capacitor connected in parallel with the load resistor. 
• The output pulsating DC voltage from a rectifier circuit passes through the inductor or choke coil.
• It is also called an L-section filter because the inductor and capacitor are connected in the shape of an inverted L. 
• The inductor has low DC resistance and extremely high AC reactance. 
• Thus, ripples get filtered through the choke coil. Some of the residual ripples if present in the filtered signal from the inductor coil will get bypassed through the capacitor. 

Frequency Response of Band Reject filter is given by:

Now above magnitude response can be obtained by the following scheme:

The combination of the frequency response of LPF with frequency f₁ with HPF with frequency f₂, for  f₂ > f₁ will be:

Explanation:

Low pass filter

The low pass filter only allows low-frequency signals from 0Hz to its cut-off frequency, ƒc point to pass while blocking those any higher.

This is designed with passive components in two different structures.

RC Low pass filter

RL Low pass filter

The ideal frequency response of the LPF is:

The frequency-domain analysis of the low pass filter is shown below and this is for the first order.

If the system functions contain the one pole then the change in slope will be 20 dB/decade, but in the given question the gain decreases at the rate of 60dB/decade so it means it contains the order three and poles also will be three.

Critical Frequency:

It is often known as “Cut-off frequency”, and it is also known as the “Corner frequency” or “break frequency”

It is defined as the frequency at which the ratio of output to input has a magnitude of 0.707 of the maximum amplitude.

When converted in decibels it is equal to – 3 dB.

\(Magnitude = 20\log \left( {\frac{{output}}{{input}}} \right)\)

The cut-off frequency is a characteristic of the filtering devices, such as RC circuits.

After this cut-off frequency point, the amount of attenuation due to the filter begins to increase rapidly.

The below figure shows the cut-off frequency of a filter.

∴ At the – 3 dB point, the response drops from the passband.

An active filter is a type of analog circuit implementing an electronic filter using active components, typically an amplifier. Amplifiers included in a filter design can be used to improve the cost, performance and predictability of a filter.

Types of active filters. 

• Butterworth 
• Chebyshev 
• Bessel
• Elliptic filters.

Butterworth Filter:

• This filter is also called as maximally flat or flat flat filter. This class of filters approximates the ideal filtted in the pass band. 
• The Butterworth filter has an essentially flat amplitude-fre­quency response upto the cut­off frequency. 
• Butterworth filters has the sharpest attenuation, their phase-shift as a function of frequency is non-linear.
• It has a monotonic drop in gain with frequency in the cut-off region and a maximally flat response below cut-off frequency.
• The Butterworth filter has characteristic somewhere be­tween those of Chebyshev and Bessel filters.
• It has a moderate roll-off of the skirt and a slightly non­linear phase responses.

Chebyshev Filter:

• It is also called a equal rip­ple filter.
• It gives a sharper cut-off than Butterworth filter in the passband.
• Both Butterworth and Chebyshev filters exhibit large phase shifts near the cut-off frequency.
• A drawback of the Chebyshev fil­ter is the appearance of gain maxima and minima below the cut-off frequency.
• This gain ripple, ex­pressed in db, is an adjustable parameter in filter design.
• A Chebyshev filter is used where very sharp roll-off is required. However, this is achieved at the expense of a gain ripple in the lower frequency passband.

Bessel Filter:

• The Bessel filter provides ideal phase characteristics with an approximately linear phase re­sponse upto nearly cut-off frequency.
• Though it has a very linear phase response but a fairly gentle skirt slope.
• For applications where the phase characteristic is important, the Bessel filter is used.
• It is a minimal phase shift filter even though its cut-off characteristics are not very sharp.
• It is well suited for pulse applica­tions.

Elliptic Filter:

• This filter has the sharpest roll-off of all filters in the transition region but has ripples in both the pass band and stop band regions.
• The elliptic filter can be designed to have very high attenuation for certain frequencies in the stop band, which reduces the attenuation for other frequencies in the stop band.

Waveguide:

• A waveguide is an electromagnetic feed line that guides waves, such as electromagnetic waves or sound, with minimal loss of energy by restricting the transmission of energy to one direction.
• Without the physical constraint of a waveguide, wave intensities decrease according to the inverse square law as they expand into three-dimensional space.
• Waveguides only allow frequencies above the cut-off frequency and do not pass below the cut-off frequency. Hence it acts as a high pass filter.

The frequency response of the waveguide is:

where, ω_c is the cut-off frequency

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:

For a wide band pass filter if f_L and f_H are the lower and higher cut-off frequencies respectively then

Center frequency (f_R) = √(f_L × f_H)

Bandwidth (BW) = (f_H - f_L)

Quality factor (Q) = f_R/ BW 

Calculation:

Given; f_L = 200 Hz , f_H = 1KHz 

Center frequency (fR) = √(200 × 1000) = 447.2 Hz

Bandwidth (BW) = (1000 - 200) = 800 Hz

Quality factor (Q) = 447.2/ 800 = 0.559