**Band Pass Filters**

The cut-off frequency or ƒc point in a simple RC passive filter can be accurately controlled using just a single resistor in series with a non-polarized capacitor, and depending upon which way around they are connected, we have seen that either a Low Pass or a High Pass filter is obtained.

One simple use for these types of passive filters is in audio amplifier applications or circuits such as in loudspeaker crossover filters or pre-amplifier tone controls. Sometimes it is necessary to only pass a certain range of frequencies that do not begin at 0Hz, (DC) or end at some upper high frequency point but are within a certain range or band of frequencies, either narrow or wide.

By connecting or “cascading” together a single **Low Pass Filter** circuit with a **High Pass Filter** circuit, we can produce another type of passive RC filter that passes a selected range or “band” of frequencies that can be either narrow or wide while attenuating all those outside of this range. This new type of passive filter arrangement produces a frequency selective filter known commonly as a **Band Pass Filter** or **BPF** for short.

Unlike a low pass filter that only pass signals of a low frequency range or a high pass filter which pass signals of a higher frequency range, a **Band Pass Filters** passes signals within a certain “band” or “spread” of frequencies without distorting the input signal or introducing extra noise. This band of frequencies can be any width and is commonly known as the filters **Bandwidth**.

Bandwidth is commonly defined as the frequency range that exists between two specified frequency cut-off points ( ƒc ), that are 3dB below the maximum centre or resonant peak while attenuating or weakening the others outside of these two points.

Then for widely spread frequencies, we can simply define the term “bandwidth”, BW as being the difference between the lower cut-off frequency ( ƒc_{LOWER} ) and the higher cut-off frequency ( ƒc_{HIGHER} ) points. In other words, BW = ƒ_{H} – ƒ_{L}. Clearly for a pass band filter to function correctly, the cut-off frequency of the low pass filter must be higher than the cut-off frequency for the high pass filter.

The “ideal” **Band Pass Filter** can also be used to isolate or filter out certain frequencies that lie within a particular band of frequencies, for example, noise cancellation. Band pass filters are known generally as second-order filters, (two-pole) because they have “two” reactive component, the capacitors, within their circuit design. One capacitor in the low pass circuit and another capacitor in the high pass circuit.

The **Bode Plot** or frequency response curve above shows the characteristics of the band pass filter. Here the signal is attenuated at low frequencies with the output increasing at a slope of +20dB/Decade (6dB/Octave) until the frequency reaches the “lower cut-off” point ƒ_{L}. At this frequency the output voltage is again 1/√2 = 70.7% of the input signal value or **-3dB** (20 log (Vout/Vin)) of the input.

The output continues at maximum gain until it reaches the “upper cut-off” point ƒ_{H} where the output decreases at a rate of -20dB/Decade (6dB/Octave) attenuating any high frequency signals. The point of maximum output gain is generally the geometric mean of the two -3dB value between the lower and upper cut-off points and is called the “Centre Frequency” or “Resonant Peak” value ƒr. This geometric mean value is calculated as being ƒr^{ 2} = ƒ_{(UPPER)} x ƒ_{(LOWER)}.

A band pass filter is regarded as a second-order (two-pole) type filter because it has “two” reactive components within its circuit structure, then the phase angle will be twice that of the previously seen first-order filters, ie, **180 ^{o}**. The phase angle of the output signal

The upper and lower cut-off frequency points for a band pass filter can be found using the same formula as that for both the low and high pass filters, For example.

Then clearly, the width of the pass band of the filter can be controlled by the positioning of the two cut-off frequency points of the two filters.

A second-order **band pass filter** is to be constructed using RC components that will only allow a range of frequencies to pass above 1kHz (1,000Hz) and below 30kHz (30,000Hz). Assuming that both the resistors have values of 10kΩ´s, calculate the values of the two capacitors required.

The value of the capacitor C1 required to give a cut-off frequency ƒ_{L} of 1kHz with a resistor value of 10kΩ is calculated as:

Then, the values of R1 and C1 required for the high pass stage to give a cut-off frequency of 1.0kHz are: R1 = 10kΩ´s and C1 = 15nF.

The value of the capacitor C2 required to give a cut-off frequency ƒ_{H} of 30kHz with a resistor value of 10kΩ is calculated as:

Then, the values of R2 and C2 required for the low pass stage to give a cut-off frequency of 30kHz are, R = 10kΩ´s and C = 510pF. However, the nearest preferred value of the calculated capacitor value of 510pF is 560pF so this is used instead.

With the values of both the resistances R1 and R2 given as 10kΩ, and the two values of the capacitors C1 and C2 found for both the high pass and low pass filters as 15nF and 560pF respectively, then the circuit for our simple passive **Band Pass Filter** is given as.

We can also calculate the “Resonant” or “Centre Frequency” (ƒr) point of the band pass filter were the output gain is at its maximum or peak value. This peak value is not the arithmetic average of the upper and lower -3dB cut-off points as you might expect but is in fact the “geometric” or mean value. This geometric mean value is calculated as being ƒr^{ 2} = ƒc_{(UPPER)} x ƒc_{(LOWER)} for example:

- Where, ƒ
_{r}is the resonant or centre frequency - ƒ
_{L}is the lower -3dB cut-off frequency point - ƒ
_{H}is the upper -3db cut-off frequency point

and in our simple example above, the calculated cut-off frequencies were found to be ƒ_{L} = 1,060 Hz and ƒ_{H} = 28,420 Hz using the filter values.

Then by substituting these values into the above equation gives a central resonant frequency of:

A simple passive **Band Pass Filter** can be made by cascading together a single **Low Pass Filter** with a **High Pass Filter**. The frequency range, in Hertz, between the lower and upper -3dB cut-off points of the RC combination is know as the filters “Bandwidth”.

The width or frequency range of the filters bandwidth can be very small and selective, or very wide and non-selective depending upon the values of R and C used.

The centre or resonant frequency point is the geometric mean of the lower and upper cut-off points. At this centre frequency the output signal is at its maximum and the phase shift of the output signal is the same as the input signal.

The amplitude of the output signal from a band pass filter or any passive RC filter for that matter, will always be less than that of the input signal. In other words a passive filter is also an attenuator giving a voltage gain of less than 1 (Unity). To provide an output signal with a voltage gain greater than unity, some form of amplification is required within the design of the circuit.

A **Passive Band Pass Filter** is classed as a second-order type filter because it has two reactive components within its design, the capacitors. It is made up from two single RC filter circuits that are each first-order filters themselves.

If more filters are cascaded together the resulting circuit will be known as an “n^{th}-order” filter where the “n” stands for the number of individual reactive components and therefore poles within the filter circuit. For example, filters can be a 2^{nd}-order, 4^{th}-order, 10^{th}-order, etc.

The higher the filters order the steeper will be the slope at n times -20dB/decade. However, a single capacitor value made by combining together two or more individual capacitors is still one capacitor.

Our example above shows the output frequency response curve for an “ideal” band pass filter with constant gain in the pass band and zero gain in the stop bands. In practice the frequency response of this Band Pass Filter circuit would not be the same as the input reactance of the high pass circuit would affect the frequency response of the low pass circuit (components connected in series or parallel) and vice versa. One way of overcoming this would be to provide some form of electrical isolation between the two filter circuits as shown below.

One way of combining amplification and filtering into the same circuit would be to use an Operational Amplifier or Op-amp, and examples of these are given in the Operational Amplifier section. In the next tutorial we will look at filter circuits which use an operational amplifier within their design to not only to introduce gain but provide isolation between stages. These types of filter arrangements are generally known as **Active Filters**.

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what about the applications of the three filters

it was nice

The freq. response of your BPF will be dependent on the source and load impedances, as well as the freqs., Rs and the Cs. You have not taken those impedances into account, so your analysis is not correct.

Please help on how to estimate the rate of attenuation of a band pass filter

Dave legend

would you help me to design no insertion loss band pass filter which has fc=10KHz and fh=2MHz

i need a project based on second order band pass filter, can any one help me out in this

Yaa,I think i can help you out.

Basicslly,I am a professional in making project related to electrical engineering.

You can mail me on my ID for some guidance on how to make one

can you tell me how to calculate the resonant frequency it would be amazing if you could give me the formula

series resonance is given as: Series Resonance

Q is fc/Bandwidth

what do you need?

I will use this lab for my physics 2e class 🙂

So, We have tested the band pass filter circuit as shown in the first figure, we were baffled as to why this was organised the way it’s shown, c1 and r1 in series. Please correct me if i am wrong but is there a mistake there? The second to last circuit works as expected, but the first doesn’t. Why has this (first circuit) been drawn as shown? Was this a mistake?

Cheers

Dave

Hello Dave, It makes no difference, because with regards to frequency both circuits are exactly the same, one reactance in series with the input signal and one reactance in parallel with the input signal. There is still a high-pass and low-pass connected together, its just that the components are arranged differently. The only difference would be in the magnitude of the output, set up by the different resistive divider network combinations but the center frequency will still be the same. But I see your point that it may be confusing to some, so I have redrawn it to make them the same and remove any confusion.

Hello, I am in my second circuits class for EE and we are just now covering filters and transfer functions. In the first figure, where the first filter is in series and the second is parallel, can you explain to me how to calculate the center frequency? I started by turning C into Z. Then combined the first series of inpedance and called it Z1. Then combined the parallel impedance and called it Z2. This gave me a nice design to use voltage divider to get Vout/Vin (which is transfer function). In my book there is a chart that explains for a Bandpass filter H(w)=1. Would I take the magnitude of the transfer function that I can up with and set it equal to 1? And then solve for w? This gives me multiple complex answers.

Hello Clay, as explained above under the heading of “Centre Frequency Equation”, the BPF has a lower corner point and an upper corner point so the filters center frequency is the mean of the two.

Nice job,but In your calculation of the capacitors,you used values of resistance of resistor in place of fragrance,why?

Sorry, I meant reactance and not fragrance

Because we are finding the value of capacitance for the required corner frequency with a given resistance, as fc = 1/(2piRC). Capacitive reactance, Xc is not needed here.