Synopsis of Frequency Response & Filter Circuits.

Hello Everyone, hope that you all are doing good in your lives. As a matter of fact, the response of a linear AC circuit is examine when excited with an AC signal of constant amplitude but varying frequency hence the term frequency response. These signals are common in everyday applications such as radio, television, and telephone can be studied using frequency response. However, in addition, we are going to study types of filter circuits their introduction and their output response is observe.

Frequency Response

However to understand the concept of frequency response. Consider a linear circuit with input and output voltage signals represented by Vi and Vo respectively as below.Two Port Linear Network

So if the input amplitude Vi is constant while its frequency is varying, then it is observe that the amplitude and the phase of the output signal(Vo) also changes. However, the input and output frequencies remain unchange. This means that Vo and ϕ become a function of frequency(ω) represented by Vo(ω) and ϕ(ω) with Vi as reference.

The ratio of the output to input voltage signal is denoted by H(jω). Which is a complex function and can be express as

H(jw) = Vo/Vi

However in polar form;

H(jw) = A∠ϕ

Whenever the voltage gain A = |H(jω)| and ϕ = arg H(jω) are both functions of frequency, ω.

Amplitude Response:- The variation observed in the amplitude gain, A with respect to w is certainly the amplitude response of the network.

Phase Response:- The variation observed in the phase, ϕ of the network with respect to w is certainly the phasor response of the network.

However, the amplitude and the phase output together define the frequency response of the network.

Next, we are going to get introduce the concept of filters and their response to an input signal with changing frequency. Besides this will also help further to clarify the concept of frequency response analysis.

Filter Circuits

Filters form an important part in electrical network special where particularly studying about frequency responses and its analysis. For instance, a radio station which is broadcasting a transmission at a frequency of 100 MHz. This means that we need to design a receiver i.e. a receiving system such that it obstructs all frequencies other than 100 MHz frequency. An ideal filter will attenuate all signal frequencies less than and greater than 100 MHz. Thus providing the best channel in short the best sound quality ever possible.

The Following are the headline or say the sections of the most commonly used filter nowadays. Apart from these, we are also going to focus on their respective frequency response.

Types of Filter Circuits

Low Pass Filter Circuits

A low pass filter circuit permits the low-frequency signals to flow through the circuit. However, in these circuits, the high-frequency signals are usually block or restricted. The figure below shows the construction of a low pass filter using a simple RC network. The output voltage, Vo is certainly taken across the capacitor.Construction of low Pass Filter using RC and RL network

Note that a low pass filter can also be constructed easily by using an RL network as shown above.

However here we are going to only go through the analysis of the RC network.

Analysis using RC network

The reactance of the capacitor is 1/jwC which will be in use for the analysis below. To find Vo, the voltage divider rule can be applied as the R and C both are in series with each other. Therefore;

    \[V_O = V_i  \frac{ \frac {1} {jwC} }{R + \frac {1} {jwC} } \end.\]

    \[V_O = V_i  \frac{1} {1 + jwRC} \end.\]

assume T = RC be the time constant, then

    \[H(jw) = \frac {V_O} {V_i}  = \frac{1} {1 + jwT} \end.\]

However at low frequencies, w ≈ 0. ∴ Vo ≈ Vi i.e. Vi is present at the output across the capacitor. When w → ∞, Vo → 0, hence output voltage amplitude is very small of Vi appear at Vo at high frequencies. Therefore these circuits act as a low pass filter. From the above equation the amplitude response equation of a low pass can be obtain as;

    \[A = |H(jw)| = \frac{1} {(1 + (jwT)^2)^\frac {1} {2}} \end.\]

and the phase is given by;

ϕ = arg(H(jw)) = tan-1(-wT)

For w ≈ 0, A ≈ 1 and ϕ ≈ 0

For w ≈ ∞, A ≈ 0 and ϕ ≈ -π/2

Lastly for w ≈ 1/T, A ≈ 1/2 = 0.7071 and ϕ ≈ -π/4 = -45°

These response plots are shown below;amplitude and phase response


The frequency w = 1/T is the cutoff frequency or bandwidth of the low pass filter. It is defined as the range of frequency for which the amplitude A ≥ 1/√2. However numerically, the bandwidth of an RC low pass filter is,

    \[Bandwidth = \frac{1} {T} = \frac {1} {RC}} \end.\]

Clearly, the bandwidth is user and problem-dependent and can be adjusted simply by turning the R and C parameters. So it is more common to express the gain, A of a circuit in Decibels (dB) units by;

    \[A_d_B = 20log_1_0A \end.\]

However which was originally developed as the logarithmic unit of power ratio i.e.

    \[dB Power Gain = 10log_1_0 \frac {P_o} {P_i} \end.\]

However, the amplitude plot shown above is redrawn below with gain converted to dB versus the logarithm of the frequency. But the phase response remains unchange.

Some of the important observation made by comparison is mention below;

  1. Zero dB corresponds to a unity gain i.e. A = 1.
  2. Negative dB corresponds to attenuation i.e. A < 1.
  3. Positive dB corresponds to amplification i.e. A > 1amplitude and phase response of a low pass filter circuits in dB

Besides the bandwidth of a low pass filter is the frequency range for which the gain, A ≥ -3dB. The = -3dB point is called the half-power point since,

    \[10log_1_0 \frac {P_o} {P_i} = 10log_1_0 \frac {V_o^2} {V_i^2} = 10log_1_0 (A)^2\end.\]

    \[= 10log_1_0 (\\frac{1} {2}^\frac{1}{2})^2\end.\]

    \[= 10log_1_0 \frac {1} {2} = -3 dB \end.\]

  1. Sources where low frequency signals are require for the working of circuit i.e. Oscillators.
  2. development of power lines.
  3. It is also used in smoothing out the ripple which is present in the output waveform of the rectifier.

High Pass Filter Circuits

A high pass filter, as the name suggests, allows high frequencies to pass through the circuit while low frequencies are attenuated. The cut-off point or bandwidth theory is the same as in that of the low pass filter.

In practice, the same circuit used for the low pass filter can be adopted as a high pass filter with the output taken across the resistor this time. The resulting figure is;High pass filter circuits using RC and RL networks

As before, the Voltage divider rule can be use to determine the output voltage across the resistor.

    \[V_o = V_i \frac {R} {R + \frac {1} {jwC}}\end.\]

    \[V_o = V_i \frac {1} {1 + \frac {1} {jwRC}}\end.\]

Clearly, when w ≈ 0, Vo ≈ 0, and for w → ∞, Vo → Vi. Thus at high frequencies, most of the input voltage appears at the output creating this circuit as a high pass filter.

The figures below are the frequency response showing low gain at low frequencies with gain approaching 1 for high frequencies. The phase plot varies from 90° to 0° with the cutoff frequency, ꞷ = 1/T, crossing at 45°Amplitude and phase response of High pass filter circuits

  1. Audio system.
  2. Image processing.

Band pass Filter Circuits

A bandpass filter allows a particular band of frequencies to pass through the network which is adjustable by the designer. It is simply a combination of a low pass and a high pass filter. If the circuit diagrams mention above for low pass filter and high pass filter are join in cascade, a bandpass filter can be found. In this case, there are two resistors and two capacitors whose values are allow to change so that the desired band can be tune.Schematic of a Band pass filter

The response for amplitude and phase of a bandpass filter is shown below. These figures however highlight the two cutoff frequencies w1 and w2. All frequencies within this range (i.e. w1 and w2 ) in an input signal will be allow to pass unchange through the circuit. However, frequencies outside these limits are blocked.Amplitude and phase response of Band pass filter circuit

  1. use in Optics.
  2. telephone services.


Subsequently, here we are at the end of the blog. I hope that this article helped you to enhance your knowledge and get all of your doubts clear. If you like the content then please do share it with others and also mention the part which you like the most. Besides if still have any doubt then feel free to ask it down below in the comment section. And please do suggest the topic which you will like to read next.

Have a nice day 🙂


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  • Harshita Singh April 18, 2021 at 5:41 pm Reply


  • Yash April 19, 2021 at 4:18 am Reply

    Great content😁

  • Aayushi Singh April 20, 2021 at 8:57 am Reply

    Nice work

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