Department of Engineering Science and Physics
College of Staten Island
Filters are an integral part of a communication system. Filters enable us to keep the signal with useful information and discard the signal with irrelevant information or noise. There are four main categories of filters:
(1) Low-pass filters (LPF): This kind of filter allows the low frequency signal to pass through the circuit and blocks the high frequency signal.
(2) High-pass filter (HPF): This kind of filter allows the high frequency signal to pass through the circuit and blocks the low frequency signal.
(3) Band-pass filter (BPF): This kind of filter allows the signal in a specific frequency range to pass through the circuit and blocks the signal with all other frequencies.
(4) Band-stop filter (BSF): This kind of filter blocks the signal in a specific frequency range and allows the signal with all other frequencies to pass through the circuits.
The range of frequencies that can be passed through the circuit (or blocked) is specified by a cutoff frequency (also known as the break frequency of half-power frequency). The output voltage at the cutoff frequency is reduced to 70.7% of its maximum value (which corresponds to a reduction of output power by 50%). In the case of LPF, the desired frequencies are less than the cutoff frequency. In the case of
HPF, the desired frequencies are above the cutoff frequency. The BPF and BSF filters are, however, are defined by two cutoff frequencies: (1) lower cutoff frequency and (2) upper cutoff frequency. In the case of BPF, the desired frequencies are between the lower and upper cutoff frequencies. In the case of BSF, the desired frequencies are below the low –cutoff frequency or above the upper cutoff frequency.
The response of a filter is measured either as the voltage gain or power gain and is defined using a logarithmic scale. The logarithmic unit of gain used is called decibels (dB). Mathematically,
𝐺𝑎𝑖𝑛 = 20 log (
𝑉 𝑜
𝑉 𝑖
) = 10 log (
𝑃
𝑃 𝑖 𝑜
) 𝑑𝐵 (1) where,
V
O
= output voltage
V i
= input voltage
P
O
= output power
P i
= input power
In this set of experiments you’ll learn to design a filter with a specific frequency response. The design is four-step process.
Step 1: Determine the pole(s) and/or zero(s) from the given frequency response
Step 2: Write the transfer function of the filter
Step 3: Determine the circuit configuration of the transfer function in Step 2
Syed A. Rizvi ENS 466/ELT 466

Department of Engineering Science and Physics
College of Staten Island
Step 4: Magnitude scaling if needed: R new
= K m
R old
and C new
= (1/K m
) C old
.
There are four commonly used circuit configurations as shown in Fig. 1.
Figure 1: Four basic circuit configurations and their impedances (in “s” domain).
A pole causes a decrease in gain of the filter at the rate of 20 dB/decade. A zero, on the other hand, causes the gain to increase at the rate of 20 dB/decade. Let’s consider the design of a low-pass filter with a cutoff frequency of 1592 Hz (see Fig. 2).
Step 1: Location of poles and zeros
The break frequency in Hz is first converted to radians/second, which, in this case, is approximately
10,000 radians/second (use ω = 2πf). Since the gain decreases at the rate of 20 dB/decade after the cutoff frequency, it indicates that this filter has a single pole at s = -10,000 and no zeros.
Step 2: Writing the transfer function
With no zeros and a single pole at s = -10,000, we can the transfer function as follows:
𝐾
𝑇(𝑠) = 𝑠 + 10000
= 𝑠
𝐾 +
1
10000
𝐾
(2)
Step 3: Circuit configuration
We’ll use op-amp in inverting amplifier configuration (see Fig. 3). The transfer function of the op-amp circuit is given by
Syed A. Rizvi ENS 466/ELT 466

Department of Engineering Science and Physics
College of Staten Island
𝑇(𝑠) =
𝑍 𝑓
𝑍 𝑖
(3)
Comparing Eqs. (2) and (3) yields and
𝑍 𝑓
= 𝑠
𝐾 +
1
10000
𝐾
𝑍 𝑖
= 1.
(4)
(5)
Figure 3: Frequency response of a low-pass filter.
Comparing Z f
and Z i
with the circuit configurations in Fig. 1 we find that Z f
and Zi can be represented by the circuits 1 and 4, respectively. Using these circuits for Z f
and Z i
yields
Ri = 1
Syed A. Rizvi ENS 466/ELT 466

Department of Engineering Science and Physics
College of Staten Island
C f
= 1/K
1/R f
= 10000/K
Choosing R f
= 1 gives us K = 10000 and C f
= 10
-4
F. Since 1 Ω is not a suitable value for this filter (it will draw a large current) let’s choose a scaling factor K m
= 1000. That gives us R i
= R f
= 1k Ω and C f
= 10
-7
F
(0.1 µF).
Figure 4: Inverting amplifier configuration.
Assignment 4.1: Design a low-pass filter with a cut-off frequency of 3.4 KHz with the gain decreasing at the rate of 40 dB/decade after the cut-off frequency. Compute and plot the frequency response.
Assignment 4.2: Implement the following transfer function:
𝑇(𝑠) =
𝐾 (𝑠 + 6280)
(𝑠 + 62800)(𝑠 + 87920)
Compute and plot the frequency response.
(6)
Syed A. Rizvi ENS 466/ELT 466