MATH/MTHE 225 - Single Frequency AM Receiver Design Project

This project focuses on the design of a single-frequency AM receiver using a series RLC circuit, modeled by a non-homogeneous linear differential equation. The goal is to maximize signal amplitude at 1200 kHz while minimizing interference from other commercial radio stations. The assignment involves selecting appropriate values for inductance (L), resistance (R), and capacitance (C) to achieve the desired resonant frequency and bandwidth, using provided component values. The student calculates LC values, evaluates design options, and analyzes frequency responses to identify the optimal component combinations that minimize signal interference and create a bandpass filter. The solution presents a detailed comparison of different design options and their performance in rejecting unwanted frequencies, using calculations of bandwidth and Q-factor. The project aims to provide a practical application of differential equations in electrical engineering, demonstrating the relationship between circuit components and frequency response.

Differential Equations
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Contents
Introduction...........................................................................................................................................3
Goal...................................................................................................................................................3
Problem Statement...............................................................................................................................3
Components used ( with reason )..........................................................................................................5
Evaluation of possible design options....................................................................................................5
Similar Result design options.................................................................................................................7
Comparison of options..........................................................................................................................8
Future designs.....................................................................................................................................11
Conclusion...........................................................................................................................................11
References...........................................................................................................................................12

Introduction
Goal
We have to choose components to construct an AM receiver that picks up
broadcasts on a single frequency with minimum interference.
Problem Statement
We can construct a simple AM radio receiver as a series RLC circuit. We can model
the circuit using a differential equation as a non homogeneous linear differential
equation with constant coefficients as follows:
L d2q/dt2 + R dq/dt + q/C = Vext
Here, L = Inductance
R = Resistance
C = Capacitance
. q = Charge
Vext = voltage of signal received in the circuit by an antenna.
i(t) = dq/dt
We obtain the steady state vibration only from 1 part of the equation’s general
solution which is Vext.
If a radio wave passes by an antenna, it creates an oscillating voltage (Vext) which
leads to the induction of an alternating current in the circuit.
Resonant frequency is the frequency at which maximum amplitude of the steady
state current is obtained. The audio data which we hear in radio is sent on the
signal. This resonant frequency is obtained for the fixed values of R, L and C [3].
We need to produce single frequency AM receivers for a network.

The purpose is to maximize the signal amplitude received from the base stations
which are broadcasting at a frequency of 1200 kHz.
We must change this frequency to the unit of rad/s. 1200 kHz = 2π x 1200 x 1000
rad / s
.ω = 7539.8 x 1000 rad/s
Let us assume Vext = sin ωt
 L d2i/dt2 + R di/dt + i/C = sin ωt
d2I/dt2 + R/L di/dt + 1/LC i = (1/L) sin ωt
This is the required differential equation.
By solving this equation, find the relationship between Current (I) and time (t).
Solution:
Complimentary function:
D2 + R/L D + 1/LC = 0
D = α , β
I = A eαt + B eβt
Particular Integral :
I = 1 / D2 + R/L D + 1/LC ((1/L) sin ωt )
Put D2 = - ω2
Another need is to minimise the signal amplitude received from other commercial
radio stations broadcasting across the region. Their signals are in the frequency
range of 600 – 1150 kHz and 1220 – 1600 kHz.
. fl1 = 600 + 1150 / 2 = 875 kHz
fl2 = 1220 + 1600 / 2 = 1410 kHz

Components used ( with reason )
The parts available are as follows:
R = 1 Ώ
L = 0.1 μH, 0.47 μH, 1 μH, 4.7 μH, 10 μH, 10 μH, 47 μH, 100 μH, 470 μH
C = 100 pF, 330 pF, 680 pF, 1 nF, 3.3 nF, 6.8 nF, 10 nF, 33 nF, 68 nF, 100
nF.
Maximum amplitude of signal is obtained at a frequency of 1200 kHz. Hence,
the resonant frequency is 1200 kHz.
fr = 1200 kHz
1/√LC = 2π x 1200 x 1000 = 7539.8 x 1000
√LC = 0.133 x 10-6
LC = 1.77 x 10-14 = 17.7 x 10-15
So, we can choose those combinations of the values of L and C which give a
value closest to above equation. We find the value of LC for all the
combinations of L and C possible from the given values. The calculated
values have been shown in Table 1.
Evaluation of possible design options

Table 1. Values of LC ( in unit 10-15 HF)
L in uH
0.1 0.47 1 4.7 10 47 100 470
C in nF
0.1 0.01 0.047 0.1 0.47 1 4.7 10 47
0.33 0.033 0.155 0.33 1.55 3.3 15.5 33 155.1
0.68 0.068 0.3196 0.68 3.196 6.8 31.96 68 319.6
1 0.1 0.47 1 4.7 10 47 100 470
3.3 0.33 1.55 3.3 15.5 33 155.1 330 1551
6.8 0.68 3.196 6.8 31.96 68 319.6 680 3196
10 1 4.7 10 47 100 470 1000 4700
33 3.3 15.51 33 155.1 330 1551 3300 15510
68 6.8 31.96 68 319.6 680 3196 6800 31960
100 10 47 100 470 1000 4700 10000 47000
LC = 17.7 x 10-15 (calculated value)
For the given equation as follows:
 L d2i/dt2 + R di/dt + i/C = sin ωt
We define the resonant frequency as the frequency at which we obtain the
maximum value of current. The value of impedance offered by the circuit is
minimum at the resonant frequency.
In case of a series RLC circuit, at a frequency ω, the impedance offred by the
circuit may be defined as follows:
Z = √ R2 + ( ωL – 1 / ωC)2
We find that the value of Z is minimum when Z = R or ωL – 1 / ωC = 0
This gives us ω = 1 / √LC
But from the given table, we can see that the value of the frequency obtained
by calculation does not match with any of the values exactly. Hence, we
choose a value of frequency which is nearest to the value of the given
frequency.

Ideally, the value of Impedance offered by the circuit must be R. So, in this
case, ideally the value of impedance at resonant frequency must be R ,ie, 1 Ώ
in this case.
Similar Result design options
From the table, we observe that the nearest value to the calculated value of
LC is 15.51 x 10-15. This value can be found at 3 place in the table formed.
The 3 cases are as follows:
Case 1 : L = 47 μH, C = 0.33 nF
Case 2 : L = 4.7 μH, C = 3.3 nF
Case 3 : L = 0.47 μH, C = 33 nF
In all these 3 cases the resonant frequency obtained is nearest to the value of the
calculated value of resonant frequency.
Our first purpose in the designing of the circuit was : to maximize the signal
amplitude received from the base stations which are broadcasting at a frequency of
1200 kHz. This need is met in all the three cases mentioned above.
All these 3 cases give the frequency as approximately equal to 1200 kHz. This
maximizes the signal amplitude received from the base stations. At this particular
frequency, the impedance offered by the series RLC circuit is the minimum. Also, as
the impedance offered is minimum, the current flowing is maximum at a given
voltage [2]. As the current is showing maximum amplitude at this frequency, we can
say that it maximizes the signal amplitude.

Comparison of options
Another concern in the design of the circuit is to minimise the signal amplitude
received from other commercial radio stations broadcasting across the region. Their
signals are in the frequency range of 600 – 1150 kHz and 1220 – 1600 kHz. It
means that when we are receiving the signal at 12 kHz, then at the same time, other
radio stations are also broadcasting at some other frequencies. Our receiver must
not receive these frequencies. In other words, it must reject these frequencies. This
is done so that any 2 frequencies do not interfere with each other [1]. If 2 or more
frequencies are received by the receiver, then there is a phenomenon called
interference. Due to the interference of 2 or more signals, there is a distortion in the
output signal. This affects the quality of the signal which we are receiving. Hence,
apart from receiving the signal at a frequency of 12 kHz, our circuit must be capable
of rejecting the signals that are in the frequency range of 600 – 1150 kHz and 1220 –
1600 kHz.
In order to implement this condition, we can check all the 3 cases possible. We can
study the frequency response of the series RLC circuit. We can plot a graph between
the current and the frequency. From the graph, it can be observed in the range of
600 – 1150 kHz and 1220 – 1600 kHz whether the current values are low or high.
The current must hold a low value at a frequency in case we want to reject that
frequency. Rejecting a frequency means we are not allowing that frequency to pass
through the circuit. In other words we can say that we are designing a band pass
filter. A band pass filter allows only a set of frequencies to pass through it. It rejects
all other frequencies.
A method to do this is to use a tool such as MATLAB to simulate the circuits with the
given values of R, L and C. We can obtain the frequency response of the 3 circuits.
Using the current versus frequency plots, we can analyse which circuit is appropriate
for us. This can be done by finding out the frequencies at which the amplitude of the
current is high. The frequencies at which current value is high are the ones which
can be readily passed through the circuit. We have to identify the cases in which the
range of frequencies given by 600 – 1150 kHz and 1220 – 1600 kHz show a low
value of current. This shows that at these frequencies, the signal will not be able to

pass through the circuit and these frequencies will be rejected. These values of
frequencies at which the current flowing is less are the frequencies at which
impedance offered by the circuit is also minimum.
We can find the most appropriate circuit out of the 3 cases obtained as follows:
It is given :
R = 1 Ώ , fr = 1200 kHz
. ω = 7539.8 x 1000 rad/s = 7.5398 x 106 rad/s
Table 2. Calculation of Band width
L XL =ωL Q=XL/R BW=fr/Q
In kHz
47 μH 354.37 354.37 3.386
4.7 μH 35.437 35.437 33.86
0.47 μH 3.5437 3.5437 338.6
We calculate the bandwidth for all the 3 cases in Table 2.
For different values of inductance, we have calculated the value of XL ,ie, the
reactance.
XL = ω x L
We have the value of ω = 7539.8 x 1000 rad/s = 7.5398 x 106 rad/s
Hence, XL = 7.5398 x 106 rad/s x L
So, we obtain the various values of XL.
Next, we find the value of Q-factor. Q = XL / R
In this case R = 1 Ώ
Hence, Q = XL / 1 = XL
Next, we calculate the bandwidth given by BW = fr / Q

These values of bandwidth obtained are very important and they concern our design
requirements. Bandwidth defines the difference of the frequencies ( f2 - f1 ) at which
the current amplitude value falls to 0.707 times the peak value of current obtained at
resonant frequency. These 2 frequency points are also called half power points.
BW1 = 3.386 kHz
BW2 = 33.86 kHz
BW3 = 338.6 kHz
The bandwidth is equally divided around the resonant frequency. Hence, we find the
value of BW/2 [4].
We can say f1 = fr – BW/2 and f2 = fr + BW/2
Fr = 1200 kHz
Table 3. Calculation of f1 and f2
BW=fr/Q BW/2 F2 F1
In kHz
3.386 1.693 1201.693 1198.3
33.86 16.93 1216.93 1183.07
338.6 169.3 1369.3 1030.7
The frequencies f1 and f2 define the range of frequencies which can be readily
accepted by the circuit and hence they are not rejected. They are shown in Table 3.
Range for case 1: 1198.3 kHz to 1201.693 kHz
Range for case 2: 1183.07 kHz to 1216.93 kHz
Range for case 3: 1030.7 kHz to 1369.3 kHz
Our requirement was that our circuit must be capable of rejecting the signals that are
in the frequency range of 600 – 1150 kHz and 1220 – 1600 kHz.
Hence, case 3 cannot represent our circuit because it is passing some of the
frequencies which must be rejected. Circuits in the cases 1 and 2 can be considered.
But if a comparison is made between the 2, then we can say that Case 1 is better as
the range of frequencies it is passing is more appropriate and close to our need.
Case 1 has a smaller bandwidth and hence it is more selective to select a single

frequency. Hence, we can say that case 1 is the most appropriate choice’ We can
design a series RLC circuit with R = 1Ώ, L = 47 μH and C = 0.33 nF.
Future designs
Another possible design is that of a super heterodyne receiver. A super heterodyne
receiver is a special kind of receiver which is designed based on the concepts of
image frequency and intermediate frequency. A super heterodyne receiver can also
be used to reject some frequency bands from passing through the circuit.
Apart from this, other devices can also be used to design the required circuits. We
can make use of CMOS technology in the design. The use of digital AM receivers is
also coming into picture with the growing advancement in the field of digital
electronics and embedded systems.
Conclusion
We have chosen the components to construct an AM receiver that picks up
broadcasts on a single frequency with minimum interference successfully. We have
made use of the given components only. We have used the concept of resonant
frequency in order to ensure that only a single frequency passes through the circuit.
Also, we have used the concept of bandwidth in order to determine the best circuit
for prevention of any kind of interference.

References
[1] Mngesh Borage, Sunil Tiwari and S.Kotaiah, “Common- mode Noise Source and its
Passive Cancellation in Full-bridge Resonant Converter,” in Proc. of INCEMIC 2003, 2003,
pp.9-14.
[2] Mitoshi Fujimoto, Taketo Matsuoka, Toshikazu Hori, Takanobu Tabata and Satoshi Hori,
Inverter Noise Canceller for AM radio reception using PI Algorithm, IEICE Communication
Express, Vol., No.6, pp.234-239, Sep. 2012.
[3] Tomohisa Harada, Yoshiyuki Hattori, Shinya Ito, Mitoshi Fujimoto, Toshikazu Hori,
“Noise Suppression System for AM radio Receiver Using Quadrature Component of
Receiving Signal,” SAE Paper No.2016-01-0079, 2016.
[4] K. Henney, Editor, Radio Engineering Handbook, New York :Mc Graw Hill , 1959, pp. 3-
18.