What is AM Transmission and Reception?

Hello everyone, hope so you all are doing good in your lives. Have you ever thought that what is AM modulation, AM transmission, and reception? What is the need for AM transmission and reception? And moreover, how do AM transmission and reception take place? if you are having a hard time tackling these questions then sit back and relax. Because the only thing you have to do is go through the article and all your doubts will get cleared.

Introduction to AM Transmission and Reception

Certainly, the main purpose of a communication system is to send data through a medium or channel separating the transmitter from the receiver. Representation of these data is often in terms of a baseband signal i.e. a signal which extends from 0 to some max frequency. One of the most common families of continuous-wave modulation systems is Amplitude modulation

In AM modulation the magnitude of the carrier signal is varied in accordance to the change in the instantaneous value of the message / data signal.

the carrier use in this method is a Sinusoidal Carrier.

AM Transmission and Reception

Take a sinusoidal carrier wave c(t) define by;

    \[c(t) = Ac Cos[2(pie)fct]\end.\]

Most certainly Ac is the carrier amplitude and the Fc is the carrier frequency. However, to simplify the exposition without affecting the result obtain and the conclusion reached, we have to assume that the phase of the carrier wave is zero. Certainly, the source of carrier wave c(t) is independent of the source responsible for present m(t). An amplitude-modulated wave thus can be described, in its most general form, as a function of time as follows:

    \[S(t) = Ac[ 1 + KaM(t) ] * Cos[2(pie)fct]\end.\]

where Ka is a constant know as Amplitude sensitivity of the modulator responsible for the generation of the modulated signal s(t). Generally, the carrier amplitude Ac and the message signal m(t) are measure in volts. Whereas the amplitude sensitivity is measure in volt-1 .wave form of am transmission and reception

Single Tone Transmission and reception

However, consider a modulating wave m(t) that consists of a single tone or a frequency component that is;

    \[m(t) = Am  Cos[2(pie)fmt]]\end.\]

where Am is that the amplitude of the modulating wave and fm is the frequency of the carrier. The sinusoidal carrier wave has an amplitude Ac and a frequency fc . The corresponding AM wave is therefore as follows;

    \[s(t) = Ac[ 1+uCos[2(3.14)fmt] ] * Cos[2(3.14)fct]\end.\]


μ = KaAm

μ is a dimensionless constant known as, modulation factor, or known as percentage modulation when there is a need to express numerically as a percentage factor. However to avoid envelope distortion the μ should be below unity.

s(t) = Ac[ 1+μCos(2πfmt) ] Cos(2πfct)

let Amax and Amin denote the maximum and the minimum value of the envelope of the modulated wave. Then, as a result, we get;


that is,time domain and frequency domain characteristic of single tone modulation in AM transmission and reception

    \[u=\frac{Amax - Amin}{Amax + Amin}\end.\]

Therefore, showing the product of two cosines as the sum of two sinusoidal waves, one having frequency fc + fm and the other having frequency fc – fm.

    \[s(t)=Ac.Cos[2(pie)fc] + \frac{1}{2}uAc.Cos[2π(fc-fm)t] + \frac{1}{2}uAc.Cos[2π(fc+fm)t ]\end.\]

However, in practice, the AM wave s(t) is a voltage or a current wave. In either case, the average power deliver to 1-ohm resistor by s(t) comprises of three keys:

  • Carrier power = 1/2 Ac2
  • Upper side band = 1/8 μ2Ac2
  • Lower-side band = 1/8 μ2Ac2

Whereas for a load resistance R different from 1 ohm, which is usually the case in practice, the above equations are merely scaled by the factor 1/R or R. This modulation certainly depends on whether the wave m(t) is current voltage.

Limitations of AM transmission and reception

Amplitude modulation is that the oldest method of acting modulation. its biggest character is the ease with which it is generated and reversed. However, its two main limitations are:

  • Amplitude modulation is wasteful of power. the carrier wave of c(t) is completely independent of the information bearing signal of baseband signal m(t). The transmission of the carrier wave therefore, represents a waste of power which means that in am modulation only a fraction of total transmitted power is affected by m(t).
  • Amplitude is a wasteful of bandwidth. The higher and lower sideband of AM wave are uniquely associated to each other by their character of symmetry concerning their carrier frequency. Thus amplitude and phase spectra of either sidebands, we can uniquely find the otherpercentage modulation of am transmission and reception

To overcome these limitations, we must make certain changes, which results in increased system complexity of the amplitude modulation process. In effect, we tend to trade off system complexness for improved utilization of communication resources. Starting with amplitude modulation as the standard, we can distinguish three modified forms of amplitude transmission and reception.

Double Sideband-suppressed Carrier Modulation

However, DSB-SC is a modulation in which the transmitted wave consists of only the upper and the lower sideband. The power transmitted is saved through the suppression of the carrier wave, however, the channel width demand is the same as before.

Basically, DSB-SC modulation consists of the product of the message signal m(t) and the carrier wave c(t) as follows;

    \[s(t) = c(t)m(t)\end.\]

    \[      = Ac Cos[2(pie)fct] m(t)\end.\]

Consequently, the modulated signal s(t) undergoes a phase reversal whenever the message signal m(t) crosses zero. The envelope of a DSB-SC modulated signal is therefore different actual message signal.

The Fourier transform of s(t) is obtained as;formation of double sideband suppressed carrier

    \[s(f) = \frac{1}{2} Ac[M(f-fc) + M(f+fc)]\end.\]

for the case when baseband signal m(t) is limited to the interval -W ≤ f ≤ W, we thus find that the spectrum S(f) of the DSB-SC wave s(t) is illustrated in the figure above. Aside from a change in multiplier factor, the modulation method simply interprets the spectrum of the baseband signal by fc. Moreover, the transmission information measure need by DSB-SC modulation is that the same as that for amplitude modulation specifically,am transmission and reception of double sideband suppressed carrier

Single Sideband Modulation

Ideally, the generation of an SSB signal is a simple and clear process. We firstly generate a double-sideband signal and then apply an ideal band-pass filter to the result with the cut-off frequencies of fc and fc+W for the upper sideband for instance. Practically, the approx. construction of an ideal filter is very tough.

Where SSB finds its greatest application is in the transmission of analog voice signals. Analog voice has very little energy at low frequency(<300Hz) that is, there is an energy gap in the spectrum near the origin as depicted in figure (a), the ideal SSB filter shown in figure(b), and the resulting band-pass spectra shown in fig(c). In particular, must only satisfy the following needs;am transmission and reception of single sideband suppressed carrier

  • The desired sideband should lie inside the passband of the filter.
  • The unwanted sideband should lie inside the stopband of the filter.

However, this shows that the filter’s transition band, separating the passband from the stopband, is twice the lowest frequency (2fa) of the message signal. This non-zero transition bandwidth greatly simplifies the design of the SSB filter. Likewise to a DSB signal, coherent demodulation is required to detect an SSB signal. The synchronization information require to perform coherent demodulation is often obtained in one of the two ways;

  • Transmitting a low power pilot carrier in addition to the certain sideband or,
  • using highly stable oscillator in both the transmitter and receiver for generating the carrier frequency.

Vestigial Sideband Modulation

The practical application of the idea of single-sideband transmission to a signal that does not have an energy gap at the origin leads to Vestigial-sideband transmission. With VSB all of the one sidebands are transmitted and a small amount of the other sideband is transmitted as shown in the figure below. With VSB, the filter is allowed to have a nonzero transition band but the question remains that what restrictions are placed on the filter? if any then why?


Let H(f) denote the transfer function of the filter following the product modulator shown below. The spectrum of the modulated signal s(t) produced bypassing the frequency-shifted signal u(t) through the filter H(f) is;

    \[s(t) = U(f)H(f)\end.\]

    \[      = \frac{Ac}{2}[M(f-fc) + M(f+fc)]H(f)\end.\]

Whereas M(f) is the F.T of the baseband signal m(t) and U(f) is the F.T of u(t). However, the problem which we wish to address is to determine the particular H(f) require to produce a modulated signal s(t) with the desire spectral characteristics. such that the original baseband signal m(t) may be recovering from s(t) by coherent detection.

The first step of the process is multiplying the modulated signal s(t) by a locally obtain sine wave Ac’Cos(2πfct) which is in sync with the Ac Cos(2πfct), in both frequency and phase.

    \[v(t) = Ac' Cos[2(pie)fct]\end.\]

Transforming this relation into the frequency domain gives the F.T of v(t) as;

    \[v(f) = \frac{Ac'}{2}[S(f-fc) + S(f+fc)]\end.\]

Therefore, by substitution

    \[v(f) = \frac{AcAc'}{4}M(f)[H(f-fc) + H(f+fc)] + \frac{AcAc'}{4}[M(f-2fc)H(f-fc) + M(f+2fc)H(f+fc)]\end.\]

The high-frequency component of v(t) is given by the second term are cancel by the low pass filter to give an output Vo(t)

    \[v(f) = \frac{AcAc'}{4}M(f)[H(f-fc) + H(f+fc)]\end.\]

To get a linear and original baseband signal at the output, we require the Vo(f) to be multiple of M(f). Therefore, this means that the transfer function H(f) must satisfy the condition,

    \[H(f-fc) + H(f+fc) = 2H(fc)\end.\]

where H( fc ) the value of H(f) at f = fc is a constant. whenever the baseband signal M(f) is zero outside the range -W ≤ f ≤ W. We need to only satisfy the above equation for the values of f in the range. Also for ease we set H(fc) = 1/2. Thus we need that H(f) satisfy the condition

    \[H(f-fc) + H(f+fc) = 1\end.\]

Thus if the T.F of the filter satisfies the above equation. we can recover the original signal easily and without any cracks or holes.


However here we are here at the last part of the blog. I hope that all of your questions are clear now, and if you liked the content then please do share it with others and also mention the part which was your fav. Above all, if have any questions related to the topic feel free to ask down below. I would really happy to know which topic you would like to read next on

Have a nice day 🙂


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1 Comment

  • Dhruv March 20, 2021 at 2:30 pm Reply

    Aapka naam lucky haina aap ladki ho ya ladka 😂🤣😂

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