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Analog Communication

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Amplitude Shift Keying
The simplest digital modulation technique is amplitude-shift keying (ASK), where a binary information signal directly modulates the amplitude of an analog carrier. ASK is similar to standard amplitude modulation except there are only two output amplitudes possible. Amplitude-shift keying is sometimes called digital amplitude modulation (DAM). In ASK, both frequency and phase remain constant while the amplitude changes. The level of amplitude can be used to represent binary logic 0s and 1s. In the other word, the carrier can be either “on” or “off” switch. In the modulated signal, logic 0 is represented by the absence of a carrier. That is why amplitude-shift keying sometimes can referred to as on-off keying (OOK).

Bandwidth of ASK :
The bandwidth of ASK is proportional to the signal rate S.
B = (1+d) X S
Where B is the bandwidth S is the signal rate d is between 0 and 1 (depend on modulation and filtering)
Baud rate is the number of signal units per second that are required to represent those bits while bit rate is the number of bits per second.

The block diagram of ASK transmitter:

In the transmitter, the precoder performs level conversion and then encodes the incoming data into groups of bits that modulate an analog carrier. The modulated carrier is shaped (filtered), amplified, and then transmitted through the transmission medium to the receiver. The transmission medium can be a metallic cable, optical fiber cable, Earth's atmosphere, or a combination of two or more types of transmission systems.
The block diagram of ASK receiver :

In the receiver, the incoming signals are filtered, amplified, and then applied to the demodulator and decoder circuits, which extracts the original source information from the modulated carrier. The clock and carrier recovery circuits recover the analog carrier and digital timing (clock) signals from the incoming modulated wave since they are necessary to perform the demodulation process.

The figure above shows the rectified and low pass filtered results of the ASK modulated signal. The waveform is shown for 8-bit data word.

The modulation circuit of the ASK :
The message signal is a unipolar (ON/OFF) format message signal, and the sinusoidal carrier signal is multiplied with each other. The resultant is an ASK signal. Whenever the message signal is "0" (represented by 0 volts) the output at the collector of the transistor is 0. When the message signal is "1", the output at the collector of the transistor is a sinusoidal signal of amplitude = (amplitude of carrier)*(amplitude of message).
The demodulation circuit of the ASK :
The first part of the demodulation circuit is a half wave rectifier, followed by a low pass filter and a comparator. The diode will remove any unwanted negative amplitudes introduced due to noise. The low pass filter will remove the high frequency carrier. Then a comparator circuit is used, with the threshold voltage= (amplitude of message signal)/2 to estimate the transmitted pulses. If the input pulse is greater than the threshold voltage, a new "1" is generated, if the input is lesser than the threshold voltage then a new "0" is generated at the output.
Equation Used In The ASK
Mathematically, equation of amplitude-shift keying is Vask (t) = [ 1 + Vm (t) ] [A/2 cos (wct) ] where vask (t) = amplitude-shift keying wave vm(t) = digital information (modulating) signal (volts) A/2 = unmodulated carrier amplitude (volts) Wc = analog carrier radian frequency (radians per second, 2πfct)
In the equation above, the modulating signal [vm (t)] is a normalized binary waveform, where + 1 V = logic 1 and -1 V = logic 0. Therefore, for a logic 1 input, vm (t) = + 1 V, the equation reduces to
Vask (t) = [1 + 1 ] [ A/2 cos (wct) ] = A cos (wct) and for a logic 0 input, vm (t) = -1 V, the equation reduces to
Vask (t) = [1 - 1 ] [A/2 cos (wct) ] = 0
Thus, the modulated wave vask(t), is either A cos(ωc t) or 0.

ADVANTAGE AND DISADVANTAGE OF ASK :
The advantage of ASK is its simplicity. You just need an oscillator which you can just turn on and off. The disadvantage of ASK is highly susceptible to noise interference because ASK relies on amplitude to differentiate between 1 and 0. To overcome this problem, a great gap between amplitude values is needed so that noise can be detected and removed.

Question for ASK:
There is an available bandwidth of 200 kHz which spans from 300 to 400 kHz. What are the carrier frequency and the bit rate if we modulated our data by using ASK with d = 1?
Solution :
The middle of the bandwidth is located at 350 kHz. This means that our carrier frequency can be at fc = 350 kHz. We can use the formula for bandwidth to find the bit rate (with d = 1 and r = 1).
B = (1+d) X S
200k =2 X N X (1/r)
200k =2 X N
N = 100 kbps

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