Digital Communication

Digital Modulation Techniques

AMPLITUDE SHIFT KEYING (ASK)

ASK:

Amplitude shift keying is a type of digital modulation technique which assigns a sinusoidal carrier signal of high amplitude $(A_H)$ for logic $’1’$ and a sinusoidal carrier signal of low amplitude $(A_L)$ for logic $’0’$ (frequency and phase of the carrier signal  remains constant).

🌾Mathematically, $$S(t)=\bigg\{\begin{matrix}A_H\;sin(2\pi ft)\;\;\;\; for \;logic\;’1’\\A_L\;sin(2\pi ft)\;\;\;\;for \;logic\;’0’\end{matrix}$$

🌾For example, for digital data $1011001$, the ASK signal will be

ASK Modulation:

🌾Block diagram of ASK Modulator:

🌾Digital Data:

  • Digital data is nothing but $1’s$ and $0’s$, which is a digitized version of any analog signal. $$…1011001…$$
  • Here digital data is represented using Unipolar NRZ encoding.

🌾Unipolar NRZ Encoding:

  • Unipolar – Only one polarity (positive or negative).
  • NRZ – Non Return to Zero.
  • UNRZ encoding assigns a rectangular pulse of amplitude ‘A’ and duration equal to bit duration $(T_b)$ for logic $’1’$ and nothing for logic $’0’$.

🌾Carrier Signal Oscillator:

  • This oscillator generates high a frequency carrier signal of amplitude $A_H$. $$C(t)=A_H\;sin(2\pi ft)$$

🌾Attenuator:

  • Attenuator is used to attenuate the carrier signal in order to get the signal of amplitude $A_L$. $$C'(t)=A_L\;sin(2\pi ft)$$

🌾Path$1$:

  • Path$1$ is dedicated for logic $’1’$.
  • Multiplier$1$ in path $1$, multiplies digital data with carrier signal of amplitude $A_H$. $$Data\;×\;A_H\;sin(2\pi ft)$$ or $$U(t)×C(t)$$

🌾Path$2$:

  • Path$2$ is dedicated for logic $’0’$.
  • Inverter in path$2$, inverts the digital data (i.e.,) converts logic $’1’$ to logic $’0’$ and vice-versa.
  • Multiplier$2$ in path$2$, multiplies inverted digital data with attenuated carrier signal of amplitude $A_L$.  $$\overline{Data}\;×\;A_L\;sin(2\pi ft)$$ or $$U'(t)×C'(t)$$

🌾Summer:

  • Summer gives us the ASK signal by adding the signals from path $1$ and path $2$.  $$S(t)=Data×A_H sin(2\pi ft)+\overline{Data}×A_L sin(2\pi ft)$$ Hence ASK signal, $$S(t)=\bigg\{\begin{matrix}A_H\;sin(2\pi ft)\;\;\;\; for \;logic\;’1’\\A_L\;sin(2\pi ft)\;\;\;\;for \;logic\;’0’\end{matrix}$$ or $$S(t)=U(t)×C(t)+U'(t)×C'(t)$$

ASK Transmission Bandwidth:

🌾ASK signal, $$S(t)=\bigg\{\begin{matrix}A_H\;sin(2\pi ft)\;\;\;\; for \;logic\;’1’\\A_L\;sin(2\pi ft)\;\;\;\;for \;logic\;’0’\end{matrix}$$,

Generation of ASK signal can be written as, $$S(t)=Data\;×\;A_H\;sin(2\pi ft)+\overline{Data}×A_L\;sin(2\pi ft)$$

Since data is represented using UNRZ coding, we can rewrite the above equation as, $$S(t)=U(t)×C(t)+U'(t)×C'(t)$$

🌾Spectrum of UNRZ pulse:

  • Since UNRZ pulse is a rectangular pulse of $T_b$ duration, we need to find the Fourier transform of rectangular pulse.
  • Rectangular pulse in time domain is given by, $$r(t)=\bigg\{\begin{matrix}1 \;\;\;for \;-T_b/2\leq t\leq T_b/2\\0\;\;\;otherwise\end{matrix}$$
  • In general,  The Fourier transform of a function $x(t)$ is given by, $$X(\omega)=\int_{-\infty}^{+\infty}x(t) e^{-j\omega t}dt$$ 
  • Now, Fourier transform of rectangular pulse be, $$R(\omega)=\int_{-T_b/2}^{T_b/2}1\; e^{-j\omega t}dt$$ $$=\Bigg[\frac{e^{-j\omega t}}{-j\omega}\Bigg]_{-T_b/2}^{T_b/2}$$$$=\frac{\big(e^{j\omega T_b/2}\big)-\big(e^{-j\omega T_b/2}\big)}{j\omega}$$$$=\frac{2}{2}\frac{\big(e^{j\omega T_b/2}\big)-\big(e^{-j\omega T_b/2}\big)}{j\omega}$$$$=\frac{2}{\omega}sin(\omega T_b/2)$$$$=\frac{T_b}{T_b}\frac{2}{\omega}sin(\omega T_b/2)$$$$=T_b\frac{sin(\omega T_b/2)}{\omega T_b/2}$$ $$=T_b\;sinc(\omega T_b/2)$$
  • Hence Fourier transform of a rectangular pulse is, $$T_b\;sinc(\omega T_b/2)$$
  • In frequency scale, $$=T_b\frac{sin(\omega T_b/2)}{\omega T_b/2}$$ $$=T_b\frac{sin(2\pi f T_b/2)}{2\pi f T_b/2}$$ $$=T_b\frac{sin(\pi f T_b)}{\pi f T_b}$$ $$=T_b\;{sinc( f T_b)}$$
  • From the above figure, we can say, we need infinite bandwidth to transmit a perfect rectangular pulse.
  • But unfortunately no channel can support infinite bandwidth.