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A 480×640 digital image is transmitted along a MIMO wireless communication system with 1.2 Mb/s bit rate for each stream. The stream is the sub-data transmitted or received through one antenna. Each pixel of this image is digitally represented by 24 bits/pixel. By neglecting the length of wireless channel, determine the required time in seconds to complete receiving this image at the receiver; considering the following techniques are used separately: i. SISO technique. ii. 2×2 MIMO technique. iii. 3×3 MIMO technique. iv. 2×4 MIMO technique.

Question

A 480×640 digital image is transmitted along a MIMO wireless communication system with 1.2 Mb/s bit rate for each stream. The stream is the sub-data transmitted or received through one antenna. Each pixel of this image is digitally represented by 24 bits/pixel. By neglecting the length of wireless channel, determine the required time in seconds to complete receiving this image at the receiver; considering the following techniques are used separately: i. SISO technique.

ii. 2×2 MIMO technique.

iii. 3×3 MIMO technique.

iv. 2×4 MIMO technique.

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Solution

Para resolver este problema, sigamos los siguientes pasos:

  1. Calcular el tamaño total de la imagen en bits:

    • La imagen tiene una resolución de 480×640 píxeles.
    • Cada píxel está representado por 24 bits.
    • Tamaño total de la imagen en bits = 480 × 640 × 24.
  2. Calcular el tiempo requerido para transmitir la imagen usando diferentes técnicas:

    • La tasa de bits es de 1.2 Mb/s por stream.
    • Convertimos la tasa de bits a bits por segundo: 1.2 Mb/s = 1.2 × 10^6 bits/s.

i. SISO (Single Input Single Output) technique:

  • En SISO, solo hay un stream.
  • Tiempo requerido = Tamaño total de la imagen en bits / Tasa de bits por stream.

ii. 2×2 MIMO technique:

  • En 2×2 MIMO, hay 2 streams.
  • Tasa de bits total = 2 × 1.2 × 10^6 bits/s.
  • Tiempo requerido = Tamaño total de la imagen en bits / Tasa de bits total.

iii. 3×3 MIMO technique:

  • En 3×3 MIMO, hay 3 streams.
  • Tasa de bits total = 3 × 1.2 × 10^6 bits/s.
  • Tiempo requerido = Tamaño total de la imagen en bits / Tasa de bits total.

iv. 2×4 MIMO technique:

  • En 2×4 MIMO, hay 2 streams (el número de streams es el menor de los dos números).
  • Tasa de bits total = 2 × 1.2 × 10^6 bits/s.
  • Tiempo requerido = Tamaño total de la imagen en bits / Tasa de bits total.

Ahora, realicemos los cálculos:

  1. Tamaño total de la imagen en bits: 480×640×24=7,372,800 bits 480 \times 640 \times 24 = 7,372,800 \text{ bits}

  2. Tasa de bits por stream: 1.2×106 bits/s 1.2 \times 10^6 \text{ bits/s}

i. SISO technique:

Tiempo requerido=7,372,800 bits1.2×106 bits/s=6.144 segundos \text{Tiempo requerido} = \frac{7,372,800 \text{ bits}}{1.2 \times 10^6 \text{ bits/s}} = 6.144 \text{ segundos}

ii. 2×2 MIMO technique:

Tasa de bits total=2×1.2×106 bits/s=2.4×106 bits/s \text{Tasa de bits total} = 2 \times 1.2 \times 10^6 \text{ bits/s} = 2.4 \times 10^6 \text{ bits/s} Tiempo requerido=7,372,800 bits2.4×106 bits/s=3.072 segundos \text{Tiempo requerido} = \frac{7,372,800 \text{ bits}}{2.4 \times 10^6 \text{ bits/s}} = 3.072 \text{ segundos}

iii. 3×3 MIMO technique:

Tasa de bits total=3×1.2×106 bits/s=3.6×106 bits/s \text{Tasa de bits total} = 3 \times 1.2 \times 10^6 \text{ bits/s} = 3.6 \times 10^6 \text{ bits/s} Tiempo requerido=7,372,800 bits3.6×106 bits/s=2.048 segundos \text{Tiempo requerido} = \frac{7,372,800 \text{ bits}}{3.6 \times 10^6 \text{ bits/s}} = 2.048 \text{ segundos}

iv. 2×4 MIMO technique:

Tasa de bits total=2×1.2×106 bits/s=2.4×106 bits/s \text{Tasa de bits total} = 2 \times 1.2 \times 10^6 \text{ bits/s} = 2.4 \times 10^6 \text{ bits/s} Tiempo requerido=7,372,800 bits2.4×106 bits/s=3.072 segundos \text{Tiempo requerido} = \frac{7,372,800 \text{ bits}}{2.4 \times 10^6 \text{ bits/s}} = 3.072 \text{ segundos}

Por lo tanto, los tiempos requeridos para cada técnica son:

  • SISO: 6.144 segundos.
  • 2×2 MIMO: 3.072 segundos.
  • 3×3 MIMO: 2.048 segundos.
  • 2×4 MIMO: 3.072 segundos.

This problem has been solved

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