Answer:
Time required is 0.007 s
Explanation:
As per the question:
Analog signal to digital bit stream conversion by Host A =64 kbps
Byte packets obtained by Host A = 56 bytes
Rate of transmission = 2 Mbps
Propagation delay = 10 ms = 0.01 s
Now,
Considering the packets' first bit, as its transmission is only after the generation of all the bits in the packet.
Time taken to generate and convert all the bits into digital signal is given by;
t = [tex]\frac{Total\ No.\ of\ packets}{A/D\ bit\ stream\ conversion}[/tex]
t = [tex]\frac{56\times 8}{64\times 10^{3}}[/tex] (Since, 1 byte = 8 bits)
t = 7 ms = 0.007 s
Time Required for transmission of the packet, t':
[tex]t' = \frac{Total\ No.\ of\ packets}{Transmission\ rate}[/tex]
[tex]t' = \frac{56\times 8}{2\times 10^{6}} = 2.24\times 10^{- 4} s[/tex]