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pashok25 [27]
3 years ago
14

4. In this problem, we consider sending real-time voice from Host A to Host B over a packet-switchednetwork (VoIP). Host A conve

rts analog voice to a digital 64 kbps bit stream on the y. Host A thengroups the bits into 56-byte packets. There is one link between Hosts A and B; its transmission rate is2 Mbps and its propagation delay is 10 msec. As soon as Host A gathers a packet, it sends it to HostB. As soon as Host B receives an entire packet, it converts the packet's bits to an analog signal. Howmuch time elapses from the time a bit is created (from the original analog signal at Host A) until thebit is decoded and played back as part of the analog signal at Host B?
Computers and Technology
1 answer:
mina [271]3 years ago
8 0

Answer:

The time elapsed is 0.017224 s

Solution:

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 = \frac{Total\ No.\ of\ packets}{A/D\ bit\ stream\ conversion}

t = \frac{56\times 8}{64\times 10^{3}}          (Since, 1 byte = 8 bits)

t = 7 ms = 0.007 s

Time Required for transmission of the packet, t':

t' = \frac{Total\ No.\ of\ packets}{Transmission\ rate}

t' = \frac{56\times 8}{2\times 10^{6}} = 2.24\times 10^{- 4} s

Now, the time elapse between the bit creation and its decoding is given by:

t + t'  + propagation delay= 0.007 + 2.24\times 10^{- 4} s + 0.01= 0.017224 s

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