Data Communications Networking Chapter 3, Problem 48E

Question: What is the total delay (latency) for a frame of size 5 million bits that is being sent on a link with 10 routers each having a queuing time of 2 μs and a processing time of 1 μs. The length of the link is 2000 Km. The speed of light inside the link is 2 × 108 m/s. The link has a bandwidth of 5 Mbps. Which component of the total delay is dominant? Which one is negligible?

Solution:

Given data:

Distance (length of the link) =2000 Km

=2000

103m

Propagation speed=2

108m/s

Frame size=5million bits

=5000000bits

The Bandwidth=5Mbps

=5

106 bps (1Mbps=106 bps)

Queuing time=10

2 (10 routers each having a queuing time of 2)

=20

Processing delay=10

1 (10 routers each having a processing time of 1)

=10

Total delay (Latency) = Propagation time + Transmission time + Queuing time

+ Processing delay.

Propagation time=

=

= 0.01s

Therefore, Propagation time=0.01s

Transmission time=

=

= 1s

Therefore, Transmission time=1s

Queuing time=10

2

=10

210-6s (1=10-6s)

=20

10-6s

= 0.00002s

Therefore, Queuing time=0.00002s

Processing delay=10

1

=10

10-6s (1=10-6s)

Therefore, Processing delay=0.000010s

Therefore, Total delay (Latency) =0.01+1+0.000020+0.000010

=1.01003s

Signals and Systems Chapter 7, Problem 41AP

Question: In many practical situations a signal is recorded in the presence of an echo, which we would like to remove by appropriate processing. For example, in Figure P7.41 (a), we illustrate a system in which a receiver simultaneously receives a signal x(t) and an echo represented by an attenuated delayed replication of x(t). Thus, the receiver output is

Assume that x(t) is band limited [i.e., X(jω) = 0 for

and that | α | < 1.

(a) If

, and the sampling period is taken to be equal to determine the difference equation for the digital filter hire] such that yc(t) is proportional to x(t).

(b) With the assumptions of part (a), specify the gain A of the ideal lowpass filter such that

(c) Now suppose that

. Determine a choice for the sampling period T, the lowpass filter gain A, and the frequency response for the digital filter h[n] such that is proportional to x(t).

Solution:

(a) The received signal is,

It is ideally sampled with an impulse train of period,

.

Sampling with impulse functions is ideal sampling.

Where,

.

Translate the received signal to frequency domain.

The sampled signal in time domain is (or) the impulse train is,

,

It is a sequence of impulses with amplitudes equal to the sample value.

Remove the sampling period, as they simply correspond to a sample.

Convert the impulse train to a sequence.

Therefore, the sequence is,

The final output must be proportional only to

.

That implies the signal

must be a baseband signal and must be proportional to .

For the final output to be proportional to be only

; the frequency domain translation of the received signal is,

That means in the delayed version of

, needs to be eliminated, to get output exactly proportional to .

In frequency domain, this translates to elimination of the

term.

So the digital filter

must eliminate term.

The signal

is applied to the digital filter, .

The expression for output

is,

Convolution in time domain translates to multiplication in frequency domain.

Substitute for (since the output is proportional to input in time domain), for .

Thus, the frequency domain description of the digital filter is,

Determine the difference equation.

…… (1)

Apply inverse Fourier transform in equation (1).

Thus, the difference equation of the filter is

.

(b)

The output sequence

is converted to an impulse train.

That is, the sequence itself is an impulse train.

To recover the signal form this impulse train, the signal is passed through a low pass filter.

The output of the digital filter is taken to be exactly equal to .

Thus, the input to the low pass filter is

.

The low pass filter simply acts as a reconstruction filter.

The input is

, for the output of the ideal low pass filter to be exactly ,

Substitute

for .

Thus, the pass band gain of the ideal low pass filter is

.

c)

The impulse

Note that in this case,

.

The minimum sampling frequency, to avoid aliasing is,

.

The delay of the echo has a range of,

The sampling frequency of the sum of two signals is twice the bandwidth of the signal with the highest bandwidth.

The discrete sequence is,

…… (1)

The sampling frequency does not change if the signal is shifted.

The received signal is the sum of the signal

and the its shifted version .

The sampling frequency of sum of two signals is the twice the bandwidth of the signal with the highest bandwidth.

Thus, the sampling frequency is,

Thus, the sampling period must be ,

.

Apply Fourier transform to equation (1).

Determine the frequency response of the digital filter,

.

For the final output to be proportional to,

.

The only possibility is altering the function using

.

The input signal is,

Since the input is directly applied to the digital filter,

The frequency response of the output is multiplication of

and .

Substitute

for and for .

Thus, the gain of frequency response of the digital filter is,

The impulse train

is a sampled version of the signal ,due to the effect of digital filter.

For the final output to be exactly

, we need to pass it through a low pass filter.

The input to the ideal low pass filter is a sampled version of

.

So get the exact replica of the signal

, the gain of the ideal low pass filter must be .

Thus, the gain of the ideal low pass filter is

.