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

.

Consider the differential equation x” + 8x’ + 25x = 10u with zero initial conditions

Question: Consider the differential equation x” + 8x’ + 25x = 10u with zero initial conditions (i.e. x(0)=x'(0)=0) and u(t) is a unit step. Determine the solution y(t) analytically and verify with MATLAB by co-ploting analytic solution and the step response obtained with the step function. (Report should include: Analytic solution, MATLAB Code and outputs, Figure 2. Analytic solution and step response (2 subplots in one image), and conclusion about the system response)

Solution:

Q3. A program P running on a single-processor system takes time T to complete.

Q3. A program P running on a single-processor system takes time T to complete. Let us assume that 40% of the program’s code is associated with “data management housekeeping” (according to Amdahl) and, therefore, can only execute sequentially on a single processor. Let us further assume that the rest of the program (60%) is “embarrassingly parallel” in that it can easily be divided into smaller tasks executing concurrently across multiple processors (without any interdependencies or communications among the tasks).

1. Calculate T2, T4, T8, which are the times to execute program P on two-,four-,eight-processor system respectively.

2. Calculate T∞on a system with an infinite number of processors. Calculate the speedup of the program on this system, where speedup is defined as T/T∞. What does this correspond to?

Solution: Assume single instruction takes 1 time unit to execute. As we know 40% of instructions are dependent then minimum 40 time unit will be required.

1.T2 (Time for system having 2 processors)
Processor P1 and P2 can execute program instructions simultaneously which are not independent.
P1 executes 40% instructions which are dependent. It takes 40 time unit and same time P2 can execute 40% instructions which are independent. Now removing 20% independent instructions can execute on both the processors.
T2 = 40 max(10,10)
= 50 time units (minimum)

T4 (Time for system having 4 processors)
T4 = 40 + max(3,3,2,2)
= 43 time units (minimum)

T8 (Time for system having 8 processors)
T8 = 40 + max(2,2,1,1,1,1,1,1)
= 42 time units (minimum)

2. T∞ = 40 time units
Speed up = T/T∞ defines how much the new (T∞) system is efficient S= T2/T∞ = 50/40 = 1.25 (means T∞ system type is 1.25 times efficient than T2).