HW3, MAE 171. Spring 2005. UCI

Nasser Abbasi





Problem 3-1

part a

Give the definition of the starred transform.

MATH is the Laplace transform of MATH, where MATH which is the time domain representation of the input signal $e\left( t\right) $ after passing though an ideal sampler.

Mathematically, MATH where $T$ is the sampling period, and $k$ is the sample number and MATH is the $Z$ transform the sequence $e\left[ n\right] $ .

Given the standard definition of the $Z$ transform, then we can now write MATH

Part b

Give the definition of the $Z$ transform.

Given a sequence MATH, then the one sided $Z$ transform of $e$ is defined as

MATH

Here we assumed that the number sequence $e$ has its first element at index $k=0$. i.e. MATH for $k<0$

Part c

For a function $e\left( t\right) $ derive a relationship between its starred transform MATH and its $Z$ transform MATH

MATH

but MATH

Hence (1) becomes

MATH

But given the property that MATH , then MATH and (2) can be now be written as

MATH

Now, given the one-sided definition of the Z transform of sequence $e\left( kT\right) $ as$\ $MATHthen by comparing (3) and (4) we see that (3) can be be written as

MATH

Which is the required relation to show.


Problem 3-4.

part c

Find the starred transform for MATH

MATH

so need to first find the Z transform of $e\left[ kT\right] $ , the discrete time representation of the ideal sampled version of $e\left( t\right) $

MATH

so we need to first find $e\left( t\right) $

MATH

Use partial fraction to find the inverse Laplace transform.

MATH

Hence

MATH

ROC is MATH and MATH

Hence, MATH

part f

Find the starred transform for MATH

MATH

so need to first find the Z transform of $e\left[ kT\right] $ , the discrete time representation of the ideal sampled version of $e\left( t\right) $

MATH

so we need to first find $e\left( t\right) $

MATH

roots of denominator are MATH

Let $r_{1}=-1+2j$, and $r_{2}=-1-2j$, so now $E\left( s\right) $ can be written as

MATH

Use partial fraction to find the inverse Laplace transform.

MATH

Hence

MATH

From Tables,

MATH

Hence

MATH


Problem 3-15.

Given the signal MATH

part a

List all frequencies less than $\omega =50$ rad/sec that are present in $e\left( t\right) $

$\sin 4t$ contains a frequency of $4$ rad/sec.

$\sin 7t$ contains a frequency of $7$ rad/sec.

Hence the answer is

MATH rad/sec

To illustrate, this is a small Mathematica script.


Figure

Part b

$e\left( t\right) $ is sampled at $\omega _{s}=22$ rad/sec. List all frequencies present in MATH less than $\omega =50$ rad/sec.

The frequency spectrum of the sampled signal will contain frequencies that are centered around multiples of $\omega _{s}$

Hence the list of generated frequencies are

MATH

substitute for $\omega _{s}=22$ we get

MATH

MATH

Sort and consider only frequencies less than $50$ rad/sec, we get

MATH

Part c

The signal MATH is applied to a ZOH device. List all frequencies present in MATH less than $\omega =50$ rad/sec.

Since a ZOH is an LTI device and the input is a set of eigenfunctions, then the output will contain the same set of input frequencies but possibly scaled and phase shifted. Hence this is the same answer as part (b) above.


Problem 3-16

MATH is applied to a samples/ZOH device, $\omega _{s}=4\ $rad/sec.

part a

what is the frequency component in the output that has the largest amplitude?

This will be the component with the frequency closest to the DC frequency.

Looking at the sequence of frequencies generated, which are

MATH

MATH

MATH

sort, and consider only positive frequencies, we obtain

MATH

Hence the component with the largest amplitude has a frequency of

$1$ rad/sec

part b

Find amplitude/phase of that component.

MATH

But MATHTo find MATH use the relation that MATH has the Fourier transform MATHThen, the Fourier transform of MATH is (where here MATH)

MATHsubstitute (3) into (2), then (1) can be written as

MATHbut MATHHence (4) becomes

MATHEvaluate at $\omega =1$ rad/sec, and given that $\omega _{s}=4$ rad/sec, we get

MATHNow to find the phase of this component.

MATHsince MATH then, since MATH then MATHthis is because MATH at frequency $1$ rad/sec which is smaller than $\omega _{s}=4$ rad/sec, so we are inside the first hub of the sinc function. i.e. MATH in this case.

Now MATH but since MATH when $\omega =1$, then $\theta =0$ (using equation 3-33 page 105 in book).

Hence MATH

Hence, (5) now becomes

MATH

part c

sketch the input signal and the component of part(b) vs. time.


Figure

part d

Find the ratio of the amplitude in part(b) to that of the frequency component in the output at $\omega =7$ rad/sec.

MATH

but

MATH

and

MATH

Hence

MATH

Hence ratio is MATH

MATH

This is expected, as at $\omega =7$ we are in the second hub of the sinc function, which will have much less amplitude.


Figure





Figure