ece3084hw03

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GEORGIA INSTITUTE OF TECHNOLOGY SCHOOL of ELECTRICAL and COMPUTER ENGINEERING ECE 3084 Fall 2012 Problem Set #3 Assigned: 21-Sept-12 Due Date: 2-Oct-12 Your homework will is due at the start of class on Tuesday, October 2. You may turn in your homework up to one day late, by 4:30 PM the following day. A 30% penalty will be assessed on late homeworks (even homeworks turned in later the day it is due, although the penalty might be slightly less at my discretion). We understand that sometimes multiple assignments hit at once, or other life events intervene, and hence you have to make some tough choices. We’d rather let you turn something in late, with some points off, than have a “no late assignments accepted at all” policy, since the former encourages you to still do the assignment and learn something from it, while the latter just grinds down your soul. The somewhat aggressive late penalty is not intended to be harsh – it’s intended to encourage you to get things in relatively on time (or just punt if you have to and not leave it hanging over you all semester) so that you can move on to assignments for your other classes. Also, there is the practical matter that we cannot accept homeworks after solutions are posted, and we would like to post solutions shortly after both sections have submitted their homework. From http://xkcd.com/26

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Transcript of ece3084hw03

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GEORGIA INSTITUTE OF TECHNOLOGYSCHOOL of ELECTRICAL and COMPUTER ENGINEERING

ECE 3084 Fall 2012Problem Set #3

Assigned: 21-Sept-12Due Date: 2-Oct-12

Your homework will is due at the start of class on Tuesday, October 2.You may turn in your homework up to one day late, by 4:30 PM the following day.

A 30% penalty will be assessed on late homeworks (even homeworks turned in later theday it is due, although the penalty might be slightly less at my discretion). We understand thatsometimes multiple assignments hit at once, or other life events intervene, and hence you have tomake some tough choices. We’d rather let you turn something in late, with some points off, thanhave a “no late assignments accepted at all” policy, since the former encourages you to still do theassignment and learn something from it, while the latter just grinds down your soul. The somewhataggressive late penalty is not intended to be harsh – it’s intended to encourage you to get thingsin relatively on time (or just punt if you have to and not leave it hanging over you all semester)so that you can move on to assignments for your other classes. Also, there is the practical matterthat we cannot accept homeworks after solutions are posted, and we would like to post solutionsshortly after both sections have submitted their homework.

From http://xkcd.com/26

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PROBLEM 3.1*:

The impulse response of a continuous-time linear time-invariant system is

h(t) = δ(t)− 0.2e−0.2t step(t).

(a) Find the frequency response H(jω) of the system. Express your answer as a single rationalfunction with powers of (jω) in the numerator and denominator.

(b) Plot the magnitude squared, |H(jω)|2 = H(jω)H∗(jω), versus ω. You may use MATLAB tocheck your answer if you want to, but try it by hand first, i.e. plot a few points on paper andsketch a smooth curve.

(c) At what frequency ω does the magnitude squared of the frequency response have it largestvalue? (Your answer might be something like “infinity.”) At what frequency is the magnitudesquared of the frequency response equal to one half of its peak value? (This is referred toas the 3dB point of the filter since the frequency response magnitude measured in decibels,10 log |H(jω)|2, is 3.01dB smaller at this frequency compared to its peak value when measuredin decibels.)

(d) Suppose that the input to this system is

f(t) = 40 + 20 cos([0.2√

3]t) + 3δ(t− 1.6).

Use superposition to find the output y(t). Note that the argument of the cosine really is(0.2√

3)t, not (0.2√

3)πt. Hint: To find the response of each term, use the easiest method,i.e., impulse response or frequency response. Also, if I set this up right, the computationsshould work out elegantly.

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PROBLEM 3.2*:

Consider the LTI system below:

- -LTI Systemh(t)

f(t) y(t)

The input to this system is the periodic pulse wave f(t) depicted below:

-

6

-10 -8 -6 -4 -2 -1 0 1 2 4 6 8 10 t

f(t)

6

· · ·· · ·

The input can be represented by the Fourier series

f(t) =∞∑

k=−∞ake

jkω0t where ak =6 sin(πk/4)

πk.

Note we can make sense of the ak formula for k = 0 using L’Hopital’s rule.

(a) Determine ω0 in the Fourier series representation of f(t).

(b) Plot F (jω), the Fourier transform of the signal f(t), for −4ω0 ≤ ω ≤ 4ω0. (See Equation4.55 on p. 144 of Chen.)

(c) If the frequency response of the system is the ideal highpass filter

H(jω) =

{0 |ω| < π/81 |ω| > π/8

plot the output of the system, y(t), when the input is f(t) as plotted above. Hint: Firstdetermine what frequency is removed by the filter, and then determine what effect this willhave on the waveform.

(d) If the frequency response of the system is an ideal lowpass filter

H(jω) =

{1 |ω| < ωc

0 |ω| > ωc

where ωc is the cutoff frequency, for what values of ωc will the output of the system have theform

y(t) = A+B cos(ω0t+ φ)

where A and B are nonzero?

(e) Find A, B, and φ in part (d).

(f) If the frequency response of the LTI system is H(jω) = 1 − e−j2ω, plot the output of thesystem, y(t), when the input is f(t) as plotted above. Hint: In this case it will be easiestto determine the impulse response h(t) corresponding to H(jω) and from h(t) you can easilyfind an equation that relates y(t) to f(t). This will allow you to plot y(t).

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PROBLEM 3.3*:

(a) The “ramp filter” is often used in medical imaging applications such as X-ray computer-aidedtomography. It has a frequency response given by

H(jω) =

{jω for |ω| < ω0

0 otherwise

Find the impulse response, h(t), of this filter. Don’t use integration by parts to do thisproblem; use Fourier transform pairs and properties, such as the “Time Derivative” propertyfrom this website:

http://fourier.eng.hmc.edu/e101/lectures/handout3/node2.html

You may need to go back to your old calculus text if you’ve forgotten how to take derivativesof quotients. Also make a plot of the impulse response (this is pretty complicated, so youshould feel free to use MATLAB or one of those fancy graphing calculators to help you makethe plot.) If you’ve ever had a CAT scan done, you’ve run across this filter in practice!

A hint on making the plot: We’re interested in the overall shape of the curve; you canpick whatever ω0 you find the most interesting. I found that the following bit of MATLABcode made a nice plot. You should feel free to use it:

omega_0 = 2*pi;

period = 2*pi/omega;

t = -5*period:period/100:5*period;

h = [fill in code that goes here]

h([t == 0]) = [fill in value that goes here]

plot(t,h);

You may ask yourself what the h([t == 0]) = business is all about. Well, it turns out thath(t) is indeterminite at t = 0, i.e. we wind up dividing zero by zero. To find a meaningfulvalue for h(0), use L’Hopital’s rule. Alternatively, you can probably guess what h(0) shouldbe by looking at h(t) for t near zero.

(b) Chemists using nuclear magnetic resonance and radiologists using magnetic resonance imag-ing1 spend a lot of time thinking about free induction decay signals, which take the form ofa decaying sinusoid, for instance:

f(t) = A exp

(− t

T2

)sin(ω0t) step(t)

The nonnegative constant T2 is called the spin-spin relaxation time. Different kinds of bodytissue have different T2 times; for instance, by determining T2 times, doctors can identify andlocate some kinds of cancer. ω0 is called the resonant frequency; in an MRI scheme, this tellsyou what part of the body the signal is coming from. A is called the spin density, which alsodepends on tissue type.

If you don’t find anything in the previous paragraph at all interesting, don’t worry about it;all I want you to do is find F (jω), the Fourier transform of f(t), in terms of the constants A,T2, and ω0.

1To be consistent, what we know now as MRI should be called NMRI. However, many of these developmentsoccured during the height of the cold war, and using the word “nuclear” was considered unwise from a marketingperspective. Hence, the “N” got dropped.

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PROBLEM 3.4*:

Consider the Fourier transform pair

e−a|t|F↔ 2a

a2 + ω2(1)

You may have noticed that Fourier transform pairs have a kind of symmetry. For instance,rectangular pulses in the time domain transform to sincs in the frequency domain, and sincs inthe time tomain transform to rectangular windows in the frequency domain. Similarly, (1) has arelated Fourier transform pair:

a

π(a2 + t2)

F↔ e−a|ω|

(a) Suppose f(t) = a/[π(a2 + t2)] and h(t) = b/[π(b2 + t2)].

(i) Compute the convolution y(t) = f(t) ∗ h(t). Theoretically, you could find the result bydirectly cranking on the convolution integral:

y(t) =1

π2

∫ ∞−∞

a

π(a2 + τ2)

b

π(b2 + (t− τ)2)dτ

However, that way lies madness. There is an easier path!

(ii) When you convolve two similar functions, you usually don’t get a function that looksanything like the functions you’re convolving. For instance, if you convolve two rect-angular pulses, you get a triangle or a trapezoid. However, there are a few families offunctions – for instance, sinc functions – that give you a member of the same family ifyou convolve two functions from that family. Such families of functions are relativelyrare and usually have quite interesting properties in many applications, particularlyprobability theory.

Looking at your answer to part (i), does it seem very different in form than f(t) or y(t)?Or does it look similar?

(b) Find the inverse Fourier transform of exp(−6|ω − 5|) + exp(−6|ω + 5|).

(c) Find the inverse Fourier transform of exp(−6|ω|+ 5jω).

(d) Evaluate the integral ∫ ∞−∞

1

2500 + c2e−j4cdc

This will not yield to a frontal assault using the standard methods you learned in your calculusclasses – attack it with Fourier theory instead. You may have to twiddle with some constantsto get things to line up right.

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PROBLEM 3.5*:In class, we presented a double side band suppressed carrier (DSB-SC) amplitude modulationscheme. As presented, that scheme requires that the frequency and phase of the sinusoidal oscillatorused to demodulate the signal be be “locked” with the frequency and phase of the sinusoidaloscillator used to modulate the signal. This problem explores what happens when there is amismatch in the phases or the frequencies.

⊗- - ⊗- -

LTILowpass

FilterH(jω)

-

66

cos(ωct+ φ)cos(ωct)

y(t)y(t)f(t) w(t) v(t)

H(jω) =

{2 |ω| < ωco

0 |ω| > ωco

We have shown that if f(t) has a bandlimited Fourier transform such that F (jω) = 0 for |ω| ≥ ωb

and ωc > ωb and φ = 0 and ωb < ωco < 2ωc − ωb, then the AMDSB/SC signal y(t) = f(t) cos(ωct)can be demodulated by the above system. That is, for precise adjustment of the demodulatorfrequency and phase, v(t) = f(t). In the following parts, assume that the input signal f(t) has abandlimited Fourier transform represented by the following plot:

-

6

�����

@@

@@@

−ωb ωb ω0

1

F (jω)

Now suppose that φ 6= 0.

(a) Use a well known trignometric identity to show that

w(t) = y(t) cos(ωct+ φ) = f(t) cos(ωct) cos(ωct+ φ) = 12f(t) cos(φ) + 1

2f(t) cos(2ωct+ φ).

(b) Use your equation from part (a) to obtain an equation for W (jω), the Fourier transform ofw(t), in terms of F (jω).

(c) Use your equation from part (b) to make a plot of W (jω) for the given F (jω).

(d) Use your plot in part (c) to make a plot of V (jω), the Fourier transform of v(t).

(e) Use your plot in part (d) to obtain an equation for v(t) in terms of f(t) and φ.