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EE102 Systems and SignalsFall Quarter 2010 Jin Hyung Lee

Homework #6

Due: Wednesday Nov. 17

1. Each of these signals can be written as a sum of scaled and shifted unit rectangles andtriangles,

2

-1

1

0

-2

1 2-2 -1

x(t)a)

2

-1

1

0

-2

1 2-2 -1t

x(t)b)

Find a simple expression for each signal, and then compute the Fourier transform.

Solution:

(a) One representation for the first signal is

x(t) = ∆(t+ 1/2)−∆(t+ 1/2)

The Fourier transform is then

X(jω) = ejω/2sinc2(ω/2π)− e−jω/2sinc2(ω/2π)

= (ejω/2 − e−jω/2)sinc2(ω/2π)

= 2j sin(ω/2)sinc2(ω/2π)

(b) One representation of the second signal is

x(t) = 2∆(t/2)−∆(t) + rect(t/2)

The Fourier transform is then

X(jω) = 4 sinc2(ω/π)− sinc2(ω/2π) + 2 rect(ω/π)

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2. Determine whether the assertions are true or false, and provide a supporting argument.

(a) If (f ∗ g)(t) = f(t), then g(t) must be an impulse, δ(t).

Solution:Time-domain convolution

(f ∗ g)(t) = f(t)

corresponds to frequency domain multiplication

F (jω)G(jω) = F (jω).

If g(t) = δ(t), then G(jω) = 1, so this is sufficient. However, it isn’t necessary. All werequire is that G(jω) = 1 when F (jω) 6= 0. For example, if F (jω) is bandlimited to±ωc, then G(jω) = rect(ω/(2ωc)) would also work. Hence the assertion is False.

F (j )

G (j )1

c− c

This is fortunate. That means that if f(t) is a signal, and g(t) is the impulse responseof a system, we can exactly reproduce f(t) provided f(t) is bandlimited, and the fre-quency response of g(t) is unity across this bandwidth.

(b) If the convolution of two functions f1(t) and f2(t) is identically zero,

(f1 ∗ f2)(t) = 0

then either f1(t) or f2(t) is identically zero, or both are identically zero.

SolutionAgain, the convolution corresponds to frequency domain multiplication, so

F1(jω)F2(jω) = 0.

A zero signal corresponds to a zero transform, so the question is whether F1(jω) orF2(jω) or both must be identically zero.

Certainly, if one or both are zero, the convolution is zero. This is sufficient. However,it is not necessary. All we need is that each be zero where the other is non-zero. Forexample, if f1(t) = sinc(t) and f2(t) = sinc(t) cos(4πt), then Then

F1(jω) = rect(ω/2π)

F2(jω) =1

2(rect((ω + 4π)/2π) + rect((ω − 4π)/2π)) .

The spectrum looks like:

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F1(j )F2(j ) F2(j )

0 2− 2− 4 4

Neither are zero, but the non-zero frequency bands don’t overlap.

The assertion is then False.

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3. Two signals f1(t) and f2(t) are defined as

f1(t) = sinc(2t)

f2(t) = sinc(t) cos(2πt).

Let the convolution of the two signals be

f(t) = (f1 ∗ f2)(t)

(a) Find the Fourier transform F [f(t)] = F (jω).

Solution:

The Fourier transform of f1(t) is

F [f1(t)] = F [sinc(2t)]

=

[1

2rect((ω/2)/2π)

]=

1

2rect(ω/4π)

This is a rect of width 4π centered at ω = 0. The signal f2(t) consists of sinc(t) multi-plied by cos(2πt). The Fourier transform of the sinc is

F [sinc(t)] = rect(ω/2π).

The Fourier transform of f2(t) can then be found using the modulation theorem,

F [sinc(t) cos(2πt)] =1

2[rect((ω − 2π)/2π) + rect((ω + 2π)/2π)]

These are two shifted rect’s of width 2π, centered at ω = ±2π

By the convolution theorem

F [(f1 ∗ f2)(t)] = F1(ω)F2(ω)

=1

4rect(ω/4π) (rect((ω − 2π)/2π) + rect((ω + 2π)/2π)) .

This can be simplified, but we’ll leave that to (b).

(b) Find a simple expression for f(t).

Solution:

To simply F1(jω)F2(jω) it is useful to plot the two functions, and their product,

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1/2

1/2

1/2F1( jω)F2( jω)

F1( jω)

F2( jω)

ω0 2π!2π!4π 4π

ω0 2π!2π!4π 4π

ω0 2π!2π!4π 4π

The latter can be written as two shifted rects, each of width π and amplitude 1/4,centered at ±3π/2

F1(jω)f2(jω) =1

4[rect((ω + 3π/2)/π) + rect((ω − 3π/2)/π)] .

This is similar to F2(ω) except the width of the rects and their shifts are different.We will expect the answer to be a modulated sinc similar to f2(t), but with differentscaling factors.

We know the Fourier transform pair sinc(t)⇔ rect(ω/2π), so by scaling,

1

2sinc(t/2)⇔ rect(ω/π)

The inverse transform is found by the modulation theorem, applied in the inversedirection,

F−1 [F1(ω)F2(ω)] =

(1

4

)(1

2

)sinc(t/2) (2 cos(3πt/2))

=1

4sinc(t/2) cos(3πt/2).

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4. Generalized Parseval’s Theorem

(a) Given two possibly complex signals f1(t) and f2(t) with Fourier transforms F1(jω)

and F2(jω), show that∫ ∞−∞

f1(t)f∗2 (t)dt =

1

2π

∫ ∞−∞

F1(jω)F ∗2 (jω)dω

This is another form of Parseval’s theorem, which reduces to the form we discussedin class if f1(t) = f2(t).

Solution:

This proceeds exactly at the proof of Paresval’s theorem did in class∫ ∞−∞

f1(t)f∗2 (t) dt =

∫ ∞−∞

f1(t)

(1

2π

∫ ∞−∞

F2(jω)ejωtdω

)∗dt

=1

2π

∫ ∞−∞

f1(t)

∫ ∞−∞

F ∗2 (jω)e−jωtdω dt

=1

2π

∫ ∞−∞

F ∗2 (jω)

∫ ∞−∞

f1(t)e−jωtdt dω

=1

2π

∫ ∞−∞

F ∗2 (jω)F1(jω)dω

(b) If f1(t) and f2(t) are real, show∫ ∞−∞

f1(t)f2(t)dt =1

2π

∫ ∞−∞

F1(jω)F2(−jω)dω

Solution:

If f2(t) is real, then f2(t) = f∗2 (t). Also, F2(jω) Hermitian symmetric, so

F ∗2 (jω) = F2(−jω)

Applying these to the results of (a),∫ ∞−∞

f1(t)f2(t)dt =1

2π

∫ ∞−∞

F1(jω)F2(−jω)dω

(c) Use this result to show that

∫ ∞−∞

sinc(t− n)sinc(t− k) dt =

{1 k = n

0 k 6= n

where n and k are integers. This shows that shifted sincs are orthogonal functions!This will be very important when we discuss reconstructing a continuous signal fromits samples.

Hint: Recall that the integral of a complex exponential over an integer number ofperiods is zero.

Solution:

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Using the results from the previous section, the fact that the transform of sinc(t) isrect(ω/2π), and the shift theorem, we get∫ ∞

−∞sinc(t− n)sinc(t−m)dt =

1

2π

∫ ∞−∞

rect(ω/2π)ejnωrect(−ω/2π)e−jmωdω

=1

2π

∫ ∞−∞

rect(ω/2π)ej(n−m)ωdω

=1

2π

∫ π

−πej(n−m)ωdω

If n = m, ∫ ∞−∞

sinc(t− n)sinc(t−m)dt =1

2π

∫ π

−πdω

=1

2π2π

= 1

If n 6= mwe are integrating over an integer number of cycles of a complex exponential,which results in 0. The result is∫ ∞

−∞sinc(t− n)sinc(t−m)dt =

{1 n = m

0 n 6= m

Hence, the family of sincs shifted by integer delays are an orthogonal set of signals.

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5. Spectral Widths Let x(t) be a signal whose spectrum is identically zero outside the range−ω0 ≤ ω ≤ ω0. An example of such a spectrum is shown below.

X (jω)

ω0−ω0 ω

For the following signals, determine the range over which their spectrum is non-zero.

a) y(t) = x(t) + x(t− 1).

Solution:By the shift theorem and linearity,

F [x(t) + x(t− 1)] = X(jω) +X(jω)e−jω = X(jω)(1 + e−jω)

The factor 1 + e−jω is never zero except at the isolated points ω = (2n + 1)π. Hence,the spectral width is determined by X(jω).

b) y(t) =dx(t)

dt.

Solution:By the derivative theorem

F[d

dtx(t)

]= jωX(jω)

Again jω is never zero except for the isolated point ω = 0, so the spectral width isdetermined by X(jω).

c) y(t) = x(t) cos(ω0t).

Solution:By the modulation theorem

F [x(t) cos(ω0t)] =1

2X(j(ω + ω0)) +

1

2X(j(ω − ω0))

The first term on the left is centered at −ω0 and has a width 2ω0, so it goes from−2ω0 < ω < 0. The term on the right is centered at ω0 and has a width 2ω0, so it goesfrom 0 < ω < 2ω0. Both terms together then go from −2ω0 < ω < 2ω0.

d) y(t) = x(t)ejb0t, where b0 is a positive real constant.

Solution:By the exponential modulation theorem

F[x(t)ejb0t

]= X(j(ω − b0))

The spectrum is shifted to the right by b0.

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e) y(t) = x3(t) ∗ x2(t).

Solution:Multiplcation in the time domain is convolution in the frequency domain. F

[x2(t)

]is then twice the width of F [x(t)], and goes from −2ω0 < ω < 2ω0. F

[x3(t)

]is three

times the width of F [x(t)], and goes from −3ω0 < ω < 3ω0.

By the convolution theorem

F[x2(t) ∗ x3(t)

]= F

[x2(t)

]F[x3(t)

]This will be non-zero only when the narrower of the two terms is non-zero. This isF[x2(t)

], so the width of the result is then −2ω0 < ω < 2ω0.

6. Invertibility of Systems

a) Consider two LTI systems with impulse responses h(t) and g(t), respectively, and sup-pose that these systems are inverses of one another. Suppose also that the systems havefrequency responses denoted by H(jω) and G(jω), respectively. What is the relationshipbetween H(jω) and G(jω)?

SolutionIf two systems are inverses of each other,

h(t) ∗ g(t) = δ(t)

In frequency domain:H(jω)G(jω) = 1

That meansG(jω) =

1

H(jω)

b) Consider the continuous time LTI system with frequency response

H(jω) =

{1 if 2 < |ω| < 3

0 otherwise

Is it possible to find an input x(t) to this system such that the output y(t) is given by:

y(t) =

{t if 0 ≤ t ≤ 1

0 otherwise

If so, find x(t). If not, explain why.

SolutionWe show that the system is uncapable of producing this y(t), so there can not exist any x(t)

such that x(t) ∗ h(t) = y(t).

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Note that H(jω) eliminates all frequencies outside of the range 2 < |ω| < 3. However, wesee that y(t) contains frequencies outside of this range. For example,

Y (j(0)) =

∫ ∞−∞

y(t)dt =1

2

Any output of the system will have X(j(0))H(j(0)) = 0, hence y(t) is not a legal output ofthis system.

c) Is the system invertible? Explain your answer.

SolutionThe system is not invertible. Information outside of the range 2 < |ω| < 3 is permanentlydestroyed. Another way to see this is to note that 1

H(jω) is not defined for all ω (because ofdivision by zero).

7. Amplitude Modulation Considerations

Let x(t) be a signal for whichX(jω) = 0 when |ω| > 1000π. Supposing that y(t) = x(t)ejωct,answer the following questions:

a) What constraint should be placed on ωc to ensure that x(t) is recoverable from y(t)?

SolutionSpectrum of y(t) is not distorted by modulation. Reconstruction is possible for any ωc, bymodulating by e−jωct. No constraint is necessary.

b) What constraint should be placed on ωc to ensure that x(t) is recoverable from real{y(t)}?

Solution

real{y(t)} =1

2(y(t) + y∗(t))↔ 1

2(Y (jω) + Y ∗(−jω))

When ωc < 1000π, spectrum of real{y(t)} is distorted on |ω| < 1000π − ωc.Accurate recovery is impossible because of this distortion. To eliminate it, ωc must begreater than 1000π.

8. Response of compound systems

H1(jω)

H2(jω)

G1(jω)

G2(jω)

F1(jω)

F2(jω)

+ -

-

+

-

+ +

+

α1

α2

x(t) y(t)

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Derive the overall system response H(jω) in frequency domain.Express it in terms of α1, α2, H1(jω), H2(jω), G1(jω), G2(jω), F1(jω), and F2(jω).

SolutionWe define intermediate signals v(t) and w(t) (input and output of small feedback loop).

H1(jω)

H2(jω)

G1(jω)

G2(jω)

F1(jω)

F2(jω)

+ -

-

+

-

+ +

+

α1

α2

x(t) y(t)v(t) w(t)

We can write:W (jω) = F1(jω)V (jω)− F1(jω)F2(jω)W (jω)

W (jω) =F1(jω)

1 + F1(jω)F2(jω)V (jω)

Also,V (jω) = X(jω)− α1V (jω)H1(jω)− α2V (jω)H2(jω)

V (jω) = X(jω)1

1 + α1H1(jω) + α2H2(jω)

Write the expression for output:

Y (jω) = W (jω) + α1G1(jω)V (jω) + α2G2(jω)V (jω)

Write W (jω) in terms of V (jω):

Y (jω) =F1(jω)

1 + F1(jω)F2(jω)V (jω) + α1G1(jω)V (jω) + α2G2(jω)V (jω)

Finally, write V (jω) in terms of X(jω):

Y (jω) =

F1(jω)1+F1(jω)F2(jω)

+ α1G1(jω) + α2G2(jω)

1 + α1H1(jω) + α2H2(jω)

9. Uncertainty Principle (based on Vetterli and Kovacevic Ch. 2)You are not required to submit the solution for this problem

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We’ve seen that when time-domain signal x(t) is narrow/short, X(jω) is extended/broad.Sharpness in time has to be traded off for sharpness in frequency (and vice versa). Thereis no way to get arbitrarily sharp in both domains. In this problem we will derive lowerbounds on resolution in both domains.

Consider time domain signal x(t) with unit energy:∫ ∞−∞|x(t)|2dt = 1

which is also centered at the origin in time and frequency∫ ∞−∞

t|x(t)|2dt = 0

∫ ∞−∞

ω|X(jω)|2dω = 0

This can always be obtained by appropriate translation/scaling - we don’t lose generaliza-tion by enforcing these properties.

Define the time and frequency widths by

∆t2 =

∫ ∞−∞

t2|x(t)|2dt

∆ω2 =

∫ ∞−∞

ω2|X(jω)|2dω

Consider real-valued x(t) that vanishes faster than 1√t

as |t| → ∞.That is, 1√

tx(t) = 0 for t→∞ and t→ −∞.

We first show that ∆t2∆ω2 ≥ π2 - that is, derive a lower bound on sharpness of signal in

time-domain and frequency.

Consider the integral of tx(t)x′(t):

By Cauchy-Schwarz inequality:∣∣∣ ∫ ∞−∞

tx(t)x′(t)dt∣∣∣2 ≤ ∫ ∞

−∞|tx(t)|2dt

∫ ∞−∞|x′(t)|2dt

a) Use Parseval relation to express∫∞−∞ |x′(t)|2dt in terms of frequency domain signal. Use

also time-derivative property of Fourier transform and finally express the result in termsof ∆ω2. Also express

∫∞−∞ |tx(t)|2dt in terms of time-width ∆t2.

SolutionSince x(t)↔ X(jω), x′(t)↔ jωX(jω). Then∫ ∞

−∞|x′(t)|2dt =

1

2π

∫ ∞−∞|jωX(jω)|2dω =

1

2π

∫ ∞−∞

ω2|X(jω)|2dω

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∫ ∞−∞|x′(t)|2dt =

1

2π∆ω2

b) Show that x(t)x′(t) = 12ddt [x

2(t)], substitute the result into the integral and integrate∫∞−∞ tx(t)x′(t)dt by parts.

Use the fact that∫∞−∞ |x(t)|2dt = 1 and tx2(t) = 0 when t→ ±∞ (by assumptions).

You should arrive at a constant.

Solution

1

2

d

dt[x2(t)] = 2x(t)x′(t)

∫ ∞−∞

1

2td

dt[x2(t)]dt =

1

2tx2(t)

∣∣∣∞−∞− 1

2

∫ ∞−∞

x2(t)dt = −1

2

c) Having evaluated∫∞−∞ tx(t)x′(t)dt, express the result∣∣∣ ∫ ∞−∞

tx(t)x′(t)dt∣∣∣2 ≤ ∫ ∞

−∞|tx(t)|2dt

∫ ∞−∞|x′(t)|2dt

only in terms of ∆t2, ∆ω2, and scaling factors.You should arrive at the original lower bound which we were trying to prove.

Solution

∣∣∣− 1

2

∣∣∣2 ≤ ∆t2∆ω2 1

2ππ

2≤ ∆t2∆ω2

For Cauchy-Schwarz inequality to hold with equality, functions must be equal within amultiplicative factor of each other, that is x′(t) = ktx(t). Thus, x(t) is of the form

x(t) = cekt2

2

Specifically,

x(t) =

√α

πe−αt

2

This result says that Gaussian functions are maximally concentrated in both time and fre-quency.