6.003 (spring 2020) may 5, 2020 · practice problems for final exam / 6.003: signal processing...

6.003 (Spring 2020) May 5, 2020

Practice Problems for Final Exam

The questions were taken from multiple previous exams.

Practice Problems for Final Exam / 6.003: Signal Processing (Spring 2020) 2

1 2D ConvolutionFor this problem, wewill consider the following 2D signals, labeled x0 through x10, each of whichis 9 rows × 13 columns. Note that the color scale is different between some of the signals.

= 0= +1

x0

−6 0 64

0

−4= 0= +1

x1

−6 0 64

0

−4= 0= +1

x2

−6 0 64

0

−4= 0= +2

x3

−6 0 64

0

−4

= −1.7= 0= +2

x4

−6 0 64

0

−4= −1.7= 0= +2

x5

−6 0 64

0

−4= 0= +1.1

x6

−6 0 64

0

−4= −1.0= 0= +1.0

x7

−6 0 64

0

−4

= −1.0= 0= +1.0

x8

−6 0 64

0

−4= 0= +1.2

x9

−6 0 64

0

−4= −1= 0= +0.4

x10

−6 0 64

0

−4

For each circular convolution below, indicate which of the graphs on the facing page most closelymatches the result by entering a single letter in each box. Note that, for each graph on the facingpage, black corresponds to the lowest value in the signal (not necessarily 0), andwhite correspondsto the highest value in the signal (not necessarily 1).

x1 ~ x0 F x2 ~ x0 D x3 ~ x0 L

x4 ~ x0 C x5 ~ x0 G x6 ~ x0 K x7 ~ x0 I

x8 ~ x0 A x9 ~ x0 E x10 ~ x0 J


Graph A

−6 0 64

0

−4Graph B

−6 0 64

0

−4Graph C

−6 0 64

0

−4Graph D

−6 0 64

0

−4

Graph E

−6 0 64

0

−4Graph F

−6 0 64

0

−4Graph G

−6 0 64

0

−4Graph H

−6 0 64

0

−4

Graph I

−6 0 64

0

−4Graph J

−6 0 64

0

−4Graph K

−6 0 64

0

−4Graph L

−6 0 64

0

−4

Graph M

−6 0 64

0

−4Graph N

−6 0 64

0

−4Graph O

−6 0 64

0

−4Graph P

−6 0 64

0

−4


2 Transforms

1. Find the Fourier series coefficients of the signal x1(·), analyzed with T chosen to be the funda-mental period of x1(·).

x1(t) = 2 cos(π2t)+ 4 cos

(π3t)

In the box below, write a simple, closed-form answer for X1[k]:

X1[k] = δ[k+ 3] + 2δ[k+ 2] + 2δ[k− 2] + δ[k− 3]

2. Find the Fourier series coefficients of the signal x2[·], shown below, which is periodic inN = 10.

0 10 20n

1x2[n]

In the box below, write a simple, closed-form answer for X2[k]:

X2[k] =1

10

(1+ e−j

2π5 k)


3. Find the Fourier transform of the signal x3(·) as defined below:

x3(t) =1 if −1 6 t 6 20 otherwise

In the box below, write a simple, closed-form answer for X3(ω):

X3(ω) =

(2 sin

(32ω)

ω

)e−j

ω2

4. Find the Fourier transform of the signal x4[·], defined below:

x4[n] = δ[n+ 3] + δ[n+ 1] − δ[n− 1] + δ[n− 3]

In the box below, write a simple, closed-form answer for X4(Ω):

X4(Ω) = 2 cos(3Ω) + 2j sin(Ω)


3 Filtering

3.1 Part 1Sketch the magnitude and phase of the frequency response of a linear, time-invariant system withthe following unit-sample response:

h1[n] = δ[n] + δ[n− 1]

Label all important magnitudes, angles, and frequencies.

−π π−2π 2π

2∣∣∣cos(

Ω2

)∣∣∣

magnitude

Ω

−π π−2π 2π

π

π/2

−π

−π/2


3.2 Part 2Consider an LTI system with a unit sample response h2[·] given by:

h2[n] =1

2δ[n] −

1

2δ[n− 2]

For each of the input signals xi[·] below, assume that the response of the system to that input isgiven by yi[·]. For each, is it possible to represent yi[n] as a single pure sinusoid of the formAi cos(Ωin + φi)? If so, specify the appropriate values of Ai, Ωi, and φi by entering a singlenumber in each box (square roots, π, and fractions are all OK). Otherwise, write none in all threeboxes. If a box is irrelevant, write any in that box.

If x1[n] = 3, is y1 expressible as y1[n] = A1 cos(Ω1n+ φ1)?

A1 = 0 Ω1 = any φ1 = any

If x2[n] = cos(πn/2), is y2 expressible as y2[n] = A2 cos(Ω2n+ φ2)?

A2 = 1 Ω2 = π/2 φ2 = 0

If x3[n] = cos(4πn/3), is y3 expressible as y3[n] = A3 cos(Ω3n+ φ3)?

A3 =√3/2 Ω3 = 4π/3 φ3 = π/6

If x4[n] = sin(πn/3) + cos(πn/3), is y4 expressible as y4[n] = A4 cos(Ω4n+ φ4)?

A4 =√6/2 Ω4 = π/3 φ4 = −π/12

If x5[n] = (−1)n, is y5 expressible as y5[n] = A5 cos(Ω5n+ φ5)?

A5 = 0 Ω5 = any φ5 = any


3.3 Part 3Let x0[n] = δ[n+ 1] + δ[n] + δ[n− 1], and let x1[·] be a scaled and periodically-extended versionof x0[·], with repetitions everyN0 samples:

x1[n] =

∞∑m=−∞

Ax0[n−mN0] =

∞∑m=−∞

Aδ[n−mN0+1]+Aδ[n−mN0]+Aδ[n−mN0−1]

As an example of the general shape of this function, here is an example withN0 = 17 (though youshould not assume thatN0 = 17 throughout the problem):

n

x1[n]

0 N0−N0 2N0−2N0

A

Also let x2[n] = Bcos(Ω0n) for some valueΩ0, and let x3[n] = x1[n] + x2[n].

Each of the plots on the facing page shows the (purely real) DTFT of some function. Which of thegraphs (1-6) corresponds to X3(Ω), the DTFT of x3[n] = x1[n] + x2[n]? And what are the valuesof A,N0, B, andΩ0? Enter a single number in each box below:

Graph number: 5

A = 83π

N0 = 8

B = 2π

Ω0 = π3


Graph 1

Ωπ−π

2π3

−2π3

π2

−π2

π3

−π3

5Graph 2

Ωπ

−π 2π3

−2π3

π3

−π3

π2

−π2

5

−5

Graph 3

Ωπ−π

2π3

−2π3

π2

−π2

π3

−π3

6Graph 4

Ωπ

−π 2π3

−2π3

π3

−π3

π2

−π2

6

−6

Graph 5

Ωπ−π

2π3

−2π3

π2

−π2

π3

−π3

2Graph 6

Ωπ

−π 2π3

−2π3

π3

−π3

π2

−π2

2

−2

Graph 7

Ωπ−π 2π

3−2π

3π2

−π2

π3

−π3

3Graph 8

Ωπ

−π

2π3

−2π3

π3

−π3

π2

−π2

3

−3


4 DFTFor both parts of this problem, we’ll consider a signal x[·], whose values are shown in the tablebelow:

n −3 −2 −1 0 1 2 3 4 5 6 7 8 9 10 11

x[n] 0 0 0 3 −1 0 2 1 0 −2 −1 2 0 0 0

You may assume that x[n] = 0 for all values of n not explicitly represented in the table above.

4.1 Part 1

Let X(·) represent this function’s DTFT, and let R[k] = X(2πk

4

)Fill in the table below with the values of r[n], the inverse DFT of R[k] (computed withN = 4).

n −3 −2 −1 0 1 2 3 4 5 6 7 8 9 10 11

r[n] −4 −8 4 24 −4 −8 4 24 −4 −8 4 24 −4 −8 4

4.2 Part 2Let X[·] represent the DFT of x[·] (computed with N = 9), and let H[·] be the DFT ofh[n] = δ[n] − δ[n− 4] (also computed withN = 9).

If we define Y[·] such that Y[k] = 9×X[k]×H[k], fill in the table below with the values of y[n], theinverse DFT of Y[k] (also computed withN = 9)

n −3 −2 −1 0 1 2 3 4 5 6 7 8 9 10 11

y[n] −2 −3 1 3 1 1 0 −2 1 −2 −3 1 3 1 1


5 SurveyingFarmland in the midwestern United States is often divided into “sections” of uniform size. Whenviewed from above, this land often looksmuch like a grid, as can be seen in the photograph below:

The following shows the magnitudes of a portion of the DFT coefficients of a different satellitephotograph of farmland, taken from directly overhead.

40 20 0 20 40

40

30

20

10

0

10

20

30

40 0.0

0.2

0.4

0.6

0.8

1.0

In order to answer the questions on the following page, you may assume that the sections in theimage are nearly identical, nearly rectangular (but not necessarily square), and laid out on a nearlyperfect grid. Given this information, answer the questions on the following page about the imagewhose DFT coefficients are shown above.


Approximately howmany sections of land are there in the imagewhoseDFT coefficients are shown

on the previous page? ≈ 112

Explain your method.

We can view this DFT as something like the sum of two DFTs, where one is given by, approx-imately:

F1[kx, ky] =

∞∑m=−∞

δ[kx − 14m]δ[ky]

and the other is given by:

F2[kx, ky] =

∞∑m=−∞

δ[ky − 8m]δ[kx]

The first tells us we have something that is repeating around 14 times over the course of theimage horizontally, and the second tells us that we have something that is repeating around8 times vertically. Given this, we can infer that there are around 14 sections horizontally, andaround 8 sections vertically. In total, this gives us around 112 sections.

If the image represented an area that was 10.5 miles wide and 4.0 miles tall, what are the approx-imate dimensions of each of the sections of land, in miles?

Height: ≈ 0.5 mi Width: ≈ 0.75 mi

Explain your method.

Using the intermediate results from above, the width of a section is10.5 mi14 sections = 0.75

misection .

Similarly, the height of a section is around 4 mi8 sections = 0.5

misection


6 Convolutions

Part 1Let x1 represent the discrete-time signal given by the following expression.

x1[n] = 0.9nu[n]

The result of convolving x1 with an unknown signal h1 is the unit sample signal,

(x1 ∗ h1)[n] = δ[n]

as illustrated in the following figure.

n

x1[n]

* ? =n

δ[n]

Determine an expression (not necessarily in closed form) for h1[n].

h1[n] = δ[n] − 0.9δ[n− 1]

On the axes below, plot the magnitude and phase of H1(Ω). In each sketch, label all importantvalues.

Ω

|H1(Ω)|

Ω

∠H1(Ω)1.9

0.1−π π −π π

π/4

−π/4

π/2

−π/2

Key features of magnitude plot:• Maximum values of |H(Ω)| atΩ = ±π.• Minimum value of |H(Ω)| atΩ = 0.• Maximum value of 1.9, minimum value of 0.1.• Smooth curve.

Key features of angle plot:• ∠H(Ω) = 0 atΩ = integer multiples of π.• Maximum ∠H2(Ω) closer to π/4 than π/2.• Steeper slope atΩ = 0 than atΩ = π.


Part 2Let x2 represent the discrete-time signal given by the following expression.

x2[n] = δ[n] − 0.8δ[n− 2]

The result of convolving x2 with an unknown signal h2 produces a new signal (y2[n])

(x2 ∗ h2)[n] = y2[n] =(−0.8)n/2u[n] for n even0 otherwise

as illustrated below.

n

x2[n]

* ? =n

y2[n]

Determine an expression (not necessarily in closed form) for h2[n].

h2[n] =

∞∑i=0

0.82iδ[n− 4i]

On the axes below, plot the magnitude and phase of H2(Ω). In each sketch, label all importantvalues.

Ω

|H2(Ω)|

Ω

∠H2(Ω)1

0.36 ≈ 3,

11.64 ≈ 3/5

−π −3π4

−π2

−π4

π3π4

π2

π4

π/4

−π/4−π

−3π4

−π2

−π4 π

3π4

π2

π4

Key features of magnitude plot:• Maximum values of |H(Ω)| atΩ = integer multiples of π/2.• Minimum values of |H(Ω)| atΩ = π/4, 3π/4.• Maximum value around 3, minimum value around 3/5.• Smooth curve.

Key features of angle plot:• ∠H(Ω) = 0 atΩ = integer multiples of π/4.• ∠H2(Ω) never reaches π/4 or −π/4.• Steeper slope at even integer multiples of π/4 than at odd integer multiples of π/4.


7 ScalingH(ω) ×x(t) y(t)

w(t)Let x(t) represent an infinite sequence of pulses, which is passed through an LTI filter H(ω) andthen multiplied byw(t) to produce y(t), as shown on the right.

x(t)t

w(t)t

|H(ω)|ω

|Y (ω)|ω

Only one of the previous signals is changed in each of the following parts. Identify the corre-sponding result from the list on the next page (original |Y(ω)| shown for reference).

Part a. The peak of H(ω) is shifted to a higher frequency. 1, 2, ... 10, or none: 5

|H(ω)|ω

Part b. The duration of each pulse in x(t) is doubled. 1, 2, ... 10, or none: 1

x(t)t

Part c. The period of x(t) is halved. 1, 2, ... 10, or none: 2

x(t)t

Part d. The duration ofw(t) is doubled. 1, 2, ... 10, or none: 4

w(t)t


|Y (ω)|ω

|Y1(ω)|ω

|Y2(ω)|ω

|Y3(ω)|ω

|Y4(ω)|ω

|Y5(ω)|ω

|Y6(ω)|ω

|Y7(ω)|ω

|Y8(ω)|ω

|Y9(ω)|ω

|Y10(ω)|ω


8 Fourier Transforms

8.1 Part 1The diagrams on the facing page show five DT signals (x1 through x5), six DTFT magnitude plots(labeled A through F), and six DTFT angle plots (labeled a through f).

In the boxes below, enter the letter corresponding to the graph that matches the magnitude andphase of each signal’s DTFT. If none of the graphs matches, write none.

Which graph, if any, shows |X1(Ω)|? (A-F, or none) D

Which graph, if any, shows ∠ (X1(Ω))? (a-f, or none) b

X1(Ω) = 2 cos(Ω) − 1

Which graph, if any, shows |X2(Ω)|? (A-F, or none) F

Which graph, if any, shows ∠ (X2(Ω))? (a-f, or none) c

X2(Ω) = 2 cos(2Ω) + 1

Which graph, if any, shows |X3(Ω)|? (A-F, or none) F

Which graph, if any, shows ∠ (X3(Ω))? (a-f, or none) none

x3[n] = x2[n− 1], so X3(Ω) = X2(Ω)e−jΩ. Thus, |X3(Ω)| = |X2(Ω)|.

We also have ∠X3(Ω) = ∠X2(Ω)−Ω. So we are looking for a phase that is linearly decreasingas Ω increases. Graph f almost looks right (the phase is linear in Ω), but it doesn’t havejumps by ±π at the same points that graph c does, so the correct answer is none.

Which graph, if any, shows |X4(Ω)|? (A-F, or none) C

Which graph, if any, shows ∠ (X4(Ω))? (a-f, or none) e


X4(Ω) = 1− 2 cos(2Ω)

Which graph, if any, shows |X5(Ω)|? (A-F, or none) C

Which graph, if any, shows ∠ (X5(Ω))? (a-f, or none) a

x5[n] = −x4[n], so X1(Ω) = −X3(Ω). Thus, its magnitude should be the same as |X2(Ω)|,and its phase graph should have the same rough shape as X4’s, but with 0 and π swapped.


n

x1[n]

Ω

A

−π 0 πΩ

a

−π π

−π

π

n

x2[n]

Ω

B

−π 0 πΩ

b

−π π

−π

π

n

x3[n]

Ω

C

−π 0 πΩ

c

−π π

−π

π

n

x4[n]

Ω

D

−π 0 πΩ

d

−π π

−π

π

n

x5[n]

Ω

E

−π 0 πΩ

e

−π π

−π

π

Ω

F

−π 0 πΩ

f

−π π

−π

π


8.2 Part 2

8.2.1 f1(·)Let f1(·) be the signal represented in the graph below, which is assumed to be 0 outside the rangeshown:

0 6

1

t

f1(t)

Enter a closed-form expression for the Fourier transform of f1(·) below.

F1(ω) =

(2 sin (3ω)

ω

)e−j3ω

We have previously solved for the Fourier transform of a rectangular pulse centered on t = 0with a height of 1 and a half-width of h, and we found its Fourier transform to be 2sin(hω)

ω .The above can be thought of as a time-shifted version of that kind of rectangular pulse.

Specifically, let a1(t) =1 if −3 6 t 6 30 otherwise

We know, then, that A1(ω) = 2sin(3ω)ω .

We can then define f1(t) in terms of a1(·):

f1(t) = a1(t− 3)

So, by the time-shift property,

F1(ω) = A1(ω)e−j3ω =

(2 sin (3ω)

ω

)e−j3ω

We could also have solved by integrating directly:

F1(ω) =

∫∞−∞

f1(t)e−jωtdt =

∫60

e−jωtdt =1

−jω

(e−j6ω − 1

)If we wanted to express this in the same form as above (using sin), we could do so by pullinga factor of e−j3ω from each of the terms in the parentheses above.



−1

1

5

−5

t

f2(t)

Enter a closed-form expression for the F2(ω) in terms of F1(·).

F2(ω) =5

6

(F1

(−ω

6

)− F1

(ω6

))

f2(t) = 5 (f1(−6t) − f1(6t))

Applying the properties of linearity, time scaling, and time reversal yields the form above.



−1 1

5

t

f3(t)


F3(ω) =F2(ω)

jω

We can express f3(t) as the integral of f2(t):

f3(t) =

∫t−∞

f2(u)du.

It follows from the derivative property of the Fourier transform that F2(ω) can be computedfrom F3(ω):

F2(ω) = jωF3(ω)

and this suggests that F3(ω) = 1jωF2(ω). However, dividing byω raises questions about the

validity of this expression atω = 0. If F2(0) = 0, then the F3(0) might converge, and we caneasily test for convergence using L’Hopital’s rule. However if F2(0) is not 0, what happensthen?

If F2(0) is not zero, then f2(t) has a DC component: i.e., the integral of f2(t) over all t isnon-zero. Since f3(t) is the integral of f2(t), the DC component of f3(t) contains the integralof the constant DC component in f2(t). As a result, F3(0) is infinite: i.e., F3(ω) contains animpulse atω = 0.

Fortunately, this complication does not occur in this problem. Here f2(t) has no DC compo-nent, so there is no impulse in F3(ω). Furthermore, L’Hopital’s rule shows that F2(ω)

jω is finite.It follows that F3(ω) = F2(ω)

jω .


8.2.4 f4(·)Let f4(·) be the signal represented in the graph below, which continues the pattern shown for allt:

−10

(67)5

−9 −8

(67)4

−7 −6

(67)3

−5 −4

(67)2

−3 −2

(67)1

−1 0 1 2

(67)1

3 4

(67)2

5 6

(67)3

7 8

(67)4

9 10

(67)5

1

t

f4(t)


F4(ω) =

(1

1− 67e

−j2ω+

1

1− 67ej2ω

− 1

)F3(ω)

5

There aremultipleways to approach this problem; we’ll walk through one below. Imporantly,notice that this signal consists of a sum of scaled and shifted copies of f3(t).

Let’s define an intermediate signal:

a1(t) =

∞∑m=0

(6

7

)mf3(t− 2m)

5

By linearity and the time-shift property, we then have:

A1(ω) =F3(ω)

5

∞∑m=0

(6

7

)me−j2ωm =

F3(ω)

5

(1

1− 67e

−j2ω

)

We can then define f4(t) in terms of a1: f4(t) = a1(t) + a1(−t) −f3(t)5

Note that we have to subtract f3(t)/5 off to avoid "double-counting" the triangle in themiddle(which would be a part of both a1(t) and a1(−t).

From here, applying linearity and the time-reversal property, we know that:

F4(ω) = A1(ω) +A1(−ω) − F3(ω)

=F3(ω)

5

(1

1− 67e

−j2ω

)+F3(−ω)

5

(1

1− 67ej2ω

)−F3(ω)

5

Since f3(·) is a symmetric and real function of t, we expect F3(·) to be a symmetric and realfunction of ω, so we know that F3(−ω) = F3(ω). Making that substitution in the equationabove and factoring out the F3(ω)

5 gives us the form in the answer box above.


9 FiltersLet H represent a system whose output y[n] is determined from a linear combination of currentand previous values of the input x[n] and output y[n].

Hx[n] y[n]

For each of the following parts, determine the frequency response for the given linear combination.Then identify which of plots A to L or none on the following page corresponds to the magnitudeand which of plots M to U or none on that same page corresponds to the angle of the frequencyresponse.

magnitude anglelinear combination (A-L or none) (M-U or none)

Part 1. y[n] = 12x[n] +

12y[n− 1] F N

Part 2. y[n] = 12x[n] −

12y[n− 2] D P

Part 3. y[n] = 14x[n− 1] + 1

2x[n] +14x[n+ 1] C R

Part 4. y[n] = −13x[n− 1] + 13x[n] −

13x[n+ 1] E O

Part 5. y[n] = 13x[n− 2] + 1

3x[n] +13x[n+ 2] K T


Ω

A

−π πΩ

B

−π πΩ

C

−π π

Ω

D

−π πΩ

E

−π πΩ

F

−π π

Ω

G

−π πΩ

H

−π πΩ

I

−π π

Ω

J

−π πΩ

K

−π πΩ

L

−π π

Ω

M

−π π

π

Ω

N

−π π

π

Ω

O

−π π

π

Ω

P

−π π

π

Ω

Q

−π π

π

Ω

R

−π π

π

Ω

S

−π π

π

Ω

T

−π π

π

Ω

U

−π π

π


10 Find the FundamentalLet x[n] represent a periodic, discrete-time signal with a period of 12 as shown below.

n

x[n] = x[n+ 12]

0 12

1

2

A new signal y[n] is derived from x[n] by inserting 2 zeros between adjacent samples of x[n] sothat the new sequence is given by

y[n] =x[n/3] if n is an integer multiple of 30 otherwise

Since y[n] is periodic, it can be described by a Fourier series of the following form

y[n] = c0 +

N/2∑k=1

ck cos2πkn

N+

N/2∑k=1

dk sin2πkn

N

Find c1, d1, and the constantN.

c1 = 0

d1 = 16

N = 36

The period of y[n] is 3 times the period of x[n]. ThusN = 36.

Let ak represent the Fourier series coefficients of y[n].

ak =1

36

∑n=〈36〉

x[n]e−j 2π36nk =1

36e−j

2π3k36 +

2

36e−j

2π9k36 +

1

36e−j

2π15k36

a1 =1

36

(e−j

2π336 + 2e−j

2π936 + e−j

2π1536)=1

36

(e−j

π6 + 2e−j

π2 + e−j

5π6)= −j

1

12

a−1 = a∗1 = j

1

12

a1ej 2πn36 +a−1e

−j 2πn36 = −j1

12

(cos

2πn

36+j sin

2πn

36

)+j1

12

(cos

2πn

36−j sin

2πn

36

)=2

12sin

2πn

36

c1 = 0

d1 =1

6

6.003 (spring 2020) may 5, 2020 · practice problems for final exam / 6.003: signal processing...

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