6.003(Spring2019) March 7, 2019 · 6.003(Spring2019) March 7, 2019 Quiz#1 Name: KerberosUsername: Enterallanswersintheboxesprovided. Other work on pages with QR codes may …

6.003 (Spring 2019) March 7, 2019

Quiz #1

Name:Kerberos Username:

Enter all answers in the boxes provided.Other work on pages with QR codes may be considered when assigningpartial credit.

You have one hour and fifty minutes.This quiz is closed book, but you may use one 8.5× 11 sheet of paper (two sides).No calculators, computers, cell phones, music players, or other electronic devices.

Quiz #1 / 6.003: Signal Processing (Spring 2019) 2

1. Series [20 points]Part 1.Consider a signal x1[n] = 3(−j)n + 7ejπn/4

What are the Fourier series coefficients of x1[·], when analyzed with N equal to thefundamental period of x1? Simplify your answer completely so that it does not containany expressions involving e, sin, cos.

X1[k] = 7 if n = −7, 1, 9, 17, . . .

3 if n = −2, 6, 14, . . .0 otherwise

The DTFS synthesis equation expresses a time-domain signal in terms of a sum ofcomplex exponentials:

x[n] =∑n=〈N〉

X[k]ej2πkN n

In our case, we have N = 8, so summing over one period looks like:

x1[n] =X[−3]ej−6π

8 n +X[−2]ej−4π

8 n +X[−1]ej−2π

8 n

+X[0] +X[1]ej2π8 n +X[2]ej

4π8 n + +X[3]ej

6π8 n +X[4]ej

8π8 n

The key thing to notice is that, if we rewrite (−j)n as e−j π2n, the given form of x1[·] isalready in this form:

x1[n] = 3e−jπ2n + 7ej

π4n

So we need to find the appropriate X[k] to make the form in the first line of this solutionlook like the above. In order to do this, we need X[−2] to be 3, and X[1] to be 7.The solution above also includes the notion that the DTFS coefficients are periodic inN = 8.


Part 2.Consider a function x2(·) that is periodic in T = 2 seconds. Its CTFS coefficients X2[·]are shown below:

X2[k] = 1/2 if k ∈ 0, 2

1 if k = 10 otherwise

k

X2[k]

0

You may assume that any coefficients that are not pictured above are 0.

On the axes below, plot the magnitude and phase of x2(t) as functions of t, and label allkey values:

0

∣∣x2(t)∣∣

t

2

−2 −1 1 2

∠x2(t)

t

π

π

−2 2

Description of method on following page.


We can start by considering a slightly different signal x′2(·), whose FS coefficients aresymmetric about k = 0:

X ′2[k] = 1/2 if k = ±1

1 if k = 00 otherwise

This corresponds to a pattern we’ve seen before: the delta at k = 0 gives us a DC offsetof 1, and the two coefficients at ±1 give us a cosine. All things considered, this signal’stime-domain representation is:

x′2(t) = 1 + 2 cos(πt)

The signal we’re considering differs from the above in that its coefficients have beenshifted to the right by one. The equivalent time-domain operation is multiplying by apure phase ejπt, so we have:

x2(t) = (1 + 2 cos(πt))ejπt

Since multiplying by ejπt does not affect the magnitude, |x2(t)| is the same as |x′2(t)|,which, because x′2(t) is strictly real and positive, is simply 1 + 2 cos(πt).Because of how complex numbers multiply, ∠x2(t) = ∠x′2(t) + πt. And since x′2(t) isstrictly positive and real, its angle is always 0, we we know that ∠x2(t) = πt.


2. Steps [26 points]Part 1. Let x[n] represent the following discrete-time signal

x[n] =

0 for n < 0a0 for n = 0, 1, 2a1 for n = 3, 4, 5a2 for n = 6, 7, 8· · ·

where a is a real number between 0 and 1, as shown in the plot below.

a0

a1a2

n

x[n]

0 3 6 9 12 15 18 21

Determine a closed form expression for X(Ω), which is the discrete-time Fourier trans-form of x[n].

X(Ω) = 1 + e−jΩ + e−j2Ω

1− ae−3Ω

Solution on following page.


Consider a related, but simpler signal:

x′[n] =a(n/3) if nmod 3 ≡ 00 otherwise

The Fourier transform of this signal is:

X ′(Ω) =∞∑m=0

ame−j3Ωm = 11− ae−j3Ω

x[n] = x′[n] + x′[n− 1] + x′[n− 2], so, by linearity and the time shift property:

X(Ω) =(

1 + e−jΩ + e−j2Ω)X ′(Ω) = 1 + e−jΩ + e−j2Ω

1− ae−j3Ω

Alternatively, apply the formula directly and simplify:

X(Ω) =∞∑

n=−∞x[n]e−jΩn

=∞∑m=0

am(e−jΩ3m + ejΩ(3m+1) + ejΩ(3m+2)

)=∞∑m=0

ame−jΩ3m(

1 + ejΩ + ej2Ω)

= 1 + e−jΩ + e−j2Ω

1− ae−j3Ω


Part 2. Let x(t) represent the following continuous-time signal

x(t) =

0 for t < 0a0 for 0 ≤ t < 3a1 for 3 ≤ t < 6a2 for 6 ≤ t < 9· · ·

where a is a real number between 0 and 1, as shown in the plot below.

a0

a1a2

t

x(t)

0 3 6 9 12 15 18 21

Determine a closed-form expression forX(ω), which is the continuous-time Fourier trans-form of x(t).

X(ω) =(

1jω

)(1− e−j3ω

1− ae−j3ω

)



Consider a simpler signal: x′(t) = 1 if 0 < n < 3

0 otherwiseThen we have x(t) = x′(t) + ax′(t− 3) + a2x′(t− 6) + a3x′(t− 9) + . . ., so, by linearityand the time shift property:

X(ω) = X ′(ω) + ae−j3ωX ′(ω) + a2e−j6ωX ′(ω) + . . .

= X ′(ω)(1 + ae−j3Ω + a2e−j6ω + . . .)

= X ′(ω)(

11− ae−j3ω

)Solving for X ′(ω), we find:

X ′(ω) =∫ 3

0a0e−jωtdt = e−jωt

−jω

∣∣∣∣30

= 1− e−j3ωjω

So we have X(ω) = X ′(ω)(

11− ae−j3ω

)=(

1− e−j3ωjω

)(1

1− ae−j3ω

)

Alternatively, apply the formula directly and simplify:

X(ω) =∫ ∞−∞

x(t)e−jωt dt

=∫ 3

0a0e−jωt dt+

∫ 6

3a1e−jωt dt+

∫ 9

6a2e−jωt dt+ · · ·

= a0[e−jωt

−jω

]3

0+ a1

[e−jωt

−jω

]6

3+ a2

[e−jωt

−jω

]9

6+ · · ·

= − 1jω

(a0 (e−jω3−e−jω0)+ a1e−jω3 (e−jω3−e−jω0)+ a2e−jω6 (e−jω3−e−jω0))

=(

1− e−j3ωjω

)( ∞∑m=0

(ae−j3ω

)m)

=(

1− e−j3ωjω

)(1

1− ae−j3ω

)


3. Related Signals [20 points]For this problem, let x[·] be a DT signal that is periodic in N = 6. One period of theFourier series coefficients X[·] is 1, 0, j, 0,−j, 0, i.e.,

X[k] = X[k + 6] =

1 if k = 0j if k = 2−j if k = 40 otherwise

Part 1.Consider a new signal y1[·], described by: y1[n] = 7 − 3x[n − 1]. What are the DTFScoefficients of y1[·]?

Y1[0] = 4 Y1[1] = 0 Y1[2] = −3je−j 2π3

Y1[3] = 0 Y1[4] = 3je−j 4π3 Y1[5] = 0

By linearity and the time shift property, Y1[k] = X1[k]− 3X[k]e−j 2πk6 , where x1[n] = 7

for all n, and x[n] is defined as above.If x1[n] = 7, then X1[k] = 7δ[k], so we have:

Y1[k] = 7δ[k]− 3X[k]e−j2πk

6

Substituting for each of the values of k in one period, we find the answers above.


Part 2. Consider another new signal y2[n] = 5(−1)nx[n]. What are the DTFS coeffi-cients of y2[·]?

Y2[0] = 0 Y2[1] = -5j Y2[2] = 0

Y2[3] = 5 Y2[4] = 0 Y2[5] = 5j

We can rewrite the expression above in a slightly different form that makes it easier tosee what is happening:

y2[n] = 5ejπnx[n]

This phase multiplication in the time domain corresponds to a frequency shift by half aperiod in the frequency domain. And we also scale by 5, so:

Y2[k] = 5X[k − 3]

Solving for the particular values of k, we find the answers above.


4. Angular Trends [14 points]Each of the expressions on the left corresponds to a plot on the right. Enter the letterfor that plot in the correponding box.

A x−π π

π

−π

B x−π π

π

−π

C x−π π

π

−π

D x−π π

π

−π

E x−π π

π

−π

F x−π π

π

−π

G x−π π

π

−π

H x−π π

π

−π

I x−π π

π

−π

J x−π π

π

−π

K x−π π

π

−π

∠e−jx : B

∠(1 + 0.8ejx) : E

∠(

1+0.4ejx

2+0.8ejx

): I

∠(1 + ejx) : C

∠(1 + 0.8ej2x) : G

∠(0.9ejx+0.8e−jx) : J

∠(

11+0.8ejx

): F


• ∠e−jx: A complex exponential of the form ejθ has magnitude 1 and angle θ. There-fore, the angle of e−jx is −x, as shown in plot B.

• ∠(1 + 0.8ejx): The number 1 + 0.8ejx is the sum of 1 with a vector of magnitude0.8 and angle x as shown in the following plot.

Re

Im

x

When x is small, the angle of the sum is zero. As x increases, the angle increasesuntil x reaches about 3π/4. At this point, the angle of the sum is on the order ofπ/3. As x increases above 3π/4, the angle of the sum quickly decreases, returningto zero when x = π. From the symmetry of the figure, it follows that the angle ofthe sum is an antisymmetric function of x. Thus the answer is plot E.

• ∠(

1+0.4ejx2+0.8ejx

): Since the denominator is twice the numerator, this is just the angle

of a real number (1/2), which is zero – plot I.

• ∠(1 + ejx):

1 + ejx = e−jx2(ej

x2 + e−j

x2)

= e−jx2 2 cos

(x2

)Thus the angle of 1 + ejx is x/2 for −π < x < π. At x = π the sign of the cosineflips so that angle jumps by π. Thus the answer is plot C.

• ∠(1 + 0.8ej2x): This expression looks like part 2 (above) except x is replaced by 2x.Therefore the answer the same as that for part 2 except that the x-axis is compressedby a factor of 2 – generating plot G.

• ∠(0.9ejx+0.8e−jx): The expression 0.9ejx+0.8e−jx can be simplified by convertingto Cartesian form:

0.9 cos(x) + j0.9 sin(x) + 0.8 cos(x)− j0.8 sin(x) = 1.7 cos(x) + j0.1 sin(x)

The angle is therefore arctan(

0.1 sin(x)1.7 cos(x)

)= arctan

( 117 tan(x)

)which is plot J.

• ∠(

11+0.8ejx

): The expression 1 + 0.8ejx was evaluated in part 2 (above). Here the

expression is in the denominator, so the answer is the negative of the answer to part2 – which yields plot F.


5. CT and DT [20 points]Part 1.Consider taking a periodic signal x(·) (with a period of T = 1 second) and sampling ata sampling rate of 8 samples per second to obtain DT signal x[·] that is also periodic.Analyzing this signal, you find that:

• x[·] is a symmetric function of n.

• x[n] is positive for all values of n.

• the sum of all x[n] in one period is 2.

• if you sum and subtract alternating samples in one period, you get 1; that is,x[0]− x[1] + x[2]− . . .+ x[N − 2]− x[N − 1] = 1.

• most of the Fourier series coefficients X[·] are 0; only two (per period) are nonzero.

What are two distinct CT functions x(·) that could have produced the results shownabove?

x(t) = 14 + 1

8 cos(8πt)

or

x(t) = 14 + 1

8 cos(24πt)



We’ll start by thinking about the DTFS coefficients of the sampled signal. The factthat

∑n=〈N〉

x[n] = 2 tells us that the DC component is:

X[0] = 18

∑n=〈N〉

x[n]

= 18 (2) = 1

4

The other important fact has to do with the alternating sum. We are given that∑n=〈N〉

x[n](−1)n = 1 This tells us about the component at k = 4:

X[4] = 18∑n=〈N〉

x[n]e−jπn = 18

∑n=〈N〉

x[n](−1)n = 1

8 (1) = 18

Since those are the only two non-zero DTFS coefficients, we can determine an expressionfor the discretized signal:

x[n] = 14 + 1

8 cos(πn)

Now we need to determine a CT signal that, when sampled at a rate of 8 samples persecond, gives us the expression above.The DC component is unaffected by sampling, so we just need to find a CT cosine that,when sampled appropriately, gives us cos(πn).We want a DT cosine where Ω = π (rad/sample), and we are given that fs = 8(samples/second). So we need ω (rad/sec) = Ωfs = 8π.Thus, one CT signal that has these properties is given by x(t) = 1

4 + 18 cos(8πt).

We can find another CT signal that aliases down to the proper value by adding 2π toΩ. In order to make Ω = 3π, we need ω = 24π. Thus, another CT signal that has thegiven properties is x(t) = 1

4 + 18 cos(24πt)


Part 2.When computing the DTFS coefficients of a purely real signal with a fundamental periodN , where N is an even number, two coefficients in particular (k = 0 and k = N/2) arealways real-valued (whereas other values can be complex). Why is this the case? Explainbriefly (1-3 sentences).

X[0] = 1N

∑n=〈N〉

x[n]

Since all of the x[n] values are real, their average (and, thus, X[0]) must also be real.

X[N/2] = 1N

∑n=〈N〉

x[n]ejπn

Since all of the x[n] values are real, and ejπn is also real (for integer n, this will alwaysbe −1 or 1), this sum (and, thus, X[N/2]) must also be real.

All other values of k can potentially lead to complex values in the sum.

When computing the DTFS coefficients of a purely real signal with a fundamental periodN where N is odd, are there any coefficients that are similarly guaranteed to be real? Ifso, specify the values of k. Explain briefly (1-3 sentences).

As above, we know that:

X[0] = 1N

∑n=〈N〉

x[n]

Since all of the x[n] values are real, their average (and, thus, X[0]) must also be real.

However, now N/2 is no longer an integer (and, thus, is not represented in the series),so we do not have any other coefficients that are guaranteed to be purely real.


When computing the CTFS coefficients of a purely real CT signal, are there any coeffi-cients that are similarly guaranteed to be real? If so, specify the values of k. Does youranswer depend on the value of T? Explain briefly (1-3 sentences).

Similarly to the DT case, here X[0] still represents the average value over one period:

X[0] =∫Tx(t)dt

Since we’re integrating a purely real function, we must get a purely real answer.

For all other values, since t is a continuous variable, integrating∫T x(t)e

−j2πktT dt could

lead to complex values. Even with k = T/2, the continuous nature of the variable tmeans we are not guaranteed a purely real result.


Worksheet (intentionally blank)













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6.003(Spring2019) March 7, 2019 · 6.003(Spring2019) March 7, 2019 Quiz#1 Name: KerberosUsername: Enterallanswersintheboxesprovided. Other work on pages with QR codes may …