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Digital Signal Processing 1: Basic Concepts and Algorithms Quiz Answers
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Digital Signal Processing 1: Basic Concepts and Algorithms Week 01 Quiz Answers
Quiz 1: Homework for Module 1.1 Quiz Answers
Q1. (Difficulty: \star⋆) What are the advantages of using digital signals over analog ones? Choose the correct answer(s).
[expand title=View Answer]
Digital signals are more robust to noise.
Processing of digital signals can be easily implemented in modern computers.
Digital signals can be easily stored.
[/expand]
.
Q2. (Difficulty: \star⋆) Amongst the signals listed below, select those that are in digital format.
[expand title=View Answer]
1.JPEG image on a website.
2.Music recorded on a CD.
[/expand]
Q3. (Difficulty: \star⋆ \star⋆) Consider the signal
x[n] = \delta[n] + 2\delta[n − 1] + 3\delta[n − 2].x[n]=δ[n]+2δ[n−1]+3δ[n−2].
Consider now its moving average, i.e. the signal
- y[n]=\frac{1}{2}(x[n]+x[n-1])y[n]=
- (x[n]+x[n−1])
Select the correct expressions from the options below.
y[n]=0.5\delta[n] + 1.5\delta[n − 1] + 2.5\delta[n − 2]+1.5\delta[n-3]y[n]=0.5δ[n]+1.5δ[n−1]+2.5δ[n−2]+1.5δ[n−3].
The output for n\ge 0n≥0 is always zero.
[expand title=View Answer] The output for n ≥ 0 is always zero. [/expand]
Q4. (Difficulty: \star⋆) A music song recorded in a studio is stored as a digital sequence on a CD. The analog signal representing the music is 2 minutes long and is sampled at a frequency f_s=44100\;s^{-1}f
s
=44100s
−1
. How many samples should be stored on the CD? Write the number of samples without commas or dots. Assume that the audio file is mono, or in other words, single channel.
[expand title=View Answer] 2 minutes * 60 seconds/minute * 44100 samples/second = 5,292,000 samples [/expand]
Q5. (Difficulty: \star \star \star⋆⋆⋆) Given the following filter
What is the input-output relationship?
- y[n]=b(ax[n]+x[n-1])-(c+1)x[n-3]y[n]=b(ax[n]+x[n−1])−(c+1)x[n−3].
- y[n]=abx[n]+bx[n-1]+cx[n-3]-x[n-4]y[n]=abx[n]+bx[n−1]+cx[n−3]−x[n−4].
- y[n]=(bax[n]+x[n-1])+(cx[n-3]+x[n-4])y[n]=(bax[n]+x[n−1])+(cx[n−3]+x[n−4]).
- y[n]=b(ax[n]+x[n-1])-(cx[n-3]+x[n-4])y[n]=b(ax[n]+x[n−1])−(cx[n−3]+x[n−4]).
[expand title=View Answer] y[n] = b(ax[n] + x[n-1]) – (c+1)x[n-3][/expand]
Q6. (Difficulty: \star \star⋆⋆) What is the minimum period PP (in samples) of the signal e^{j(M/N)2\pi n}e
j(M/N)2πn
for M=1,N=3M=1,N=3.
[expand title=View Answer] P = N = 3 samples [/expand]
Week 2 Quiz Answers
Quiz 1: Homework for Module 1.2 Quiz Answers
Q4. Which of the following sets form a basis of \mathbb R^4R?
[expand title=View Answer]
{\mathbf{y},\mathbf{v}_0,\mathbf{v}_1,\mathbf{v}_2}{y,v
{\mathbf{y},\mathbf{v}_0,\mathbf{v}_2,\mathbf{v}_3}{y,v
{\mathbf{y},\mathbf{v}_1+\mathbf{v}_2,\mathbf{v}_2,\mathbf{v}_3}{y,v
{\mathbf{y},\mathbf{v}_0,\mathbf{v}_1,\mathbf{v}_3}{y,v
[/expand]
Q5. (Difficulty: \star⋆) If we represent finite-length signals as vectors in Euclidean space, many operations on signals can be encoded as a matrix-vector multiplication. Consider for example a circular shift in \mathbb{C}
: a delay by one (i.e. a right shift) transforms the signal \mathbf{x} = [x_0 \, x_1 \, x_2]^Tx=[x
into \mathbf{x}’ = [x_2 \, x_0 \, x_1]^Tx
and it can be described by the matrix
D= ⎡⎣010001100⎤⎦
so that \mathbf{x}’ = D\mathbf{x}x
′
=Dx.
Determine the matrix FF that implements the one-step-difference operator in {C}
i.e. the operator that transforms a signal \mathbf{x}x into [(x_0-x_2) \,\, (x_1-x_0) \,\, (x_2-x_1)]^T[(x
Write the 9 integer matrix coefficients one after the other, row by row and separated by spaces.
[expand title=View Answer]
The one-step-difference operator can be implemented using the following matrix F:
F =
[1 0 -1]
[-1 1 0]
[0 -1 1]
So, when you write the 9 integer matrix coefficients one after the other, row by row and separated by spaces, you get:
1 0 -1 -1 1 0 0 -1 1
[/expand]
Q6. (Difficulty \star⋆) Given the matrix
A= ⎡⎣⎢⎢0100001000011000⎤⎦⎥⎥
compute the matrix A^4A
(i.e. the fourth power of AA).
(Hint: there is a simple way to do that and, if you’ve solved the previous question, it should be obvious).
Write the 16 integer matrix coefficients one after the other, row by row, and separated by spaces.
[expand title=View Answer]
[0 0 0 0]
[0 1 0 0]
[0 0 1 0]
[1 0 0 0]
[/expand]
Week 3 Quiz Answers
Quiz 1: Homework for Module 1.3 Quiz Answers
Q1. (Difficulty: \star⋆) Write out the phase of the complex numbers a_1=1-\mathrm{j}a
1
=1−j and a_2=-1-\mathrm{j}a
2
=−1−j.
Express the phase in degrees and separate the two phases by a single white space. Each phase should be a number in the range [-180,180][−180,180].
Enter answer here
Q2. (Difficulty: \star⋆) Let W_N^k=e^{-\mathrm{j}\frac{2\pi}{N}k}W
N
k
=e
−j
N
2π
k
and N>1N>1. Then W^{N/2}_NW
N
N/2
is equal to…
-1
1
-\mathrm{j}−j
e^{-\mathrm{j}(2\pi/N)+N}e
−j(2π/N)+N
Q3. (Difficulty: \star⋆) Which of the following signals (continuous- and discrete-time) are periodic signals?
Note that t\in\mathbb{R}t∈R and n\in\mathbb{Z}n∈Z.
1 point
x[n]=1x[n]=1.
x(t)=\cos(2\pi f_0 t+\phi)x(t)=cos(2πf
0
t+ϕ) with f_0\in\mathbb{R}f
0
∈R.
x(t)=t-\operatorname{floor}(t)x(t)=t−floor(t).
x[n]=e^{-\mathrm{j} f_0 n}+e^{+\mathrm{j} f_0 n}x[n]=e
−jf
0
n
+e
+jf
0
n
, where f_0 = \sqrt{2}f
0
=
2
.
x[n] = \sin(n)x[n]=sin(n).
Q4. (Difficulty: \star\star\star⋆⋆⋆) Choose the correct statements from the choices below.
Consider the length-NN signal x[n]=(-1)^nx[n]=(−1)
n
with NN even. Then X[k]=0X[k]=0 for all kk except k=N/2k=N/2
If we apply the DFT twice to a signal x[n]x[n], we obtain the signal itself scaled by NN, i.e. Nx[n]Nx[n].
Consider the length-NN signal x[n]=\cos(\frac{2\pi}{N}Ln+\phi),x[n]=cos(
N
2π
Ln+ϕ), where NN is even and L = N/2L=N/2. Then X[k]=
⎧⎩⎨N2ejϕ0 for k=L otherwise
.X[k]={
2
N
e
jϕ
0
for k=L
otherwise
.
Q5. (Difficulty: \star⋆) Consider the Fourier basis {\mathbf{w}^k}_{k=0,\ldots,N-1}{w
k
}
k=0,…,N−1
, where \mathbf{w}^k[n]=e^{-j \frac{2\pi}{N}nk}w
k
[n]=e
−j
N
2π
nk
for 0\le n\le N-10≤n≤N−1.
Select the correct statement below.
The elements of the basis are orthonormal:
\langle \mathbf{w}^i,\mathbf{w}^j\rangle=
{1 for i=j0 otherwise.
\quad⟨w
i
,w
j
⟩={
1 for i=j
0 otherwise.
The orthogonality of the vectors depends on the length NN of the elements of the basis.
The elements of the basis are orthogonal:
\langle \mathbf{w}^i,\mathbf{w}^j\rangle=
{N for i=j0 otherwise.
\quad⟨w
i
,w
j
⟩={
N for i=j
0 otherwise.
Q6. (Difficulty: \star\star⋆⋆) Consider the three sinusoids of length N=64N=64 as illustrated in the above figure; note that the signal values are shown from n=0n=0 to n=63n=63.
Call y_1[n]y
1
[n] the blue signal, y_2[n]y
2
[n] the green and y_3[n]y
3
[n] the red. Further, define x[n] = y_1[n] + y_2[n] + y_3[n]x[n]=y
1
[n]+y
2
[n]+y
3
[n].
Choose the correct statements from the list below. Note that the capital letters indicate the DFT vectors.
Y_2[k]=
⎧⎩⎨16j16j0 for k=8 for k=56 otherwise
Y
2
[k]=
⎩
⎪
⎪
⎨
⎪
⎪
⎧
16j
16j
0
for k=8
for k=56
otherwise
Y_1[k]=
{N0 for k=4,60 otherwise
Y
1
[k]={
N
0
for k=4,60
otherwise
Y_3[k]=
⎧⎩⎨32320 for k=0 for k=64 otherwise
Y
3
[k]=
⎩
⎪
⎪
⎨
⎪
⎪
⎧
32
32
0
for k=0
for k=64
otherwise
|x|_2^2=|X|_2^2=12800∥x∥
2
2
=∥X∥
2
2
=12800
Q7. (Difficulty: \star\star\star⋆⋆⋆) Consider the length-NN signal
x[n]=\cos\left(2\pi\frac{L}{M}n\right)x[n]=cos(2π
M
L
n)
where MM and LL are integer parameter with 0 \lt L \le N-10<L≤N−1, 0 \lt M \le N0<M≤N.
Choose the correct statements among the choices below.
If N=MN=M and N\neq 2LN
=2L, it is easier to compute the norm of the signal |\mathbf{x}|_2∥x∥
2
in the Fourier domain, using Parseval’s Identity.
Consider the circularly shifted signal y[n]=x[(n-D)\mod N].y[n]=x[(n−D)modN]. In the Fourier domain, since the DFT operator is shift invariant, it is Y[k]=X[k]Y[k]=X[k].
The signal has always exactly LL periods 0 \le n \lt N0≤n<N
The DFT X[k]X[k] has two elements different from zero if N=MN=M and N\neq 2lN
=2l.
Q8. (Difficulty: \star⋆) Consider an orthogonal basis {\phi_i}_{i=0,\dots,N-1}{ϕ
i
}
i=0,…,N−1
for \mathbb{R}^NR
N
. Select the statements that hold for any vector \mathbf{x} \in \mathbb{R}^Nx∈R
N
.
\displaystyle \Vert \mathbf{x}\Vert_2^2 = \frac{1}{P}\sum_{i=0}^{N-1} \vert \langle x , \phi_i \rangle \vert^2∥x∥
2
2
=
P
1
i=0
∑
N−1
∣⟨x,ϕ
i
⟩∣
2
if and only if |\phi_i|_2^2=P∥ϕ
i
∥
2
2
=P \forall i∀i.
\displaystyle \Vert \mathbf{x}\Vert_2^2 = \frac{1}{P}\sum_{i=0}^{N-1} \vert \langle x , \phi_i \rangle \vert^2∥x∥
2
2
=
P
1
i=0
∑
N−1
∣⟨x,ϕ
i
⟩∣
2
if and only if |\phi_i|_2=P∥ϕ
i
∥
2
=P \forall i∀i.
\displaystyle \Vert \mathbf{x}\Vert_2^2 = \sum_{i=0}^{N-1} \vert \langle x , \phi_i \rangle \vert^2 . ∥x∥
2
2
=
i=0
∑
N−1
∣⟨x,ϕ
i
⟩∣
2
.
\displaystyle \Vert \mathbf{x}\Vert_2^2 = \sum_{i=0}^{N-1} \vert \langle x , \phi_i \rangle \vert^2∥x∥
2
2
=
i=0
∑
N−1
∣⟨x,ϕ
i
⟩∣
2
if and only if |\phi_i|_2=1∥ϕ
i
∥
2
=1 \forall i∀i.
Q9. (Difficulty: \star\star⋆⋆) Pick the correct sentence(s) among the following ones regarding the DFT \mathbf{X}X of a signal \mathbf{x}x of length NN, where NN is odd.
Remember the following definitions for an arbitrary signal (asterisk denotes conjugation):
hermitian-symmetry: x[0]x[0] real and x[n]=x[N-n]^*x[n]=x[N−n]
∗
for n=1,\dots,N-1n=1,…,N−1.
hermitian-antisymmetry: x[0]=0x[0]=0 and x[n]=-x[N-n]^*x[n]=−x[N−n]
∗
for n=1,\dots,N-1n=1,…,N−1.
- If the signal \mathbf{x}x is hermitian antisymmetric, then its DFT \mathbf{X}X is purely imaginary.
- If the signal \mathbf{x}x is hermitian-symmetric, then the DFT \mathbf{X}X is also hermitian-symmetric.
- If the signal \mathbf{x}x is purely real, then the DFT \mathbf{X}X is purely imaginary.
- If the signal \mathbf{x}x is hermitian-symmetric, then its DFT is real.
Week 4 Quiz Answers
Quiz 1: Homework for Module 1.4 Quiz Answers
Q1. (Difficulty: \star⋆) Consider the magnitude DTFTs |X(e^{j\omega})|∣X(e
jω
)∣ and |Y(e^{j\omega})|∣Y(e
jω
)∣ shown below, where vertical lines represent Dirac deltas:
Both underlying signals x[n]x[n] and y[n]y[n] are periodic. Find their periods and write them below, separated by a space. Please write the smallest period, i.e. a 5-periodic signal is also obviously 15-periodic but we’re interested in 5.
Enter the period of x[n]x[n] and y[n]y[n] with a unique white space in between.
Enter answer here
Q2. (Difficulty: \star⋆) We will see in later lectures that communication systems must fulfill what is called the “bandwidth constraint”, that is, the energy of the signals that they transmit must strictly fit into pre-defined frequency bands. In this problem we will look at the bandwidth constraint in the discrete-time domain.
The signal x[n]x[n] is real-valued and its spectrum is nonzero only over the [-\pi/8,\ \pi/8][−π/8, π/8] interval. Due to the bandwidth constraint we need to “fit” the signal over the bands indicated in green in the following figure
To this aim, we need to design a processing block \mathcal{H}H in order to convert x[n]x[n] into a sequence s[n]s[n] satisfying the following requirements:
The support of the DTFT of s[n] must be limited to [−3π/4, −π/2]∪[π/2, 3π/4]
The sequence s[n] must be real-valued (x[n] is real-valued)
Which of the following input/output relationships for \mathcal{H}H meet both requirements? (check all correct answers) :
s[n]=\sin(\frac{21\pi}{8}n)\cdot x[n]s[n]=sin(
8
21π
n)⋅x[n]
s[n] = e^{j\frac{5\pi}{8}n}\cdot x[n] s[n]=e
j
8
5π
n
⋅x[n]
s[n]=\cos(\frac{5\pi}{8}n)\cdot x[n]s[n]=cos(
8
5π
n)⋅x[n]
s[n]=\mathrm{IDTFT}{X(e^{j(\omega-5\pi/8)})}s[n]=IDTFT{X(e
j(ω−5π/8)
)}
Q3. (Difficulty: \star⋆) Consider the length-LL signal
x[n]=
{100≤n≤M−1M≤n≤L−1
\; ,x[n]={
1
0
0≤n≤M−1
M≤n≤L−1
,
Write out the closed-form analytical expression for its DFT coefficients X[k]X[k].
Be careful with your typing since the regular-expression parser can be a bit picky. Check Coursera help to enter math expression. In particular, remember that in the Coursera platform the symbols are different:
II (capital i) is used for the imaginary unit instead of jj
Euler’s number is EE instead of ee
you can also use the exponential function \exp(\cdot)exp(⋅)
\piπ is defined as pipi
The only other symbols you’ll need for the answer are the case-sensitive variables k, M, Lk,M,L.
Finally, do not forget to validate your syntax by clicking “Preview” before submitting your answer.
For instance, the expression e^{j(\pi L + 3\pi)}/(k + M)e
j(πL+3π)
/(k+M) should be entered as
E^(I(piL + 3 * pi))/(k + M)
Preview will appear here…
Enter math expression here
Q4. The real and imaginary parts of X(e^{j\omega})X(e
jω
) are:
After examining the plots, check all the correct statements below.
x[n]x[n] is Hermitian-symmetric x[n]=x^\ast[-n]x[n]=x
∗
[−n].
x[n]x[n] is 0-mean, i.e. \sum_{n\in\mathbb Z} x[n]=0∑
n∈Z
x[n]=0.
x[n]x[n] is real valued.
Q5. (Difficulty: \star \star⋆⋆) Consider a signal x[n]x[n] and its DTFT X(e^{j\omega})X(e
jω
). Assume X(e^{j\omega})X(e
jω
) is differentiable. Compute the inverse DTFT of
j\frac{d}{d\omega}X(e^{j\omega})j
dω
d
X(e
jω
).
You should write your answer in terms of x[n]x[n] and elementary functions and constants, for example \frac{\pi}{2}x[n]
2
π
x[n] would be written :
pi/2*x[n]
Enter answer here
Q6. (Difficulty: \star⋆) Which property of the DTFT allows you to easily compute the inverse DTFT of 4X(e^{j\omega})/\pi -24X(e
jω
)/π−2 once you know x[n]x[n]? Just type the name of the property.
Enter answer here
Q7. (Difficulty: \star⋆) Take a length-NN signal x[n]x[n] and its DFT X[k]X[k], with 0 \le n,k, \le N-10≤n,k,≤N−1. Next, consider its periodized version \tilde x[n]=x[n\ \mathrm{mod}\ N]
x
~
[n]=x[n mod N] with its DFS \tilde X[k]
X
~
[k] where now n,k \in \mathbb{Z}n,k∈Z.
Which of the following statements are true?
\tilde X[l] = X[l\ \mathrm{mod}\ N]
X
~
[l]=X[l mod N], for all l\in\mathbb Zl∈Z
\tilde X[-2] = X[2]
X
~
[−2]=X[2] for all x[n]x[n] and N>2N>2
\tilde X[k+lN] = X[k]
X
~
[k+lN]=X[k], for all l\in\mathbb Zl∈Z and k=0,…,N-1k=0,…,N−1.
Q8. (Difficulty: ⋆) In the class, we learned how the modulation theorem can help us tune a musical instrument. Martin showed us an example with a bass but of course the same works with a classical guitar. Listen carefully to these two samples (with earphones, if possible); each audio clip is the recording of two notes played together:
Audio clip A
Audio Clip B
Select the correct options below.
- The notes are in tune in audio clip A and out of tune in audio clip B
- The notes are in tune in audio clip B and out of tune in audio clip A
- The notes are out of tune in both audio clips
- The notes are in tune in both audio clips
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