### Get All Weeks Digital Signal Processing 2: Filtering Coursera Quiz Answers

Digital Signal Processing is the branch of engineering that, in the space of just a few decades, has enabled unprecedented levels of interpersonal communication and of on-demand entertainment. By reworking the principles of electronics, telecommunication, and computer science into a unifying paradigm, DSP is the heart of the digital revolution that brought us CDs, DVDs, MP3 players, mobile phones, and countless other devices.

The goal, for students of this course, will be to learn the fundamentals of Digital Signal Processing from the ground up. Starting from the basic definition of a discrete-time signal, we will work our way through Fourier analysis, filter design, sampling, interpolation, and quantization to build a complete DSP toolset to analyze a practical communication system in detail. Hands-on examples and demonstrations will be routinely used to close the gap between theory and practice.

### Digital Signal Processing 2: Filtering Coursera Quiz Answers

### Week 1 Quiz Answers

#### Quiz 1: Homework for Module 2.1

Q1. (Difficulty: \star⋆) Among the choices below, select all the linear systems. (Please note that some of the choices use functions rather than discrete-time signals; the concept of linearity is identical in both cases).

The DTFT, i.e. transform a sequence \mathbf xx into \mathrm{DTFT}{\mathbf x}DTFT{x}.

AM radio modulation, i.e. multiply a signal x[n]x[n] by a cosine at the carrier frequency :

y[n]= x[n]\cos(2\pi\omega_cn)y[n]=x[n]cos(2πω

c

n)

Time-stretch, i.e. y(t)=x(\alpha t),y(t)=x(αt), e.g. if you play an old LP-45 vinyl disc at 33 rpm, the time-stretch coefficient would be \alpha = 33/45α=33/45.

Envelope detection (via squaring), i.e. y[n] = |x[n]|^2 \ast h[n]y[n]=∣x[n]∣

2

∗h[n] ,where h[n]h[n] is the impulse response of a lowpass filter such as the moving average filter.

Second derivative, i.e.

y(t) = \frac{d^2}{dt^2}x(t)y(t)=

dt

2

d

2

x(t)

Clipping, i.e. enforce a maximum signal amplitude MM,e.g.:

y[n]=\left{

x[n]M, x[n]≤M, otherwise

\right.y[n]={

x[n]

M

, x[n]≤M

, otherwise

Scrambling, i.e. a permutation to the input sequence, e.g. :

Q2. (Difficulty: \star⋆) Among the choices below, select all the time-invariant systems. (Please note that some of the choices use functions rather than discrete-time signals; the concept of time invariance is identical in both cases).

Time-stretch, i.e. y(t)=x(\alpha t),y(t)=x(αt), e.g. if you play an old LP-45 vinyl disc at 33 rpm, the time-stretch coefficient would be \alpha = 33/45α=33/45

Second derivative, i .e.

y(t) = \frac{d^2}{dt^2}x(t)y(t)=

dt

2

d

2

x(t)

The DTFT, i.e. transform a sequence \mathbf xx into \mathrm{DTFT}{\mathbf x}DTFT{x}.

Scrambling : apply a permutation to the input sequence, e.g. :

AM radio modulation, i.e. multiply a signal x[n]x[n] by a cosine at the carrier frequency :

y[n]= x[n]\cos(2\pi\omega_cn)y[n]=x[n]cos(2πω

c

n)

Clipping, i.e. enforce a maximum signal amplitude MM,e.g.:

y[n]=\left{

x[n]M, x[n]≤M, otherwise

\right.y[n]={

x[n]

M

, x[n]≤M

, otherwise

Envelope detection (via squaring), i.e. y[n] = |x[n]|^2 \ast h[n]y[n]=∣x[n]∣

2

∗h[n] ,where h[n]h[n] is the impulse response of a lowpass filter such as the moving average filter.

Q3. (Difficulty: \star⋆) The impulse response of a room can be recorded by producing a sharp noise (impulsive sound source) in a silent room, thereby capturing the scattering of the sound produced by the walls.

The impulse response h[n]h[n] of Lausanne Cathedral was measured by Dokmanic et al. by recording the sound of balloons being popped ( hear it! ).

The balloon is popped at time n=0n=0 and after a number of samples NN, the reverberations die out, i.e. h[n]=0h[n]=0 for n\lt 0n<0 or n\gt Nn>N.

The acoustic of this large space can then be artificially recreated by convolving any audio recording with the impulse response, e.g. this cello recording becomes this.

What are the properties of h[n]h[n] ? (tick all the correct answers)

BIBO stable

Anticausal

FIR

Q4. (Difficulty: \star⋆) Let

h[n]=\delta[n]-\delta[n-1]h[n]=δ[n]−δ[n−1]

x[n]=\left{

10, n≥0,, else.

\right.x[n]={

1

0

, n≥0,

, else.

y[n]=x[n]\ast h[n]y[n]=x[n]∗h[n].

Compute y[-1]y[−1], y[0]y[0], y[1]y[1], y[2]y[2] and write the result as space-separated values. E.g.: If you find y[-1]=-2y[−1]=−2, y[0]=-1y[0]=−1, y[1]=0y[1]=0, y[2]=1y[2]=1, you should enter

1

-2 -1 0 1

Enter answer here

Q5. (Difficulty: \star\star⋆⋆) Consider the filter h[n]=\delta[n]-\delta[n-1]h[n]=δ[n]−δ[n−1],

the signal x[n]=\left{

n0, n≥0,, else.

\right.x[n]={

n

0

, n≥0,

, else.

and the output y[n]=x[n]\ast h[n]y[n]=x[n]∗h[n].

Compute y[-1]y[−1], y[0]y[0], y[1]y[1], y[2]y[2] and write the result as space-separated values. E.g.: If you find y[-1]=-2y[−1]=−2, y[0]=-1y[0]=−1, y[1]=0y[1]=0, y[2]=1y[2]=1, you should enter

1

-2 -1 0 1

Enter answer here

Q6. (Difficulty: \star\star\star⋆⋆⋆) Which of the following filters are BIBO-stable?

Assume N\in\mathbb{N}N∈N and 0 \lt \omega_c \lt \pi0<ω

c

<π.

The ideal low pass filter with a cutoff frequency \omega_cω

c

: H(e^{j\omega})=

{10|ω|≤ωcotherwise

H(e

jω

)={

1

0

∣ω∣≤ω

c

otherwise

.

The moving average: h[n]=\frac{\delta[n]+\delta[n-1]}{2}.h[n]=

2

δ[n]+δ[n−1]

.

The following smoothing filter: h[n]=\sum_{k=0}^\infty \frac{1}{k+1}\delta[n-k].h[n]=∑

k=0

∞

k+1

1

δ[n−k].

The filter h[n]=\sum_{k=0}^{N-1}\delta[n-k]\sin\left(2\pi \frac{k}{N}\right)h[n]=∑

k=0

N−1

δ[n−k]sin(2π

N

k

).

Q7. (Difficulty: \star\star⋆⋆) Consider an LTI system \mathcal{H}H. When the input to \mathcal{H}H is the following signal

then the output is

Assume now the input to \mathcal{H}H is the following signal

Which one of the following signals is the system’s output?

- Output
- Output
- Output

Q8. (Difficulty: \star⋆) Consider the system shown below, consisting of a cosine modulator at frequency \omega_0ω

0

followed by an ideal bandpass filter h[n]h[n] whose frequency response is also shown in the figure; assume that the input to the system is the signal x[n]x[n], whose spectrum is shown below.

Determine the value of \omega_0\in[0,2\pi]ω

0

∈[0,2π] that maximizes the energy of the output y[n]y[n] when the input is x[n]x[n].

Remember that \piπ must be entered in the answer box as pi.

Preview will appear here…

Enter math expression here

Q9. (Difficulty: \star⋆) Consider a lowpass filter with the following frequency response.

What is the output y[n]y[n] when the input to this filter is x[n]=\cos(\frac{\pi}{5}n)+\sin(\frac{\pi}{4}n)+0.5\cos(\frac{3\pi}{4}n)x[n]=cos(

5

π

n)+sin(

4

π

n)+0.5cos(

4

3π

n)?

Preview will appear here…

Enter math expression here

Q10. (Difficulty: \star⋆) Consider a filter with real-valued impulse response h[n]h[n]. The filter is cascaded with another filter whose impulse response is h'[n] = h[-n]h

′

[n]=h[−n], i.e. whose impulse response is the time-reversed version of h[n]h[n]:

The cascade system can be seen as a single filter with impulse response g[n].

What is the phase of G(ejω)?

Preview will appear here…

Enter math expression here

Q11. (Difficulty: \star⋆) Let x[n]=\cos(\frac{\pi}{2} n)x[n]=cos(

2

π

n) and h[n]=\frac{1}{5}\text{sinc}(\frac{n}{5})h[n]=

5

1

sinc(

5

n

). Compute the convolution y[n]=x[n]*h[n]y[n]=x[n]∗h[n], and write the value of y[5]y[5].

Hint: First find the convolution result in the frequency domain.

Enter answer here

Q12. (Difficulty: \star\star⋆⋆) Consider the system below, where H(e^{j\omega})H(e

jω

) is an ideal lowpass filter with cutoff frequency \omega_c=\pi/4ω

c

=π/4:

Consider two input signals to the system:

x_1[n]x

1

[n] is bandlimited to [-\pi/4, \pi/4][−π/4,π/4]

x_2[n]x

2

[n] is band-limited to [-\pi, -3\pi/4] \cup [3\pi/4, \pi][−π,−3π/4]∪[3π/4,π].

Which of the following statements is correct?

Both x_1[n]x

1

[n] and x_2[n]x

2

[n] are eliminated by the system.

Both x_1[n]x

1

[n] and x_2[n]x

2

[n] are not modified by the system.

x_2[n]x

2

[n] is not modified by the system while x_1[n]x

1

[n] is eliminated.

x_1[n]x

1

[n] is not modified by the system while x_2[n]x

2

[n] is eliminated.

Q13. (Difficulty: \star⋆) x[n]x[n] and y[n]y[n] are two square-summable signals in \ell_2(\mathbb{Z})ℓ

2

(Z); X(e^{j\omega})X(e

jω

) and Y(e^{j\omega})Y(e

jω

) are their corresponding DTFTs.

We want to compute the value.

\sum_{n=-\infty}^{\infty}x[n]y^*[n]∑

n=−∞

∞

x[n]y

∗

[n].

in terms of X(e^{j\omega})X(e

jω

) and Y(e^{j\omega})Y(e

jω

). Select the correct expression among the choices below.

X(e^{j\omega})*Y(e^{-j\omega})X(e

jω

)∗Y(e

−jω

)

\frac{1}{2\pi}\int_{-\pi}^{\pi}X(e^{j\omega})Y(e^{-j\omega})d \omega

2π

1

∫

−π

π

X(e

jω

)Y(e

−jω

)dω

\frac{1}{2\pi}\int_{-\pi}^{\pi}X(e^{j\omega})Y^*(e^{j\omega})d \omega

2π

1

∫

−π

π

X(e

jω

)Y

∗

(e

jω

)dω

X(e^{j\omega})Y(e^{-j\omega})X(e

jω

)Y(e

−jω

)

\frac{1}{2\pi}X(e^{j\omega})Y^*(e^{j\omega})

2π

1

X(e

jω

)Y

∗

(e

jω

)

\frac{1}{2\pi}X(e^{j\omega})Y(e^{-j\omega})

2π

1

X(e

jω

)Y(e

−jω

)

Q14. (Difficulty: \star\star\star⋆⋆⋆) h[n]h[n] is the impulse response of an ideal lowpass filter with cutoff frequency \omega_c\lt \frac{\pi}{2}ω

c

<

2

π

. Select the correct description for the system represented in the following figure?

Hint: Use the trigonometric identity \cos(x)^2 = \frac{1}{2}(1+\cos(2x))cos(x)

2

=

2

1

(1+cos(2x)).

- A lowpass filter with gain 1 and cutoff frequency \omega_c/2ω
- A highpass filter with gain \frac{1}{4}

and cutoff frequency \omega_cω - A highpass filter with gain 1 and pass band [\omega_c,\pi-\omega_c][ω
- A lowpass filter with gain 1 and cutoff frequency c
- A lowpass filter with gain
- A highpass filter with gain 21

and cutoff frequency cπ−ω c

Q15. (Difficulty:r⋆⋆⋆) Consider the following system, where jω

) is a half-band filter, i.e. an ideal lowpass with cutoff frequenω =π/2:

Assume the input to the system is x[n]=\delta[n]x[n]=δ[n]. Compute

∑ n=−∞

y[n]

Hint: Perform the derivations in the frequency domain.

Enter answer here

### Week 2 Quiz Answers

#### Quiz 1: Homework for Module 2.2

Q1. (Difficulty: \star⋆) Consider the following causal CCDE

y[n] + 2 y[n-1] = 3\ x[n] + 2.5\ x[n-1].y[n]+2y[n−1]=3 x[n]+2.5 x[n−1].

Which of the following statements are correct?

The system is stable.

Its ROC contains the unit circle.

It has two poles at -2−2 and \frac{-5}{6}

6

−5

.

If the input signal is \delta[n]-\delta[n+1]δ[n]−δ[n+1], then the z-transform of the output would be (-3z+1/2+5/2z^{-1})/(1+2z^{-1})(−3z+1/2+5/2z

−1

)/(1+2z

−1

).

Q2. (Difficulty: \star\star⋆⋆) Suppose that the ROC of the sequence x[n]x[n] is r_L \lt |z| \lt r_Ur

L

<∣z∣<r

U

. What is the ROC of x^*[-n]x

∗

[−n]?

\frac{1}{r_U} \lt |z| \lt \frac{1}{r_L}

r

U

1

<∣z∣<

r

L

1

\frac{1}{r_L} \lt |z| \lt \frac{1}{r_U}

r

L

1

<∣z∣<

r

U

1

r_U \lt |z| \lt r_Lr

U

<∣z∣<r

L

r_L \lt |z| \lt r_Ur

L

<∣z∣<r

U

Q3. (Difficulty: \star⋆) Consider an LTI system h[n]h[n], whose transfer function’s ROC is R_hR

h

. Consider a second LTI system g[n]g[n] with ROC R_gR

g

. Now consider the cascade of the two filters.

What is the ROC of the cascade?

It contains R_g \cup R_hR

g

∪R

h

.

It is only R_hR

h

.

It contains R_g \cap R_hR

g

∩R

h

.

It is only R_gR

g

.

Q4. (Difficulty: \star⋆) Consider the following CCDE

y[n] – \frac{1}{2} y[n-1] = 2 x[n] – 5 x[n-1] – x[n-2]\,.y[n]−

2

1

y[n−1]=2x[n]−5x[n−1]−x[n−2].

Let H(e^{j\omega})H(e

jω

) denote the transfer function of this system. What is H(e^{j\pi})H(e

jπ

)?

Enter answer here

Q5. (Difficulty: \star\star⋆⋆) Write some code in your preferred programming language that implements the following CCDE:

y[n] + 2 y[n-1] = x[n+1] – \frac{1}{2} x[n]y[n]+2y[n−1]=x[n+1]−

2

1

x[n]

Use y[n] = 0y[n]=0 for n \lt 0n<0 as initial conditions and run the algorithm for x[n] = \delta[n] + \frac{1}{2}\delta[n-1]x[n]=δ[n]+

2

1

δ[n−1].

y[4]=2y[4]=2

The output shows a diverging oscillation around zero: as nn grows, it assumes always larger values with alternated signs.

The filter is stable but the output y[n]y[n] diverges because of the chosen input x[n]x[n].

y[5]=1y[5]=1

The filter is mathematically unstable. Even in practice, you can see the output diverging |y[50]|>10^{13}∣y[50]∣>10 13

.

Q6. (Difficulty: \star\star\star⋆⋆⋆) A filter H(z)H(z) has the following pole-zero plot:

Which of the following figures shows the magnitude response of the filter?

Q7. (Difficulty: \star\star⋆⋆) Let h[n]h[n] represent the impulse response of the following system.

Select the correct statement about the poles and zeros of H(z)H(z).

H(z)H(z) has one zero at z_1=1/4z

1

=1/4 and two poles at z_3=5/4z

3

=5/4 and z_2=3/8z

2

=3/8.

It has one zero at z_1=3/2z

1

=3/2 and one pole at z_2=5/6z

2

=5/6.

It has one zero at z_1=-3/4z

1

=−3/4.

H(z)H(z) has two zeros at z_1=5/4z

1

=5/4 and z_2=3/8z

2

=3/8 and one pole at z_3=1/4.

It has one pole at z_1= =3/2

Q8. (Difficulty: \star\star⋆⋆) The following bit of Python code implements a discrete-time filter (assume x[n]x[n] and y[n]y[n] are suitably defined arrays):

`f = 0;`

g = 0;

for n in range(0, L-1):

y[n] = x[n] + f;

g = -f;

f = -x[n] + 0.5 * y[n] + g;

What is the minimum number of delays necessary to implement this filter efficiently?

Enter answer here

Q9. (Difficulty: \star\star⋆⋆) Which of the following statements describes the system in this figure?

- This is a resonator at 0=π/2.
- None of these statements describe this system.
- The system is a hum removal filter with 0=3π/4.
- The system is a DC notch.
- The system is a hum removal filter with 0= =π/2.
- The system is a resonator at 0=3π/4.

### Week 3 Quiz Answers

#### Quiz 1: Homework for Module 2.3

Q1. (Difficulty: \star ⋆) Assume x[n]x[n] is a WSS random process with power spectral density

- P_x(e^{j\omega})
*Px*(*ejω*) is constant - P_x(e^{j\omega})
*Px*(*ejω*) is real-valued. - P_x(e^{j\omega})
*Px*(*ejω*) is symmetric around \omega = 0*ω*=0 - P_x(e^{j\omega}) \geq 0
*Px*(*ejω*)≥0

Q2. (Difficulty: \star⋆) Consider a white noise process w[n]w[n] with variance

to produce a WSS process x[n]x[n]

Compute the value of the power spectral density of x[n]x[n] in \omega=\pi/2ω=π/2

Enter answer here

Q3. (Difficulty: \star⋆) Consider the stochastic process defined as

y[n] = x[n] + \beta x[n-1]y[n]=x[n]+βx[n−1]

where \beta\in\mathbb Rβ∈R and x[n]x[n] is a zero-mean wide-sense stationary process with autocorrelation

y[n]y[n] can also be expressed as filtered version of x[n]x[n] where the filter’s impulse response h[n]h[n] is:

- h[n]=\delta[n] – \beta\delta[n-1]h[n]=δ[n]−βδ[n−1]
- h[n]=\delta[n] + \beta\delta[n+1]h[n]=δ[n]+βδ[n+1]
- h[n]=\delta[n] + \beta\delta[n-1]h[n]=δ[n]+βδ[n−1]
- h[n]=\delta[n] – \beta\delta[n+1]h[n]=δ[n]−βδ[n+1]
- h[n]=\delta[n+1] + \beta\delta[n]h[n]=δ[n+1]+βδ[n]

Q4. (Difficulty: \star \star \star⋆⋆⋆) Using the same setup as in the previous question, select the correct expression for the power spectral density P_y2βcos(ω)

*Py*(*ejω*)=1+*α*2−2*αcos*(*ω*)1+*β*2+2*βcos*(*ω*)- P_y(e^{j\omega}) = \alpha(1-\sigma)\frac{1 – \beta^2 + 2\beta cos(\omega)}{1 + \alpha^2 -2\alpha cos(\omega)}
*Py*(*ejω*)=*α*(1−*σ*)1+*α*2−2*αcos*(*ω*)1−*β*2+2*βcos*(*ω*) - P_y(e^{j\omega}) = \sigma(1-\alpha)\frac{1 + \beta + 2 cos(\omega)}{1 + \alpha -2 cos(\omega)}
*Py*(*ejω*)=*σ*(1−*α*)1+*α*−2*cos*(*ω*)1+*β*+2*cos*(*ω*) - P_y(e^{j\omega}) = \alpha^2\frac{\beta^2 + 2\beta cos(\omega)}{\alpha^2 -2\alpha cos(\omega)}
*Py*(*ejω*)=*α*2*α*2−2*αcos*(*ω*)*β*2+2*βcos*(*ω*) - P_y(e^{j\omega}) = \sigma^2(1-\alpha^2)\frac{1 + \beta^2 + 2\beta cos(\omega)}{1 + \alpha^2 -2\alpha cos(\omega)}
*Py*(*ejω*)=*σ*2(1−*α*2)1+*α*2−2*αcos*(*ω*)1+*β*2+2*βcos*(*ω*) - P_y(e^{j\omega}) = \sigma^2\frac{1 + \alpha^2 + 2\alpha cos(\omega)}{1 + \beta^2 -2\beta cos(\omega)}
*Py*(*ejω*)=*σ*21+*β*2−2*βcos*(*ω*)1+*α*2+2*αcos*(*ω*)

Q5. (Difficulty: \star \star⋆⋆) Using the same setup as in the previous questions, assume that the output y[n]y[n] turns out to be a white noise sequence. Which of the following statements are necessarily true ?

- The samples of y[n]y[n] must be correlated.
- The samples of y[n]y[n] must be uncorrelated.
- The power spectrum P_y(e^{j\omega})P

y

(e

jω

) must be constant. - \beta=-\alphaβ=−α
- \beta=\alphaβ=α
- \beta= 2 \alphaβ=2α

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