### Get All Weeks Digital Signal Processing 3: Analog vs Digital Quiz Answers

### Digital Signal Processing 3: Analog vs Digital Coursera Quiz Answers

### Week 1 Quiz Answers

#### Quiz 1: Homework for Module 3.1

Q1. (Difficulty: \star⋆) Select the correct statement among the ones below:

In continuous time, there is always a maximum positive frequency FF.

An interpolated discrete-time signal is always bandlimited.

The total energy of a continuous time signal is always infinite.

A signal that is F_sF

-bandlimited is also ff-bandlimited if f \ge F_sf≥F

Q2. (Difficulty: \star⋆) A piecewise constant function in continuous time can always be expressed as a linear combination of scaled and translated unit step functions. Recall the definition of the unit step is:

u(t)={10for t>0otherwise Consider the function shown in the figure below:

Which is the correct expression for the function in terms of units steps?

u(t-1) – u(t+\frac{1}{2})u(t−1)−u(t+

u(t+\frac{1}{2}) – u(t+1)u(t+

\frac{1}{2} \cdot u(t) + \frac{1}{2} \cdot u(t)

u(t+\frac{1}{2}) – u(t-\frac{1}{2})u(t+

\frac{1}{2} \cdot u(t) – \frac{1}{2} \cdot u(t)

Q3. (Difficulty: \star⋆) The figures below show the result of filtering the signal x(t) = \text{rect}(t)x(t)=rect(t) with four ideal lowpass filters with different cutoff frequencies F_iF

i = 0, 1, 2, 3i=0,1,2,3.

Check the correct statements below.

All the signals in the figures are bandlimited.

The cutoff frequencies F_iF

are decreasing from plot (a) to plot (d).

Q4. (Difficulty: \star\star⋆⋆) Consider the interpolator

i(t)=⎧⎩⎨1−2|t|0for |t|≤12otherwise

The interpolator has a triangular shape in the time domain. Select the correct expression for I(f)I(f), the Fourier transform of the interpolator. (Hint: a triangle can be obtained by convolving two rectangles).

I(j\Omega) = \frac{1}{4}{\mathrm{sinc}}\left( \frac{\Omega}{4 \pi}\right)I(jΩ)=

4

1

sinc(

4π

Ω

)

I(f) = \frac{1}{4}{\mathrm{sinc}}^2\left( \frac{f}{2}\right)I(f)=

Q5. (Difficulty: \star\star\star⋆⋆⋆) Consider a discrete-time signal x[n]x[n] whose spectrum between -\pi−π and \piπ is

X(ejω)={10for |ω|≤2π3otherwise

with 2\pi2π-periodicity over the entire frequency axis.

x[n]x[n] is interpolated to the continuous-time signal x(t) = \sum_{n=-\infty}^{\infty} x[n] i\left(\frac{t-nT_s}{T_s}\right)x(t)=∑

n=−∞

i(t)=⎧⎩⎨1−2|t|0for |t|≤12otherwise Which graph represents the resulting spectrum of x(t)x(t)?

(Note that the graphs use the normalized frequency notation \Omega_N = \pi/T_sΩ N =π/T s )

Q6. (Difficulty: \star\star⋆⋆) When using a first-order interpolator (such as the one described in the previous question) to interpolate a finite-support sequence, which of the following statements are true?

The interpolated signal is bandlimited.

The interpolated signal has finite length in time because of the limited support of the interpolating function i(t)i(t).

The periodic copies of X(e^{j 2\pi f/F_s})X(e

j2πf/F

) outside of [-F_s/2,F_s/2][−F

/2,F

/2] are not eliminated by the interpolation filter, since it is not an ideal lowpass.

The spectrum between [-F_s/2,F_s/2][−F

/2] (the baseband) is distorted by the non-flat response of the interpolating function over the baseband.

Q7. (Difficulty: \star⋆) Select the correct statement(s).

The sampling of a bandlimited signal x(t)x(t) (with maximum frequency F_s/2F

/2) with a sampling frequency F \ge F_sF≥F

will result in no information loss.

Increasing the interpolation interval, T_sT

results in a wider spectrum of the interpolated signal.

The sampling theorem implies that the space of bandlimited functions is a Hilbert Space.

Q8. (Difficulty: \star\star\star⋆⋆⋆) Consider a local interpolation scheme based on Lagrange polynomials. Suppose you want to compute the approximate value of s_c((n+\tau)T), |\tau| \le 1/2s c

((n+τ)T),∣τ∣≤1/2 using the discrete-time version of the signal s[n] = s_c(nT)s[n]=s c

(nT). For simplicity, let’s set T=1T=1.

Lagrange interpolation of order NN will fit an order-NN polynomial through the N+1N+1 samples that are closest to s_c(n+\tau)s c

(n+τ). For instance, for N=1N=1, Lagrange interpolation will fit a straight line between s_c(n-1)s

(n−1) and s_c(n)s c

(n) if \tau < 0τ<0 and between s_c(n)s c (n) and s_c(n+1)s c (n+1) if \tau > 0τ>0. For N=2N=2, Lagrange interpolation will fit a parabola through s_c(n-1)s c

(n−1), s_c(n)s c

(n) and s_c(n+1)s c

(n+1) for all values of \tauτ.

Consider the following signal, showing 5 values of s_c(t)s c

(t) sampled around t=nt=n with T_s=1T s

=1; the gray area represents the range of the local approximation that we want to perform. Below you will find six plots each one of which shows, in red, a polynomial interpolator passing through s_c(n)s c

(n). Select the interpolators that are valid Lagrange interpolators for the gray interval.

- 2
- 3
- 1
- 6
- 5
- 4

Q9. (Difficulty: \star\star⋆⋆) Consider the same interpolation setup as in the previous question:

The numeric values of the five samples are

s[n−2]s[n−1]s[n]s[n+1]s[n+2]=−4=4=−2=−3=3 Using Lagrange interpolation, compute the interpolated value s_c(n+\tau)s c

(n+τ) for \tau=\frac{1}{4}τ= N=1N=1.

Hint: the easiest way to solve the exercise is to write a short program.

Enter answer here

### Week 2 Quiz Answers

#### Quiz 1: Homework for Module 3.2

Q1. (Difficulty: \star⋆) Consider a real-valued, continuous-time signal x(t)x(t). All you know about the signal is that x(t) = 0x(t)=0 for |t| \gt t_0∣t∣>t

so that when you sample x(t)x(t), there is no aliasing?

Select the correct statement(s).

Yes, since F_SF

does not depend on the support of x(t)x(t), but on the highest frequency contained within it.

No. The signal is time-limited, so it is not bandlimited. There will always be a certain amount of aliasing in the sampled version.

No, since the exact numerical value for t_0t

Yes. The signal is time-limited and therefore it is bandlimited. Consequently, there exists a sampling frequency F_SF

Q2. (Difficulty: \star\star⋆⋆) Listen to the sound of a triangle (the percussive musical instrument):

Below you are given 4 processed version of the original sound. In which one can you hear aliasing artifacts due to downsampling?

Try to find the answer just by listening. If you are stuck, you may use a numerical package to study the spectrum.

Sound A

Sound B

Sound C

Sound D

Q3. (Difficulty: \star⋆) Consider a real-valued, continuous-time signal x_c(t)x

(t) with the following spectrum (note that the frequency axis uses normalized frequencies \Omega = 2\pi fΩ=2πf):

What is the maximum sampling period T_sT

that we can use to sample x_c(t)x

(t) so that the spectral copies caused by sampling do not overlap?

Q4. (Difficulty: \star⋆) Assume x(t)x(t) is a continuous-time pure sinusoid at 10 kHz. The signal is raw-sampled at 8 kHz and then interpolated back to a continuous-time signal with an interpolator at 8 kHz. What is the perceived frequency in kHz of the interpolated sinusoid?

Enter answer here

Q5. (Difficulty: \star\star\star⋆⋆⋆) Consider a real-valued continuous-time signal x_c(t)x

(t) whose bandlimited Fourier transform is

Xc(f)={cos2(π2fFs)0for |f|<Fsotherwise

(f) is shown here between -F_s−F

The signal x(t)x(t) is now sampled with T_s = 1/F_sT

this defines the periodized spectrum

Which of the following pictures depicts \tilde{X}_c(f)

Q6. (Difficulty: \star⋆) A chirp is a sinusoidal signal whose frequency increases over time. Consider the linear chirp defined as

x(t) = \cos(2 \pi f_0 t + \alpha \pi t^2)x(t)=cos(2πf

By setting \alpha = \frac{f_1 – f_0}{t_{1}}α=

is the inital frequency of the chirp at time t=0t=0, and \frac{f_0+f_1}{2}

is the instantaneous frequency at time t=t_{1}t=t

and the frequency increases linearly with time.

Write a program in your favorite programming language that computes a discrete-time version of the chirp at a sampling frequency F_s = 8000\text{Hz}F

=8000Hz for 0\leq t\leq 20≤t≤2 seconds. Set f_0=0\text{Hz}f

=0Hz and f_1=10\text{Hz}f

=10Hz and t_1 = 2t

=2 seconds.

Use the program to count how many times the signal crosses the abscissa. How many zero-crossings can you count in the generated chirp?

Enter the number of zero-crossings as an integer.

Enter answer here

Q7. (Difficulty: \star\star⋆⋆) Consider the same chirp signal as in the previous question. Which of the following statements are true for the signal?

Tick all the correct statements.

- The instantaneous frequency of the chirp is linearly increasing with time.
- If we sample the signal with a very low sampling frequency, the resulting signal will have an instantaneous frequency that varies between 0 Hz and 2\pi2π
- A linear chirp needs the same time to go from 100Hz to 200Hz as to go from 1000Hz to 2000Hz.
- If the runtime for the chirp generation is not limited, we cannot set a sampling frequency such that there will be no aliasing.

### Week 3 Quiz Answers

#### Quiz 1: Homework for Module 3.4

Q1. Check all the statements that are true

- Donwsampling is a time-invariant operation.
- Upsampling can always be undone.
- Downsampling can always be undone
- Upsampling by 6 followed by downsampling by 4 is equivalent to downsampling by 4 followed by upsampling by 6
- Upsampling by 3 followed by downsampling by 2 is equivalent to downsampling by 2 followed by upsampling by 3

Q2. Consider the signal x = \cos((2\pi/3)n)x=cos((2π/3)n). The signal is downsampled by a factor of two. Indicate the frequency of the resulting output, normalized by 2\pi2π.

(E.g., if the frequency is \pi/2π/2, write 1/4)

Preview will appear here…

Enter math expression here

Q3. Consider the signal x = \cos((\pi/3)n)x=cos((π/3)n). The signal is upsampled by a factor of two. Check the true statements about x_{2U} x

- x_{2U} = \cos((\pi/6)n) + \cos((5\pi/6)n))x
- x_{2U} = \cos((\pi/6)n)x
- x_{2U} = \cos((\pi/3)n) + \cos((\pi/6)n))x

Q4. Consider a discrete-time signal $x $ with the following spectrum, shown here between 00 and \piπ and symmetric around 00:

The signal is processed by the following system in which L(z)L(z) is an ideal lowpass filter with cutoff frequency \pi/2π/2 and H(z)H(z) is an ideal highpass filter with cutoff frequency \pi/2π/2:

Indicate which output (1, 2, 3, or 4) has the following spectrum:

Enter answer here

Q5. Consider a system implementing a rational sampling rate change by 5/75/7: for this, we cascade upsampler by 5, a lowpass filter with cutoff frequency \pi/7π/7 and a downsampler by 7. The lowpass filter is a 99-tap FIR.

Assume that the input works at a rate of 1000 samples per second. What is the number of multiplications per second required by the system? Assume that multiplications by zero do not count and round the number of operations to the nearest integer.

Enter answer here

Q6. Consider the same setup as in the previous exercise but now the lowpass filter is a 4rd-order Butterworth filter with transfer function

H(z) = \frac{b_0 + b_1z^{-1} + b_2z^{-2} + b_3z^{-3} + + b_4z^{-4}}{1 – a_1z^{-1} – a_2z^{-2} – a_3z^{-3} – a_4z^{-4}}H(z)=

Assume that the input works at a rate of 1000 samples per second. What is the number of multiplications per second required by the system? Assume that multiplications by zero do not count.

Enter answer here

### Week 4 Quiz Answers

#### Quiz 1: Homework for Module 3.4

Q1. (Difficulty: \star⋆) What is the minimum number of bits per sample you must use in order to sample and uniformly-quantize an analog signal with at least 80 dB of SNR?

Hint : you can use the "rule of thumb" from the lecture notes.

Q2. (Difficulty: \star \star\star⋆⋆⋆) A uniformly-distributed, zero mean stochastic signal with power spectral density P_x(e^{j\omega})=\sigma_x^2*P**x*(*e**j**ω*)=*σ**x*2 is quantized by means of a uniform linear quantizer with input range from -2\sqrt{3}\sigma_x−23*σ**x* to +2\sqrt{3}\sigma_x+23*σ**x* and resolution of R*R* bits per sample.

What is the SNR at the output of the quantizer?

- \frac{\sigma_x^2}{2^{2R}}22
*Rσx*2 - \frac{2^{2R}}{6}622
*R* - \frac{2^{2R}}{4}422
*R* - \frac{2^{R}}{2}22
*R* - 6R6
*R*

Q3. Consider a D/A converter for audio signals consisiting of a zero-order-hold interpolator followed by a continuous-time lowpass filter with positive passband between 0 and 20KHz and stopband starting at f_a=40*f**a*=40KHz.

Assume we want to convert a digital signal originally sampled at 16KHz. What is the minimum oversampling factor that we need to use?

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