Oxford Revise AQA A Level Physics | Chapter P30 answers

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P30: Analogue and digital signals

Question

Answers

Extra information

Mark

AO

Spec reference

01.1

 \(\textrm{Time for one sample } = \frac{1}{500} = 0.002\ \rm{s}\)

Time for one sample correct

Remaining points plotted correctly

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3.13.2.1

01.2

The sampling rate is too low and high-frequency sections of the signal will be lost

There will be aliasing (spurious low signals)

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3.13.2.1

01.3

Number of samples = 3 × 60 × 500 = 90 000

Bits per sample = 4

Total number of bits = 360 000

Number of bytes = 45 000B = 45 kB

Number of samples

Number of bits

Answer

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3.13.2.1

01.4

Increase the bits per sample

The song would increase the storage space required/take longer to send/download

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3.13.2.1

01.5

The output of the ADC is in parallel form because each sample contains 4 (simultaneous) bits

These bits have to be converted to a serial stream in order to be sent

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3.13.2.1

02.1

Circuit with capacitor and inductor in parallel

Aerial and earth connected at opposite points

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3.13.2.2

02.2

 \(
\begin{array}{l}
f_0 = \frac{1}{2\pi \sqrt {LC}}\\
C = \frac{1}{\left( 2\pi f_0 \right)^2 L}\\
=\frac{1}{\left( 2\pi \times 1.1 \times 10^6 \right)^2 \times 2.3 \times 10^{-3}}\\
= 9.1 \times 10^{-12}\ \rm{F} = 9.1\ \rm{pF}
\end{array}
\)

Capacitor A would work

Substitution

Answer

Capacitor identified

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3.13.2.2

02.3

 \(
\begin{array}{l}
Q = \frac{f_0}{f_{\rm{B}}}\\
{f_{\rm{B}}} = \frac{f_0}{Q} = \frac{1\,100\,000}{50}\\
= 22\,000\ \rm{Hz} = 22\ \textrm{kHz so the peak at}\ 0.7\ \rm{is}\ 22\ \rm{kHz\ wide}\\
\end{array}
\)

Bandwidth

Peak at 1100 and 1.0 relative gain

22 kHz wide at 0.71 relative gain

Approximate shape

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3.13.2.2

02.4

The mark scheme gives some guidance as to what statements are expected to be seen in a 1 or 2 mark (L1), 3 or 4 mark (L2), and 5 or 6 mark (L3) answer

Mark

Criteria

QoWC

6

A thorough and well-communicated discussion using most of the statements in bullets 1 and 2

The student presents relevant information coherently, employing structure, style, and SP&G to render meaning clear. The text is legible

5

An explanation that includes discussion using most of the statements in bullets 1 and 2 but may contain minor errors or omissions

4

The response includes a well-presented discussion of two from bullets 1 and two from bullet 2

The student presents relevant information and in a way which assists the communication of meaning. The text is legible.

SP&G are sufficiently accurate not to obscure meaning

3

The response includes a discussion of one comment from each bullet

2

The response makes comments about two bullet points (This is likely to be from bullets 2 and 3)

The student presents some relevant information in a simple form. The text is usually legible.

SP&G allow meaning to be derived although errors are sometimes obstructive

1

Makes relevant comment from the list

0

No relevant coverage of the likely statements

The student’s presentation, SP&G seriously obstruct understanding

The following statements are likely to be present:

Bullet point 1 in question

(Analogy between components)

1. The mass is analogous to the inductance

2. The spring constant is analogous to \(\frac{1}{\rm{capacitance}}\)

3. Energy can be stored in the mass–spring system, and in the LC system

4. The mass–spring system oscillates with a characteristic/ natural frequency, and so does the LC circuit

5. Energy can be transferred more easily at low frequencies, but there is increasing resistance to motion as the acceleration, which is analogous to the LC circuit

Bullet point 2 in question (Explaining resonant frequency)

5. For a mass–spring system, the time period is \(T = 2\pi \sqrt {\frac{m}{k}}\), so in the LC circuit \(T = 2\pi \sqrt {LC}\)

6. \(f\frac{1}{T}\), so \(f = \frac{1}{{2\pi \sqrt {LC} }}\)

7. When a mass–spring system is forced to oscillate at its natural frequency, it will resonate, so an LC circuit will oscillate when an alternating pd at the natural frequency is applied

8. In a mass–spring system, energy is transferred between being stored kinetically and being stored potentially. In an LC circuit, energy is transferred from being stored in the electric field in the capacitor and the magnetic field of the inductor

6

1

3.13.2.2

03.1

The output depends on the difference between the inputs to the inverting and non-inverting input

The gain of a comparator is infinite, but the gain of a difference amplifier can be controlled by the values of the resistors connected to the amplifier

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3.13.4.2

03.2

If the operational amplifier was being used as a comparator then the output would be ±24 V, and not values in between

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3.13.4.2

03.3

 \(
\begin{array}{l}
\rm{Gain}\ = \frac{{{V_{{\rm{out}}}}}}{{{V_ + } – {V_ -}}} = 4 = \frac{{{R_{\rm{f}}}}}{{{R_{{\rm{in}}}}}}\\
R_{in} = 4 \times 10\ \rm{k}\Omega = 40\ \rm{k}\Omega
\end{array}
\)

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3.13.4.3

03.4

The third electrode ensures that the ‘noise’ picked up by the mains would be the same for each electrode

Because the amplifier is amplifying the difference, the noise would not be amplified

The only p.d. that is amplified is that produced by differences between the signals produced by the electrodes, making it suitable for detecting irregularities in heart rhythms

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3.13.4.3

04.1

1101

Do not accept 1011

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3.13.2.1

04.2

 \(
\begin{array}{l}
{V_{{\rm{out}}}} = -10\,{\rm{k}} \times \left( {\frac{5}{{10\,{\rm{k}}}} + \frac{5}{{20\,{\rm{k}}}} + \frac{0}{{40\,{\rm{k}}}} + \frac{5}{{80\,{\rm{k}}}}} \right)\\
= -8.125\ \rm{V}
\end{array}
\)

Substitution

Answer (must be negative)

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1

3.13.4.3

04.3

Connect the output to an inverting amplifier to change to a positive value

\(
\begin{array}{l}
\frac{{{V_{{\rm{out}}}}}}{{{V_{{\rm{in}}}}}} = \frac{13}{{-8.125}} = -1.6\\
-\frac{{{R_{\rm{f}}}}}{{{R_{{\rm{in}}}}}} = – 1.6\\
{R_{{\rm{in}}}} = \frac{{{R_{\rm{f}}}}}{{1.6}} = \frac{{10\,{\rm{k}}\Omega }}{{1.6}} = 6.25\,{\rm{k}}\Omega
\end{array}
\)

Change to positive

Substitution

Answer

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3.13.4.1

04.4

Left-hand end of Rf is connected to virtual earth/inverting input is at 0‘V

Current in the feedback resistor is the same as the current in the input resistor/\({I_{{R_{\rm{f}}}}} = {I_{{R_{{\rm{in}}}}}}\)

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3.13.4.1

05.1

 \(
(\textrm{A}\ \bar{\textrm{B}}) + (\textrm{C})
\)

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3.13.5.1

05.2

A and NOT B into AND gate

Output and C into OR gate

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3.13.5.1

05.3

A

B

C

D

E

F

0

0

0

1

0

0

0

1

0

0

0

0

1

0

0

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1

0

0

0

0

0

0

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0

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0

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0

0

1

Column D correct 1 mark

Column E correct 1 mark

Column F correct 1 mark

Max 3

3

3.13.5.1

05.4

A BCD counter

The output of the logic circuit triggers the counter to start

The carry-out is used to input a circuit operating a switch to open the door

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3

3.13.5.2

06.1

Power supply, resistor, and capacitor in series

Voltmeter across the capacitor

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3.7.4.4

06.2

Time constant = RC = approximately 2 – 5 seconds

Capacitance = 6 – 15 × 10–6 F, or 6–16 mF

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3.7.4.4

06.3

 \(
\begin{array}{l}
\textrm{The time for the lights to be on and off}\ {t_{\rm{p}}} = \frac{10}{4} = 2.5\ \rm{s}\\
\textrm{Clock rate}\ = \frac{1}{{{t_{\rm{p}}}}} = \frac{1}{{2.5}} = 0.4\ \rm{Hz}\\
\textrm{Duration that the lights are on}\ = 1.25\ \rm{s}\\
\textrm{Duty cycle } = \frac{{{t_{{\rm{on}}}}}}{{{t_{\rm{p}}}}} \times 100 = 50\%
\end{array}
\)

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3.13.5.3

06.4

 \(
\begin{array}{l}
\textrm{The clock rate}\ = \frac{1}{{{\rm{period}}}},\ \rm{so}\ 1.3\ RC = \frac{1}{{{0}{.4}}} = 2.5\ \rm{s}\\
C = \frac{{2.5}}{{1.3 \times 330\,000}} = 5.8(7) \times 10^{-6}\ \rm{F}
\end{array}
\)

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3.13.5.3

07.1

The original length, the cross-sectional area

The extension for different loads/forces applied to the wire

\(\textrm{Calculate the strain } = \frac{{{\rm{extension}}}}{{{\rm{original}}\,{\rm{length}}}}\ \textrm{for each force}\) \(\textrm{Calculate the stress } = \frac{{{\rm{force}}}}{{{\rm{cross\ -\ sectional}}\,{\rm{area}}}}\ \textrm{for each force}\)

Plot a graph of stress (y-axis) against strain (x-axis)

The gradient of the initial linear portion of the graph is the Young modulus

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3.4.2.2

07.2

The output voltage is zero when there is no difference between the inputs to the operational amplifier

This happens when the potentials at P and Q are the same

P and Q are each the centre point of a potential divider, so they are at the same potential when the resistance of the wire is equal to the resistance of the variable resistor

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3.5.1.5

3.13.4.3

07.3

 \(
\begin{array}{l}
{V_{{\rm{out}}}} = \left( {{V_ + } – {V_ – }} \right)\frac{{{R_{\rm{f}}}}}{{{R_{{\rm{in}}}}}}\\
\left( {{V_ + } – {V_ – }} \right) = \frac{{{R_{{\rm{in}}}}}}{{{R_{\rm{f}}}}}{V_{{\rm{out}}}} = \frac{{10\,{\rm{k}}\,{\rm{\Omega\ }}}}{{410\,{\rm{k\Omega\ }}}} \times 12.5\\
= 0.305\ \rm{V}
\end{array}
\)

Substitution

Answer

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3.13.4.3

07.4

 \(
\begin{array}{l}
\textrm{The new potential at Q}\ = 5.688 + 0.305 = 5.993\ \rm{V}\\
\frac{50}{{50 + R}} \times 6 = 5.993\\
R = 0.056\ \Omega\\
\textrm{Original resistance } = \frac{50}{50 + R} \times 6 = 5.688\\
R = 2.743\ \Omega\\
\textrm{Change in resistance } = 2.743 – 0.056 = 2.687\ \Omega
\end{array}
\)

New potential at Q

Resistance now

Resistance before

Change

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3.5.1.5

08.1

The pressure due to the sound wave produces a force that moves the (diaphragm and) coil in and out

The magnetic flux that the coil cuts changes and induces a p.d. that matches the sound wave

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3.7.5.4

08.2

Transmitting device: laser/infrared LED

Transmission path: optical fibre

Receiving device: photodiode

This is more secure than copper wire/free space (electromagnetic waves)

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3.13.6.1

3.13.6.2

08.3

The signal travels long distances and the satellites have only a certain amount of electrical power, so the down-link signals received require much amplification

The up-link transmission frequency must be different from the down-link frequency to prevent the high-power down-link signal from the satellite from overwhelming the weak up-link signal

This would de-sensitise the high-gain up-link receiver

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3.13.6.2

08.4

 \(\lambda = \frac{v}{f} = \frac{{3 \times {{10}^8}}}{{1548 \times {{10}^3}}} = 193\,{\rm{m}}\ \approx\ 200\,{\rm{m}}\)

Yes, there is appreciable diffraction when the wavelength approximately equals the diameter/size of the obstacle

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3.3.1.1

3.3.2.2

08.5

Speech and music on the AM radio band have a frequency range of 4 kHz

There is a range of side frequencies/sidebands about the carrier frequency and there has to be a gap between them

The bandwidth for AM = 2 × fm where fm = the maximum frequency in the signal, which here is 4 kHz, so the spacing is a minimum of 8 kHz

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3.13.6.4

Skills box answers

Question

Answer

1

 \({f_{\rm{o}}} = \frac{1}{{2\pi \sqrt {\left( {LC} \right)} }} = \frac{1}{{2\pi \sqrt {\left( {8.0 \times 1.2 \times {{10}^{-6}}} \right)} }} = 51\,{\rm{Hz}}\)

2

 \(C = \left( {\frac{1}{L}} \right) \times {\left( {\frac{1}{{2\pi {f_{\rm{o}}}}}} \right)^2} = \left( {\frac{1}{{0.100}}} \right) \times {\left( {\frac{1}{{320\pi }}} \right)^2} = 10\,\mu {\kern 1pt} {\rm{F}}\)

3

 \(L = \left( {\frac{1}{C}} \right) \times {\left( {\frac{1}{{2\pi {f_{\rm{o}}}}}} \right)^2} = \left( {\frac{1}{{1.0 \times {{10}^{-6}}}}} \right) \times {\left( {\frac{1}{{8.2 \times {{10}^3}\pi }}} \right)^2} = 1.5\,{\rm{mH}}\)
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