Practice Problems (KEY)

Practice Problems (KEY)#

Problem 1#

An accelerometer used to measure the G loading on an aircraft outputs values between \(100\ \text{mV}\) and \(500\ \text{mV}\). We are interested in acceleration frequencies below \(4500\ \text{Hz}\).

Gain and Bias for Signal Conditioning#

We want to map the accelerometer range \(0.1\ \text{V}\) to \(0.5\ \text{V}\) into the ADC range \(2\ \text{V}\) to \(8\ \text{V}\).

Set up the linear interface:

\[ v_{\text{out}} = K v_{\text{in}} + B \]

Minimum condition:

\[ (0.1\ \text{V})K + B = 2\ \text{V} \]

Maximum condition:

\[ (0.5\ \text{V})K + B = 8\ \text{V} \]

Solving:

\[ K = 15 \qquad B = 0.5\ \text{V} \]

Therefore,

\[ v_{\text{out}} = 15v_{\text{in}} + 0.5\ \text{V} \]

Can the accelerometer be used with this ADC?#

Yes. The signal can be properly conditioned to match the ADC dynamic range.

Does the ADC meet the Nyquist rate?#

Yes. The sampling frequency satisfies:

\[ f_s > 2f_{\text{High}} \]

If \(v_{in}(t)=400\ \text{mV}\), what is the ADC output?#

First compute the ADC resolution (\(b=10\) bits):

\[ \Delta V=\frac{V_{\max}-V_{\min}}{2^b} =\frac{8\ \text{V}-2\ \text{V}}{2^{10}} =\frac{6\ \text{V}}{1024} =5.859375\ \text{mV/level} \]

Conditioned voltage:

\[ v_{\text{out}}=15(0.4\ \text{V})+0.5\ \text{V}=6.5\ \text{V} \]

Expected level:

\[ EL=\frac{V_{in}-V_{\min}}{\Delta V} =\frac{6.5\ \text{V}-2\ \text{V}}{5.859375\ \text{mV/level}} =768 \]

Quantized level:

\[ QL=768 \]

Binary encoding (10-bit):

\(2^9\)

\(2^8\)

\(2^7\)

\(2^6\)

\(2^5\)

\(2^4\)

\(2^3\)

\(2^2\)

\(2^1\)

\(2^0\)

512

256

128

64

32

16

8

4

2

1

1

1

0

0

0

0

0

0

0

0

Digital output:

\[ 1100000000 \]

Bit Rate into the Flight Data Recorder#

\[ \text{bit rate} = f_s b = (10\ \text{kHz})(10\ \text{bits}) = 100\ \text{kbps} \]

Time Until the FDR Fills (1 MB card)#

\[ \text{Time} =\frac{1\ \text{MB}\cdot 2^{20}\ \text{Bytes/MB}\cdot 8\ \text{bits/Byte}}{100\,000\ \text{bits/s}} =83.9\ \text{s} \]

Problem 2#

How can a low-pass filter (LPF) be used to prevent aliasing?

What must the cutoff frequency, \(f_c\), satisfy relative to the sampling frequency, \(f_s\)?

A low-pass filter (LPF) can be used to limit the highest frequency component, \(f_{\text{High}}\), of the input signal before it enters the ADC.

The filtered signal is then applied to the ADC, which samples at frequency \(f_s\). To prevent aliasing, the cutoff frequency of the LPF must satisfy:

\[ f_c \le \frac{f_s}{2} \]

This ensures that no frequency components above the Nyquist frequency enter the ADC.

Problem 3#

T / F: An ADC’s maximum quantization error cannot be greater than its resolution.

A low-pass filter (LPF) can be used to limit the highest frequency component, \(f_{\text{High}}\), of the input signal before it enters the ADC.

The filtered signal is then applied to the ADC, which samples at frequency \(f_s\). To prevent aliasing, the cutoff frequency of the LPF must satisfy:

\[ f_c \le \frac{f_s}{2} \]

This ensures that no frequency components above the Nyquist frequency enter the ADC.

Problem 4#

The following ADC is to be used to convert a music signal into a digital output:

a. To prevent aliasing, what must the highest frequency of the input signal be limited to by the low-pass filter?

\[ f_{\text{High}} \le \frac{f_s}{2} \]

b. What is the resolution of this ADC?

\[ \Delta V = \frac{V_{\max} - V_{\min}}{2^b} \]

c. What is the digital bandwidth of this system (i.e., how many bits per second does the ADC output)?

\[ \text{bit rate} = b f_s \]

a. To prevent aliasing, what must the highest frequency of the input be limited to?#

Given the sampling frequency:

\[ f_s = 25\ \text{kHz} \]

From the Nyquist criterion:

\[ f_s \ge 2f_{\text{High}} \quad\Rightarrow\quad f_{\text{High}} \le \frac{f_s}{2} \]

Therefore,

\[ f_{\text{High}} \le \frac{25\ \text{kHz}}{2} = 12.5\ \text{kHz} \]

The low-pass filter must limit the highest input frequency to \(12.5\ \text{kHz}\) or lower.

b. What is the resolution for this ADC?#

Use the resolution equation:

\[ \Delta V = \frac{V_{\max} - V_{\min}}{2^b} = \frac{10\ \text{V} - (-10\ \text{V})}{2^5} = \frac{20\ \text{V}}{32} = 0.625\ \text{V/level} = 625\ \text{mV/level} \]

Problem 5#

Given the two analog-to-digital converters below, answer the following questions and justify your answers.

Use the resolution equation,

\[ \Delta V=\frac{V_{\max}-V_{\min}}{2^b}, \]

to solve for each ADC.

a. Determine the resolution for each ADC#

ADC A:

\[ \Delta V_A=\frac{5\ \text{V}-0}{2^6} =\frac{5\ \text{V}}{64} =0.078125\ \text{V/level} =78.125\ \text{mV/level} \]

ADC B:

\[ \Delta V_B=\frac{10\ \text{V}-(-2\ \text{V})}{2^7} =\frac{12\ \text{V}}{128} =0.09375\ \text{V/level} =93.75\ \text{mV/level} \]

ADC A: 78.125 mV/level
ADC B: 93.75 mV/level

b. Which one has the better resolution?#

A smaller resolution is a better resolution. Since

\[ \Delta V_A < \Delta V_B, \]

ADC A has the better resolution.

c. Which one will take more samples of the input in 10 seconds?#

The ADC with the higher sampling frequency takes more samples. Since

\[ f_{s,A}=12\ \text{kHz} > f_{s,B}=10\ \text{kHz}, \]

ADC A will take more samples in 10 seconds.

d. If connected to a 2 Gbit thumb drive, which ADC would fill it faster?#

Compare bit rates:

\[ \text{bit rate}=f_s b \]

ADC A:

\[ \text{bit rate}_A=(12\ \text{kHz})(6)=72\ \text{kbps} \]

ADC B:

\[ \text{bit rate}_B=(10\ \text{kHz})(7)=70\ \text{kbps} \]

Since \(\text{bit rate}_A > \text{bit rate}_B\), ADC A will fill the thumb drive faster.#

Problem 6#

An accelerometer is used to measure the G loading on an aircraft.
It generates a \(+20\ \text{mV}\) signal at \(+8\ \text{G}\) and a \(-10\ \text{mV}\) signal at \(-2\ \text{G}\).
This signal must be conditioned before entering the ADC shown below.
The conditioning must also prevent aliasing.

a. Design a transducer interface (gain \(K\) and bias \(B\)) to map the accelerometer output across the full ADC dynamic range.

\[ v_{\text{out}} = K v_{\text{in}} + B \]

b. What is the resolution of this ADC?

\[ \Delta V = \frac{V_{\max} - V_{\min}}{2^b} \]

To cover the full dynamic range of the ADC, we should use the entire operating voltage range. Map the accelerometer maximum output to \(V_{\max}\) and the accelerometer minimum output to \(V_{\min}\).

Assume a linear transducer interface:

\[ v_{\text{out}} = K v_{\text{in}} + B \]

Minimum condition (\(-2\ \text{G}\)):

\[ (-10\ \text{mV})K + B = -2\ \text{V} \]

Maximum condition (\(8\ \text{G}\)):

\[ (20\ \text{mV})K + B = 12\ \text{V} \]

Solving the system gives:

\[ K = 466.7 \qquad B = 2.67\ \text{V} \]

To prevent aliasing, add a low-pass filter (LPF) before the ADC. A common design choice is to set the cutoff frequency to the Nyquist frequency:

\[ f_c = \frac{f_s}{2} \]

Given \(f_s = 10\ \text{kHz}\):

\[ f_c = \frac{10\ \text{kHz}}{2} = 5\ \text{kHz} \]

b. What is the resolution of this ADC?

\[ \Delta V=\frac{V_{\max}-V_{\min}}{2^b} =\frac{12\ \text{V}-(-2\ \text{V})}{2^4} =\frac{14\ \text{V}}{16} =0.875\ \text{V/level} =875\ \text{mV/level} \]

Problem 7#

An accelerometer signal is amplified with a gain of \(1000\), filtered to prevent aliasing, and then digitized using the system below.

a. If the flight data recorder has a storage capacity of \(275\ \text{MB}\), how long can it record data?

\[ \text{bit rate} = b f_s \]

Convert storage to bits and determine recording time.

The first step is to determine the data (bit) rate:

\[ \text{bit rate} = f_s b = (10\ \text{kHz})(4\ \text{bits}) = 40\ \text{kbps} \]

Now use unit conversions to compute the total stored data for \(320\ \text{min}\):

\[ \left(40\ \frac{\text{kb}}{\text{s}}\right) \left(\frac{1000\ \text{b}}{1\ \text{kb}}\right) \left(\frac{1\ \text{B}}{8\ \text{b}}\right) \left(\frac{1\ \text{MB}}{2^{20}\ \text{B}}\right) \left(\frac{60\ \text{s}}{1\ \text{min}}\right) \left(320\ \text{min}\right) = 91.6\ \text{MB} \]
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