Practice Problems (KEY)

Practice Problems (KEY)#

Problem 1#

What is aliasing and how can it be prevented?

Aliasing is distortion caused by sampling a signal at a rate that is too low. It can be prevented by selecting a sampling frequency, \(f_s\), that is greater than the Nyquist rate, which is

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

Aliasing can also be prevented by pre-filtering the signal prior to the analog-to-digital conversion process. The cutoff frequency of the low-pass filter must satisfy

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

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

Problem 2#

What are the three steps for converting an analog signal into digital?

  1. Sampling — Taking discrete measurements of the signal in time at the sampling frequency, \(f_s\).

  2. Quantizing — Assigning each sampled value to one of a finite number of discrete voltage levels.

  3. Encoding — Converting the quantized level into a binary number for storage or transmission.

Problem 3#

Given the following spectra, determine the minimum sampling frequency.

\[ f_s = 2f_{\text{High}} = 2(300\ \text{MHz}) = 600\ \text{MHz} \]
\[ f_s = 2f_{\text{High}} = 2(50\ \text{kHz}) = 100\ \text{kHz} \]

Problem 4#

Find the number of levels and resolution for an 8-bit ADC with

\(V_{\max} = 6\ \text{V}\) and \(V_{\min} = -4\ \text{V}\).

The number of levels is related to the number of bits by:

\[ \text{levels} = 2^b \]

Substituting \(b = 8\):

\[ \text{levels} = 2^8 = 256\ \text{levels} \]

Resolution is:

\[ \Delta V = \frac{V_{\max} - V_{\min}}{2^b} = \frac{6\ \text{V} - (-4\ \text{V})}{256} = \frac{10\ \text{V}}{256} = 0.0390625\ \text{V/level} = 39.0625\ \text{mV/level} \]

Problem 5#

Given a cosine input

\[ v_{in}(t) = 1 + 3\cos\!\left(360^\circ (50\,\text{kHz})t\right)\ \text{V}, \]

answer the following questions:

a. What is the Nyquist sampling frequency for this input signal?#

\[ f_s = 2f_{\text{High}} = 2(50\ \text{kHz}) = 100\ \text{kHz} \]

b. Which of these sampling frequencies would you use: 75 kHz, 90 kHz, or 120 kHz?#

You must use the \(120\ \text{kHz}\) sampling frequency. The other two options (\(75\ \text{kHz}\) and \(90\ \text{kHz}\)) would result in aliasing because they are below the Nyquist requirement.

c. What are the minimum and maximum values of the input signal?#

Given $\( v_{in}(t)=1+3\cos(\cdot)\ \text{V}, \)\( the cosine term ranges from \)-1\( to \)1$.

Minimum value: $\( V_{\min}=1+3(-1)=-2\ \text{V} \)$

Maximum value: $\( V_{\max}=1+3(1)=4\ \text{V} \)$

d. Which of the following ADCs would work with this signal?#

An ADC will work if its input range contains the signal range: $\( V_{\min,\text{ADC}} \le V_{\min,\text{sig}} \quad\text{and}\quad V_{\max,\text{ADC}} \ge V_{\max,\text{sig}}. \)$

  • \(V_{\max}=5\ \text{V}\) and \(V_{\min}=-3\ \text{V}\): Works (covers \(-2\ \text{V}\) to \(4\ \text{V}\)).

  • \(V_{\max}=5\ \text{V}\) and \(V_{\min}=-1\ \text{V}\): Does not work (clips below \(-1\ \text{V}\)).

  • \(V_{\max}=3\ \text{V}\) and \(V_{\min}=-3\ \text{V}\): Does not work (clips above \(3\ \text{V}\)).

e. Using the \(V_{\max}\) and \(V_{\min}\) from part (d), how many bits are required for a resolution of 600 mV or better?#

Start with the resolution equation and solve for \(2^b\):

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

Using \(V_{\max}=5\ \text{V}\) and \(V_{\min}=-3\ \text{V}\) and \(\Delta V=600\ \text{mV/level}=0.600\ \text{V/level}\):

\[ 2^b=\frac{5-(-3)}{0.600}=\frac{8}{0.600}=13.333\ \text{levels} \]

We need at least 14 levels, so choose the next power of 2:

\[ 2^4=16 \ge 14 \quad\Rightarrow\quad b=4\ \text{bits} \]

f. Given the number of bits from part (e), what is the actual resolution?#

\[ \Delta V=\frac{V_{\max}-V_{\min}}{2^b} =\frac{5-(-3)}{2^4} =\frac{8}{16} =0.5\ \text{V/level} =500\ \text{mV/level} \]

Problem 6#

An input signal

\[ v_{in}(t) = 6 + 5\cos\!\left(360^\circ (6\,\text{kHz})t\right) + 3\cos\!\left(360^\circ (15\,\text{kHz})t\right)\ \text{V} \]

is to be digitized using the ADC below.

Note: The 15 kHz portion of the signal is mainly noise and does not carry any useful information.

Is this a good ADC for this signal? Why or why not?
If not, what could you do to make this ADC work with this signal?

  1. Check for aliasing

Minimum sampling frequency (Nyquist), based on the highest frequency present:

\[ f_{s,\min} = 2f_{\text{High}} = 2(15\ \text{kHz}) = 30\ \text{kHz} \]

Given the ADC sampling frequency \(f_{s,\text{ADC}}=20\ \text{kHz}\):

\[ f_{s,\text{ADC}} < f_{s,\min} \quad\Rightarrow\quad \text{aliasing will occur.} \]

Since we are told the \(15\ \text{kHz}\) component is noise, we can place a low-pass filter (LPF) before the ADC to remove it. The cutoff frequency should pass the useful \(6\ \text{kHz}\) content while attenuating the \(15\ \text{kHz}\) noise. A reasonable requirement is:

\[ 6\ \text{kHz} < f_c < 10\ \text{kHz} \]

With this LPF in place, the highest frequency entering the ADC is below the Nyquist limit of \(f_{s,\text{ADC}}/2=10\ \text{kHz}\), so aliasing is avoided.

  1. Check for clipping

For the full signal (including the \(15\ \text{kHz}\) term), use the fact that \(\cos(\cdot)\) ranges from \(-1\) to \(1\).

Minimum value:

\[ V_{\min} = 6 + 5(-1) + 3(-1) = -2\ \text{V} \qquad\Rightarrow\qquad \text{min clipping will occur.} \]

Maximum value:

\[ V_{\max} = 6 + 5(1) + 3(1) = 14\ \text{V} \qquad\Rightarrow\qquad \text{max clipping will occur.} \]

If the \(15\ \text{kHz}\) term is filtered out (LPF applied), the signal becomes:

\[ v_{in}(t)=6+5\cos\!\left(360^\circ(6\,\text{kHz})t\right)\ \text{V} \]

Minimum value:

\[ V_{\min}=6+5(-1)=1\ \text{V} \qquad\Rightarrow\qquad \text{min clipping will occur.} \]

Maximum value:

\[ V_{\max}=6+5(1)=11\ \text{V} \qquad\Rightarrow\qquad \text{max clipping will occur.} \]

  1. Conclusion

In both cases (with or without the LPF), clipping occurs. To fix this, the signal must be conditioned with a transducer interface (gain and offset) to map the signal range into the ADC input range. Using the filtered-signal case from Step 2, the required interface is:

\[ K=0.7 \qquad\text{and}\qquad B=2.3\ \text{V} \]

so that the conditioned signal limits match the ADC limits.

Problem 7#

What are the advantages of digital signals?

Digital signals have several key advantages:

  • They are less susceptible to noise.

  • They can be easily stored and recovered.

  • They allow for easier encryption and signal processing.

Problem 8#

T / F. Improving the resolution is always better.

When you improve resolution, you typically increase the number of bits, \(b\) (assuming \(V_{\max}\) and \(V_{\min}\) remain fixed).

Increasing \(b\) increases the number of levels:

\[ \text{levels} = 2^b \]

While this improves resolution,

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

it also increases memory usage and required bandwidth. Therefore, resolution improvement involves an engineering tradeoff that must be carefully considered.

Problem 9#

T / F. Increasing the sampling rate improves resolution.

Sampling rate and resolution are independent of one another.

The sampling rate, \(f_s\), determines how finely the signal is divided in time (x-axis), while resolution, \(\Delta V\), determines how finely the signal is divided in amplitude (y-axis):

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

Changing \(f_s\) does not affect resolution, and changing the number of bits \(b\) does not affect the sampling rate.

Problem 10#

T / F. A smaller resolution is a better resolution.

Yes, a smaller resolution means less quantization error, since

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

However, although a smaller \(\Delta V\) is technically a better resolution, it may not be better at the system level. Achieving a smaller resolution typically requires increasing the number of bits, \(b\), which increases memory usage and required bandwidth.

Therefore, improving resolution involves an engineering tradeoff between accuracy and system resources.

Problem 11#

You want to convert \(v(t)\) to binary numbers to be stored. Your ADC has

\(V_{\max} = 0.5\ \text{V}\),
\(V_{\min} = -0.5\ \text{V}\),
and a 3-bit output.

\[ v(t) = 200 + 400\cos\!\left(360^\circ (200\,\text{kHz})t\right)\ \text{mV} \]
\[ v(t)=200+400\cos\!\left(360^\circ(200\,\text{kHz})t\right)\ \text{mV} \]

Draw the amplitude spectrum of \(v(t)\)#

a. How would you prepare the signal? Design the transducer interface.#

First, find the input signal limits (since \(\cos(\cdot)\in[-1,1]\)):

\[ v_{\min}=200-400=-200\ \text{mV} \qquad v_{\max}=200+400=600\ \text{mV} \]

We want a linear interface of the form:

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

so that the ADC range is matched:

\[ (-200\ \text{mV})K + B = -0.5\ \text{V} \]
\[ (600\ \text{mV})K + B = 0.5\ \text{V} \]

Solving gives:

\[ K=1.25 \qquad B=-0.25\ \text{V} \]

Therefore,

\[ v_{\text{out}} = 1.25\,v_{\text{in}} - 0.25\ \text{V} \]

b. What sampling rate would you choose?#

Highest frequency is \(f_{\text{High}}=200\ \text{kHz}\), so the Nyquist sampling rate is:

\[ f_s = 2f_{\text{High}} = 2(200\ \text{kHz}) = 400\ \text{kHz} \]

c. What is the resolution?#

For a 3-bit ADC, \(2^b=2^3=8\) levels, and \(V_{\max}=0.5\ \text{V}\), \(V_{\min}=-0.5\ \text{V}\):

\[ \Delta V=\frac{V_{\max}-V_{\min}}{2^b} =\frac{0.5\ \text{V}-(-0.5\ \text{V})}{8} =0.125\ \text{V/level} =125\ \text{mV/level} \]

d. What is the encoded output if the input is 315 mV?#

First, apply the transducer interface:

\[ v_{\text{out}} = 1.25(0.315\ \text{V}) - 0.25\ \text{V} = 0.14375\ \text{V} \]

Compute expected level:

\[ EL=\frac{V_{in}-V_{\min}}{\Delta V} =\frac{0.14375\ \text{V}-(-0.5\ \text{V})}{0.125\ \text{V/level}} =5.15 \]

Quantize (truncate):

\[ QL=\lfloor EL \rfloor = 5 \]

Encode \(QL=5\) in 3-bit binary:

\(2^2\)

\(2^1\)

\(2^0\)

4

2

1

1

0

1

So the encoded output is:

\[ 101 \]

Problem 12#

What is the resolution of a 10-bit ADC with

\(V_{\max} = 5\ \text{V}\) and \(V_{\min} = -3\ \text{V}\)?

\[ \Delta V=\frac{V_{\max}-V_{\min}}{2^b} =\frac{5\ \text{V}-(-3\ \text{V})}{1024} =\frac{8\ \text{V}}{1024} =0.0078125\ \text{V/level} =7.8125\ \text{mV/level} \]

Problem 13#

Given the following signal, create an ADC to capture it with a 5-bit ADC and determine its resolution:

\[ V(t) = \cos\!\left(360^\circ (8\,\text{kHz})t\right) + 2\cos\!\left(360^\circ (15\,\text{kHz})t\right)\ \text{V} \]

Signal limits#

Using \(\cos(\cdot)\in[-1,1]\):

\[ V_{\min}=1(-1)+2(-1)=-3\ \text{V} \]
\[ V_{\max}=1(1)+2(1)=3\ \text{V} \]

Sampling rate (Nyquist)#

Highest frequency is \(15\ \text{kHz}\), so:

\[ f_s = 2f_{\text{High}} = 2(15\ \text{kHz}) = 30\ \text{kHz} \]

Resolution (5-bit ADC)#

\[ \Delta V=\frac{V_{\max}-V_{\min}}{2^b} =\frac{3\ \text{V}-(-3\ \text{V})}{2^5} =\frac{6\ \text{V}}{32} =0.1875\ \text{V/level} =187.5\ \text{mV/level} \]

b) Using this ADC, encode 1.62 V#

Expected level:

\[ EL=\frac{V_{in}-V_{\min}}{\Delta V} =\frac{1.62\ \text{V}-(-3\ \text{V})}{0.1875\ \text{V/level}} =24.64 \]

Quantized level (truncate):

\[ QL=\lfloor EL \rfloor = 24 \]

Encode \(QL=24\) as a 5-bit binary number:

\(2^4\)

\(2^3\)

\(2^2\)

\(2^1\)

\(2^0\)

16

8

4

2

1

1

1

0

0

0

So the encoded output is:

\[ 11000 \]

Problem 14#

Given

\[ v_{in}(t)=4+4\cos\!\left(360^\circ(2\,\text{kHz})t\right)\ \text{V}, \]

and \(\text{max } QE = 200\ \text{mV}\), find the unknowns in the figure.

Signal limits#

\[ V_{\max}=4+4(1)=8\ \text{V} \]
\[ V_{\min}=4+4(-1)=0\ \text{V} \]

Sampling rate (Nyquist)#

Highest frequency is \(2\ \text{kHz}\), so:

\[ f_s = 2f_{\text{High}} = 2(2\ \text{kHz}) = 4\ \text{kHz} \]

Bits required from max quantization error#

Using the (given) design requirement \(\Delta V \le 200\ \text{mV/level}\):

\[ \Delta V=\frac{V_{\max}-V_{\min}}{2^b} \quad\Rightarrow\quad 2^b=\frac{V_{\max}-V_{\min}}{\Delta V} =\frac{8\ \text{V}-0}{0.200\ \text{V/level}} =40\ \text{levels} \]

We need \(2^b \ge 40\), so:

\[ 2^5=32<40,\qquad 2^6=64\ge 40 \quad\Rightarrow\quad b=6 \]

If a value of 1.6 V is measured, find QE and the encoded value#

First, compute the actual resolution for \(b=6\):

\[ \Delta V=\frac{8\ \text{V}-0}{2^6} =\frac{8}{64} =0.125\ \text{V/level} =125\ \text{mV/level} \]

Expected level:

\[ EL=\frac{V_{in}-V_{\min}}{\Delta V} =\frac{1.6\ \text{V}-0}{0.125\ \text{V/level}} =12.8 \]

Quantized level:

\[ QL=\lfloor EL \rfloor = 12 \]

Quantization error:

\[ QE=\Delta V\left(EL-\lfloor EL\rfloor\right) =0.125(12.8-12) =0.100\ \text{V} =100\ \text{mV} \]

Encode \(QL=12\) as a 6-bit binary number:

\(2^5\)

\(2^4\)

\(2^3\)

\(2^2\)

\(2^1\)

\(2^0\)

32

16

8

4

2

1

0

0

1

1

0

0

So the encoded output is:

\[ 001100 \]
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