Chapter 02 Periodic Sampling

2.1

Suppose you have a mechanical clock that has a minute hand, but no hour hand. Next, suppose you took a photograph of the clock when the minute hand was pointed at 12:00 noon and then took additional photos every 55 minutes. Upon showing those photos, in time order, to someone:

(a) What would that person think about the direction of motion of the minute hand as time advances?

Solution: He or she will think it is going counter-clock wise

(b) With the idea of lowpass sampling in mind, how often would you need to take photos, measured in photos/hour, so that the successive photos show proper (true) clockwise minute-hand rotation?

Solution: He or she just needs to take photos > photos/hour. Since we need to have $f_s > 2 B$.

$\square$

2.2

Assume we sampled a continuous $x(t)$ signal and obtained 100 $x(n)$ time- domain samples. What important information (parameter that we need to know in order to analyze $x(t)$) is missing from the $x(n)$ sequence?

Solution: The frequency of $x(t)$ is missing.

$\square$

2.3

National Instruments Corporation produces an analog-to-digital (A/D) converter (Model #NI-5154) that can sample (digitize) an analog signal at a sample rate of fs=2.0 GHz (gigahertz).

(a) What is the $t_s$ period of the output samples of such a device?

Solution: The $t_s = 500 \text{ ps}$.

(b) Each A/D output sample is an 8-bit binary word (one byte), and the converter is able to store 256 million samples. What is the maximum time interval over which the converter can continuously sample an analog signal?

Solution: It can sample $2 \times 10^9$ samples per second. Then it can only sample for

$$\frac{256 \cdot 10^6}{2 \times 10^9} = 0.128 \text{ second}$$

$\square$

2.4

Consider a continuous time-domain sinewave, whose cyclic frequency is 500 Hz, defined by

$$x(t) = \cos [2 \pi (500) t + \pi / 7]$$

Write the equation for the discrete $x(n)$ sinewave sequence that results from sampling $x(t)$ at an $f_s$ sample rate of 4000 Hz.

Solution:

$f_o = 500, f_s = 4000$, so $t_s = 1/4000$.

Then

$$x(n) = \cos [2 \pi \frac{n}{8} + \pi / 7]$$

$\square$

2.5

If we sampled a single continuous sinewave whose frequency is $f_o$ Hz, over what range must $t_s$ (the time between digital samples) be to satisfy the Nyquist criterion? Express that $t_s$ range in terms of $f_o$.

Solution:

We need to have $f_s \geq 2 f_o$, i.e. $\frac{1}{t_s} \geq 2 f_o$. So

$$t_s \leq \frac{1}{2 f_o}$$

$\square$

2.6

Suppose we used the following statement to describe the Nyquist criterion for lowpass sampling: “When sampling a single continuous sinusoid (a single analog tone), we must obtain no fewer than $N$ discrete samples per continu- ous sinewave cycle.” What is the value of this integer $N$?

Solution $N = 2$.

$\square$

2.7

The Nyquist criterion, regarding the sampling of lowpass signals, is sometimes stated as “The sampling rate fs must be equal to, or greater than, twice the highest spectral component of the continuous signal being sampled.” Can you think of how a continuous sinusoidal signal can be sampled in accordance with that Nyquist criterion definition to yield all zero-valued discrete samples?

Solution: Consider

$$x(t) = \sin 2 \pi \frac{f_s}{2} t$$

Then

$$x(n) = \sin 2 \pi \frac{f_s}{2} n \frac{1}{f_s} = \sin n \pi = 0$$

$\square$

2.8

Stock market analysts study time-domain charts (plots) of the closing price of stock shares. A typical plot takes the form of that in Figure P2–8, where in- stead of plotting discrete closing price sample values as dots, they draw straight lines connecting the closing price value samples. What is the $t_s$ period for such stock market charts?

Solution:

The $t_s$ is one day, 24 hours.

$\square$

2.9

Consider a continuous time-domain sinewave defined by

$$x(t) = \cos (4000 \pi t)$$

that was sampled to produce the discrete sinewave sequence defined by

$$x(n) = \cos (n \pi / 2)$$

What is the $f_s$ sample rate, measured in Hz, that would result in sequence $x(n)$?

Solution:

$x(t) = \cos (2 f_o \pi t) = \cos (4000 \pi t)$, so $f_o = 2000$ Hz. Then

$$x(n) = \cos 2 \pi f_o n \frac{1}{f_s} = \cos 4000 \pi n \frac{1}{f_s}$$

So

$$4000 \pi n \frac{1}{f_s} = n \pi / 2 \\ \Rightarrow f_s = 8000 \text{ Hz}$$

$\square$

2.10

Consider the two continuous signals defined by

$$a(t) = \cos (4000 \pi t) \text{ and } b(t) = \cos (200 \pi t)$$

whose product yields the $x(t)$ signal shown in Figure P2–10. What is the minimum $f_s$ sample rate, measured in Hz, that would result in a sequence $x(n)$ with no aliasing errors (no spectral replication overlap)?

Solution:

Use the trigonometric identities

$$\cos (x) \cos (y) = \frac{1}{2} [\cos (x-y) + \cos (x+y)]$$

So

$$x(t) = a(t) \times b(t) = \frac{1}{2} [ \cos (4200 \pi t) + \cos (3800 \pi t)]$$

So the max frequency component is $f_o = 2100$ Hz. Then we need to have $f_s \geq 2 \times 2100 = 4200$ Hz.

$\square$

2.11

Consider a discrete time-domain sinewave sequence defined by

$$x(n) = \sin (n \pi / 4)$$

that was obtained by sampling an analog $x(t) = \sin(2π f_o t)$ sinewave signal whose frequency is $f_o$ Hz. If the sample rate of $x(n)$ is $fs = 160$ Hz, what are three possible positive frequency values, measured in Hz, for $f_o$ that would result in sequence $x(n)$?

Solution:

$$x(n) = \sin 2 \pi (f_o + k f_s) n \frac{1}{f_s} = \sin (n \pi / 4) \\ \Rightarrow \\ 8 (f_o + k f_s) = f_s \\ \Rightarrow \\ 8 (f_o + k f_s) = f_s \\ \Rightarrow \\ f_o + k \cdot 160 = 20 \\ $$

Let $k = 0, -1, -2$ we can get $f_o = 20, 180, 340$ Hz.

$\square$

2.12

In the text we discussed the notion of spectral folding that can take place when an $x_a(t)$ analog signal is sampled to produce a discrete $x_d(n)$ sequence. We also stated that all of the analog spectral energy contained in $X_a(f)$ will reside within the frequency range of $\pm f_s/2$ of the $X_d(f)$ spectrum of the sampled $x_d(n)$ sequence.

Given those concepts, consider the spectrum of an analog signal shown in Figure P2–12(a) whose spectrum is divided into the six segments marked as 1 to 6. Fill in the following table showing which of the A-to-F spectral segments of $X_d(f)$, shown in Figure P2–12(b), are aliases of the 1-to-6 spectral segments of $X_a(f)$.

Solution:

$$A \rightarrow 3 \\ B \rightarrow 4 \\ C \rightarrow 5 \\ D \rightarrow 2 \\ E \rightarrow 1 \\ F \rightarrow 6 \\ $$

$\square$

2.13

Consider the simple analog signal defined by $x(t)=\sin (2π700t)$ shown in Figure P2–13. Draw the spectrum of $x(n)$ showing all spectral components, labeling their frequency locations, in the frequency range $[-2 f_s, 2 f_s]$

Solution: We have alias for $700$ at $[-1300, 300, 700, 1700]$ and alias for $-700$ Hz at $[-1700, -700, 300, 1300]$.

$\square$

2.14

The Nançay Observatory, in France, uses a radio astronomy receiver that generates a wideband analog $s(t)$ signal whose spectral magnitude is represented in Figure P2–14. The Nançay scientists bandpass sample the analog $s(t)$ signal, using an analog-to-digital (A/D) converter to produce an $x(n)$ discrete sequence, at a sample rate of $f_s = 56$ MHz.

(a) Draw the spectrum of the $x(n)$ sequence, $X(f)$, showing its spectral energy over the frequency range –70 MHz to 70 MHz.

Solution:

We have alias for $[70-7, 70+7]$ at $[14-7, 14+7], [-42-7,-42+7]$. We have alias for $[-70-7, -70+7]$ at $[-14-7, -14+7], [42-7,42+7]$.

(b) What is the center frequency of the first positive-frequency spectral replication in $X(f)$?

Solution: It's 14 MHz.

(c) How is your solution to Part (b) related to the $f_s$ sample rate?

Solution:

$f_s = 56$ MHz, then it's $f_s / 4$.

$\square$

2.15

Think about the continuous (analog) signal $x(t)$ that has the spectral magnitude shown in Figure P2–15. What is the minimum $f_s$ sample rate for lowpass sampling such that no spectral overlap occurs in the frequency range of 2 to 9 kHz in the spectrum of the discrete $x(n)$ samples?

Solution:

If $f_s = 20$ kHz, then there will be no overlap at all. Since we can tolerate the overlap in the range of 9 to 10, we can reduce $f_s$ further to $f_s = 19$ kHz. Then there is overlap for $[9,10]$, but no overlap between $[2,9]$.

$\square$

2.16

If a person wants to be classified as a soprano in classical opera, she must be able to sing notes in the frequency range of 247 Hz to 1175 Hz. What is the minimum $f_s$ sampling rate allowable for bandpass sampling of the full audio spectrum of a singing soprano?

Solution:

In this case,

$$f_c = 711, B = 1175 - 247 = 928$$

So

$$f_s = \frac{2 f_c+B}{1} = 2350 \text{ Hz}$$

2.17

This problem requires the student to have some knowledge of electronics and how a mixer operates inside a radio. (The definition of a bandpass filter is given in Appendix F.) Consider the simplified version of what is called a su- perheterodyne digital radio depicted in Figure P2–17.

(a) For what local oscillator frequency, $f_{\text{LO}}$, would an image (a copy, or duplication) of the $w(t)$ signal’s spectrum be centered at 15 MHz (megahertz) in signal $u(t)$?

Solution: Since the analog bandpass filter #1's Center frequency = 50 MHz and bandwidth = 10 MHz, then we can assume

$$w(t) = \sin (2 \pi 40 t) + \sin (2 \pi 30 t)$$

Then since $m(t) = \sin (2 \pi l t)$, then

$$u(t) = w(t) m(t) = \frac{1}{2} [ \cos (2 \pi t (40 - l)) - \cos (2 \pi t (40 + l)) ] + \frac{1}{2} [ \cos (2 \pi t (30 - l)) - \cos (2 \pi t (30 + l)) ]$$

So if $f_{\text{LO}} = 20$, we can have an image of the $w(t)$ signal’s spectrum be centered at 15 MHz in signal $u(t)$.

$\square$

(b) What is the purpose of the analog bandpass filter #2?

Solution: filter out all the signals beyond $[12.5, 17.5]$ MHz.

(c) Fill in the following table showing all ranges of acceptable fs bandpass sampling rates to avoid aliasing errors in the discrete $x(n)$ sequence. Also list, in the rightmost column, for which values of $m$ the sampled spectrum, centered at 15 MHz, will be inverted.

Solution:

Since $f_c = 15, B = 5$

$$\left[ \frac{2 f_c + B}{m+1}, \frac{2 f_c - B}{m} \right] = \begin{cases} [17.5, 25.0] &\text{if } m = 1 \text{ inverted }\\ [11.7, 12.5] &\text{if } m = 2 \text{ not inverted }\\ [8.75, 8.33] &\text{if } m = 3 \text{ invalid!!} \\ \end{cases}$$

Note here, when $m = 3$, $f_s < 2B$, then it's not valid.

(d) In digital receivers, to simplify AM and FM demodulation, it is advantageous to have the spectrum of the discrete $x(n)$ sequence be centered at one-quarter of the sample rate. The text’s Eq. 2–11 describes how to achieve this situation. If we were constrained to have $f_s$ equal to 12 MHz, what would be the maximum $f_{\text{LO}}$ local oscillator frequency such that the spectra of $u(t)$, $x(t)$, and $x(n)$ are centered at $f_s/4$? (Note: In this scenario, the $f_c$ center frequency of analog bandpass filter #2 will no longer be 15 MHz.)

Solution:

$$f_s = \frac{4 f_c}{2k - 1} \\ \Rightarrow \\ f_c = \frac{2k-1}{4} f_s = 6k - 3 \\ \Rightarrow \\ f_c = 3, 9, 15, \cdots$$

The $f_c$ cannot be $3$, since $w(t)$ then bandwidth is $10$. $9$ is fine. In this case, the $f_{\text{LO}} = 26$ MHz.

$\square$

2.18

Think about the analog anti-aliasing filter given in Figure P2–18(a), having a one-sided bandwidth of $B$ Hz. A wideband analog signal passed through that filter, and then sampled, would have an $|X(m)|$ spectrum as shown in Figure P2–18(b), where the dashed curves represent spectral replications.

Suppose we desired that all aliased spectral components in $|X(m)|$ over our $B$ Hz bandwidth of interest must be attenuated by at least $60$ dB. Determine the equation, in terms of $B$ and the $f_s$ sampling rate, for the frequency at which the anti-aliasing filter must have an attenuation value of –60 dB. The solution to this problem gives us a useful rule of thumb we can use in specifying the desired performance of analog anti-aliasing filters.

Solution: let the width of small area be $c$, then we have $2B + c = f_s$, so

$$B+c = f_s - B$$

So the anti-aliasing filter needs to have an attenuation value of –60 dB at $f_s - B$ Hz.

$\square$

2.19

This problem demonstrates a popular way of performing frequency down-conversion (converting a bandpass signal into a lowpass signal) by way of bandpass sampling. Consider the continuous 250-Hz-wide bandpass $x(t)$ signal whose spectral magnitude is shown in Figure P2–19. Draw the spectrum, over the frequency range of –1.3$f_s$ to +1.3$f_s$, of the $x(n)$ sampled sequence obtained when $x(t)$ is sampled at $f_s=1000$ samples/second.

Solution:

The results is exactly like the following 2-9 (c).

2.20

Here’s a problem to test your understanding of bandpass sampling. Think about the continuous (analog) signal $x(t)$ that has the spectral magnitude shown in Figure P2–20.

(a) What is the minimum $f_c$ center frequency, in terms of $x(t)$’s bandwidth $B$, that enables bandpass sampling of $x(t)$? Show your work.

Solution: To be able to do the bandpass sampling, use the figure 2-9 (a) (b) you need at least to have

$$f_c = B + B/2 = \frac{3}{2}B$$

$\square$

(b) Given your results in Part (a) above, determine if it is possible to perform bandpass sampling of the full spectrum of the commercial AM (amplitude modulation) broadcast radio band in North America. Explain your solution.

Solution:

According to the wiki, the commercial AM broadcast radio band in North America occupies the frequency range 535 to 1705 kHz.

Then $B = 1170, B/2 = 585, f_c = 1120$, so $f_c < \frac{3}{2}B$.

It is not possible.

$\square$

2.21

Suppose we want to perform bandpass sampling of a continuous 5 kHz-wide bandpass signal whose spectral magnitude is shown in Figure P2–21.

Fill in the following table showing the various ranges of acceptable $f_s$ band-pass sampling rates, similar to the text’s Table 2–1, to avoid aliasing errors. Also list, in the rightmost column, for which values of m the sampled spectrum in the vicinity of zero Hz is inverted.

Solution:

$f_c = 25, B = 5$ $$\left[ \frac{2 f_c + B}{m+1}, \frac{2 f_c - B}{m} \right] = \begin{cases} [27.5, 45.0] &\text{if } m = 1 \text{ inverted }\\ [18.3, 22.5] &\text{if } m = 2 \text{ not inverted }\\ [13.75, 15.0] &\text{if } m = 3 \text{ inverted } \\ [11, 11.25] &\text{if } m = 4 \text{ not inverted } \\ [9.17, 9] &\text{if } m = 5 \text{ invalid!! } \\ \end{cases}$$

$\square$

2.22

I recently encountered an Internet website that allegedly gave an algorithm for the minimum fs bandpass sampling rate for an analog bandpass signal centered at $f_c$ Hz, whose bandwidth is $B$ Hz. The algorithm is

$$f_{s, \min } = \frac{4 f_c}{2Z - 1}$$

where

$$Z = \left\lfloor \frac{4 f_c + 2B}{4B} \right\rfloor$$

Here’s the problem: Is the above $f_{s,\min}$ algorithm correct in computing the absolute minimum possible nonaliasing $f_s$ bandpass sampling rate for an analog bandpass signal centered at $f_c$ Hz, whose bandwidth is $B$ Hz? Verify your answer with an example.

Solution:

In ex 2.21, $f_c = 25, B = 5, Z = 5$, then

$$f_{s, \min } = \frac{100}{9} = 11.11 > 11$$

So this algorithm is not correct.

$\square$