5.5 FREQUENCY RESPONSE OF [DISCRETE-TIME](#page-11-0) SYSTEMS

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5.5 FREQUENCY RESPONSE OF DISCRETE-TIME SYSTEMS

For (asymptotically or BIBO-stable) continuous-time systems, we showed that the system response to an input ejω*^t* is H(jω)ejω*^t* and that the response to an input cos ωt is |H(jω)| cos[ωt + H(jω)]. Similar results hold for discrete-time systems. We now show that for an (asymptotically or BIBO-stable) LTID system, the system response to an input ejⁿ is H[ej]ejⁿ and the response to an input cos n is |H[ej]| cos(n+ H[ej]).

The proof is similar to the one used for continuous-time systems. In Sec. 3.8-2, we showed that an LTID system response to an (everlasting) exponential zn is also an (everlasting) exponential H[z]zⁿ. This result is valid only for values of z for which H[z], as defined in Eq. (5.11), exists (converges). As usual, we represent this input–output relationship by a directed arrow notation as

$$z^n \Longrightarrow H[z]z^n \tag{5.30}$$

Setting z = ej in this relationship yields

$$e^{i\Omega n} \Longrightarrow H[e^{i\Omega}]e^{i\Omega n} \tag{5.31}$$

Noting that cos n is the real part of ejⁿ, use of Eq. (3.34) yields

$$\cos \Omega n \Longrightarrow \text{Re}\{H[e^{i\Omega}]e^{i\Omega n}\}\tag{5.32}$$

Expressing H[ej] in the polar form

$$H[e^{i\Omega}] = |H[e^{i\Omega}]|e^{i\angle H[e^{i\Omega}]}$$

Eq. (5.32) can be expressed as

$$\cos \Omega n \Longrightarrow |H[e^{i\Omega}]|\cos(\Omega n + \angle H[e^{i\Omega}])$$

In other words, the system response y[n] to a sinusoidal input cos n is given by

$$y[n] = |H[e^{j\Omega}]|\cos(\Omega n + \angle H[e^{j\Omega}])$$

Following the same argument, the system response to a sinusoid cos(n+θ ) is

$$y[n] = |H[e^{j\Omega}]\cos(\Omega n + \theta + \angle H[e^{j\Omega}])\tag{5.33}$$

This result is valid only for BIBO-stable or asymptotically stable systems. The frequency response is meaningless for BIBO-unstable systems (which include marginally stable and asymptotically unstable systems). This follows from the fact that the frequency response in Eq. (5.31) is obtained by setting z=ej in Eq. (5.30). But, as shown in Sec. 3.8-2 [Eqs. (3.38) and (3.39)], the relationship of Eq. (5.30) applies only for values of z for which H[z] exists. For BIBO-unstable systems, the ROC for H[z] does not include the unit circle where z = ej. This means, for BIBO-unstable systems, that H[z] is meaningless when z = ej. †

This important result shows that the response of an asymptotically or BIBO-stable LTID system to a discrete-time sinusoidal input of frequency is also a discrete-time sinusoid of the same frequency. The amplitude of the output sinusoid is |H[ej]| times the input amplitude, and the phase of the output sinusoid is shifted by H[ej] with respect to the input phase. Clearly, |H[ej]| is the amplitude gain, and a plot of |H[ej]| versus is the amplitude response of the discrete-time system. Similarly, H[ej] is the phase response of the system, and a plot of H[ej] versus shows how the system modifies or shifts the phase of the input sinusoid. Note that H[ej] incorporates the information of both amplitude and phase responses and therefore is called the frequency responses of the system.

STEADY-STATE RESPONSE TO CAUSAL SINUSOIDAL INPUT

As in the case of continuous-time systems, we can show that the response of an LTID system to a causal sinusoidal input cos n u[n] is y[n] in Eq. (5.33), plus a natural component consisting of the characteristic modes (see Prob. 5.5-9). For a stable system, all the modes decay exponentially, and only the sinusoidal component in Eq. (5.33) persists. For this reason, this component is called the sinusoidal steady-state response of the system. Thus, yss[n], the steady-state response of a system to a causal sinusoidal input cos n u[n], is

$$y_{ss}[n] = |H[e^{j\Omega}]\cos{(\Omega n + \angle H[e^{j\Omega}])}u[n]$$

SYSTEM RESPONSE TO SAMPLED CONTINUOUS-TIME SINUSOIDS

So far we have considered the response of a discrete-time system to a discrete-time sinusoid cos n (or exponential ejⁿ). In practice, the input may be a sampled continuous-time sinusoid cos ωt (or an exponential ejω*^t* ). When a sinusoid cos ωt is sampled with sampling interval T, the resulting

^† This may also be argued as follows. For BIBO-unstable systems, the zero-input response contains nondecaying natural mode terms of the form cos0n or γ ⁿ cos0n (γ > 1). Hence, the response of such a system to a sinusoid cosn will contain not just the sinusoid of frequency but also nondecaying natural modes, rendering the concept of frequency response meaningless. Alternately, we can argue that when ^z=ej, a BIBO-unstable system violates the dominance condition ^|γi^{| &}lt; ^|ej^| for all ⁱ, where ^γi represents *ⁱ*th characteristic root of the system.

528 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM

signal is a discrete-time sinusoid cos ωnT, obtained by setting t = nT in cosωt. Therefore, all the results developed in this section apply if we substitute ωT for :

$$\Omega = \omega T \tag{5.34}$$

EXAMPLE 5.10 Sinusoidal Response of a Difference Equation System

For a system specified by the equation

$$y[n+1] - 0.8y[n] = x[n+1]$$

find the system response to the inputs

(a)

$$1^n = 1$$

(b) $\cos\left[\frac{\pi}{6}n - 0.2\right]$

(c) a sampled sinusoid cos 1500t with sampling interval T = 0.001

The system equation can be expressed as

$$(E - 0.8)y[n] = Ex[n]$$

Therefore, the transfer function of the system is

$$H[z] = \frac{z}{z - 0.8} = \frac{1}{1 - 0.8z^{-1}}$$

The frequency response is

$$H[e^{i\Omega}] = \frac{1}{1 - 0.8e^{-i\Omega}} = \frac{1}{(1 - 0.8\cos\Omega) + i0.8\sin\Omega}$$

Therefore,

$$|H[e^{i\Omega}]| = \frac{1}{\sqrt{(1 - 0.8 \cos \Omega)^2 + (0.8 \sin \Omega)^2}} = \frac{1}{\sqrt{1.64 - 1.6 \cos \Omega}}$$

(5.35)

and

$$\angle H[e^{i\Omega}] = -\tan^{-1}\left[\frac{0.8\sin\Omega}{1 - 0.8\cos\Omega}\right]$$

\n(5.36)

The amplitude response |H[ej]| can also be obtained by observing that |H| ² = HH∗. Since our system is real, we therefore see that

$$|H[e^{i\Omega}]|^{2} = H[e^{i\Omega}]H^{*}[e^{i\Omega}] = H[e^{i\Omega}]H[e^{-i\Omega}]$$

\n(5.37)

Substituting for H[ej], it follows that

$$|H[e^{i\Omega}]|^2 = \left(\frac{1}{1 - 0.8e^{-i\Omega}}\right)\left(\frac{1}{1 - 0.8e^{i\Omega}}\right) = \frac{1}{1.64 - 1.6\cos\Omega}$$

which matches the result found earlier.

Figure 5.14 shows plots of amplitude and phase response as functions of . We now compute the amplitude and the phase response for the various inputs.

Figure 5.14 Frequency response of the LTID system.

(a) Since 1*ⁿ* = (ej)ⁿ with = 0, the amplitude response is H[ej⁰]. From Eq. (5.35) we obtain

$$H[e^{i0}] = \frac{1}{\sqrt{1.64 - 1.6 \cos(0)}} = \frac{1}{\sqrt{0.04}} = 5 = 5 \angle 0$$

Therefore,

$$|H[e^{j0}]| = 5 \quad \text{and} \quad \angle H[e^{j0}] = 0$$

These values also can be read directly from Figs. 5.14a and 5.14b, respectively, corresponding to = 0. Therefore, the system response to input 1 is

$$y[n] = 5(1^n) = 5 \qquad \text{for all } n$$

(b) For x[n] = cos[(π/6)n−0.2], = π/6. According to Eqs. (5.35) and (5.36),

$$|H[e^{j\pi/6}]| = \frac{1}{\sqrt{1.64 - 1.6 \cos \frac{\pi}{6}}} = 1.983$$ $$\angle H[e^{j\pi/6}] = -\tan^{-1} \left[ \frac{0.8 \sin \frac{\pi}{6}}{1 - 0.8 \cos \frac{\pi}{6}} \right] = -0.916 \text{ rad}$$

These values also can be read directly from Figs. 5.14a and 5.14b, respectively, corresponding to = π/6. Therefore,

$$y[n] = 1.983 \cos\left(\frac{\pi}{6}n - 0.2 - 0.916\right) = 1.983 \cos\left(\frac{\pi}{6}n - 1.116\right)$$

Figure 5.15 shows the input x[n] and the corresponding system response.

Figure 5.15 Sinusoidal input and the corresponding output of the LTID system.

(c) A sinusoid cos 1500t sampled every T seconds (t = nT) results in a discrete-time sinusoid

$$x[n] = \cos 1500nT$$

For T = 0.001, the input is

x[n] = cos(1.5n)

In this case, = 1.5. According to Eqs. (5.35) and (5.36),

$$|H[e^{j1.5}]| = \frac{1}{\sqrt{1.64 - 1.6 \cos(1.5)}} = 0.809$$ $$\angle H[e^{j1.5}] = -\tan^{-1} \left[ \frac{0.8 \sin(1.5)}{1 - 0.8 \cos(1.5)} \right] = -0.702 \text{ rad}$$

These values also could be read directly from Fig. 5.14 corresponding to = 1.5. Therefore,

$$y[n] = 0.809 \cos(1.5n - 0.702)$$

FREQUENCY RESPONSE PLOTS USING MATLAB

MATLAB makes it easy to compute and plot magnitude and phase responses directly using a system’s transfer function. As the following code demonstrates, there is no need to derive separate expressions for the magnitude and phase responses.

Omega = linspace(-pi,pi,400); H = @(z) z./(z-0.8);

subplot(1,2,1); plot(Omega,abs(H(exp(1j*Omega))),‘k’); axis tight;

xlabel(‘\Omega’); ylabel(‘|H[e^{j \Omega}]|’);

subplot(1,2,2); plot(Omega,angle(H(exp(1j*Omega))*180/pi),‘k’); axis tight;

>> xlabel('\Omega'); ylabel('\angle H[e^{j \Omega}] [deg]');

The resulting plots, shown in Fig. 5.16, confirm the earlier results of Fig. 5.14.

532 CHAPTER 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE Z-TRANSFORM

Comment: Figures 5.14 and 5.16 show amplitude and phase response plots as functions of . These plots as well as Eqs. (5.35) and (5.36) indicate that the frequency response of a discrete-time system is a continuous (rather than discrete) function of frequency . There is no contradiction here. This behavior is merely an indication of the fact that the frequency variable is continuous (takes on all possible values) and therefore the system response exists at every value of .

DR ILL 5.18 Frequency Response of Difference Equation

For a system specified by the equation

$$y[n+1] - 0.5y[n] = x[n]$$

find the amplitude and the phase response. Find the system response to sinusoidal input cos[1000t −(π/3)] sampled every T = 0.5 ms.

ANSWER

$$|H[e^{i\Omega}]| = \frac{1}{\sqrt{1.25 - \cos \Omega}}$$

\n

$$\angle H[e^{i\Omega}] = -\tan^{-1} \left[ \frac{\sin \Omega}{\cos \Omega - 0.5} \right]$$

\n

$$y[n] = 1.639 \cos \left( 0.5n - \frac{\pi}{3} - 0.904 \right) = 1.639 \cos (0.5n - 1.951)$$

DR ILL 5.19 Frequency Response of an Ideal Delay System

Show that for an ideal delay (H[z] = 1/z), the amplitude response |H[ej]| = 1, and the phase response H[ej]=−. Thus, a pure time delay does not affect the amplitude gain of sinusoidal input, but it causes a phase shift (delay) of radians in a discrete sinusoid of frequency . Thus, for an ideal delay, the phase shift of the output sinusoid is proportional to the frequency of the input sinusoid (linear phase shift).

5.5-1 The Periodic Nature of Frequency Response

In Ex. 5.10 and Fig. 5.14, we saw that the frequency response H[ej] is a periodic function of . This is not a coincidence. Unlike continuous-time systems, all LTID systems have periodic frequency response. This is seen clearly from the nature of the expression of the frequency response of an LTID system. Because e±j2π*^m* = 1 for all integer values of m [see Eq. (B.10)],

$$H[e^{j\Omega}] = H[e^{j(\Omega + 2\pi m)}] \qquad m \text{ integer}$$

Therefore, the frequency response H[ej] is a periodic function of with a period 2π. This is the mathematical explanation of the periodic behavior. The physical explanation that follows provides a much better insight into the periodic behavior.

NON-UNIQUENESS OF DISCRETE-TIME SINUSOID WAVEFORMS

A continuous-time sinusoid cos ωt has a unique waveform for every real value of ω in the range 0 to ∞. Increasing ω results in a sinusoid of ever-increasing frequency. Such is not the case for the discrete-time sinusoid cos n because

$$\cos[(\Omega \pm 2\pi m)n] = \cos \Omega n \qquad m \text{ integer}$$

and

$$e^{j(\Omega \pm 2\pi m)n} = e^{j\Omega n} \qquad m \text{ integer}$$

This shows that the discrete-time sinusoids cos n (and exponentials ejⁿ) separated by values of in integral multiples of 2π are identical. The reason for the periodic nature of the frequency response of an LTID system is now clear. Since the sinusoids (or exponentials) with frequencies separated by interval 2π are identical, the system response to such sinusoids is also identical and, hence, is periodic with period 2π.

This discussion shows that the discrete-time sinusoid cos n has a unique waveform only for the values of in the range −π to π. This band is called the fundamental band. Every frequency , no matter how large, is identical to some frequency, a, in the fundamental band (−π ≤ ^a < π), where

$$\Omega_a = \Omega - 2\pi m \qquad -\pi \le \Omega_a < \pi \quad \text{and} \quad m \text{ integer} \tag{5.38}$$

The integer m can be positive or negative. We use Eq. (5.38) to plot the fundamental band frequency ^a versus the frequency of a sinusoid (Fig. 5.17a). The frequency ^a is modulo 2π value of .

All these conclusions are also valid for exponential ejⁿ.

ALL DISCRETE-TIME SIGNALS ARE INHERENTLY BANDLIMITED

This discussion leads to the surprising conclusion that all discrete-time signals are inherently bandlimited, with frequencies lying in the range −π to π radians per sample. In terms of frequency F = /2π, where F is in cycles per sample, all frequencies F separated by an integer number are identical. For instance, all discrete-time sinusoids of frequencies 0.3, 1.3, 2.3, … cycles per sample are identical. The fundamental range of frequencies is −0.5 to 0.5 cycles per sample.

Any discrete-time sinusoid of frequency beyond the fundamental band, when plotted, appears and behaves, in every way, like a sinusoid having its frequency in the fundamental band. It is impossible to distinguish between the two signals. Thus, in a basic sense, discrete-time frequencies beyond || = π or |F| = 1/2 do not exist. Yet, in a “mathematical” sense, we must admit the existence of sinusoids of frequencies beyond = π. What does this mean?

Figure 5.17 (a) Actual frequency versus (b) apparent frequency.

A MAN NAMED ROBERT

To give an analogy, consider a fictitious person Mr. Robert Thompson. His mother calls him Robby; his acquaintances call him Bob, his close friends call him by his nickname, Shorty. Yet, Robert, Robby, Bob, and Shorty are one and the same person. However, we cannot say that only Mr. Robert Thompson exists, or only Robby exists, or only Shorty exists, or only Bob exists. All these four persons exist, although they are one and the same individual. In a same way, we cannot say that the frequency π/2 exists and frequency 5π/2 does not exist; they are both the same entity, called by different names.

It is in this sense that we have to admit the existence of frequencies beyond the fundamental band. Indeed, mathematical expressions in the frequency domain automatically cater to this need by their built-in periodicity. As seen earlier, the very structure of the frequency response is 2π-periodic. We shall also see later, in Ch. 9, that discrete-time signal spectra are also 2π-periodic.

Admitting the existence of frequencies beyond π also serves mathematical and computational convenience in digital signal-processing applications. Values of frequencies beyond π may also originate naturally in the process of sampling continuous-time sinusoids. Because there is no upper limit on the value of ω, there is no upper limit on the value of the resulting discrete-time frequency = ωT either.†

The highest possible frequency is π and the lowest frequency is 0 (dc or constant). Clearly, the high frequencies are those in the vicinity of = (2m + 1)π and the low frequencies are those in the vicinity of = 2πm for all positive or negative integer values of m.

^† However, if goes beyond π, the resulting aliasing reduces the apparent frequency to ^a < π.

FURTHER REDUCTION IN THE FREQUENCY RANGE

Because cos(−n + θ ) = cos(n − θ ), a frequency in the range −π to 0 is identical to the frequency (of the same magnitude) in the range 0 to π (but with a change in phase sign). Consequently the apparent frequency for a discrete-time sinusoid of any frequency is equal to some value in the range 0 to π. Thus, cos(8.7πn + θ ) = cos(0.7πn + θ ), and the apparent frequency is 0.7π. Similarly,

$$\cos(9.6\pi n + \theta) = \cos(-0.4\pi n + \theta) = \cos(0.4\pi n - \theta)$$

Hence, the frequency 9.6π is identical (in every respect) to frequency −0.4π, which, in turn, is equal (within the sign of its phase) to frequency 0.4π. In this case, the apparent frequency reduces to |a| = 0.4π. We can generalize the result to say that the apparent frequency of a discrete-time sinusoid is |a|, as found from Eq. (5.38), and if ^a <0, there is a phase reversal. Figure 5.17b plots versus the apparent frequency |a|. The shaded bands represent the ranges of for which there is a phase reversal, when represented in terms of |a|. For example, the apparent frequency for both the sinusoids cos(2.4π + θ ) and cos(3.6π + θ ) is |a| = 0.4π, as seen from Fig. 5.17b. But 2.4π is in a clear band and 3.6π is in a shaded band. Hence, these sinusoids appear as cos(0.4π +θ ) and cos(0.4π −θ ), respectively.

Although every discrete-time sinusoid can be expressed as having frequency in the range from 0 to π, we generally use the frequency range from −π to π instead of 0 to π for two reasons. First, exponential representation of sinusoids with frequencies in the range 0 to π requires a frequency range −π to π. Second, even when we are using a trigonometric representation, we generally need the frequency range −π to π to have exact identity (without phase reversal) of a higher-frequency sinusoid.

For certain practical advantages, in place of the range −π to π, we often use other contiguous ranges of width 2π. The range 0 to 2π, for instance, is used in many applications. It is left as an exercise for the reader to show that the frequencies in the range from π to 2π are identical to those in the range from −π to 0.

EXAMPLE 5.11 Apparent Frequency

Express the following signals in terms of their apparent frequencies: (a) cos(0.5πn + θ ), (b) cos(1.6πn+θ ), (c) sin(1.6πn+θ ), (d) cos(2.3πn+θ ), and (e) cos(34.699n+θ ).

(a) =0.5π is in the reduced range already. This is also apparent from Fig. 5.17a or 5.17b. Because ^a = 0.5π, there is no phase reversal, and the apparent sinusoid is cos(0.5πn+θ ).

(b) We express 1.6π = −0.4π + 2π so that ^a = −0.4π and |a| = 0.4. Also, ^a is negative, implying sign change for the phase. Hence, the apparent sinusoid is cos(0.4πn−θ ). This fact is also apparent from Fig. 5.17b.

(c) We first convert the sine form to cosine form as sin(1.6πn+θ ) = cos(1.6πn − (π/2) + θ ). In part (b), we found ^a = −0.4π. Hence, the apparent sinusoid is cos(0.4πn + (π/2)−θ ) = −sin(0.4πn−θ ). In this case, both the phase and the amplitude change signs.

(d) 2.3π = 0.3π +2π so that ^a = 0.3π. Hence, the apparent sinusoid is cos(0.3πn+θ ).

(e) We have 34.699 = −3+6(2π ). Hence, ^a = −3, and the apparent frequency |a| = 3 rad/sample. Because ^a is negative, there is a sign change of the phase. Hence, the apparent sinusoid is cos(3n−θ ).

DR ILL 5.20 Apparent Frequency

Show that the sinusoids having frequencies of (a) 2π, (b) 3π, (c) 5π, (d) 3.2π, (e) 22.1327, and (f) π + 2 can be expressed, respectively, as sinusoids of frequencies (a) 0, (b) π, (c) π, (d) 0.8π, (e) 3, and (f) π −2. Show that in cases (d), (e), and (f), phase changes sign.

5.5-2 Aliasing and Sampling Rate

The non-uniqueness of discrete-time sinusoids and the periodic repetition of the same waveforms at intervals of 2π may seem innocuous, but in reality it leads to a serious problem for processing continuous-time signals by digital filters. A continuous-time sinusoid cosωt sampled every T seconds (t = nT) results in a discrete-time sinusoid cosωnT, which is cosn with = ωT. The discrete-time sinusoids cosn have unique waveforms only for the values of frequencies in the range <π or ωT < π. Therefore, samples of continuous-time sinusoids of two (or more) different frequencies can generate the same discrete-time signal, as shown in Fig. 5.18. This phenomenon is known as aliasing because through sampling, two entirely different analog sinusoids take on the same “discrete-time” identity. †

Aliasing causes ambiguity in digital signal processing, which makes it impossible to determine the true frequency of the sampled signal. Consider, for instance, digitally processing

Figure 5.18 Demonstration of the aliasing effect.

^† Figure 5.18 shows samples of two sinusoids cos 12πt and cos 2πt taken every 0.2 second. The corresponding discrete-time frequencies ( = ωT = 0.2ω) are cos 2.4π and cos 0.4π. The apparent frequency of 2.4π is 0.4π, identical to the discrete-time frequency corresponding to the lower sinusoid. This shows that the samples of both these continuous-time sinusoids at 0.2-second intervals are identical, as verified from Fig. 5.18.

a continuous-time signal that contains two distinct components of frequencies ω¹ and ω2. The samples of these components appear as discrete-time sinusoids of frequencies ¹ = ω1T and ² = ω2T. If ¹ and ² happen to differ by an integer multiple of 2π (if ω² − ω¹ = 2kπ/T), the two frequencies will be read as the same (lower of the two) frequency by the digital processor.‡ As a result, the higher-frequency component ω² not only is lost for good (by losing its identity to ω1), but also it reincarnates as a component of frequency ω1, thus distorting the true amplitude of the original component of frequency ω1. Hence, the resulting processed signal will be distorted. Clearly, aliasing is highly undesirable and should be avoided. To avoid aliasing, the frequencies of the continuous-time sinusoids to be processed should be kept within the fundamental band ωT ≤ π or ω ≤ π/T. Under this condition the question of ambiguity or aliasing does not arise because any continuous-time sinusoid of frequency in this range has a unique waveform when it is sampled. Therefore, if ω*^h* is the highest frequency to be processed, then, to avoid aliasing,

$$\omega_h < \frac{\pi}{T}$$

If fh is the highest frequency in hertz, fh = ωh/2π, and we avoid aliasing if

$$f_h < \frac{1}{2T} \qquad \text{or} \qquad T < \frac{1}{2f_h} \tag{5.39}$$

This shows that discrete-time signal processing places the limit on the highest frequency fh that can be processed for a given value of the sampling interval T. Fortunately, we can process a signal of any frequency (without aliasing) by choosing a suitably small value of T. Since the sampling frequency fs is the reciprocal of the sampling interval T, we can also express Eq. (5.39) as

$$f_s = \frac{1}{T} > 2f_h$$

or $f_h < \frac{f_s}{2}$ (5.40)

This result is a special case of the well-known sampling theorem (to be proved in Ch. 8). It states that for a discrete-time system to process a continuous-time sinusoid, the sampling rate must be greater than twice the frequency (in hertz) of the sinusoid. In short, a sampled sinusoid must have a minimum of two samples per cycle. † For sampling rates below this minimum value, the output signal will be aliased, which means it will be mistaken for a sinusoid of lower frequency.

ANTI-ALIASING FILTER

If the sampling rate fails to satisfy Eq. (5.40), aliasing occurs, causing the frequencies beyond fs/2 Hz to masquerade as lower frequencies to corrupt the spectrum at frequencies below fs/2. To avoid such a corruption, a signal to be sampled is passed through an anti-aliasing filter of bandwidth fs/2 prior to sampling. This operation ensures the condition of Eq. (5.40). The drawback of such a filter is that we lose the spectral components of the signal beyond frequency fs/2, which is preferable to the aliasing corruption of the signal at frequencies below fs/2. Chapter 8 presents a detailed analysis of the aliasing problem.

^‡ In the case shown in Fig. 5.18, ^{ω1 = 12}π, ^{ω2 = 2}π, and ^{T =} 0.2. Hence, ^{ω2 − ω1 = 10}π*^T* ^{= 2}π, and the two frequencies are read as the same frequency = 0.4π by the digital processor.

^† Strictly speaking, we must have more than two samples per cycle.

EXAMPLE 5.12 Maximum Sampling Interval

Determine the maximum sampling interval T that can be used in a discrete-time oscillator that generates a sinusoid of 50 kHz.

Here the highest significant frequency fh = 50 kHz. Therefore from Eq. (5.39),

$$T < \frac{1}{2f_h} = 10\,\mu\,\mathrm{s}$$

The sampling interval must be less than 10µs. The sampling frequency is fs = 1/T > 100 kHz.

EXAMPLE 5.13 Maximum Frequency Without Aliasing

A discrete-time amplifier uses a sampling interval T = 25µs. What is the highest frequency of a signal that can be processed with this amplifier without aliasing?

From Eq. (5.39)

$$f_h < \frac{1}{2T} = 20 \, \text{kHz}$$