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[9.4 LTI DISCRETE-TIME](#page-14-0) SYSTEM ANALYSIS BY DTFT

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9.4 LTI DISCRETE-TIME SYSTEM ANALYSIS BY DTFT

Consider a linear, time-invariant, discrete-time system with the unit impulse response h[n]. We shall find the (zero-state) system response y[n] for the input x[n]. Let

x[n] ⇐⇒ X() y[n] ⇐⇒ Y() and h[n] ⇐⇒ H()

Because y[n] = x[n] βˆ— h[n], it follows from Eq. (9.34) that

Y(Ξ©)=X(Ξ©)H(Ξ©)(9.37)Y(\Omega) = X(\Omega)H(\Omega) \tag{9.37}

This result is similar to that obtained for continuous-time systems. Let us examine the role of H(), the DTFT of the unit impulse response h[n].

Equation (9.37) holds for BIBO-stable systems and also for marginally stable systems if the input does not contain the system’s natural mode(s). In other cases, the response grows with n and is not Fourier-transformable. Moreover, the input x[n] also has to be DTF-transformable. For cases where Eq. (9.37) does not apply, we use the z-transform for system analysis.

Equation (9.37) shows that the output signal frequency spectrum is the product of the input signal frequency spectrum and the frequency response of the system. From this equation, we obtain

|Y()|=|X()||H()| and Y() = X()+ H()

This result shows that the output amplitude spectrum is the product of the input amplitude spectrum and the amplitude response of the system. The output phase spectrum is the sum of the input phase spectrum and the phase response of the system.

We can also interpret Eq. (9.37) in terms of the frequency-domain viewpoint, which sees a system in terms of its frequency response (system response to various exponential or sinusoidal components). The frequency domain views a signal as a sum of various exponential or sinusoidal components. The transmission of a signal through a (linear) system is viewed as transmission of various exponential or sinusoidal components of the input signal through the system. This concept can be understood by displaying the input–output relationships by a directed arrow as follows:

eiΩn⟹H(Ω)eiΩne^{i\Omega n} \Longrightarrow H(\Omega)e^{i\Omega n}

which shows that the system response to ejn is H()ejn, and

x[n]=12Ο€βˆ«2Ο€X(Ξ©)eiΞ©ndΞ©x[n] = \frac{1}{2\pi} \int_{2\pi} X(\Omega) e^{i\Omega n} d\Omega

which shows x[n] as a sum of everlasting exponential components. Invoking the linearity property, we obtain

y[n]=12Ο€βˆ«2Ο€X(Ξ©)H(Ξ©)eiΞ©ndΞ©y[n] = \frac{1}{2\pi} \int_{2\pi} X(\Omega) H(\Omega) e^{i\Omega n} d\Omega

which gives y[n] as a sum of responses to all input components and is equivalent to Eq. (9.37). Thus, X() is the input spectrum and Y() is the output spectrum, given by X()H().

EXAMPLE 9.13 LTID System Analysis by the DTFT

An LTID system is specified by the equation y[n] βˆ’ 0.5y[n βˆ’ 1] = x[n]. Find H(), the frequency response of this system. Determine the (zero-state) response y[n] if the input x[n] = (0.8)nu[n].

Let x[n] ⇐⇒ X() and y[n] ⇐⇒ Y(). Taking the DTFT of the system’s difference equation yields

(1βˆ’0.5eβˆ’jΞ©)Y(Ξ©)=X(Ξ©)(1 - 0.5e^{-j\Omega})Y(\Omega) = X(\Omega)

According to Eq. (9.37),

H(Ξ©)=Y(Ξ©)X(Ξ©)=11βˆ’eβˆ’jΞ©=ejΞ©ejΞ©βˆ’0.5H(\Omega) = \frac{Y(\Omega)}{X(\Omega)} = \frac{1}{1 - e^{-j\Omega}} = \frac{e^{j\Omega}}{e^{j\Omega} - 0.5}

Also, x[n] = (0.8)nu[n]. Hence,

X(Ξ©)=eiΞ©eiΞ©βˆ’0.8X(\Omega) = \frac{e^{i\Omega}}{e^{i\Omega} - 0.8}

and

Y(Ξ©)=X(Ξ©)H(Ξ©)=2eiΞ©(eiΞ©βˆ’0.8)(eiΞ©βˆ’0.5)Y(\Omega) = X(\Omega)H(\Omega) = \frac{2e^{i\Omega}}{(e^{i\Omega} - 0.8)(e^{i\Omega} - 0.5)}

880 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS

We can express the right-hand side as a sum of two first-order terms (modified partial fraction expansion as discussed in Sec. B.5-6) as follows† :

Y(Ξ©)eiΞ©=eiΞ©(eiΞ©βˆ’0.5)(eiΞ©βˆ’0.8)=βˆ’53eiΞ©βˆ’0.5+83eiΞ©βˆ’0.8\frac{Y(\Omega)}{e^{i\Omega}} = \frac{e^{i\Omega}}{(e^{i\Omega} - 0.5)(e^{i\Omega} - 0.8)} = \frac{-\frac{5}{3}}{e^{i\Omega} - 0.5} + \frac{\frac{8}{3}}{e^{i\Omega} - 0.8}

Consequently,

Y(Ξ©)=βˆ’(53)eiΞ©eiΞ©βˆ’0.5+(83)eiΞ©eiΞ©βˆ’0.8Y(\Omega) = -\left(\frac{5}{3}\right) \frac{e^{i\Omega}}{e^{i\Omega} - 0.5} + \left(\frac{8}{3}\right) \frac{e^{i\Omega}}{e^{i\Omega} - 0.8}

= -\left(\frac{5}{3}\right) \frac{1}{1 - 0.5e^{-i\Omega}} + \left(\frac{8}{3}\right) \frac{1}{1 - 0.8e^{-i\Omega}}

From entry 2 of Table 9.1, the inverse DTFT of this equation is

y[n]=[βˆ’53(0.5)n+83(0.8)n]u[n]y[n] = \left[ -\frac{5}{3}(0.5)^n + \frac{8}{3}(0.8)^n \right] u[n]

This example demonstrates the procedure for using the DTFT to determine an LTID system response. It is similar to the Fourier transform method in the analysis of LTIC systems. As in the case of the Fourier transform, this method can be used only if the system is asymptotically or BIBO-stable and if the input signal is DTF-transformable.‑ We shall not belabor this method further because it is clumsier and more restrictive than the z-transform method discussed in Ch. 5.

9.4-1 Distortionless Transmission

In several applications, digital signals are passed through LTI systems, and we require that the output waveform be a replica of the input waveform. As in the continuous-time case, transmission is said to be distortionless if the input x[n] and the output y[n] satisfy the condition

y[n]=G0x[nβˆ’nd]y[n] = G_0 x[n - n_d]

Here, nd, the delay (in samples), is assumed to be integer. Taking the Fourier transform yields

Y(Ξ©)=G0X(Ξ©)eβˆ’jΞ©ndY(\Omega) = G_0 X(\Omega) e^{-j\Omega n_d}

But

Y(Ξ©)=X(Ξ©)H(Ξ©)Y(\Omega) = X(\Omega) H(\Omega)

† Here, Y() is a function of variable ej. Hence, x = ej for the purpose of comparison with the expression in Sec. B.5-6.

‑ It can also be applied to marginally stable systems if the input does not contain natural mode(s) of the system.

Figure 9.13 LTI system frequency response for distortionless transmission.

Therefore,

H(Ξ©)=G0eβˆ’jΞ©ndH(\Omega) = G_0 e^{-j\Omega n_d}

This is the frequency response required for distortionless transmission. From this equation, it follows that

∣H(Ξ©)∣=G0and∠H(Ξ©)=βˆ’Ξ©nd(9.38)|H(\Omega)| = G_0 \quad \text{and} \quad \angle H(\Omega) = -\Omega n_d \tag{9.38}

Thus, for distortionless transmission, the amplitude response |H()| must be a constant, and the phase response H() must be a linear function of with slope βˆ’nd, where nd is the delay in the number of samples with respect to input (Fig. 9.13). These are precisely the characteristics of an ideal delay of nd samples with a gain of G0 [see Eq. (9.31)].

MEASURE OF DELAY VARIATION

For distortionless transmission, we require a linear phase characteristic. In practice, many systems have a phase characteristic that may be only approximately linear. A convenient way of judging phase linearity is to plot the slope of H() as a function of frequency. This slope is constant for the ideal linear phase (ILP) system, but it may vary with in the general case. The slope can be expressed as

ng(Ξ©)=βˆ’ddΩ∠H(Ξ©)(9.39)n_g(\Omega) = -\frac{d}{d\Omega} \angle H(\Omega) \tag{9.39}

If ng() is constant, all the components are delayed by ng samples. But if the slope is not constant, the delay ng varies with frequency. This variation means that different frequency components undergo different amounts of delay, and consequently, the output waveform will not be a replica of the input waveform. As in the case of LTIC systems, ng(), as defined in Eq. (9.39), plays an important role in bandpass systems and is called the group delay or envelope delay. Observe that constant nd implies constant ng. Note that H() = Ο†0 βˆ’ nd also has a constant ng. Thus, constant group delay is a more relaxed condition.

DISTORTIONLESS TRANSMISSION OVER BANDPASS SYSTEMS

As in the case of continuous-time systems, the distortionless transmission conditions can be relaxed for discrete-time bandpass systems. For lowpass systems, the phase characteristic should not only be linear over the band of interest, it should also pass through the origin [Eq. (9.38)]. For bandpass systems, the phase characteristic should be linear over the band of interest, but it need not pass through the origin (ng should be constant). The amplitude response is required to

882 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS

be constant over the passband. Thus, for distortionless transmission over a bandpass system, the frequency response for positive range of is of the form†

H(Ξ©)=G0ej(Ο•0βˆ’Ξ©ng)Ξ©β‰₯0H(\Omega) = G_0 e^{j(\phi_0 - \Omega n_g)} \qquad \Omega \ge 0

The proof is identical to that for the continuous-time case in Sec. 7.4-2 and will not be repeated. In using Eq. (9.39) to compute ng, we should ignore jump discontinuities in the phase function.

9.4-2 Ideal and Practical Filters

Ideal filters allow distortionless transmission of a certain band of frequencies and suppress all the remaining frequencies. The general ideal lowpass filter shown in Fig. 9.14 for || ≀ Ο€ allows all components below the cutoff frequency = c to pass without distortion and suppresses all components above c. Figure 9.15 illustrates ideal highpass and bandpass filter characteristics.

The ideal lowpass filter in Fig. 9.14a has a linear phase of slope βˆ’nd, which results in a delay of nd samples for all its input components of frequencies below c rad/sample. Therefore, if the input is a signal x[n] bandlimited to c, the output y[n] is x[n] delayed by nd; that is,

y[n]=x[nβˆ’nd]y[n] = x[n - n_d]

The signal x[n] is transmitted by this system without distortion, but with delay of nd samples. For this filter,

H(Ξ©)=βˆ‘m=βˆ’βˆžβˆžrect(Ξ©βˆ’2Ο€m2Ξ©c)eβˆ’jΞ©ndH(\Omega) = \sum_{m=-\infty}^{\infty} \text{rect}\left(\frac{\Omega - 2\pi m}{2\Omega_c}\right) e^{-j\Omega n_d}

The unit impulse response h[n] of this filter is obtained from pair 8 (Table 9.1) and the time-shifting property

h[n]=Ξ©cΟ€sinc⁑[Ξ©c(nβˆ’nd)]h[n] = \frac{\Omega_c}{\pi} \operatorname{sinc} \left[ \Omega_c (n - n_d) \right]

Because h[n] is the system response to impulse input Ξ΄[n], which is applied at n = 0, it must be causal (i.e., it must not start before n = 0) for a realizable system. Figure 9.14b shows h[n] for

Figure 9.14 Ideal lowpass filter: its frequency response and impulse response.

† Because the phase function is an odd function of , if H() = Ο†0 βˆ’ ng for β‰₯ 0, over the band 2*W* (centered at c), then H() = βˆ’Ο†0 βˆ’ng for < 0 over the band 2W (centered at βˆ’c).

Figure 9.15 Ideal highpass and bandpass filter frequency response.

Figure 9.16 Approximate realization of an ideal lowpass filter by truncation of its impulse response.

c = Ο€/4 and nd = 12. This figure also shows that h[n] is noncausal, hence unrealizable. Similarly, one can show that other ideal filters (such as the ideal highpass or and bandpass filters depicted in Fig. 9.15) are also noncausal and therefore physically unrealizable.

One practical approach to realize an ideal lowpass filter approximately is to truncate both tails (positive and negative) of h[n] so that it has a finite length and then delay sufficiently to make it causal (Fig. 9.16). We now synthesize a system with this truncated (and delayed) impulse response. For closer approximation, the truncating window has to be correspondingly wider. The delay required also increases correspondingly. Thus, the price of closer realization is higher delay in the output; this situation is common in noncausal systems.