7.4 SIGNAL [TRANSMISSION](#page-12-0) THROUGH LTIC SYSTEMS

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7.4 SIGNAL TRANSMISSION THROUGH LTIC SYSTEMS

If x(t) and y(t) are the input and output of an LTIC system with impulse response h(t), then, as demonstrated in Eq. (7.35),

$$Y(\omega) = H(\omega)X(\omega)$$

This equation does not apply to (asymptotically) unstable systems because h(t) for such systems is not Fourier transformable. It applies to BIBO-stable as well as most of the marginally stable systems.† Similarly, this equation does not apply if x(t) is not Fourier transformable.

In Ch. 4, we saw that the Laplace transform is more versatile and capable of analyzing all kinds of LTIC systems whether stable, unstable, or marginally stable. Laplace transform can also handle exponentially growing inputs. In comparison to the Laplace transform, the Fourier transform in system analysis is not just clumsier, but also very restrictive. Hence, the Laplace transform is preferable to the Fourier transform in LTIC system analysis. We shall not belabor the application of the Fourier transform to LTIC system analysis. We consider just one example here.

EXAMPLE 7.18 Fourier Transform to Determine the Zero-State Response

Use the Fourier transform to find the zero-state response of a stable LTIC system with frequency response

$$H(s) = \frac{1}{s+2}$$

and the input is x(t) = e−^t u(t). Stability implies that the region of convergence of H(s) includes the ω axis.

In this case,

$$X(\omega) = \frac{1}{j\omega + 1}$$

^† For marginally stable systems, if the input x(t) contains a finite-amplitude sinusoid of the system’s natural frequency, which leads to resonance, the output is not Fourier transformable. It does, however, apply to marginally stable systems if the input does not contain a finite-amplitude sinusoid of the system’s natural frequency.

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Moreover, because the system is stable, the frequency response H(jω) = H(ω). Hence,

$$H(\omega) = H(s)|_{s=j\omega} = \frac{1}{j\omega + 2}$$

Therefore,

$$Y(\omega) = H(\omega)X(\omega) = \frac{1}{(j\omega + 2)(j\omega + 1)}$$

Expanding the right-hand side in partial fractions yields

$$Y(\omega) = \frac{1}{j\omega + 1} - \frac{1}{j\omega + 2}$$

and

$$y(t) = (e^{-t} - e^{-2t})u(t)$$

DR ILL 7.10 Fourier Transform to Determine the Zero-State Response

For the system in Ex. 7.18, show that the zero-input response to the input et ^u(−t) is ^y(t) ^{= 1 3} [et u(−t)+e−2*^t u*(t)]. [Hint: Use pair 2 (Table 7.1) to find the Fourier transform of et u(−t).]

HEURISTIC UNDERSTANDING OF LINEAR SYSTEM RESPONSE

In finding the linear system response to arbitrary input, the time-domain method uses convolution integral and the frequency-domain method uses the Fourier integral. Despite the apparent dissimilarities of the two methods, their philosophies are amazingly similar. In the time-domain case, we express the input x(t) as a sum of its impulse components; in the frequency-domain case, the input is expressed as a sum of everlasting exponentials (or sinusoids). In the former case, the response y(t) obtained by summing the system’s responses to impulse components results in the convolution integral; in the latter case, the response obtained by summing the system’s response to everlasting exponential components results in the Fourier integral. These ideas can be expressed mathematically as follows:

1. For the time-domain case,

δ(t) ⇒ h(t)	shows the system response δ(t) h(t) to is the impulse response
= $ ∞ x(t) x(τ )δ(t −τ )dτ −∞	expresses x(t) as a sum of impulse components
= $ ∞ y(t) x(τ )h(t −τ )dτ −∞	expresses y(t) as a sum of responses to the impulse components of input x(t)

2. For the frequency-domain case,

$$e^{j\omega t} \implies H(\omega)e^{j\omega t} \qquad \text{shows the system response}$$

\nto $e^{j\omega t}$ is $H(\omega)e^{j\omega t}$
\n

$$x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega)e^{j\omega t} d\omega \qquad \text{expresses } x(t) \text{ as a sum}$$

\nof everyday exponential components
\n

$$y(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega)H(\omega)e^{j\omega t} d\omega \qquad \text{expresses } y(t) \text{ as a sum of responses to}$$

\nthe exponential components of input $x(t)$

The frequency-domain view sees a system in terms of its frequency response (system response to various sinusoidal components). It views a signal as a sum of various sinusoidal components. Transmission of an input signal through a (linear) system is viewed as transmission of various sinusoidal components of the input through the system.

It was not by coincidence that we used the impulse function in time-domain analysis and the exponential ejω*^t* in studying the frequency domain. The two functions happen to be duals of each other. Thus, the Fourier transform of an impulse δ(t − τ ) is e−jωτ , and the Fourier transform of ej^ω0*^t* is an impulse 2πδ(ω − ω0). This time-frequency duality is a constant theme in the Fourier transform and linear systems.

7.4-1 Signal Distortion During Transmission

For a system with frequency response H(ω), if X(ω) and Y(ω) are the spectra of the input and the output signals, respectively, then

$$Y(\omega) = X(\omega)H(\omega) \tag{7.38}$$

The transmission of the input signal x(t) through the system changes it into the output signal y(t). Equation (7.38) shows the nature of this change or modification. Here, X(ω) and Y(ω) are the spectra of the input and the output, respectively. Therefore, H(ω) is the spectral response of the system. The output spectrum is obtained by the input spectrum multiplied by the spectral response of the system. Equation (7.38), which clearly brings out the spectral shaping (or modification) of the signal by the system, can be expressed in polar form as

$$|Y(\omega)|e^{j\angle Y(\omega)} = |X(\omega)||H(\omega)|e^{j[\angle X(\omega)+\angle H(j\omega)]}$$

Therefore,

$$|Y(\omega)| = |X(\omega)| |H(\omega)| \quad \text{and} \quad \angle Y(\omega) = \angle X(\omega) + \angle H(\omega)$$

During transmission, the input signal amplitude spectrum |X(ω)| is changed to |X(ω)||H(ω)|. Similarly, the input signal phase spectrum X(ω) is changed to X(ω) + H(ω). An input signal spectral component of frequency ω is modified in amplitude by a factor |H(ω)| and is shifted in phase by an angle H(ω). Clearly, |H(ω)| is the amplitude response, and H(ω) is the phase response of the system. The plots of |H(ω)| and H(ω) as functions of ω show at a glance how the system modifies the amplitudes and phases of various sinusoidal inputs. This is the reason why H(ω) is also called the frequency response of the system. During transmission through the system, some frequency components may be boosted in amplitude, while others may be attenuated. The

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relative phases of the various components also change. In general, the output waveform will be different from the input waveform.

DISTORTIONLESS TRANSMISSION

In several applications, such as signal amplification or message signal transmission over a communication channel, we require that the output waveform be a replica of the input waveform. In such cases we need to minimize the distortion caused by the amplifier or the communication channel. It is, therefore, of practical interest to determine the characteristics of a system that allows a signal to pass without distortion (distortionless transmission).

Transmission is said to be distortionless if the input and the output have identical waveshapes within a multiplicative constant. A delayed output that retains the input waveform is also considered to be distortionless. Thus, in distortionless transmission, the input x(t) and the output y(t) satisfy the condition

$$y(t) = G_0 x(t - t_d)$$

The Fourier transform of this equation yields

$$Y(\omega) = G_0 X(\omega) e^{-j\omega t_d}$$

But

$$Y(\omega) = X(\omega)H(\omega)$$

Therefore,

$$H(\omega) = G_0 e^{-j\omega t_d}$$

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

$$|H(\omega)| = G_0 \quad \text{and} \quad \angle H(\omega) = -\omega t_d \tag{7.39}$$

This result shows that for distortionless transmission, the amplitude response |H(ω)| must be a constant, and the phase response H(ω) must be a linear function of ω with slope −td, where td is the delay of the output with respect to input (Fig. 7.29).

MEASURE OF TIME-DELAY VARIATION WITH FREQUENCY

The gain |H(ω)| = G⁰ means that every spectral component is multiplied by a constant G0. Also, as seen in connection with Fig. 7.22, a linear phase H(ω) = −ωtd means that every spectral

Figure 7.29 LTIC system frequency response for distortionless transmission.

component is delayed by td seconds. This results in the output equal to G⁰ times the input delayed by td seconds. Because each spectral component is attenuated by the same factor (G0) and delayed by exactly the same amount (td), the output signal is an exact replica of the input (except for attenuating factor G⁰ and delay td).

For distortionless transmission, we require a linear phase characteristic. The phase is not only a linear function of ω, it should also pass through the origin ω = 0. 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(ω) as a function of frequency. This slope, which is constant for an ideal linear phase (ILP) system, is a function of ω in the general case and can be expressed as

$$t_g(\omega) = -\frac{d}{d\omega} \angle H(\omega)$$

\n(7.40)

If tg(ω) is constant, all the components are delayed by the same time interval tg. But if the slope is not constant, the time delay tg varies with frequency. This variation means that different frequency components undergo different amounts of time delay, and consequently, the output waveform will not be a replica of the input waveform. As we shall see, tg(ω) plays an important role in bandpass systems and is called the group delay or envelope delay. Observe that constant td [Eq. (7.39)] implies constant tg. Note that H(ω) = φ⁰ −ωtg also has a constant tg. Thus, constant group delay is a more relaxed condition.

It is often thought (erroneously) that flatness of amplitude response |H(ω)| alone can guarantee signal quality. However, a system that has a flat amplitude response may yet distort a signal beyond recognition if the phase response is not linear (td not constant).

THE NATURE OF DISTORTION IN AUDIO AND VIDEO SIGNALS

Generally speaking, the human ear can readily perceive amplitude distortion but is relatively insensitive to phase distortion. For the phase distortion to become noticeable, the variation in delay [variation in the slope of H(ω)] should be comparable to the signal duration (or the physically perceptible duration, in case the signal itself is long). In the case of audio signals, each spoken syllable can be considered to be an individual signal. The average duration of a spoken syllable is of a magnitude of the order of 0.01 to 0.1 second. Audio systems may have nonlinear phases, yet no noticeable signal distortion results because in practical audio systems, maximum variation in the slope of H(ω) is only a small fraction of a millisecond. This is the real truth underlying the statement that “the human ear is relatively insensitive to phase distortion” [3]. As a result, the manufacturers of audio equipment make available only |H(ω)|, the amplitude response characteristic of their systems.

For video signals, in contrast, the situation is exactly the opposite. The human eye is sensitive to phase distortion but is relatively insensitive to amplitude distortion. Amplitude distortion in television signals manifests itself as a partial destruction of the relative half-tone values of the resulting picture, but this effect is not readily apparent to the human eye. Phase distortion (nonlinear phase), on the other hand, causes different time delays in different picture elements. The result is a smeared picture, and this effect is readily perceived by the human eye. Phase distortion is also very important in digital communication systems because the nonlinear phase characteristic of a channel causes pulse dispersion (spreading out), which in turn causes pulses to interfere with neighboring pulses. Such interference between pulses can cause an error in the pulse amplitude at the receiver: a binary 1 may read as 0, and vice versa.

7.4-2 Bandpass Systems and Group Delay

The distortionless transmission conditions [Eq. (7.39)] can be relaxed slightly for bandpass systems. For lowpass systems, the phase characteristics not only should be linear over the band of interest but also should pass through the origin. For bandpass systems, the phase characteristics must be linear over the band of interest but need not pass through the origin.

Consider an LTI system with amplitude and phase characteristics as shown in Fig. 7.30, where the amplitude spectrum is a constant G⁰ and the phase is φ⁰ − ωtg over a band 2W centered at frequency ωc. Over this band, we can describe H(ω) as†

$$H(\omega) = G_0 e^{j(\phi_0 - \omega t_g)} \qquad \omega \ge 0 \tag{7.41}$$

The phase of H(ω) in Eq. (7.41), shown dotted in Fig. 7.30b, is linear but does not pass through the origin.

Consider a modulated input signal z(t) = x(t) cosωct. This is a bandpass signal, whose spectrum is centered at ω = ωc. The signal cosωct is the carrier, and the signal x(t), which is a lowpass signal of bandwidth W (see Fig. 7.25), is the envelope of z(t). ‡ We shall now show that the transmission of z(t) through H(ω) results in distortionless transmission of the envelope x(t). However, the carrier phase changes by φ0. To show this, consider an input zˆ(t) = x(t)ejωct and the corresponding output yˆ(t). From Eq. (7.30), Zˆ(ω) = X(ω − ωc), and the corresponding output

Figure 7.30 Generalized linear phase characteristics.

^† Because the phase function is an odd function of ^ω, if ^H(ω) ^{= φ0 − ω}tg for ^{ω ≥} 0, over the band 2*^W* (centered at ωc), then H(ω) = −φ⁰ − ωtg for ω < 0 over the band 2W (centered at −ωc), as shown in Fig. 7.30a.

^‡ The envelope of a bandpass signal is well defined only when the bandwidth of the envelope is well below the carrier ω*^c* (W ωc).

spectrum Yˆ(ω) is given by

$$\hat{Y}(\omega) = H(\omega)\hat{Z}(\omega) = H(\omega)X(\omega - \omega_c)$$

Recall that the bandwidth of X(ω) is W so that the bandwidth of X(ω −ωc) is 2W, centered at ωc. Over this range, H(ω) is given by Eq. (7.41). Hence,

$$\hat{Y}(\omega) = G_0 X(\omega - \omega_c) e^{j(\phi_0 - \omega t_g)} = G_0 e^{j\phi_0} X(\omega - \omega_c) e^{-\omega t_g}$$

Use of Eqs. (7.29) and (7.30) yields yˆ(t) as

$$\hat{y}(t) = G_0 e^{j\phi_0} x(t - t_g) e^{j\omega_c(t - t_g)} = G_0 x(t - t_g) e^{j[\omega_c(t - t_g) + \phi_0]}$$

This is the system response to input zˆ(t) = x(t)ejωct , which is a complex signal. We are really interested in finding the response to the input z(t) = x(t) cosωct, which is the real part of zˆ(t) = x(t)ejωct . Hence, we use Eq. (2.31) to obtain y(t), the system response to the input z(t) = x(t) cosωct, as

$$y(t) = G_0 x(t - t_g) \cos [\omega_c (t - t_g) + \phi_0)]$$

\n(7.42)

where tg, the group (or envelope) delay, is the negative slope of H(ω) at ωc. † The output y(t) is basically the delayed input z(t − tg), except that the output carrier acquires an extra phase φ0. The output envelope x(t − tg) is the delayed version of the input envelope x(t) and is not affected by extra phase φ⁰ of the carrier. In a modulated signal, such as x(t) cosωct, the information generally resides in the envelope x(t). Hence, the transmission is considered to be distortionless if the envelope x(t) remains undistorted.

Most practical systems satisfy Eq. (7.41), at least over a very small band. Figure 7.30b shows a typical case in which this condition is satisfied for a small band W centered at frequency ωc.

A system in Eq. (7.41) is said to have a generalized linear phase (GLP), as illustrated in Fig. 7.30. The ideal linear phase (ILP) characteristics is shown in Fig. 7.29. For distortionless transmission of bandpass signals, the system need satisfy Eq. (7.41) only over the bandwidth of the bandpass signal.

Caution. Recall that the phase response associated with the amplitude response may have jump discontinuities when the amplitude response goes negative. Jump discontinuities also arise because of the use of the principal value for phase. Under such conditions, to compute the group delay [Eq. (7.40)], we should ignore the jump discontinuities.

$$y(t) = G_o x(t - t_g) \cos \omega_c (t - t_{ph})$$

where tph, called the phase delay at ωc, is given by tph(ωc) = (ωctg − φ0)/ωc. Generally, tph varies with ω, and we can write

$$t_{\rm ph}(\omega) = \frac{\omega t_g - \phi_0}{\omega}$$

Recall also that tg itself may vary with ω.

^† Equation (7.42) can also be expressed as

EXAMPLE 7.19 Distortionless Bandpass Transmission

(a) A signal z(t), shown in Fig. 7.31b, is given by

$$z(t) = x(t) \cos \omega_c t$$

where ω*^c* = 2000π. The pulse x(t) (Fig. 7.31a) is a lowpass pulse of duration 0.1 second and has a bandwidth of about 10 Hz. This signal is passed through a filter whose frequency response is shown in Fig. 7.31c (shown only for positive ω). Find and sketch the filter output y(t).

(b) Find the filter response if ω*^c* = 4000π.

(a) The spectrum Z(ω) is a narrow band of width 20 Hz, centered at frequency f⁰ = 1 kHz. The gain at the center frequency (1 kHz) is 2. The group delay, which is the negative of the slope of the phase plot, can be found by drawing tangents at ωc, as shown in Fig. 7.31c. The negative of the slope of the tangent represents tg, and the intercept along the vertical axis by the tangent represents φ⁰ at that frequency. From the tangents at ωc, we find tg, the group delay, as

$$t_g = \frac{2.4\pi - 0.4\pi}{2000\pi} = 10^{-3}$$

The vertical axis intercept is φ⁰ = −0.4π. Hence, by using Eq. (7.42) with gain G⁰ = 2, we obtain

$$y(t) = 2x(t - t_g)\cos[\omega_c(t - t_g) - 0.4\pi] \qquad \omega_c = 2000\pi \quad t_g = 10^{-3}$$

Figure 7.31d shows the output y(t), which consists of the modulated pulse envelope x(t) delayed by 1 ms and the phase of the carrier changed by −0.4π. The output shows no distortion of the envelope x(t), only the delay. The carrier phase change does not affect the shape of envelope. Hence, the transmission is considered distortionless.

(b) Figure 7.31c shows that when ω*^c* = 4000π, the slope of H(ω) is zero so that tg = 0. Also, the gain G⁰ = 1.5, and the intercept of the tangent with the vertical axis is φ⁰ = −3.1π. Hence,

$$y(t) = 1.5x(t)\cos(\omega_c t - 3.1\pi)$$

This, too, is a distortionless transmission for the same reasons as for case (a).