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ECE4330 Lecture 19 the Fourier Transform (Cont.) Prof ECE4330 Lecture 19 The Fourier Transform (cont.) Prof. Mohamad Hassoun This lecture covers the following topics: The significance of phase information: Signal transmission example Applications of the Fourier transform o Fourier transform in a basic communication system o Low-pass filters: ideal vs practical (Butterworth filters) o Filter Design: low-pass, high-pass, band-pass, band-reject, notch, and all-pass Analog filter design as a numerical optimization problem Appendix: Experimental demonstration of amplitude modulation. The Significance of Phase Information: Signal Transmission Example As we saw in the previous lecture, an ideal transmission system has uniform attenuation 푘 for all frequencies. Also, the transmitted signal is received delayed by 푡푑, but with no distortion; i.e., 푦(푡) = 푘푓(푡 − 푡푑). The above system has the transfer function 퐻(휔) = 푘푒−푗푡푑휔 from which the magnitude and angle are obtained as: |퐻(휔)| = 푘, ∠퐻(휔) = −푡푑휔. Example. Determine the steady-state output for the above ideal transmission system to the input, 푓(푡) = cos(푡) + cos(2푡). Here, the signal has two component frequencies (휔1 = 1, 휔2 = 2). 푦푠푠(푡) = |퐻(휔1)| cos(푡 + ∠퐻(휔1)) + |퐻(휔2)| cos(2푡 + ∠퐻(휔2)) = 푘cos(푡 − 푡푑) + 푘cos(2푡 − 2푡푑) = 푘cos(푡 − 푡푑) + 푘cos(2(푡 − 푡푑)) = 푘푓(푡 − 푡푑) ←No distortion Let us repeat this problem assuming a constant angle, ∠퐻(휔) = −푡푑. 푦푠푠(푡) = 푘cos(푡 − 푡푑) + 푘cos(2푡 − 푡푑) 푡 = 푘cos(푡 − 푡 ) + 푘cos (2 (푡 − 푑)) ≠ 푘푓(푡 − 푡 ) 푑 2 푑 Notice that the second component is delayed by half the delay of the first component. So, different parts of the signal arrive at different times! This leads to a distorted signal. The following Mathcad simulations illustrate this phase-based distortion phenomenon. Your turn: Consider the following passive linear circuit with 퐿 = 1 and 퐶 = 1. Determine (by inspection) the steady-state output for the following inputs: o 푣푖푛(푡) = 1 o 푣푖푛(푡) = cos(휔푡), as 휔 → ∞ Determine the transfer function 퐻(휔). Does the circuit distort the amplitude of its input? Does the circuit distort the phase of its input? 3 Let 푣푖푛(푡) = sin (푡). Determine the steady-state output, 푣푠푠(푡). Repeat for 푣푖푛(푡) = sin(10휋푡). Ans. 1 − 푠2 1 + 휔2 퐻(푠) = → 퐻(휔) = 1 + 푠2 1 − 휔2 푣푠푠(푡) = ∞ + 0.244 sin(3푡) = ∞ Note: The circuit resonates at 휔 = 1. This circuit is supposed to operate at 휔 ≫ 1. For those high frequencies, the frequency response of the filter is given by, 퐻(휔) = −1 = 1푒푗휋 Therefore, for 휔 ≫ 1 the filter is an all-pass filter with constant phase shift of 휋. Refer to the section entitled “All-Pass Filter” later in this lecture for additional analysis and application. Fourier Transform in a Basic Communication System Consider the following communication system. There are two main factors affecting the design of a communication system. First, the transmission antenna must be of practical size. The size of the transmission antenna (퐿) is proportional to the wave-length, 푐 2휋푐 m 퐿 ∝ 휆 = = , 푐 ≅ 3 x 108 푓 휔 s For a (talk radio channel) speech signal, 푚(푡), the frequency components are in the range 100Hz < 푓 < 20KHz. And the effective signal bandwidth for intelligible speech is about 5KHz. So, if we are to transmit speech signals directly then the length of required transmission antenna would be of the order of several kilometers: (3)108 퐿 ∝ 휆 > ≅ 15Km (20)103 Therefore, we need to shift the signal frequency to a higher range; i.e., we need to modulate 푚(푡) so that it rides on a high-frequency signal 푦(푡) in order to make the size of the antenna practical. Other requirements include the ability of the receiver to select one specific transmission channel from the superposition of several concurrent transmissions (by different stations), as illustrated in the following figure. The Fourier transform is a convenient tool for designing communication systems. The frequency spectrum representation of a signal renders it easier to understand the processing and design aspects of the system as compared to the time-domain picture. We start by Fourier transforming the signals, 푚1(푡) and 푚2(푡). Next, we need to shift the spectra of 푀1(휔) and 푀2(휔) to high frequencies, but with minimal spectral overlap. This is achieved by applying frequency division multiplexing (FDM) as illustrated in the following figure. Note: For simplicity, we assumed that 푚1(푡) and 푚2(푡) are even and real test signals, which leads to even and real Fourier transforms 푀1(휔) and 푀2(휔), respectively. The carrier frequencies for AM radio range from 540KHz to 1700KHz and are spaced at 10KHz intervals. Recall that the radian frequency 휔 is 2휋 times the Hertz frequency, 푓. From the above figure, we notice that in order to minimize spectral overlap, we should require that the sum of the signals’ (rad/sec) bandwidth to be less than the carrier frequency separation: 휔퐵2 + 휔퐵1 < 휔푐2 − 휔푐1 or 퐵2 + 퐵1 < 푓푐2 − 푓푐1 For AM radio, the intelligible signal bandwidth is about 5KHz, which would require neighboring carrier frequencies to be separated by 10KHz. The above requirement is important so that the receiving station can employ a proper narrow band-pass filter to select the spectra of the signal of interest. Modulation at the transmitter stage: Consider the Fourier transform pair 푚(푡) ↔ 푀(휔). The frequency shift of the signal can be obtained by multiplying the time-domain signal by 푒푗휔푐푡 (recall the frequency-shift property of the Fourier transform) as shown below. 푗휔푐푡 Since 푒 = cos(휔푐푡) + 푗sin(휔푐푡) is complex (non-physical), let us see what happens if the modulator used cos(휔푐푡) as the modulating signal. So, the transmitted signal has a duplicated spectrum centered at ±휔푐 1 1 푦(푡) ↔ 푀(휔 − 휔 ) + 푀(휔 + 휔 ) 2 푐 2 푐 with uniform attenuation of 50% for each component (refer to the spectral plot shown below). The following figure depicts the modulation operation in the time domain: Now, having received the modulated transmitted signal 푦(푡) at the receiver, how can one extract 푚(푡)? Let us assume that the receiver can generate a cos(휔푐푡) that is in phase with the transmitter’s modulating sinusoid. Multiplying 푦(푡) by this sinusoid generates: 1 Using the trigonometric identity: cos2(푥) = (1 + cos 2푥), we may write 2 the demodulated signal 푒(푡) as 1 1 푒(푡) = 푚(푡) + 푚(푡) cos(2휔 푡) 2 2 푐 and write its corresponding Fourier transform as 1 1 퐸(휔) = 푀(휔) + [푀(휔 + 2휔 ) + 푀(휔 − 2휔 )] 2 4 푐 푐 Here is the associated spectrum, The spectrum suggests that 푀(휔) can be extracted using a low-pass filter whose cutoff frequency is equal to the bandwidth of 푚(푡). The above analysis suggests the following receiver circuit [휔푐 represents the carrier frequency in rad/sec], Your turn: Assume a signal 푚(푡) is modulated by cos (휔푐푡) and then demodulated by cos (휔푐푡 + 휃). Determine the expression for the signal 푒(푡). Discuss the effect of 휃. The following is a Mathcad validation of the above communication system for the even signal 푚(푡) = 푒−2|푡|. It assumes that the modulation and demodulation signal is cos (휔푐푡), where 휔푐 = 40 rad/sec. A low-pass filter cut-off frequency is the frequency 휔0 at which the filter 2 attenuation is |퐻(휔 )| = √ |퐻(0)|. The cut-off frequency leads to 50% 0 2 2 2 1 power attenuation [(√ ) = ], with a dB (decibels) value of 2 2 2 2 1 10log ((√ ) ) = 10log ( ) ≅ −3dB 10 2 10 2 The negative sign means attenuation. When a non-ideal low-pass filter is used, the output signal will be distorted as can be seen from the above simulation with the first-order low-pass filter. Simulations Repeated with a 2nd-order LP filter Note that the shift 푡푑 is due to the delay introduced by the filter. This delay is about 0.17 sec and can be verified by measuring the slope of the 2nd- order phase response in the vicinity of the origin. Note how the 2nd order filter leads to less distortion but slightly more delay. We may repeat the above simulation of the communication system using Matlab’s Simulink (employing a 2nd-order filter with cut-off frequency 휔0 = 10 and dc gain of 2): Refer to the appendix for an experimental demonstration of amplitude modulation Low-Pass Filters: Ideal vs Practical (Butterworth Filters) Ideal “Brick-Wall” Low-Pass Filter (no signal distortion) The transfer function for this filter can be easily obtained from the magnitude 휔 response, |퐻(휔)| = rect ( ), and phase response, angle(퐻(휔)) = 푒−푗휔푡푑, as 2휔0 휔 퐻(휔) = rect ( ) 푒−푗휔푡푑 2휔표 The unit-impulse response ℎ(푡) for this filter can be obtained using the inverse Fourier transform of 퐻(휔) (employ Fourier Pair #18 and the time-shift property) 휔 휔표 −1 −푗휔푡푑 ℎ(푡) = 퐹 {rect ( ) 푒 } = sinc[휔표(푡 − 푡푑)] 2휔표 휋 So, from the following plot, we observe that ℎ(푡) is non-causal; the response of the filter for an impulse 훿(푡) applied at 푡 = 0 exists for 푡 < 0. The above ideal low- pass filter characteristics cannot be realize physically because ℎ(푡) ≠ 0 for 푡 < 0. In the frequency domain, the Paley-Wiener Criterion can be used to test to see if a mathematical transfer function 퐻(휔) can be realized physically. The criterion is necessary and sufficient and is mathematically stated in terms of the amplitude response as ∞ |ln|퐻(휔)|| ∫ 2 푑휔 < ∞ −∞ 1 + 휔 If |퐻(휔)| does not satisfy the above criterion then it is physically unrealizable.
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