chap07 - PROBLEM 7.1 KNOWN: FIND: SOLUTION The signal is...

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PROBLEM 7.1 KNOWN: E(t) = 5sin 2 π t mV FIND: Convert to a discrete time series and plot SOLUTION The signal is converted to a discrete time series for using N = 8 and sample time increments of 0.125, 0.30, and 0.75 s and plotted below. The time increments of 0.125 and 0.30 s produce discrete series with a period of 1s or frequency of 1 Hz. The series created with a time increment of 0.75 s, which fails the Sampling Theorem criterion, portrays a signal with a different frequency content; this frequency is the alias frequency. -6 -4 -2 0 2 4 6 0 0.2 0.4 0.6 0.8 1 1.2 TIME [s] DISCRETE SERIES dt = 0.125 s
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-6 -4 -2 0 2 4 6 0 0.5 1 1.5 2 2.5 3 TIME [s] DISCRETE SERIES dt = 0.30 s -6 -4 -2 0 2 4 6 0123456 TIME [s] dt =0.75 s
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PROBLEM 7.2 KNOWN: Repeat Problem 7.1 using N = 128 points. FIND: The discrete Fourier transform for each series. SOLUTION The DFT for the time series representations of E(t) using N = 128 and t δ = 0.125, 0.30, and 0.75 s, respectively was executed using a DFT algorithm (any such algorithm on the companion software or using the approach described in Chapter 2 will work) and shown below. The DFT returns an exact Fourier transform of the discrete time series but not necessarily the time signal from which it is based. Whether this DFT exactly represents E(t) depends on the criteria: (1) f s = 1/ t > 2f (2) m/f 1 = N t m = 1,2,3, . .. With f = 1 Hz and f 1 = f: (a) t = 0.125 s and N = 128: f s = 1/0.125s = 8 Hz > 2 Hz m/1 Hz = 128/0.125 or m = 16 an exact integer value. Another way to look at this second criterion: the DFT resolution f = 1/N t = 0.0625 Hz, to which 1 Hz is an exact multiple. Both criteria are met. Therefore, this DFT will exactly represent E(t) in both frequency and amplitude, as shown below. (b) t = 0.3 s and N = 128 f s = 1/0.3s = 3.3 Hz > 2 Hz m/1 Hz = 128/0.3 or m = 38.4 not an exact integer value. Criterion (1) is met but Criterion (2) is not met. Therefore, an alias frequency will not appear. But this DFT will not exactly represent E(t) in amplitude and spectral leakage will occur, as seen below. So we find an amplitude less than 5 at a frequency centered at 1 Hz.
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(c) t δ = 0.75 s and N = 128 f s = 1/0.75s = 1.33 Hz < 2 Hz m/1 Hz = 128/0.75 or m = 96 an exact integer value. Criterion (1) is not met but Criterion (2) is met. Therefore, this DFT will not exactly represent E(t) in frequency. However, it does represent the signal amplitude exactly. An alias frequency at 0.33Hz appears in the DFT. Note how the frequency resolution scale changes between plots. 0.0 1.0 2.0 3.0 4.0 5.0 6.0 012345 FREQUENCY [Hz] AMPLITUDE dt=0.125 0 0.5 1 1.5 2 2.5 3 3.5 4 00 . 511 . 52 FREQUENCY [Hz] dt=0.3
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0 1 2 3 4 5 6 0 0.2 0.4 0.6 0.8 FREQUENCY [Hz] AMPLITUDE dt=0.3
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PROBLEM 7.3 KNOWN: T(t) = 2 sin 4 π t o C f s = 4 and 8 Hz; N = 128 FIND: Compute the Fourier transform from the resulting series. SOLUTION The fundamental frequency of this signal is f 1 = 2 Hz. From the Sampling theorem, an appropriate sample rate is f s > 2f 1 or f s > 4Hz At f s = 4 Hz the Sampling Theorem is not upheld whereas at f s = 8 Hz the Sampling theorem is upheld. For f s = 4 Hz, see the COMMENT below.
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chap07 - PROBLEM 7.1 KNOWN: FIND: SOLUTION The signal is...

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