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Sec. 2–2 Fourier Transform and Spectra 45 where [ # ] denotes the Fourier transform of [·] and f is the frequency parameter with units of hertz (i.e., 1/s). † This defines the term frequency. It is the parameter f in the Fourier transform. W(f) is also called a two-sided spectrum of w(t), because both positive and negative frequency components are obtained from Eq. (2–26). It should be clear that the spectrum of a voltage (or current) waveform is obtained by a mathematical calculation and that it does not appear physically in an actual circuit. f is just a parameter (called frequency) that determines which point of the spectral function is to be evaluated. The FT is used to find the frequencies in w(t). One chooses some value of f, say, f=f 0 , and evaluates ƒW(f 0 )ƒ. If ƒW(f 0 )ƒ is not zero, then the frequency f 0 is present in w(t). In general, the FT integral is evaluated again and again for all possible values of f over the range - q6f 6qto find all of the frequencies in w(t). It is easy to demonstrate why the FT finds the frequencies in w(t). For example, suppose that w(t) = 1. For this DC waveform, the integrand of Eq. (2–26) is e -j2pft = cos 2pft - j sin 2pft, and consequently, the FT integral is zero (provided that f Z 0), because the area under a sinusoidal wave over multiple periods is zero. However, if one chooses f = 0, then the FT integral is not zero. Thus, as expected, Eq. (2–26) identifies f = 0 as the frequency of w(t) = 1. In another example, let w(t) = 2 sin2pf 0 t. For this case, the integrand of Eq. (2–26) is sin 2p(f 0 - f)t + sin 2p(f 0 + f)t - j cos 2p(f 0 - f)t + j cos 2p(f 0 + f)t and the integral is nonzero when f = f or when f = -f ‡ 0 0 . In this way, the FT finds the frequencies in w(t). Direct evaluation of the FT integral can be difficult, so a list of alternative evaluation techniques is very helpful. The FT integral, Eq. (2–26), can be evaluated by the use of 1. Direct integration. § See Example 2–3. 2. Tables of Fourier transforms or Laplace transforms. See Table 2–2 and Example 2–10. 3. FT theorems. See Table 2–1 and Example 2–4. 4. Superposition to break the problem into two or more simple problems. See study-aid example SA–5. 5. Differentiation or integration of w(t). See Example 2–7. 6. Numerical integration of the FT integral on the PC via MATLAB integration functions. See Example 2–6 and file Example2_06.m. 7. Fast Fourier transform (FFT) on the PC via MATLAB FFT functions. See Fig. 2–21, file Example2_20.m, and Fig. 2–22, file Example2_21.m. † Some authors define the FT in terms of the frequency parameter v = 2pf where v is in radians per second. That definition is identical to (2–26) when v is replaced by 2pf. I prefer Eq. (2–26) because spectrum analyzers are usually calibrated in units of hertz, not radians per second. ‡ The frequency of the sine wave is f 0 , but the FT finds both f 0 and its mirror image -f 0 , as explained in the discussion following Example 2–5. § Contour integration, covered in a mathematics course in complex variables, is also a helpful integration technique.
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Sec. 2–2 Fourier Transform and Spectra 45<br />
where [ # ] denotes the Fourier transform of [·] and f is the frequency parameter with<br />
units of hertz (i.e., 1/s). † This defines the term frequency. It is the parameter f in the<br />
Fourier transform.<br />
W(f) is also called a two-sided spectrum of w(t), because both positive and negative frequency<br />
components are obtained from Eq. (2–26). It should be clear that the spectrum of a voltage (or<br />
current) waveform is obtained by a mathematical calculation and that it does not appear physically<br />
in an actual circuit. f is just a parameter (called frequency) that determines which point<br />
of the spectral function is to be evaluated.<br />
The FT is used to find the frequencies in w(t). One chooses some value of f, say, f=f 0 ,<br />
and evaluates ƒW(f 0 )ƒ. If ƒW(f 0 )ƒ is not zero, then the frequency f 0 is present in w(t). In general,<br />
the FT integral is evaluated again and again for all possible values of f over the range<br />
- q6f 6qto find all of the frequencies in w(t).<br />
It is easy to demonstrate why the FT finds the frequencies in w(t). For example, suppose<br />
that w(t) = 1. For this DC waveform, the integrand of Eq. (2–26) is e -j2pft =<br />
cos 2pft - j sin 2pft, and consequently, the FT integral is zero (provided that f Z 0),<br />
because the area under a sinusoidal wave over multiple periods is zero. However, if one chooses<br />
f = 0, then the FT integral is not zero. Thus, as expected, Eq. (2–26) identifies f = 0 as the frequency<br />
of w(t) = 1. In another example, let w(t) = 2 sin2pf 0 t. For this case, the integrand of<br />
Eq. (2–26) is sin 2p(f 0 - f)t + sin 2p(f 0 + f)t - j cos 2p(f 0 - f)t + j cos 2p(f 0 + f)t<br />
and the integral is nonzero when f = f or when f = -f ‡ 0<br />
0 . In this way, the FT finds the<br />
frequencies in w(t).<br />
Direct evaluation of the FT integral can be difficult, so a list of alternative evaluation<br />
techniques is very helpful. The FT integral, Eq. (2–26), can be evaluated by the use of<br />
1. Direct integration. § See Example 2–3.<br />
2. Tables of Fourier transforms or Laplace transforms. See Table 2–2 and Example 2–10.<br />
3. FT theorems. See Table 2–1 and Example 2–4.<br />
4. Superposition to break the problem into two or more simple problems. See study-aid<br />
example SA–5.<br />
5. Differentiation or integration of w(t). See Example 2–7.<br />
6. Numerical integration of the FT integral on the PC via MATLAB integration functions.<br />
See Example 2–6 and file Example2_06.m.<br />
7. Fast Fourier transform (FFT) on the PC via MATLAB FFT functions. See Fig. 2–21, file<br />
Example2_20.m, and Fig. 2–22, file Example2_21.m.<br />
† Some authors define the FT in terms of the frequency parameter v = 2pf where v is in radians per second.<br />
That definition is identical to (2–26) when v is replaced by 2pf. I prefer Eq. (2–26) because spectrum analyzers are<br />
usually calibrated in units of hertz, not radians per second.<br />
‡ The frequency of the sine wave is f 0 , but the FT finds both f 0 and its mirror image -f 0 , as explained in the<br />
discussion following Example 2–5.<br />
§ Contour integration, covered in a mathematics course in complex variables, is also a helpful integration<br />
technique.