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86
Signals and Spectra Chap. 2
From Table 2–1, we find that the Fourier transform of this differential equation is
Thus, the transfer function for this network is
RC(j2pf)y(f) + Y(f) = X(f)
H(f) = Y(f)
X(f)
=
1
1 + j(2pRC)f
(2–145)
Using the Fourier transform of Table 2–2, we obtain the impulse response
1
e -t/t 0
,
h(t) = c t 0
t Ú 0
0, t 6 0
(2–146)
where t 0 = RC is the time constant. When we combine Eqs. (2–143) and (2–145), the power
transfer function is
G h (f) = |H(f)| 2 =
1
1 + (f/f 0 ) 2
(2–147)
where f 0 = 1/(2pRC).
The impulse response and the power transfer function are shown in Fig. 2–15. See
Example2_17.m for detailed plot. Note that the value of the power gain at f = f 0 (called
the 3-dB frequency) is G h (f 0 ) = 1 2. That is, the frequency component in the output waveform
at f = f 0 is attenuated by 3 dB compared with that at f = 0. Consequently, f = f 0 is said
to be the 3-dB bandwidth of this filter. The topic of bandwidth is discussed in more detail in
Sec. 2–9.
Distortionless Transmission
In communication systems, a distortionless channel is often desired. This implies that the
channel output is just proportional to a delayed version of the input
y(t) = Ax(t - T d )
(2–148)
where A is the gain (which may be less than unity) and T d is the delay.
The corresponding requirement in the frequency domain specification is obtained by
taking the Fourier transform of both sides of Eq. (2–148).
Y(f) = AX(f)e -j2pfT d
Thus, for distortionless transmission, we require that the transfer function of the channel be
given by
H(f) = Y(f)
X(f)
= Ae -j2pfT d
(2–149)