563489578934

Problems 117
b. Given the RC low-pass filter transfer function
where f 0 = 1 Hz, use the inverse fast Fourier transform (IFFT) of MATLAB to calculate
the impulse response h(t).
Solution.
H(f) =
(a) From Eq. (2–30), the ICFT is
Referring to the discussion leading up to Eq. (2–184), the ICFT is approximated by
But ¢t = T/N,¢f = 1/T, and f s = 1/¢t, so
q
w(t) = W(f)e j2pft df
L
-q
1
1 + j(f/f 0 )
w(k¢t) L©W(n¢f)e j2pn¢fk¢t ¢f
w(k¢t) L Nc 1 N ©W(n¢f)ej(2p/N)nk d ¢f
Using the definition of the IDFT as given by Eq. (2–177), we find that the ICFT is related to the
IDFT by
w(k¢t) L f s x(k)
(2–207)
where x(k) is the kth element of the N-element IDFT vector. As indicated in the discussion
leading up to Eq. (2–184), the elements of the X vector are chosen so that the first N/2 elements
are samples of the positive frequency components of W(f), where f = n¢f, and the
second N/2 elements are samples of the negative frequency components of W(f).
(b) Run file SA 2_9.m for a plot of h(t) as computed by using the IFFT and Eq. (2–207). Compare
this IFFT-computed h(t) with the analytical h(t) that is shown in Fig. 2–15b, where
t 0 = RC = 1/(2pf 0 ).
PROBLEMS
★ 2–1 For a sinusoidal waveform with a peak value of A and a frequency of f 0 , use the time average
operator to show that the RMS value for this waveform is A/12.
2–2 A function generator produces the periodic voltage waveform shown in Fig. P2–2.
(a) Find the value for the DC voltage.
(b) Find the value for the RMS voltage.
(c) If this voltage waveform is applied across a 100-Ω load, what is the power dissipated in the load?
2–3 The voltage across a load is given by v(t) = A 0 cos v 0 t, and the current through the load is a
square wave,
i(t) = I 0
q
a
n=-q
cßa t - nT 0
T 0 /2
b -ßa t - nT 0 - (T 0 /2)
T 0 /2
where v 0 = 2p/T 0 , T 0 = 1 sec, A 0 = 10 V, and I 0 = 5 mA.
bd