Page 287 - Schaum's Outline of Theory and Problems of Signals and Systems

FOURIER ANALYSIS OF TIME SIGNALS AND SYSTEMS [CHAP.
5

(a) From Eq. (5.137) (Prob. 5.20) we have
sin at
~(t) I4 <a
= - -X(w) =p0(w) =
~t Iwl > a
Then when a < w,, we have
Y(w) = X(w)H(w) =X(w)
Thus,
sin at
y(t) =x(t) = -
X t

(b) When a > w,, we have
Y(w) =X(w)H(w) = H(w)
Thus,
sin o,t
y(t) = h(t) = -
Xt

(c) In case (a), that is, when w, > a, y(t) =x(t) and the filter does not produce any
distortion. In case (b), that is, when w, < a, y(t) = h(t) and the filter produces distortion.

5.53. Consider an ideal low-pass filter with frequency response

The input to this filter is the periodic square wave shown in Fig. 5-27. Find the output
y( t 1.

Setting A = 10, T, = 2, and w, = 2n/To = n in Eq. (5.107) (Prob. 5.9, we get

Since the cutoff frequency o, of the filter is 47~ rad, the filter passes all harmonic components
of x(t) whose angular frequencies are less than 4n rad and rejects all harmonic components of
x(t) whose angular frequencies are greater than 477 rad. Therefore,
20 20
y(t) = 5 + - sinnt + - sin3~t
x 3n

5-54 Consider an ideal low-pass filter with frequency response

The input to this filter is

Find the value of o, such that this filter passes exactly one-half of the normalized
energy of the input signal x(t ).