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Because x(n) is a sinusoid, the response is
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where A and 4 are the magnitude and phase, respectively, of the frequency response at o = 1712. With
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it follows that I H (ejw)l = 2. We may evaluate the phase as follows:
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Therefore. which, when evaluated at o = n/2, gives
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4 , , ( ~ ) tan-' =
4,, (w)J,=,/~ = tan-' Thus, y(n) = 2sin(:[n
= tan-'
= 0.2952~
+ 0.59031)
Consider the following system consisting of an ideal D/C converter, a linear time-invariant filter, and an ideal C P converter.
The continuous-time system h , ( t ) is an ideal low-pass filter with a frequency response
Hdf) =
(a) If T I = T2 =
(b) If T I =
(a) x
find an expression relating the output y ( n ) to the input x(n). and T2 = find y(n) when
I f 1 5 1OkH.z otherwise
(a) When TI = T2,this system behaves as a linear shift-invariant discrete-time system with a frequency response
Because H,(jS2) = 1 for 152) < 2 n .
lo4,
h(n) = 6(n)
Therefore, y(n) = x(n). Another way to analyze this system is to note that the output of the D/C converter, x,(t), is bandlimited to f = 5 kHz. Because Ha(f )is an ideal low-pass filter with a cutoff frequency 10 kHz, yo(!) = xo(t). Therefore, this system is equivalent to the one shown below.
Because an ideal D/C converter followed by an ideal D/C converter is the identity system, y(n) = x(n).
(b) When T I T2,this system is, in general, no longer a linear shift-invariant system. However, we may analyze this system in the frequency domain as follows. First, note that the DTFTofx(n) is as illustrated in the following figure:
SAMPLING
[CHAP. 3
Thus, the output of the D/C converter is a bandlimited signal that has a Fourier transform as shown in the following figure:
The analog low-pass filter removes all frequencies in x a ( t ) above 10 kHz to produce a signal y,(t) that has a Fourier transform as shown below.
Because the highest frequency in y a ( t ) is 10 kHz, the Nyquist rate is 20 kHz. However, the sampling frequency of the C/D converter is 10 kHz, so y a ( t ) will be aliased. The DTFT of y ( n ) is related to Y a ( j Q ) as follows:
Summing the shifted and scaled transforms yields
Sample Rate Conversion
3.21 Suppose that a discrete-time sequence x ( n ) is bandlimited so that
This sequence is then sampled to form the sequence
where N is an integer. Find the largest value for N for which x ( n ) may be uniquely recovered from y ( n ) .
The easiest way to view this problem is as illustrated below.
Converting x ( n ) into a continuous-time signal with an ideal D/C converter with a sampling frequency f produces , a continuous-time signal x a ( t ) that is bandlimited to fo = 0.3 . f s / 2 . Therefore, xa(r) may be sampled, without
CHAP. 31
SAMPLING
aliasing, if we use a sampling frequency fsf 2 2 fo = 0 . 3 f,, or
Therefore, if T,' = 3T,,
y ( n ) = x,(3nTs) = x ( 3 n )
and x ( n ) may be uniquely recovered from y(n). Thus, N = 3.
Consider the following system:
Assume that X a ( f ) = 0 for If ( > l / T s and that
How is the output of the discrete-time system, y ( n ) , related to the input signal x a ( t )
In this system, the bandlimited signalx,(t) is sampled, without aliasing, to produce the sampled signal x ( n ) = x,(nT,). Up-sampling x ( n ) by a factor of L , and filtering with an ideal low.-pass filter with a cutoff frequency w, = n / L , produces the signal
that is, a signal that is sampled with a sampling frequency L f,. However, because the low-pass filter has linear phase with a group delay of one sample, the interpolated up-sampled signal is delayed by 1. Therefore, the output of the low-pass filter is
u ( n ) = w ( n - 1 ) = x,
Down-sampling by L then produces the output
Thus, y ( n ) corresponds to samples of x,(c - to) where ro = T , / L .
Consider the system shown in the figure below.
Assume that the input is bandlimited, X a ( j i 2 ) = 0 for ( i 2 ( > 2 n . 1000.
(a) What constraints must be placed on M, T I ,and T2 in order for y a ( t ) to be equal to x,(t)
(b) If f , = f2 = 20 kHz and M = 4, find an expression for ya(r) in terms of x , ( t ) .
SAMPLING
(a) Suppose that x,(r) has a Fourier transform as shown in the figure below.
[CHAP. 3
Because y ( n ) = x ( M n ) = x , ( n M T I ) , in order to prevent x ( n ) from being aliased, it is necessary that
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