barcode generator for ssrs (a) The Nyquist rate is equal to twice the highest frequency in x,(t). If in Software

Creating Code 128 Code Set A in Software (a) The Nyquist rate is equal to twice the highest frequency in x,(t). If

(a) The Nyquist rate is equal to twice the highest frequency in x,(t). If
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Y,(jSt) = jStX,(jQ)
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Thus, if X,(jR) = 0 for IQI > Sto, the same will be true for Y,(jR). Therefore, the Nyquist frequency is not changed by differentiation.
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(b) The signal ya(t) = xa(2t) is formed from x,(t) by compressing the time axis by a factor of 2. This results in an
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expansion of the frequency axis by a factor of 2. Specifically, note that
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Consequently, if the Nyquist frequency for x,(r) is St,, the Nyquist frequency for y&) will be 2S2,.
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(c) When two signals are multiplied, their Fourier transforms are convolved. Therefore, if
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Thus, the highest frequency in yo([) will be twice that of x,,(r), and the Nyquist frequency will be 2Q,. (d) Modulating a signal by cos(Qol) shifts the spectrum of xa(l) up and down by n o . Therefore, the Nyquist frequency for ya(t) = cos(Qot)xa(r) will be S2, 2Qo.
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[CHAP. 3
Let h , ( t ) be the impulse response of a causal continuous-time filter with a system function
Thus, H,(s) has a zero at s = -a and a pair of poles at s = -a k jh. By sampling h , ( t ) we form a discrete-time filter with a unit sample response
Find the frequency response H ( e j w )of the discrete-time filter.
To find the frequency response H(eJU), is necessary to find the impulse response of the analog filter, h,(t), sample it the impulse response, h(n) = h,(nT,) and then find the discrete-time Fourier transform,
To find the impulse response, we first perform a partial fraction expansion of Ha(s)as follows:
Ha(s) = A B s + ( a jb) + s + ( a - jb)
The constant A is
Similarly, for B we have
Therefore,
Ha(s)=
Another way to find the constants A and B would be to write Eq. (3.13)over a common denominator,
Ha(s) =
+ + b2
= A(s + a - jb)
and equate the polynomial coefficients in the numerators of H,(s):
A+B=I A(a - j b )
+ B(a + j b ) = a
Solving these two equations for A and B gives the same result as before. From the partial fraction expansion of Ha(s),the impulse response may be found using the Laplace transform pair
s +(a
+ jb)
+ (a2- jb)
+ a)2+ b2
+ B(s + a + jb)
CHAP. 31
Sampling ha(! ), we have
SAMPLING
h(n) = h,(nT,) = e - a " T ~ o s ( b n ~ s ) u ( n ) Finally, for the frequency response we have
Note that in order for these sums to converge, and for the frequency response to exist, it is necessary that le-aTs( < I or, because T, > 0, we must have a > 0. In other words, the poles of H,(s) must lie in the left-half s-plane or, equivalently, h,,(t) must be a stable filter. With a > 0 we have
which, after combining over a common denominator and simplifying, gives
A continuous-time filter has a system function
If h , ( t ) is sampled to form a discrete-time system with a unit sample response
h ( n ) = ha(nT7)
find the value for Ts so that ~ ( e . ~at w = rr/2 is down 6 dB from its maximum value at w = 0, that is, " )
10 log
I H (eJT'2)12
IH (eio)12
= -6
The impulse response of the continuous-time system is h,(t) = e-'u(t) When sampled with a sampling period T,, the resulting unit sample response is h(n) = h,,(nT,) = e-"%u(n) and the frequency response is
SAMPLING
[CHAP. 3
With
and it follows that we want
lo log
~ H ( e ~ ~ f ~ ) l ( ~ - c - & ) ~ -6 1 = = lolog I H(ej0)12 1 + ecZTs
Thus, we have
1 - 2KTs + e-2T' = 0.2512 [I
+ e-2T']
0.7488e-~~' 2KT' + 0.7488 = 0 -
which is a quadratic equation in e d . Solving for the roots of this quadratic equation, we find
Taking the natural logarithm, and selecting the positive value for T,, we have
T, = 0.7978
A continuous-time signal x a ( t ) is bandlimited with X,(jS1) = 0 for IS11 > n o . If x a ( t ) is sampled with a sampling frequency S1, 2 2Slo, how is the energy in x ( n ) ,
related to the energy in x a ( t ) ,
and the sampling period T,
Using Parseval's theorem, the energy in the analog signal x,(t) may be expressed in the frequency domain as follows:
Because x,(t) is bandlimited with X,(jQ) = 0 for Is21
Sampling x,(t) at or above the Nyquist rate results in a sequence x(n) with a discrete-time Fourier transform
CHAP. 31
SAMPLING
Therefore, the energy in x ( n ) , using Parseval's theorem, is
and we have
I E,, = - E, Ts
As a check on this result. suppose that x,(t) is a bandlimited signal with a spectrum shown in the figure below.
The energy in x,
""' ""
When sampled with a sampling frequency Q, 1 2Qo, the DTFT of the sampled signal is as shown in the following figure:
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