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barcode generator for ssrs (a) The Nyquist rate is equal to twice the highest frequency in x,(t). If in Software
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Let h , ( t ) be the impulse response of a causal continuoustime 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 discretetime filter with a unit sample response Find the frequency response H ( e j w )of the discretetime 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 discretetime 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 leaTs( < I or, because T, > 0, we must have a > 0. In other words, the poles of H,(s) must lie in the lefthalf splane 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 continuoustime filter has a system function
If h , ( t ) is sampled to form a discretetime 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 continuoustime 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 + e2T' = 0.2512 [I
+ e2T'] 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 continuoustime 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 discretetime 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:

