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Determine the characteristics of this mapping, and compare it to the bilinear transformation. Is this a good mapping to use Explain why or why not.

Unlike the harmonic sequence, the factorial sequence grows exponentially. This is reflected by Stirling s formula:

As with the bilinear transformation, the first backward difference will map a rational function of s into a rational function of z. To see how points in the s-plane map to points in the z-plane, let us write the mapping as follows,

1.000000 1.500000 1.833333 2.083333 2.283333 2.450000 2.592857 2.717857 2.828968 2.928968

+jR,

12n ,

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It is not hard to see that this series diverges. That is, its partial sums increase without bound. The partial sums of the harmonic series are called the harmonic numbers and are denoted by Hn :

.and it follows that points in the left-half s-plane (a c 0) are mapped to points inside the unit circle, lzl < I . Thus, stable analog filters are mapped to stable digital filters. Now, let us look at how the jR-axis is mapped to the z-plane. With s = j R , we see that

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The value of the variable depends upon n, but in any case it is bounded between 0 and 1. Thus, for large n, the exponent /12n will be very close to 0, making the factor e /12n very close to 1. Consequently, Stirling s formula is often expressed in this simpler approximate form: n n n! 2n -e The factorial numbers grow exponentially: n! = (2n). This fact follows from Stirling s formula. Another important consequence of Stirling s formula is that lg(n!) is asymptotically equivalent to n lgn: lg(n!) = (n lgn). FIBONACCI NUMBERS The Fibonacci numbers also frequently appear in the analysis of algorithms. They are defined by:

Since the jR-axis does not map onto the unit circle, the frequency response of the digital filter produced with this mapping will not, in general. be an accurate representation of the frequency response of the analog filter except when w is close to zero. In other words, the frequency response of a continuous-time filter will be well preserved only for low frequencies.

3 2 1

Use the impulse invariance method to design a digital filter from an analog prototype that has a system function

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0 1 2 3 4 5 6 7 8 9 10 11 12

Ha(s) =

H2 =

+ (a + jh)

0 1 1 2 3 5 8 13 21 34 55 89 144

To design a filter using the impulse invariance technique, we first expand H,,(s) in a partial fraction expansion as follows,

H3 =

HAS)=

The first 13 Fibonacci numbers are shown in Table A.3. Like the factorial sequence, the Fibonacci sequence grows exponentially, as is verified by De Moivre s formula:

+ (a - jb)

Although the harmonic numbers increase without bound, it is not obvious how fast they increase. Table A.1 suggests that they increase very slowly. The fact is that the harmonic numbers increase logarithmin n! cally: H n = (lgn). This means that they increase at about the 0 1 same rate as logarithmic numbers. More precisely, it means that 1 1 both ratios H n/ lgn and lgn /H n are bounded.

Hu(s)=

1 5 = --------------2

s+a+,jh+s+a-

2 3 4 5 6 7 8 9 2 6 24 120 720 5040 40,320 362,880

cos(bT,). Thus, the location of the zero in the discreteNote that the zero at s = -a is mapped to a zero at z = PT7 time filter depends on the position of the poles as well as the zero in the analog filter.

Thus, Fn = ( ). Here, = 1.618034 and = 0.618034. These two constants are the golden mean and its conjugate.

With the impulse invariance method, the unit sample response of a digital filter is formed by sampling the impulse response of the continuous-time filter,

STIRLING S FORMULA The factorial numbers frequently appear in the analysis of algorithms. They are defined by:

Another approach is to use the step invariance method in which the step response of the digital filter is formed by sampling the step response of the continuous-time filter.

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