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Alternate Method: Make the same substitution as above, but directly in the de nite integral, changing the limits of integration accordingly. When x = 0, u = 02 + 1 = 1; when x = 1, u = 12 + 1 = 2. Thus, the rst line of the computation above yields

1 72.

2-l)'

7.8 Finding fourth proportionals to three given numbers Find the fourth proportional to (a) 2, 4, 6; (b) 4, 2, 6; (c) 1, 3, 4; (d) b, d, c. 2

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As a check on the design, we may compute the magnitude of the frequency response at w = 0. I n , which is

7.9 Finding the mean proportional to two given numbers Find the positive mean proportional x between (a) 5 and 20; (b) 1 and 8. 2 9

I ~ ( e j ~ ) l , , o . s ,= 0.9044

1 3 1 (( 2) ( 1)3 ) = (2 2 1) 3 3

8 4 9

We would like to design a digital low-pass filter that has a passband cutoff frequency w p = 0 . 3 7 5 ~ with 6, = 0.01 and a stopband cutoff frequency w, = 0.5x with 6 = 0.01. The filter is to be designed using , the bilinear transformation. What order Butterworth, Chebyshev, and elliptic filters are necessary to meet the design specifications

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To find the required filter order, we begin by finding the discrimination factor and the selectivity factor for the analog low-pass filter prototype. With 8, = 6, = 0.01, the discrimination factor is

2 3.

For the selectivity factor, we first find the cutoff frequencies for the analog prototype. With wp = 0.375r and w, = O S r , we prewarp the frequencies as follows (T, = 2),

7.10 Changing equal products into proportions (a) Form a proportion whose fourth term is x and such that 2bx (b) Find the ratio x to y if ay

0 375r ( = 0.6682 l )

2 3/2 2 1 1/2 du = 1 u u = u3/2 2 1 2 3 3 1 2 1 2

3s2.

Therefore,

7.11 Changing proportions into new proportions Use each of the following to form a new proportion whose first term is x: 15 (a) x

R k = 2 = 0.6682 -

For the Butterworth filter, the required filter order is log d N = - - 16.25 log k or N = 17. For the Chebyshev filter,

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x 15 x (b) By Principle 5, 6 (a) By Principle 3, 4 3 8 3 x 8 x (d) By Principle 4, 2 (c) By Principle 6, 1 3 x 15 ; thus, 5 2

31.8 (a) If f is an even function (Section 7.3), show that, for any a > 0,

9 = 0.0369

15 x

With impulse invariance, a first-order pole in H,(s) at s = sk is mapped to a pole in H ( z ) at z = e s k T s :

7.12 Combining numerators and denominators of proportions Use Principle 8 to find x in each of the following proportions: (a) x 9 2 2 3 (b) x 8 y x 4 y 2 3 (c) 3x y 15 y 10 3 3 5

the impulse response is ha(t) = tesk'u(t) where u(t) is the unit step function. Sampling h,(t) with a sampling period T,, we have

(a) Adding numerators and denominators yields (b) Here we have (x y) 8 (x 4 y) (3x x 9 2 3 2 x 2 or 3 12 2 , so x 3 3x 3 or 5 30 4. 3 , so x 5 6. 2 , so x 3 8.

(b) Assume that f (x) 0 for arguments to the right of and near a. Prove:

Using the z-transform property nx(n) and the z-transform pair cunu(n) c4-t it follows that the z-transform of h(n) is

2 2x , which gives 3 12 y) 15 (y 10 3) 5 3

- az-'

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