1) Using the bilinear transform frequency mapping equation,
(𝜔𝑐 + 2𝜋𝑘)
2
2
𝜔𝑐
Ω𝑐 = tan (
) = tan ( )
𝑇
2
𝑇
2
3𝜋
2
=> T =
tan ( 5 ) = 1.46𝑚𝑠
2π(300)
2
2) For impulse invariance method, we have:
𝛺=
𝜔
,
𝑇𝑑
𝐻(𝑒 𝑗𝜔 ) = 𝐻(𝑗𝛺)
So,
(a) 89125 ≤ |𝐻(𝑗Ω)| ≤ 1 ,
|𝐻(𝑗Ω)| ≤ 0.17783,
0 ≤ |Ω| ≤
0.2𝜋
𝑇𝑑
0.3𝜋
𝜋
≤ |Ω| ≤
𝑇𝑑
𝑇𝑑
(b) Because the Butterworth frequency response is monotonic, we can solve:
𝑗0.2𝜋 2
)| =
|𝐻𝑐 (
𝑇𝑑
𝑗0.3𝜋 2
)| =
|𝐻𝑐 (
𝑇𝑑
1
0.2𝜋 2𝑁
1 + (Ω 𝑇 )
𝑐 𝑑
1
= 0.891252
= 0.177832
0.3𝜋 2𝑁
1 + (Ω 𝑇 )
𝑐 𝑑
To obtain Ω𝑐 𝑇𝑑 = 0.7047 and 𝑁 = 5.88. Rounding up to 𝑁 = 6 and yieldsΩ𝑐 𝑇𝑑 = 0.7032.
(c) We see that the poles of the magnitude-squared function are again evenly distributed around a
circle radius 0.7032. For being causal and stable, 𝐻𝑐 (𝑠) should have the left half-plane poles of
the magnitude squared function & the result is the same for any value of 𝑇𝑑 . Correspondingly,
𝐻(𝑧) does not depend on 𝑇𝑑 .
3) (a)
1 − 𝛿̂1 =
1 − 𝛿1
1 + 𝛿1
𝛿̂1 =
=>
𝛿̂2 =
2𝛿1
1 + 𝛿1
𝛿2
1 + 𝛿1
(b)
𝛿1 =
𝛿̂1
2 − 𝛿̂1
,
𝛿2 =
2𝛿̂2
2 − 𝛿̂2
For the example, we were given
𝛿̂1 = 1 − 0.89125 = 0.10875
𝛿̂2 = 0.17783
So:
𝛿1 = 0.0575
𝛿2 = 0.1881
(c) We can calculate 𝐻(𝑧) as example 7.2.
4) Using the bilinear transform frequency mapping equation,
2 𝜔𝑠
2
0.2𝜋
𝑟𝑎𝑑
Ω𝑠 = ( ) =
tan (
) = 2𝜋(51.7126)
𝑇 2
0.002
2
𝑠
2 𝜔𝑝
2
0.3𝜋
𝑟𝑎𝑑
Ω𝑝 = ( ) =
tan (
) = 2𝜋(81.0935)
𝑇 2
0.002
2
𝑠
Thus, the specification which should be used to design the prototype continuous-time filter are
|𝐻𝑐 (𝑗Ω)| < 0.04
0.995 < |𝐻𝑐 (𝑗Ω)| < 1.005
|Ω| ≤ 2𝜋(51.7126)
|Ω| ≥ 2𝜋(81.0935)