SIGNALS AND SYSTEMS II Assignment 8 Solutions

1. Compute h_{k}^{d} first,

h_{k}^{d} = \frac{1}{2π} ∫_{−π}^{π} H_{d}(e^{iω})e^{iωk}dω
= \frac{1 − (−1)^{k}}{kπ} .

(a) δ = 2%

A = −20 log_{10} δ = 33.98
α = 0.5842(A − 21)^{0.4} + 0.07886(A − 21) = 2.652

Using Matlab, we obtain

k ω_{k} ω_{k}h_{k}^{d} 0 1 0 1 0.9592 0.6106 2 0.8434 0 3 0.6714 0.1425 4 0.4704 0 5 0.2702 0.0344

(b) δ = 0.056%

A = −20 log_{10} δ = 65.04
α = 0.1102(A − 8.7) = 6.208

k ω_{k} ω_{k}h_{k}^{d} 0 1 0 1 0.8916 0.5676 2 0.6235 0 3 0.3251 0.0690 4 0.1098 0 5 0.0123 0.0016

(c) Hamming Window.

k ω_{k} ω_{k}h_{k}^{d} 0 1 0 1 0.9121 0.5807 2 0.6821 0 3 0.3978 0.0844 4 0.1679 0 5 0.08 0.0102

Note that the Hamming weights are between those of Kaiser designs 1 and 2, with the value of α = 5.4414 in a Kaiser window design corresponding very closely to that of the Hamming window design. Hence we can say, qualitatively, that Kaiser design 1 has the largest ripple but smallest transition width; Kaiser design 2 has the smallest ripple but the widest transition, while the Hamming design is in between.

2. (a)
(\frac{ω_{p}}{ω_{s}})^{2N_{c}} = \frac{ϵ^{2}}{A^{2} − 1} ⟹ \frac{ω_{p}^{2N_{c}}}{ω_{s}^{2N_{c}}} = \frac{ϵ^{2}}{A^{2} − 1}
⟹ \frac{ω_{p}^{2N_{c}}}{ϵ^{2}} = \frac{ω_{s}^{2N_{c}}}{A^{2} − 1}
⟹ ω_{1}(N_{c}) = \frac{ω_{p}}{(ϵ^{2})^{\frac{1}{2Nc}}} = \frac{ω_{s}}{(A^{2} − 1)^{\frac{1}{2Nc}}} = ω_{2}(N_{c})

Therefore, \frac{ω_{1}(N_{c})}{ω_{2}(N_{c})} = 1.

(b) Notice N ≥ N_{c} and \frac{ω_{p}}{ω_{s}} < 1. So

(\frac{ω_{p}}{ω_{s}})^{2N} ≤ (\frac{ω_{p}}{ω_{s}})^{2N_{c}} = \frac{ϵ^{2}}{A^{2} − 1}

Then,

\underbrace{\frac{ω_{p}}{(ϵ^{2})^{\frac{1}{2N}}}}_{ω_{1}(N)} ≤ \underbrace{\frac{ω_{s}}{(A^{2} − 1)^{\frac{1}{2N}}}}_{ω_{2}(N)}

i.e., ω_{1}(N) ≤ ω_{2}(N).

(c) From ω_{c} ≤ ω_{2}(N) = \frac{ω_{s}}{(A^{2}−1)^{\frac{1}{2N}}} , we have

ω_{c}^{2N}(A^{2} − 1) ≤ ω_{s}^{2N} ⇒

A^{2} ≤ (\frac{ω_{s}}{ω_{c}})^{2N} + 1.

3. Refer to the Example in Chapter 7. Originally, it was intended that you would use the second choice (ω_{1}(N) in question 2) for computing ω_{c}, but it was unfortunately not specified in the problem!

4. (a)
h_{a}(t) = \frac{1}{2π} ∫_{−∞}^{∞} H_{a}(ω′)e^{iω′t}dω′
= \frac{1}{2π} ∫_{−π}^{π} H_{a}(ω′)e^{iω′t}dω′

So

h_{k} = h_{a}(k) = \frac{1}{2π} ∫_{−π}^{π} H_{a}(ω′)e^{iω′k}dω′

Thus, the digital filter frequency response is

H(ω) = ∑_{k=−∞}^{∞} h_{k}e^{−iωk}

with h_{k} = \frac{1}{2π} ∫_{−π}^{π} H(e^{iω})e^{iωk}dω

On comparison, we see that

H(e^{iω}) = H_{a}(ω) |ω| ≤ π

(b) Using the fact that

L^{−1} (\frac{ω_{p}}{(s + ω_{p})^{2} + ω_{p}^{2}} ) = e^{−ω_{p}t} sin(ω_{p}t)u(t),

we have

L^{−1} (\frac{2ω_{p}^{2}}{(s + ω_{p})^{2} + ω_{p}^{2}} ) = 2ω_{p}e^{−ω_{p}t} sin(ω_{p}t)u(t)
= (−iω_{p}e^{(i−1)ω_{p}t} + iω_{p}e^{−(i+1)ω_{p}t})u(t),

by sin(ω_{p}t) = (e^{iω_{p}t} − e^{−iω_{p}t})/(2i). Then

h_{k} = (−iω_{p}e^{(i−1)ω_{p}k} + iω_{p}e^{−(i+1)ω_{p}k})u(k).

And

H(ω) = ∑_{k=0}^{∞} (−iω_{p}e^{(i−1)ω_{p}k} + iω_{p}e^{−(i+1)ω_{p}k})e^{−iωk}
= \frac{−iω_{p}}{1 − e^{(i−1)ω_{p}−iω}} + \frac{iω_{p}}{1 − e^{−(i+1)ω_{p}−iω}}
= \frac{2ω_{p} sin ω_{p}e^{−ω_{p}}e^{−iω}}{1 − 2 cos ω_{p}e^{−ω_{p}}e^{−iω} + e^{−2ω_{p}}e^{−2iω}}