SIGNALS AND SYSTEMS II Assignment 6 Solutions

1. (a)
X(e^{iω}) = ∑_{k=−∞}^{∞} x_{k}e^{−iωk}

= ∑_{k=−∞}^{−1} (\frac{1}{3})^{−k}e^{−iωk} + ∑_{k=0}^{∞} (\frac{1}{2})^{k}e^{−iωk}
= ∑_{k=1}^{∞} (\frac{1}{3}e^{iω})^{k} + ∑_{k=0}^{∞} (\frac{1}{2}e^{−iω})^{k}
= \frac{\frac{1}{3}e^{iω}}{1 − \frac{1}{3}e^{iω}} + \frac{1}{1 − \frac{1}{2}e^{−iω}}
= −\frac{5e^{iω}}{2(e^{iω} − 3)(e^{iω} − \frac{1}{2})}

(b) Fourier inversion is given by

x_{k} = \frac{1}{2π} ∫_{2π} X(e^{iω})e^{iωk}dω  (1)
= \frac{1}{2π} ∫_{2π} −\frac{5e^{iω(k+1)}}{2(e^{iω} − 3)(e^{iω} − \frac{1}{2})}dω  (2)

Let z = e^{iω}, dz = izdω. Equation 2 can be rewritten as

x_{k} = \frac{1}{2πi} ∮_{c} X(z)z^{k−1}dz
= \frac{1}{2πi} ∮_{c} −\frac{5z^{k}}{2(z − 3)(z − \frac{1}{2})}dz
:= \frac{1}{2πi} ∮_{c} f(z)dz

where c is the counterclockwise unit circle. Now we can use Residue Theorem to carry out the above integral. (i). k ≥ 0 Inside the unit circle, there is only one pole at z_{o} = \frac{1}{2}. Thus

x_{k} = [−\frac{5z^{k}}{2(z − 3)}]_{z= \frac{1}{2}}
= (\frac{1}{2})^{k}, k ≥ 0

(ii).k < 0 Now there are a pole at z_{o} = \frac{1}{2} and a pole of order −k at the origin, inside the unit circle. It is not easy to calculate the residue at the origin. Let v = \frac{1}{z}, dv = −\frac{1}{z^{2}} dz = −v^{2}dz. Notice that the integral path c′ in v-plane is the clockwise unit circle ^{1}.

x_{k} = \frac{1}{2πi} ∮_{c′} f(\frac{1}{v})(−v^{2})dv
= −\frac{1}{2πi} ∮_{c′} f(\frac{1}{v})v^{2}dv
= \frac{1}{2πi} ∮_{c} f(\frac{1}{v})v^{2}dv
= \frac{1}{2πi} ∮_{c} −\frac{5v^{−k}}{3(v − \frac{1}{3})(v − 2)}dv

Then by Residue Theorem, we have

x_{k} = [−\frac{5v^{−k}}{3(v − 2)}]_{v= \frac{1}{3}} = (\frac{1}{3})^{−k}, k < 0,

since there is only one pole of order 1 at v_{o} = \frac{1}{3} inside the unit circle.

2. (a)
x_{k} = \frac{1}{2πi} ∮ X(z)z^{k−1}dz
= \frac{1}{2πi} ∮ \frac{z^{k+1}}{(z − a)(z − b)}dz
= [\frac{z^{k+1}}{z − b}]_{z=a} + [\frac{z^{k+1}}{z − a}]_{z=b}
= \frac{a^{k+1} − b^{k+1}}{a − b} , k ≥ 0

(b)

X(z) = α_{1} \frac{1}{z^{−1} − a^{−1}} + α_{2} \frac{1}{z^{−1} − b^{−1}}
= \frac{−a · α_{1}}{1 − az^{−1}} + \frac{−b · α_{2}}{1 − bz^{−1}}
= \frac{(−aα_{1} − bα_{2}) + (abα_{1} + abα_{2})z^{−1}}{(1 − az^{−1})(1 − bz^{−1})}
= \frac{1}{(1 − az^{−1})(1 − bz^{−1})}

1At path c, z = e^{jω} goes counterclockwise when ω goes from 0 to 2π. Now at path c′ in v-plane, v = \frac{1}{z} = e^{−jω} must go clockwise along the unit circle.

Thus, α_{1} = \frac{1}{b−a}, α_{2} = \frac{1}{a−b}. And

x_{1}(k) = Z^{−1} [α_{1} \frac{1}{z^{−1} − a^{−1}} ]
= Z^{−1} [\frac{a}{a − b} · \frac{1}{1 − az^{−}1}]
= \frac{a}{a − b} · a^{k}
= \frac{a^{k+1}}{a − b}, k ≥ 0
x_{2}(k) = Z^{−1} [α_{2} \frac{1}{z^{−1} − b^{−1}} ]
= \frac{b^{k+1}}{b − a}, k ≥ 0

Therefore, x_{k} = x_{1}(k) + x_{2}(k) = \frac{a^{k+1}−b^{k+1}}{a−b} , k ≥ 0

3. (a) i. {kx_{k}}

X(z) = ∑_{k=0}^{∞} x_{k}z^{−k}
\frac{dX(z)}{dz} = − ∑_{k=0}^{∞} kx_{k}z^{−k−1}
−z · \frac{dX(z)}{dz} = ∑_{k=0}^{∞} kx_{k}z^{−k}
= Z[kx_{k}]

ii. {k^{2}x_{k}} Already know

−z · \frac{dX(z)}{dz} = ∑_{k=0}^{∞} kx_{k}z^{−k}

Continue the (−z) · \frac{d}{dz} operation on both sides, we have

−\frac{dX(z)}{dz} − z \frac{d^{2}X(z)}{dz^{2}} = − ∑_{k=0}^{∞} k^{2}x_{k}z^{−k−1}
z \frac{dX(z)}{dz} + z^{2} \frac{d^{2}X(z)}{dz^{2}} = ∑_{k=0}^{∞} k^{2}x_{k}z^{−k}
= Z[k^{2}x_{k}]

(b) By definition, Z[x_{k}y_{k}] = ∑_{k=0}^{∞} x_{k}y_{k}z^{−k}. If |z| > R_{y}, then Y (z) exists. Rewrite the above equation by substituting the inverse z-transform of y_{k} for y_{k}, we get

Z[x_{k}y_{k}] = ∑_{k=0}^{∞} x_{k} [\frac{1}{2πi} ∮ Y (v)v^{k−1}dv] z^{−k}

= \frac{1}{2πi} ∮ Y (v) ∑_{k=0}^{∞} (x_{k}z^{−k}v^{k−1})dv
= \frac{1}{2πi} ∮ \frac{Y (v)}{v} · ∑_{k=0}^{∞} x_{k}(\frac{z}{v})^{−k}dv

If |\frac{z}{v}| > R_{x}, then X(\frac{z}{v}) exists. We have

Z[x_{k}y_{k}] = \frac{1}{2πi} ∮ X(\frac{z}{v})Y (v)\frac{dv}{v} .

For the existence of Y (v), we need |v| > R_{y}. For the existence of X(\frac{z}{v}), we need |\frac{z}{v}| > R_{x}. Thus the ROC at z-plane is |z| > R_{x}R_{y}.

(c) The ROC for Z[k] is |z| > 1. By the result in 3(b), the ROC for Z[kx_{k}] is |z| > R_{x} · 1 = R_{x}. The ROC for Z[k^{2}] is |z| > 1. By the result in 3(b), the ROC for Z[k^{2}x_{k}] is |z| > R_{x} · 1 = R_{x}.

4. (a)
X(z) = ∑_{k=−∞}^{∞} x_{k}z^{−k}

= ∑_{k=−∞}^{−1} x_{k}z^{−k} + ∑_{k=0}^{∞} x_{k}z^{−k}

= ∑_{k=−∞}^{−1} 3^{k}z^{−k} + ∑_{k=0}^{∞} 2^{k}z^{−k}

= ∑_{k=−∞}^{−1} (\frac{z}{3})^{−k} + ∑_{k=0}^{∞} (2z^{−1})^{k}
= ∑_{k=1}^{∞} (\frac{z}{3})^{k} + ∑_{k=0}^{∞} (2z^{−1})^{k}
= \frac{\frac{1}{3}z}{1 − \frac{1}{3}z} + \frac{1}{1 − 2z^{−1}} , if |\frac{z}{3}| < 1 & |2z^{−1}| < 1
= \frac{z}{3 − z} + \frac{z}{z − 2}

The ROC is 2 < |z| < 3 from |\frac{z}{3}| < 1 & |2z^{−1}| < 1.

(b) i. |z| < 4.

X(z) = \frac{1}{z − 4}
= −\frac{1}{4} · \frac{1}{1 − \frac{z}{4}}

= −\frac{1}{4} · ∑_{k=0}^{∞} (\frac{z}{4})^{k}
= − ∑_{k=0}^{∞} (\frac{1}{4})^{k+1}z^{k}

= − ∑_{k=−∞}^{0} (\frac{1}{4})^{−k+1}z^{−k}

= ∑_{k=−∞}^{0} (−4^{k−1})z^{−k}

Therefore, from the definition of bilateral z-transform,

x_{k} = \{ \begin{matrix}−4^{k−1}, & k ≤ 0 \\ 0 & k > 0 \end{matrix}

ii. |z| > 4.

X(z) = \frac{1}{z − 4}
= \frac{\frac{1}{z}}{1 − \frac{4}{z}}
= ∑_{k=0}^{∞} \frac{1}{z}(\frac{4}{z})^{k}
= ∑_{k=0}^{∞} 4^{k}z^{−k−1}
= ∑_{k=1}^{∞} 4^{k−1}z^{−k}

Therefore, from the definition of bilateral z-transform,

x_{k} = \{ \begin{matrix}0 & k < 1 \\ 4^{k−1} & k ≥ 1 \end{matrix}