Calculate the Z transform of the sequences1) (−15)k2) (cos⁡kπ)3) 3k

1) (−15)kf(k)=(−15)kTherefore,Z{f(k)}=Z{(−15)k}=∑k=0∞(−15)kz−k=1−15z+(−15z)2−…⇒Z{(−15)k}=11−(−15z)=5z5z+1,|z|&gt;15Thus, the answer is 5z5z+1 and the region of convergence is |z|≻152) (cos⁡kπ)Z{f(k)}=Z{cos⁡kπ}=∑k=0∞cos⁡(kπ)z−k⇒{cos⁡(kπ)}=∑k=0∞[eiπk+e−iπk2]z−k=12∑k=0∞eiπkz−k+12∑k=0∞e−iπkz−k=12∑k=0∞(eiπz)k+12∑k=0∞(e−iπz)k=1211−eiπz+1211−e−iπz=1211−(−1)z+1211−(−1)z=12zz+1+12zz+1=zz+1,|z|≻1Thus, the answer is zz+1, and the region of convergence is |z|≻1

3. 3kZ-form is given by Z{f(k)}=Z{3k}=∑k=0∞3kz−kZ{f(k)}=3∑k=0∞kzk=3z(1+2z+3z2+…)=3z(1−1z)−2=3z(z−1z)−2=3z(z−1)2 (Answer)Region of convergence |z|&gt;1

1) (−15)kf(k)=(−15)kTherefore,Z{f(k)}=Z{(−15)k}=∑k=0∞(−15)kz−k=1−15z+(−15z)2−…⇒Z{(−15)k}=11−(−15z)=5z5z+1,|z|&gt;15Thus, the answer is 5z5z+1 and the region of convergence is |z|≻152) (cos⁡kπ)Z{f(k)}=Z{cos⁡kπ}=∑k=0∞cos⁡(kπ)z−k⇒{cos⁡(kπ)}=∑k=0∞[eiπk+e−iπk2]z−k=12∑k=0∞eiπkz−k+12∑k=0∞e−iπkz−k=12∑k=0∞(eiπz)k+12∑k=0∞(e−iπz)k=1211−eiπz+1211−e−iπz=1211−(−1)z+1211−(−1)z=12zz+1+12zz+1=zz+1,|z|≻1Thus, the answer is zz+1, and the region of convergence is |z|≻1

3. 3kZ-form is given by Z{f(k)}=Z{3k}=∑k=0∞3kz−kZ{f(k)}=3∑k=0∞kzk=3z(1+2z+3z2+…)=3z(1−1z)−2=3z(z−1z)−2=3z(z−1)2 (Answer)Region of convergence |z|&gt;1

Key to "Calculate the Z transform of the sequences1) (−15)k2)..."

Ella Maddox

Answered question

2022-03-18

Calculate the Z transform of the sequences
1)

{(- \frac{1}{5})}^{k}

(\cos k π)

3 k

Answer & Explanation

diesel817637dsf

Beginner2022-03-19Added 13 answers

{(- \frac{1}{5})}^{k}

f (k) = {(- \frac{1}{5})}^{k}

Therefore,

Z {f (k)} = Z {{(- \frac{1}{5})}^{k}} = \sum_{k = 0}^{\infty} {(- \frac{1}{5})}^{k} z^{- k}

= 1 - \frac{1}{5 z} + {(- \frac{1}{5 z})}^{2} - \dots

\Rightarrow Z {{(- \frac{1}{5})}^{k}} = \frac{1}{1 - (- \frac{1}{5 z})}

= \frac{5 z}{5 z + 1}, | z | > \frac{1}{5}

Thus, the answer is

\frac{5 z}{5 z + 1}

and the region of convergence is

| z | ≻ \frac{1}{5}

(\cos k π)

Z {f (k)} = Z {\cos k π} = \sum_{k = 0}^{\infty} \cos (k π) z^{- k}

\Rightarrow {\cos (k π)} = \sum_{k = 0}^{\infty} [\frac{e^{i π k} + e^{- i π k}}{2}] z^{- k}

= \frac{1}{2} \sum_{k = 0}^{\infty} e^{i π k} z^{- k} + \frac{1}{2} \sum_{k = 0}^{\infty} e^{- i π k} z^{- k}

= \frac{1}{2} \sum_{k = 0}^{\infty} {(\frac{e^{i π}}{z})}^{k} + \frac{1}{2} \sum_{k = 0}^{\infty} {(\frac{e^{- i π}}{z})}^{k}

= \frac{1}{2} \frac{1}{1 - \frac{e^{i π}}{z}} + \frac{1}{2} \frac{1}{1 - \frac{e^{- i π}}{z}}

= \frac{1}{2} \frac{1}{1 - \frac{(- 1)}{z}} + \frac{1}{2} \frac{1}{1 - \frac{(- 1)}{z}}

= \frac{1}{2} \frac{z}{z + 1} + \frac{1}{2} \frac{z}{z + 1}

= \frac{z}{z + 1}, | z | ≻ 1

Thus, the answer is

\frac{z}{z + 1}

, and the region of convergence is

| z | ≻ 1

orangepaperiz7

Beginner2022-03-20Added 9 answers

3 k

Z-form is given by

Z {f (k)} = Z {3 k} = \sum_{k = 0}^{\infty} 3 k z^{- k}

Z {f (k)} = 3 \sum_{k = 0}^{\infty} \frac{k}{z^{k}}

= \frac{3}{z} (1 + \frac{2}{z} + \frac{3}{z^{2}} + \dots)

= \frac{3}{z} {(1 - \frac{1}{z})}^{- 2}

= \frac{3}{z} {(\frac{z - 1}{z})}^{- 2}

= \frac{3 z}{{(z - 1)}^{2}}

(Answer)
Region of convergence

| z | > 1

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