mit aoe mcq z transform

Question 1 Z-transform of the sequence {f(k)} is F(z)=Option A

∑k=−∞

∞

f (k ) zk

Option B∑k=−∞

∞

f (k ) z−k

Option C∑k=−∞

∞

[ f (k ) ]−1zk

Option D∑k=−∞

∞

[ f (k ) ]−1z−k

Option ECorrect Answer BExplanation No ExplanationDifficulties 1Sub Topic Z-transform

Question 2 Z-transform of f ( k )=cos αk ;k ≥0 is

Option A z( z−cosα )z2−2 zcos α+1

;|z|>1

Option Bzsinα

z2−2 zcos α+1;|z|>1

Option Czcos α

z2−2 zcos α+1;|z|>1

Option D z (z−sinα )z2−2 zcos α+1

;|z|>1

Option E

Correct Answer A

Explanation(Optional) No Explanation

Difficulties 1

Sub Topic Z-transform

Question 3 Z-transform of f ( k )=sinαk ;k ≥0 is

Option A z( z−cosα )z2−2 zcos α+1

;|z|>1

Option B zsinα

z2−2 zcos α+1;|z|>1

Option C zcos α

z2−2 zcos α+1;|z|>1

Option D z (z−sinα )z2−2 zcos α+1

;|z|>1

Option E

Correct Answer B

Explanation No Explanation

Difficulties 1


Question 4 Z-transform of f ( k )=cosh αk ; k≥0 is

Option A zsinhα

z2−2 zcosh α+1;|z|>max {|ea|,|e−a|}

Option B zcosh α


Option C z (z−cosh α)z2−2 zcosh α+1

;|z|>max {|ea|,|e−a|}Option D z (z−sinhα)


Option E

Correct Answer C


Difficulties 1


Question 5 Z-transform of f ( k )=sinh αk ;k ≥0 is

Option A zsinhα


Option B zcosh α


Option C z (z−cosh α)z2−2 zcosh α+1

;|z|>max {|ea|,|e−a|}Option D z (z−sinhα)


Option E

Correct Answer A

Explanation No Explanation No Explanation

Difficulties 1


Question 6The inverse Z-transform of

z sinhα

z2−2 z cosh α+1is

Option Acosh αk ; k≥0

Option Bα sinhαk ;k ≥0

Option Csinhαk ;k ≥0

Option Dαcoshαk ; k≥0

Option E

Correct Answer C


Difficulties 1



z sinα

z2−2 z cosα+1is

Option A cos αk ;k ≥0

Option B k sinαk ;k ≥0

Option C kcos αk ; k≥0

Option D sinαk ;k ≥0

Option E

Correct Answer D


Difficulties 1



z( z−cosα )z2−2 zcos α+1

;|z|>1 is

Option A α sin k ;k ≥0

Option B cos αk ;k ≥0

Option C kcos α ;k ≥0

Option D αcos k ;k ≥0

Option E

Correct Answer B


Difficulties 1



z (z−cosh α)z2−2 zcosh α+1

;|z|>1 is

Option A coshα k ;k ≥0

Option B α sinhα ;k ≥0

Option C k cos α ;k ≥0

Option D α coshk ; k≥0

Option E

Correct Answer A


Difficulties

Sub Topic

Question 10 The Z-transform of unit impulse function is

Option A Z

Option B 1

Option C δ (k )

Option D 0

Option E

Correct Answer B


Difficulties 1


Question 11The Z-transform unit impulse function is δ (k ) = f ( k )={1 ,∧K=0

0 ,∧K≠0is

Option A k

Option B z

Option C 1

Option D 0

Option E

Correct Answer C


Difficulties 1


Question 12 The Z-transform of f ( k ) = cos3 k ; k ≥0 is

Option A z sin3

z2−2 zcos 3+1;|z|>1

Option B zcos 3

z2−2 zcos 3+1;|z|>1

Option C z ¿¿Option D z ¿¿Option E

Correct Answer D


Difficulties


Question 13 The Z-transform of f ( k ) = cosh 5k ; k ≥0 is

Option A zsinh 5

z2−2 zcosh 5+1;|z|>max {|e5|;|e−5|}

Option B z ¿¿Option C z cosh 5

z2−2 zcosh 5+1;|z|>max {|e5|;|e−5|}

Option D z ¿¿Option E

Correct Answer B


Difficulties 2


Question 14 The Z-transform of f ( k ) = sin 5k ; k ≥0 is

Option A zcos 5

z2−2 zcos 5+1;|z|>1

Option B z sin5

z2−2 zcos 5+1;|z|>1

Option C z ¿¿Option D

z (z−sin 3)z2−2 zcos 5+1

;|z|>1

Option E

Correct Answer B


Difficulties 2


Question 15 The Z-transform of f ( k ) = ak ; k ≥0 is

Option A zz−a

;|z|>|a|

Option B aa−z

;|z|>a

Option C azz−a

;|a|>|z|

Option D azz−a

;|z|>|a|

Option E

Correct Answer A


Difficulties 2


Question 16 The Z-transform of U (k )={0 ,∧k<0

1 ,∧k ≥0 is

Option A 11−z

;|z|<1

Option B z1−z

;|z|<1

Option C 1z−1

;|z|>1

Option D zz−1

;|z|>1

Option E

Correct Answer D


Difficulties 2


Question 17 The Z-transform of f ( k ) = ak ; k<0 is

Option A zz−a

;|z|>|a|

Option B 1z−a

;|z|>a

Option C za−z

;|z|<|a|

Option D 1a−z

;|z|<|a|

Option E

Correct Answer C


Difficulties 2


Question 18 The Z-transform of f ( k ) = 2k ; k ≥0 is

Option A 2 z2−z

;|z|<2

Option B 12−z

;|z|<2

Option C zz−2

;|z|>2

Option D 1z−2

;|z|>2

Option E

Correct Answer C


Difficulties 2


Question 19 The Z-transform of f ( k ) = 2k +1 ; k ≥0 is

Option A 2 z2−z

;|z|<2

Option B zz−2

;|z|>2

Option C 2 z2−z

;|z|>2

Option D z2−z

;|z|<2

Option E

Correct Answer C


Difficulties 2


Question 20The Z-transform of f ( k )=(2

3)k

; k≥0

Option A 2 z2 z−3

;|z|> 23

Option B z3 z−2

;|z|> 23

Option C 3 z3 z−2

;|z|> 23

Option D zz−3

;|z|>3

Option E

Correct Answer C


Difficulties

Sub Topic

Question 21 The Z-transform of f ( k )=1 ; k≥0

Option A z1−z

;|z|<1

Option B 1z−1

;|z|>1

Option C zz−1

;|z|>1

Option D 11−z

;|z|<1

Option E

Correct Answer C


Difficulties 1


Question 22 The Z-transform of f ( k )=e−ak ; k≥0

Option A z

z−e−a;|z|>|e−a|

Option B e−a z1−z

;|z|>|e−a|Option C ea

z−e−a;|z|>|ea|

Option D z

e−a−z;|z|>|e−a|

Option E

Correct Answer A


Difficulties 2


Question 23 The Z-transform of f ( k )=sin (3k+2) ; k≥0

Option A zsin 1

z2+2 zcos3+1Option B z sin (3k+2 )

z2−2 zcos(3k+2)+1Option C z(sin 1+z sin 2)

z2−2 zcos 3+1Option D z (sin 3+sin 2)

z2−2 zcos 3+1Option E

Correct Answer C


Difficulties 2


Question 24If Z (1 )= z

z−1;|z|>1 then Z-transform of f ( k )=k 2 ; k≥0 is

Option A z( z+1)(z−1)3

Option B z( z+1)(z−1)2

Option C z

(z−1)2

Option D z

(z−1)3

Option ECorrect Answer AExplanation No ExplanationDifficulties 2Sub Topic Z-transform

Question 25If Z (2¿¿k )= z

z−2;|z|>2¿ then Z-transform of f ( k )=k 2k ; k≥0 is

Option A z

(2−z)2

Option B z2

(z−2)Option C z2

(2−z)2

Option D 2 z

(2−z)2

Option ECorrect Answer DExplanation No ExplanationDifficulties 2Sub Topic Z-transform


z−5;|z|>5¿ then Z-transform of f ( k )=k 25k ; k≥0 is

Option A (z+5)(z−5)3

Option B 5 z (z+5)(z−5)3

Option C z

(z−5)4

Option D 5 z

(z−5)4



z−2;|z|>2¿ then Z-transform of f ( k )=2k

k; k≥1 is

Option Alog( z−2

z );|z|<2

Option Blog( z2 );|z|<2

Option Ctan−1( z2 );|z|<2

Option Dlog( z

z−2 );|z|>2

Option ECorrect Answer DExplanation No ExplanationDifficulties 2Sub Topic Z-transform

Question 28If Z (sin k )= z sin1

z2−2 zcos 1+1; then Z-transform of f ( k )= sin k

k; k≥0 is

Option Acot−1( sin 1

z−cos1 )Option B

cot−1( z−cos1sin 1 )

Option Ctan−1( z−cos1

sin1 )Option D

log( z−cos1sin 1 )


Question 29If Z [cos (3k+2 )]= z [ zcos 2−cos1]

z2−2 zcos3+1; then Z-transform of

[2kcos (3k+2)] isOption A z [zcos 2−2 cos1]

z2−4 zcos 3+4Option B z [cos 2−2cos1]

z2−4 z+1Option C z [zcos 2−cos1]

z2−4 zcos 3+4Option D z [zcos 2−cos1]

z2−4 z cos3+1Option ECorrect Answer AExplanation No ExplanationDifficulties 2Sub Topic Z-transform

Question 30If Z (k )= z

(z−1)2; then Z-transform of f ( k )=k e−ak; k≥0 is

Option A e−a z

( z−e−a )2

Option B z

( z−e−a )2

Option C e−a z

(e−a z−1 )2

Option D e−a z

( z+e−a )2

Option ECorrect Answer AExplanation No ExplanationDifficulties 2Sub Topic Z-transform

Question 31Option AOption BOption COption DOption ECorrect AnswerExplanation No ExplanationDifficultiesSub Topic


Question 38

Option AOption BOption COption DOption ECorrect AnswerExplanation No ExplanationDifficultiesSub Topic

Question 39Option AOption BOption C

Option DOption ECorrect AnswerExplanation No ExplanationDifficultiesSub Topic

Question 40Option AOption BOption COption DOption ECorrect Answer

Explanation No ExplanationDifficultiesSub Topic


Question 46Option AOption B

Option COption DOption ECorrect AnswerExplanation No ExplanationDifficultiesSub Topic

Question 47Option AOption BOption COption DOption E

Correct AnswerExplanation No ExplanationDifficultiesSub Topic

Question 48Option AOption BOption COption DOption ECorrect AnswerExplanation No ExplanationDifficulties

Sub Topic


Question 50 Z-transform of the sequence {f(k)} is F(z)=Option A

∑k=−∞

∞

f (k ) zk

Option B∑k=−∞

∞

f (k ) z−k

Option C∑k=−∞

∞

[ f (k ) ]−1zk

Option D∑k=−∞

∞

[ f (k ) ]−1z−k


mit aoe mcq z transform

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