Most Latest Signals and Systems MCQs – Z-Transform and Digital Filtering MCQs ( Signals and Systems ) MCQs

Most Latest Signals and Systems MCQs – Z-Transform and Digital Filtering MCQs ( Signals and Systems ) MCQs

Latest Signals and Systems MCQs

By practicing these MCQs of Z-Transform and Digital Filtering MCQs ( Signals and Systems ) MCQs – Latest Competitive MCQs , an individual for exams performs better than before. This post comprising of objective questions and answers related to Z-Transform and Digital Filtering MCQs ( Signals and Systems ) Mcqs “. As wise people believe “Perfect Practice make a Man Perfect”. It is therefore practice these mcqs of Signals and Systems to approach the success. Tab this page to check ” Z-Transform and Digital Filtering MCQs ( Signals and Systems )” for the preparation of competitive mcqs, FPSC mcqs, PPSC mcqs, SPSC mcqs, KPPSC mcqs, AJKPSC mcqs, BPSC mcqs, NTS mcqs, PTS mcqs, OTS mcqs, Atomic Energy mcqs, Pak Army mcqs, Pak Navy mcqs, CTS mcqs, ETEA mcqs and others.

Signals and Systems MCQs – Z-Transform and Digital Filtering MCQs ( Signals and Systems ) MCQs

The most occurred mcqs of Z-Transform and Digital Filtering MCQs ( Signals and Systems ) in past papers. Past papers of Z-Transform and Digital Filtering MCQs ( Signals and Systems ) Mcqs. Past papers of Z-Transform and Digital Filtering MCQs ( Signals and Systems ) Mcqs . Mcqs are the necessary part of any competitive / job related exams. The Mcqs having specific numbers in any written test. It is therefore everyone have to learn / remember the related Z-Transform and Digital Filtering MCQs ( Signals and Systems ) Mcqs. The Important series of Z-Transform and Digital Filtering MCQs ( Signals and Systems ) Mcqs are given below:

The Z-Transform

1. When do DTFT and ZT are equal?
a) When σ = 0
b) When r = 1
c) When σ = 1
d) When r = 0
Answer: b
Explanation: Discrete Time Fourier Transform, X(e-jω) = \(\sum_{n=-∞}^∞ x(n) e^{-jωn}\)
Z-Transform, X(Z) = \(∑_{n=-∞}^∞ x(n) z^{-n}\), z = r e
When r=1, z = e and hence DTFT and ZT are equal.


2. Find the Z-transform of δ(n+3).
a) z
b) z2
c) 1
d) z3
Answer: d
Explanation: Given x(n) = δ(n+3)
We know that δ(n+3) = \( \begin{cases}
1 &\text{\(n=-3\)} \\
0 &\text{otherwise} \\
\end{cases}\)
X(Z) = \(\sum\limits_{n=-\infty}^{\infty} x(n) z^{-n} = \sum\limits_{n=-\infty}^{\infty} δ(n+3) z^{-n}\) = z3.


3. Find the Z-transform of an u(n);a>0.
a) \(\frac{z}{z-a}\)
b) \(\frac{z}{z+a}\)
c) \(\frac{1}{1-az}\)
d) \(\frac{1}{1+az}\)
Answer: a
Explanation: Given x(n) = an u(n)
We know that \( u(n)=\begin{cases}
1 &\text{\(n≥0\)} \\
0 &\text{\(n<0\)} \\
\end{cases}\)
X(Z) = \(\sum\limits_{n=-\infty}^{\infty} x(n) z^{-n} = \sum\limits_{n=-\infty}^{\infty} a^n u(n) z^{-n}\)
= \(\sum\limits_{n=0}^{∞} a^n (1) z^{-n} = \sum\limits_{n=0}^{∞} (az^{-1})^n = (1-az^{-1})^{-1}\)
= \(\frac{1}{1-az^{-1}} = \frac{z}{z-a}\).


4. Find the Z-transform of cos⁡ωn u(n).
a) \(\frac{z(z+cos⁡ω)}{z^2-2z cos⁡ω+1}\)
b) \(\frac{z(z-cos⁡ω)}{z^2-2z cos⁡ω+1}\)
c) \(\frac{z(z-cos⁡ω)}{z^2+2z cos⁡ω+1}\)
d) \(\frac{z(z+cos⁡ω)}{z^2+2z cos⁡ω+1}\)
Answer: b
Explanation: Given x(n) = cos⁡ωn u(n)
We know that \( u(n)=\begin{cases}
1 &\text{\(n≥0\)} \\
0 &\text{\(n<0\)} \\
\end{cases}\)
Z[cos⁡ωn u(n)] = \(Z\Big[\frac{e^jωn+e^{-jωn}}{2} u(n)\Big] = \frac{1}{2} Z[e^{jωn} u(n)] + \frac{1}{2} Z[e^{-jωn} u(n)]\)
\(= \frac{1}{2} \left(\frac{z}{z-e^{jω}} + \frac{z}{z-e^{-jω}}\right) = \frac{1}{2} \Big[\frac{z(z-e^{-jω}) + z(z-e^{jω})}{(z-e^{jω})(z-e^{-jω})}\Big]\)
\(= \frac{1}{2} \Big\{\frac{z[2z-(e^{jω}+e^{-jω})]}{z^2-z(e^{jω}+e^{-jω})+1}\Big\} = \frac{z(z-cos⁡ω)}{z^2-2z cos⁡ω+1}\).


5. For causal sequences, the ROC is the exterior of a circle of radius r.
a) True
b) False
Answer: a
Explanation: Consider a causal sequence, x(n) = rn u(n)
X(Z) = \(\sum\limits_{n=-∞}^{∞} x(n) z^{-n} = \sum\limits_{n=-∞}^{∞} r^n u(n) z^{-n} = \sum\limits_{n=0}^{∞} r^n (1) z^{-n} = \sum\limits_{n=0}^{∞} (rz^{-1})^n\)
The above summation converges for |rz-1|<1, i.e. for |z|>r
Hence, for the causal sequences, the ROC is the exterior of a circle of radius r.


6. x(n) = an u(n) and x(n) = -an u(-n-1) have the same X(Z) and ROC.
a) True
b) False
Answer: b
Explanation: an u(n) ↔ \(\frac{1}{1-az^{-1}} = \frac{z}{z-a}\); ROC:|z|>|a|
-an u(-n-1) ↔ \(\frac{1}{1-az^{-1}} = \frac{z}{z-a}\); ROC:|z|<|a|
Hence, x(n) = an u(n) and x(n) = -an u(-n-1) have the same X(Z) and differ only in ROC.


7. Find the Z-transform of y(n) = x(n+2)u(n).
a) z2 X(Z) – z2 x(0) – zx(1)
b) z2 X(Z) + z2 x(0) – zx(1)
c) z2 X(Z) – z2 x(0) + zx(1)
d) z2 X(Z) + z2 x(0) + zx(1)
Answer: a
Explanation: Given y(n) = x(n+2)u(n)
Y(z) = Z[y(n)] = Z[x(n+2)u(n)] = \(\sum\limits_{n=0}^{∞} x(n+2)u(n) z^{-n} = \sum\limits_{n=0}^{∞} x(n+2)z^{-n}\)
Let n + 2 = p,i.e.n = p – 2
Y(z) = \(∑_{p=2}^∞ x(p)z^{-(p-2)} = z^2 ∑_{p=2}^∞ x(p)z^{-p} = z^2 ∑_{p=0}^∞ x(p)z^{-p} – x(0) – x(1) z^{-1}\)
=z2 X(Z) – z2 x(0) – zx(1).


8. Find the Z-transform of x(n) = a|n|; |a|<1.
a) \(\frac{z}{z-a} – \frac{z}{z-(1/a)}\)
b) \(\frac{z}{z-(1/a)} – \frac{z}{z-a}\)
c) \(\frac{z}{z-a} + \frac{z}{z-(1/a)}\)
d) \(\frac{1}{z-a} – \frac{1}{z-(1/a)}\)
Answer: a
Explanation: a^|n| = a^n u(n) + a-n u(-n-1) = an u(n) + \((\frac{1}{a})^n\) u(-n-1)
Z[a|n|] = Z[an u(n)] + Z[\((\frac{1}{a})^n\) u(-n-1)] = \(\frac{z}{z-a} – \frac{z}{z-(1/a)}\).


9. Find the Z-transform of u(-n).
a) \(\frac{1}{1-z}\)
b) \(\frac{1}{1+z}\)
c) \(\frac{z}{1-z}\)
d) \(\frac{z}{1+z}\)
Answer: a
Explanation: Given x(n) = u(-n)
Z[x(n)] = X(Z) = \(∑_{n=-∞}^∞ x(n) z^{-n} = ∑_{n=-∞}^∞ u(-n) z^{-n} = ∑_{n=-∞}^0 (1) z^{-n}\)
=\(∑_{n=0}^∞ z^n = \frac{1}{1-z}\).


10. For a right hand sequence, the ROC is entire z-plane.
a) True
b) False
Answer: b
Explanation: If x(n) is finite duration causal sequence (right-sided sequence),the X(z) converges for all values of z except at z = 0. Hence, the ROC is entire z-plane except at z = 0.

Properties of Z-Transforms – 1

1. The z-transform of δ[n-k]>0 is __________
a) Zk, Z>0
b) Z-k, Z>0
c) Zk, Z≠0
d) Z-k, Z≠0
Answer: d
Explanation: Performing z-transform on δ[n-k], we get,
X (z) = n=0x[n]zn
= Z-k, Z≠0.


2. The z-transform of δ[n+k]>0 is __________
a) Z-k, Z≠0
b) Zk, Z≠0
c) Z-k, all Z
d) Zk, all Z
Answer: d
Explanation: Performing z-transform on δ[n+k], we get,
X (z) = n=0x[n]zn
= Zk, all Z


3. The z-transform of u[n] is _________
a) 11z1, |Z|>1
b) 11z1, |Z|<1
c) z1z1, |Z|<1
d) z1z1, |Z|>1
Answer: a
Explanation: Performing z-transform on u[n], we get,
X (z) = n=0x[n]zn
11z1, |Z|>1.


4. The z-transform of (14)n (u[n] – u[n-5]) is __________
a) z50.255z4(z0.25), z>0.25
b) z50.255z4(z0.25), z>0
c) z50.255z3(z0.25), z<0.25
d) z50.255z4(z0.25), all z
Answer: d
Explanation: X (z) = 4n=0(14z1)4
1(14z1)51(14z1)1
z50.255z4(z0.25), all z.


5. The z-transform of (14)4 u[-n] is ___________
a) 4z4z1, |Z|>14
b) 4z4z1, |Z|<14
c) 114z, |Z|>14
d) 114z, |Z|<14
Answer: d
Explanation: X (z) = 0n=(14z1)n
n=0(4z)n
114z, |Z|<14.


6. The z-transform of 3n u[-n-1] is ___________
a) z3z, |Z|>3
b) z3z, |Z|<3
c) 33z, |Z|>3
d) 33z, |Z|<3
Answer: b
Explanation: X (z) = 1n=(3z1)n
n=1(z13)n
13z113z, |z|<3
z3z.


7. The z-transform of (23)[n] is ____________
a) 5z(2z3)(3z2), –32 < z < –23
b) 5z(2z3)(3z2)23 < |z| < 32
c) 5z(2z3)(3z2)23 < |z|
d) 5z(2z3)(3z2), –32 < z< –23
Answer: b
Explanation: X(z) = 1(32z1)n+n=0(23z1)n
1(132z1)+1(123z1)
5z(2z3)(3z2)23 < |z| < 32.

 

Laplace Transform And System Design MCQs




8. The z-transform of cos(π3 n) u[n] is __________
a) z2(2z1)(z2z+1), 0<|z|<1
b) z2(2z1)(z2z+1), |z|>1
c) z2(12z)(z2z+1), 0<|z|<1
d) z2(12z)(z2z+1), |z|>1
Answer: b
Explanation: Performing z-transform on anu[n], we get zza
∴ x[n] = 12ej(π3)nu[n]+12ej(π3)nu[n]
Hence, X (z) =0.5(11ej(π3)z1+11ej(π3)z1)
Hence, X (z) = z2(2z1)(z2z+1), |z|>1.


9. The z-transform of {3,0,0,0,0,6,1,-4} (1 as the reference variable) is ___________
a) 3z5 + 6 + z-1 – 4z-2, 0≤|z|<∞
b) 3z5 + 6 + z-1 – 4z-2, 0<|z|<∞
c) 3z5 + 6 + z – 4z-2 0<|z|<∞
d) 3z5 + 6 + z-1 – 4z-2, 0≤|z|<∞
Answer: b
Explanation: Performing z-transform on x (n+n0), we get zn0 X (z)
Now, x[n] = 3δ[n+5] + 6δ[n] + δ[n-1] – 4δ[n-2]
So, X (z) = 3z5 + 6 + z-1 – 4z-2, 0<|z|<∞.


10. The z-transform of x[n]= {2,4,5,7,0,1} (5 as the reference variable) is ___________
a) 2z2 + 4z + 5 +7z + z3, z≠∞
b) 2z-2 + 4z-1 + 5 + 7z + z3, z≠∞
c) 2z-2 + 4z-1 + 5 + 7z + z3, 0<|z|<∞
d) 2z2 + 4z + 5 + 7z-1 + z3, 0<|z|<∞
Answer: d
Explanation: Performing z-transform on x (n+n0), we get zn0 X (z)
Now, x[n] = 2δ[n+2] + 4δ[n+1] + 5δ[n] + 7δ[n-1] + δ[n-3]
So, X (z) = 2z2 + 4z + 5 + 7z-1 + z3, 0<|z|<∞.


11. The z-transform of x[n]= {1,0,-1,0,1,-1} (1st 1 as the reference variable) is __________
a) 1+2z-2 -4 z-4 + 5z-5
b) 1-z-2 + z-4 – z-5
c) 1-2z2 + 4z4 – 5z5
d) 1-z2 + z4 – z5
Answer: b
Explanation: Performing z-transform on x (n-n0), we get z-n0 X (z)
Now, x[n] = δ[n] – δ[n-2] + δ[n-4] – δ[n-5]
So, X (z) = 1-z-2 + z-4 – z-5, z≠0.


12. Given the z-transform pair
X[n]32z216, |z|<4
The z-transform of the signal x [n-2] is _________
a) z4z216
b) (z+2)2(z+2)216
c) 1z216
d) (z2)2(z2)216
Answer: c
Explanation: Performing z-transform on x (n-n0), we get zn0 X (z)
Now, z-transform of y[n] = x [n-2] is given by,
Y (z) = z-2 X (z)
1z216.


13. Given the z-transform pair
X[n]32z216, |z|<4
The z-transform of the signal y[n] = 12n x[n] is _________
a) (z+2)2(z+2)216
b) z2z24
c) (z2)2(z2)216
d) z2z264
Answer: b
Explanation: y[n] = 12n x[n] Performing z-transform on y[n], we get, Y (z) = X (2z)
∴ Y(z) = z2z24.


14. Given the z-transform pair
X[n]32z216, |z|<4
The z-transform of the signal x [-n]*x[n] is ____________
a) z216z2257z416
b) 16z2(z216)2
c) z2(257z216z416)
d) 16z2(z216)2
Answer: c
Explanation: y[n] = x [-n]*x[n] Performing z-transform on y[n], we get, Y (z) = X (1z) X (z)
∴ X(1z) ↔ x [-n].


15. Given the z-transform pair
X[n]32z216, |z|<4
The z-transform of the signal x[n]*x [n-3] is __________
a) z3(z216)2
b) z7(z216)2
c) z5(z216)2
d) z(z216)2
Answer: d
Explanation: y[n] = x[n]*x [n-3] Performing z-transform on y[n], we get, Y (z) = X (z) z-3X (z)
Or, Y (z) = z(z216)2.

Properties of Z-Transforms – 2

1.Find the Z-transform of the causal sequence x(n) = {1,0,-2,3,5,4}. (1 as the reference variable)
a) 1 – 2z-2 + 3z-3 + 5z-4 + 4z-5
b) 1 – 2z2 + 3z3 + 5z4 + 4z5
c) z-1 – 2z2 + 3z3 + 5z4 + 4z5
d) z – 2z3 + 3z4 + 5z5 + 4z6
Answer: a
Explanation: Given sequence values are :
x(0)=1, x(1)=0, x(2)=-2, x(3)=3, x(4)=5, x(5)=4.
We know that
X(Z)=n=x(n)zn
X(Z) = x(0) + x(1) z-1 + x(2) z-2 + x(3) z-3 + x(4) z-4 + x(5) z-5
X(Z) = 1 – 2z-2 + 3z-3 + 5z-4 + 4z-5.


2. Find the Z-transform of the anticausal sequence x(n) = {4,2,3,-1,-2,1}. (1 as the reference variable)
a) 4z5 + 2z4 + 3z3 – z2 – 2z + 1
b) 4z-5 + 2z-4 + 3z-3 -z-2 – 2z-1 + 1
c) -4z5 – 2z4 – 3z3 + z2 + 2z – 1
d) -4z-5 – 2z-4 – 3z-3 + z-2 + 2z-1 – 1
Answer: a
Explanation: Given sequence values are :
x(-5)=4, x(-4)=2, x(-3)=3, x(-2)=-1, x(-1)=-2, x(0)=1
We know that
X(Z)=n=x(n)zn
X(Z) = x(-5) z5 + x(-4) z4 + x(-3) z3 + x(-2) z2 + x(-1)z + x(0)
X(Z) = 4z5 + 2z4 + 3z3 – z2 – 2z + 1.


3. Find the Z-transform of x(n) = u(-n).
a) 1z1
b) 1z+1
c) 11z
d) 1z+1
Answer: c
Explanation: Given x(n) = u(-n)
Time reversal property of Z-transform states that
If x(n) ↔ X(z), then x(-n) ↔ X(1z)
Z[u(-n)] = (1z1)z=(1/z)=1/z(1/z)1=11z.


4. Find the Z-transform of x(n) = u(-n-2).
a) z2z1
b) z21z
c) z21+z
d) z22z
Answer: b
Explanation: Given x(n) = u(-n-2)
Time shifting property of Z-transform states that
If x(n) ↔ X(z), then x(n-m) ↔ z-m X(z)
Z[u(-n-2)] = Z{u[-(n+2)]}=z2 Z[u(-n)] = z21z.

 

Sampling Theorem MCQs




5. Find the Z-transform of x(n) = n2 u(n).
a) z(z1)(z1)3
b) z(z+1)(z1)3
c) z(z+1)(z+1)3
d) z(z1)(z+1)3
Answer: b
Explanation: Given x(n) = n2 u(n)
We know that X(z) = Z[x(n)] = Z[u(n)] = z1z
The multiplication of n or differentiation in z-domain property of Z-transform states that
If x(n) ↔ X(z), then nk x(n) ↔ (-1)k zk dkX(z)dzk
Z[n2 u(n)] = z2 d2X(z)dz2=z2d2dz2[z1z]=z(z+1)(z1)3.


6. Find the Z-transform of x(n) = 2n u(n-2).
a) zz2
b) zz+2
c) zz(z2)
d) 4z(z2)
Answer: d
Explanation: Given x(n) = 2n u(n-2)
Time shifting property of Z-transform states that
If x(n) ↔ X(z), then x(n-m) ↔ z-m X(z)
Z[u(n-2)] = z-2 Z[u(n)] = z2zz1=1z(z1)
The multiplication by an exponential sequence property of Z-transform states that
If x(n) ↔ X(z), then an x(n) ↔ X(z/a)
Z[2n u(n-2)] = Z[u(n-2)]|z=(z/2) = [1z(z1)]z=(z/2)
=1(z/2)[(z/2)1]=4z(z2).


7. Find the Z-transform of x(n) = n[an u(n)].
a) zz(za)
b) azz(za)
c) azz(z+a)
d) az(za)2
Answer: d
Explanation: Given x(n) = n[an u(n)]
We know that an u(n) ↔ zza
Time differentiation property states that
If x(n) ↔ X(z), then nx(n) ↔ -z dX(z)dz
Z[x(n)] = Z{n[an u(n)]} = -z dX(z)dz=zddz[zza]=az(za)2.


8. Find the Z-transform of x(n) = (12)n u(n)*(14)n u(n).
a) zz(1/2)zz(1/4)
b) zz(1/2)+zz(1/4)
c) zz+(1/2)zz(1/4)
d) zz(1/2)zz+(1/4)
Answer: a
Explanation: We know that an u(n) ↔ zza
Let x1 (n)=(12)n u(n) and x2 (n) = (14)n u(n)
∴X1 (z) = zz(1/2) and X2 (z) = zz(1/4)
Given x(n) = x1 (n) * x2 (n)
The convolution property of Z-transform states that
x1 (n) * x2 (n) ↔ X1 (z) X2 (z)
∴Z[x(n)] = X(z) = Z[x1 (n)*x2 (n)] = X1 (z) X2 (z) = zz(1/2)zz(1/4).


9. Find x(∞) if X(z) = Z+1(z0.6)2.
a) 1
b) 0
c) ∞
d) 0.6
Answer: b
Explanation: Given X(z) = Z+1(z0.6)2
The final value theorem of Z-transform states that
If x(n) ↔ X(z), then x(∞) = Ltz→1⁡ (z-1)X(z)
x(∞) = Ltz→1 (z-1)X(z) = Ltz→1 (z-1) Z+1(z0.6)2 = 0.


10. Find x(∞) if X(z) = z+3(z+1)(z+2).
a) ∞
b) -1
c) 1
d) 0
Answer: d
Explanation: Given X(z) = z+3(z+1)(z+2)=z[1+(3/z)]z2[1+(1/z)][1+(2/z)]=1z1+(3/z)[1+(1/z)][1+(2/z)]
The initial value theorem of Z-transform states that
If x(n) ↔ X(z), then x(0) = Ltz→∞⁡ X(z)
x(0) = Ltz→∞⁡ X(z) = Ltz1z[1+(3/z)][1+(1/z)][1+(2/z)] = 0.

Inverse Z-Transform

1. Given the z-transform pair 3nn2 u[n] ↔ X (z). The time signal corresponding to X(2z) is ___________
a) n23n u[2n]
b) (32)nn2u[n]
c) (32)nn2u[n]
d) 6nn2u[n]
Answer: c
Explanation: Y (z) = X (2z) ↔ y[n] = 12n x[n]
Or, y[n] = 12n n2 3n u[n]
Or, y[n] = (32)nn2u[n].


2. Given the z-transform pair 3nn2 u[n] ↔ X (z). The time signal corresponding to X(z-1) is ___________
a) n23-nu[n]
b) n23-nu[-n]
c) 1n231nu[n]
d) 1n231nu[n]
Answer: b
Explanation: Y (z) = X (1z) ↔ y[n] = X [-n]
Or, y[n] = n23-nu[-n].


3. Given the z-transform pair 3nn2 u[n] ↔ X (z). The time signal corresponding to dX(z)dz is ___________
a) (n-1)33n-1u[n-1]
b) n33nu[n-1]
c) (1-n)33n-1u[n-1]
d) (n-1)33n-1u[n]
Answer: c
Explanation: Y (z) = dX(z)dz
z1[zdX(z)dz]
Now, Y (z) ↔ y (n) = – (n-1) X [-n-1] Or, y (n) = – (n-1)33n-1u[n-1].


4. Given the z-transform pair 3nn2 u[n] ↔ X (z). The time signal corresponding to z2z22 X(z) is ___________
a) 12(x[n+2]-x[n-2])
b) (x[n+2]-x[n-2])
c) 12(x[n-2]-x[n+2])
d) (x[n-2]-x[n+2])
Answer: a
Explanation: Y (z) = z2z22 X (z) ↔ y[n] = 12(x[n+2]-x[n-2]).


5. Given the z-transform pair 3nn2 u[n] ↔ X (z). The time signal corresponding to {X(z)}2 is ___________
a) {x[n]}2
b) x[n]*x[n]
c) x[n]*x[-n]
d) x[-n]*x[-n]
Answer: b
Explanation: Y (z) = X (z) H (z)
Y (z) = X (z) X (z) ↔ y[n] Or, y [n] = x[n]*x[n]


6. The system described by the difference equation y(n) – 2y(n-1) + y(n-2) = X(n) – X(n-1) has y(n) = 0 and n<0. If x (n) = δ(n), then y (z) will be?
a) 2
b) 1
c) 0
d) -1
Answer: c
Explanation: Given equation = y (n) – 2y (n-1) + y (n-2) = X (n) – X (n-1) has y (n) = 0
For n = 0, y (0)2y (-1) + y (-2) = x (0) – x (-1)
∴ y(0) = x(0) – x(-1)
Or, y (n) = 0 for n<0
For n=1, y (1) = -2y (0) + y (-1) = x (1) – x (0)
Or, y (1) = x (1) – x (0) + 2x (0) – 2x (-1)
Or, y (1) = x (1) + x (0) – 2x (-1)
For n=2, y (2) = x (2) – x (1) + 2y (1) – y (0)
Or, y(2) = x(2) – x(1) + 2x(1) + 2x(0) – 4x(-1) – x(0) + x(-1)
∴y (2) = d (2) + d (1) + d (0) – 3d (-1).


7. The value of Z1{z2(za)(zb)} is ____________
a) an+1bn+1a+b
b) an+1bn+1ab
c) an+1+bn+1ab
d) an+1+bn+1a+b
Answer: b
Explanation: We know that, Z1zza=an andZ1zzb=bn
Z1{z2(za)(zb)}=Z1zza.zzb=anbn
nm=0am.bnm
bnnm=0amb
bn.an+1b1ab1
an+1bn+1ab.


8. Given the z-transform pair
X[n]32z216, |z|<4
The z-transform of the signal y[n] = nx[n] is _________
a) 32z2(z216)2
b) 32z2(z216)2
c) 32(z216)2
d) 32z(z216)2
Answer: a
Explanation: y[n] = n x[n] n x[n] ↔ Y (z) = zdX(z)dz
Now, zdX(z)dz=32z2(z216)2.


9. Given the z-transform pair
X[n]32z216, |z|<4
The z-transform of the signal y[n] = x[n+1] + x[n-1] is _________
a) (z+1)2(z+1)216+(z1)2(z1)216
b) z2(1+z)z216
c) z2(z1)z216
d) (z+2)2(z+2)216
Answer: b
Explanation: x (n-n0) ↔ z-n0 X (z)
Now, y[n] = x [n+1] + x [n-1] ↔ Y (z)
Y (z) = (z+z-1) X (z)
∴ Y (z) = z2(1+z)z216.


10. The value of inverse Z-transform of log(zz+1) is _______________
a) (-1)n/n for n = 0; 0 otherwise
b) (-1)n/n
c) 0, for n = 0; (-1)n/n, otherwise
d) 0
Answer: c
Explanation: Putting z = 1t, U (z) = log (1y1y+1)
= – log (1+y) = -y + 12 y2 – 13 y3 + …..
= -z-1 + 12 z-2 – 13 z-3 + …..
Thus, un = 0, for n = 0; (-1)n/n otherwise.


11. The inverse Z-transform of z/(z+1)2 is ______________
a) (-1)n+1
b) (-1)n-1 n
c) (-1)n-1
d) (-1)n+1 n
Answer: b
Explanation: U (z) = zz2+2z+1
z12+z1z2+2z+1
z12z2+2z2+3z1z2+2z+1
z12z2+3z34z2+3z3z2+2z+1
So, U (z) = n=0(1)n1nzn
Hence, un = (-1)n-1 n.

Most Latest Signals and Systems MCQs – Z-Transform and Digital Filtering MCQs ( Signals and Systems ) MCQs

Share with Friends

Leave a Reply

%d bloggers like this: