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

Latest Signals and Systems MCQs

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

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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^jω
When r=1, z = e^jω 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) Z^k, Z>0
b) Z^-k, Z>0
c) Z^k, Z≠0
d) Z^-k, Z≠0
Answer: d
Explanation: Performing z-transform on δ[n-k], we get,
X (z) = ∑∞n=0x[n]z−n
= Z^-k, Z≠0.


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


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


4. The z-transform of (14)n (u[n] – u[n-5]) is __________
a) z5–0.255z4(z−0.25), z>0.25
b) z5–0.255z4(z−0.25), z>0
c) z5–0.255z3(z−0.25), z<0.25
d) z5–0.255z4(z−0.25), all z
Answer: d
Explanation: X (z) = ∑4n=0(14z−1)4
= 1−(14z−1)51−(14z−1)1
= z5–0.255z4(z−0.25), all z.


5. The z-transform of (14)4 u[-n] is ___________
a) 4z4z−1, |Z|>14
b) 4z4z−1, |Z|<14
c) 11−4z, |Z|>14
d) 11−4z, |Z|<14
Answer: d
Explanation: X (z) = ∑0n=−∞(14z−1)n
= ∑∞n=0(4z)n
= 11−4z, |Z|<14.


6. The z-transform of 3ⁿ u[-n-1] is ___________
a) z3−z, |Z|>3
b) z3−z, |Z|<3
c) 33−z, |Z|>3
d) 33−z, |Z|<3
Answer: b
Explanation: X (z) = ∑−1n=−∞(3z−1)n
= ∑∞n=1(z13)n
= 13z1−13z, |z|<3
= z3−z.


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

Laplace Transform And System Design MCQs

8. The z-transform of cos(π3 n) u[n] is __________
a) z2(2z−1)(z2−z+1), 0<|z|<1
b) z2(2z−1)(z2−z+1), |z|>1
c) z2(1−2z)(z2−z+1), 0<|z|<1
d) z2(1−2z)(z2−z+1), |z|>1
Answer: b
Explanation: Performing z-transform on aⁿu[n], we get zz−a
∴ x[n] = 12ej(π3)nu[n]+12e−j(π3)nu[n]
Hence, X (z) =0.5(11−ej(π3)z−1+11−ej(π3)z−1)
Hence, X (z) = z2(2z−1)(z2−z+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 z^n₀ 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 z^n₀ 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]↔32z2−16, |z|<4
The z-transform of the signal x [n-2] is _________
a) z4z2−16
b) (z+2)2(z+2)2−16
c) 1z2−16
d) (z−2)2(z−2)2−16
Answer: c
Explanation: Performing z-transform on x (n-n0), we get z^n₀ X (z)
Now, z-transform of y[n] = x [n-2] is given by,
Y (z) = z^-2 X (z)
= 1z2−16.


13. Given the z-transform pair
X[n]↔32z2−16, |z|<4
The z-transform of the signal y[n] = 12n x[n] is _________
a) (z+2)2(z+2)2−16
b) z2z2−4
c) (z−2)2(z−2)2−16
d) z2z2−64
Answer: b
Explanation: y[n] = 12n x[n] Performing z-transform on y[n], we get, Y (z) = X (2z)
∴ Y(z) = z2z2−4.


14. Given the z-transform pair
X[n]↔32z2−16, |z|<4
The z-transform of the signal x [-n]*x[n] is ____________
a) z216z2−257z4−16
b) −16z2(z2−16)2
c) z2(257z2−16z4−16)
d) 16z2(z2−16)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]↔32z2−16, |z|<4
The z-transform of the signal x[n]*x [n-3] is __________
a) z−3(z2−16)2
b) z7(z2−16)2
c) z5(z2−16)2
d) z(z2−16)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(z2−16)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)z−n
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)z−n
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) 1z−1
b) 1z+1
c) 11−z
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)] = (1z−1)z=(1/z)=1/z(1/z)−1=11−z.


4. Find the Z-transform of x(n) = u(-n-2).
a) z2z−1
b) z21−z
c) z21+z
d) z22−z
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)] = z21−z.

Sampling Theorem MCQs

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


6. Find the Z-transform of x(n) = 2ⁿ u(n-2).
a) zz−2
b) zz+2
c) zz(z−2)
d) 4z(z−2)
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)] = z−2zz−1=1z(z−1)
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(z−1)]z=(z/2)
=1(z/2)[(z/2)−1]=4z(z−2).


7. Find the Z-transform of x(n) = n[aⁿ u(n)].
a) zz(z−a)
b) azz(z−a)
c) azz(z+a)
d) az(z−a)2
Answer: d
Explanation: Given x(n) = n[aⁿ u(n)]
We know that aⁿ u(n) ↔ zz−a
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[zz−a]=az(z−a)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) ↔ zz−a
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(z−0.6)2.
a) 1
b) 0
c) ∞
d) 0.6
Answer: b
Explanation: Given X(z) = Z+1(z−0.6)2
The final value theorem of Z-transform states that
If x(n) ↔ X(z), then x(∞) = Lt_z→1⁡ (z-1)X(z)
x(∞) = Lt_z→1 (z-1)X(z) = Lt_z→1 (z-1) Z+1(z−0.6)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) = Lt_z→∞⁡ X(z)
x(0) = Lt_z→∞⁡ X(z) = Ltz→∞1z[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)³3^n-1u[n-1]
b) n³3ⁿu[n-1]
c) (1-n)³3^n-1u[n-1]
d) (n-1)³3^n-1u[n]
Answer: c
Explanation: Y (z) = dX(z)dz
= −z−1[−zdX(z)dz]
Now, Y (z) ↔ y (n) = – (n-1) X [-n-1] Or, y (n) = – (n-1)³3^n-1u[n-1].


4. Given the z-transform pair 3ⁿn² u[n] ↔ X (z). The time signal corresponding to z2−z−22 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) = z2−z−22 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 Z−1{z2(z−a)(z−b)} is ____________
a) an+1–bn+1a+b
b) an+1–bn+1a−b
c) an+1+bn+1a−b
d) an+1+bn+1a+b
Answer: b
Explanation: We know that, Z−1zz−a=an andZ−1zz−b=bn
∴Z−1{z2(z−a)(z−b)}=Z−1zz−a.zz−b=an∗bn
= ∑nm=0am.bn−m
= bn∑nm=0amb
= bn.an+1b−1ab−1
= an+1–bn+1a−b.


8. Given the z-transform pair
X[n]↔32z2−16, |z|<4
The z-transform of the signal y[n] = nx[n] is _________
a) 32z2(z2−16)2
b) −32z2(z2−16)2
c) 32(z2−16)2
d) −32z(z2−16)2
Answer: a
Explanation: y[n] = n x[n] n x[n] ↔ Y (z) = −zdX(z)dz
Now, −zdX(z)dz=32z2(z2−16)2.


9. Given the z-transform pair
X[n]↔32z2−16, |z|<4
The z-transform of the signal y[n] = x[n+1] + x[n-1] is _________
a) (z+1)2(z+1)2−16+(z−1)2(z−1)2−16
b) z2(1+z)z2−16
c) z2(z−1)z2−16
d) (z+2)2(z+2)2−16
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)z2−16.


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, u_n = 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
= z−1–2+z−1z2+2z+1
= z−1–2z−2+2z−2+3z−1z2+2z+1
= z−1–2z−2+3z−3–4z−2+3z−3z2+2z+1
So, U (z) = ∑∞n=0(−1)n−1nz−n
Hence, un = (-1)^n-1 n.

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