Signals & Systems Questions and Answers – The Z-Transform

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This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “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
View Answer

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.
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2. Find the Z-transform of δ(n+3).
a) z
b) z2
c) 1
d) z3
View Answer

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) \(\frac{1}{1+az}\)
View Answer

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) \(\frac{z(z+cos⁡ω)}{z^2+2z cos⁡ω+1}\)
View Answer

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
View Answer

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.
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6. x(n) = an u(n) and x(n) = -an u(-n-1) have the same X(Z) and ROC.
a) True
b) False
View Answer

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)
View Answer

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) \(\frac{1}{z-a} – \frac{1}{z-(1/a)}\)
View Answer

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) \(\frac{z}{1+z}\)
View Answer

Answer: a
Explanation: Given x(n) = u(-n)
Z[x(n)] = X(Z) = \(∑_{n=-∞}^∞ x(n) z^{-n} = ∑_{n=-∞}^∞ u(-n) z^{-n} = ∑_{n=-∞}^0 (1) z^{-n}\)
=\(∑_{n=0}^∞ z^n = \frac{1}{1-z}\).
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10. For a right hand sequence, the ROC is entire z-plane.
a) True
b) False
View Answer

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.

Sanfoundry Global Education & Learning Series – Signals & Systems.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn