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

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

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) $$\frac{1}{1+az}$$

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

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

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

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)

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

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) $$\frac{z}{1+z}$$

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