1. Which of the following justifies the linearity property of z-transform?[x(n)↔X(z)].
a) x(n)+y(n) ↔ X(z)Y(z)
b) x(n)+y(n) ↔ X(z)+Y(z)
c) x(n)y(n) ↔ X(z)+Y(z)
d) x(n)y(n) ↔ X(z)Y(z)
Explanation: According to the linearity property of z-transform, if X(z) and Y(z) are the z-transforms of x(n) and y(n) respectively then, the z-transform of x(n)+y(n) is X(z)+Y(z).
2. What is the z-transform of the signal x(n)=[3(2n)-4(3n)]u(n)?
a) \(\frac{3}{1-2z^{-1}}-\frac{4}{1-3z^{-1}}\)
b) \(\frac{3}{1-2z^{-1}}-\frac{4}{1+3z^{-1}}\)
c) \(\frac{3}{1-2z}-\frac{4}{1-3z}\)
d) None of the mentioned
Explanation: Let us divide the given x(n) into x1(n)=3(2n)u(n) and x2(n)= 4(3n)u(n)
and x(n)=x1(n)-x2(n)
From the definition of z-transform X1(z)=\(\frac{3}{1-2z^{-1}}\) and X2(z)=\(\frac{4}{1-3z^{-1}}\)
So, from the linearity property of z-transform
X(z)=X1(z)-X2(z)
=> X(z)=\(\frac{3}{1-2z^{-1}}-\frac{4}{1-3z^{-1}}\).
3. What is the z-transform of the signal x(n)=sin(jω0n)u(n)?
a) \(\frac{z^{-1} sin\omega_0}{1+2z^{-1} cos\omega_0+z^{-2}}\)
b) \(\frac{z^{-1} sin\omega_0}{1-2z^{-1} cos\omega_0-z^{-2}}\)
c) \(\frac{z^{-1} cos\omega_0}{1-2z^{-1} cos\omega_0+z^{-2}}\)
d) \(\frac{z^{-1} sin\omega_0}{1-2z^{-1} cos\omega_0+z^{-2}}\)
Explanation: By Euler’s identity, the given signal x(n) can be written as
x(n) = sin(jω0n)u(n)=\(\frac{1}{2j}[e^{jω_0n} u(n)-e^{-jω_0n} u(n)]\)
Thus X(z)=\(\frac{1}{2j}[\frac{1}{1-e^{jω_0} z^{-1}}-\frac{1}{1-e^{-jω_0} z^{-1}}]\)
On simplification, we obtain
=> \(\frac{z^{-1} sin\omega_0}{1-2z^{-1} cos\omega_0+z^{-2}}\).
4. According to Time shifting property of z-transform, if X(z) is the z-transform of x(n) then what is the z-transform of x(n-k)?
a) zkX(z)
b) z-kX(z)
c) X(z-k)
d) X(z+k)
Explanation: According to the definition of Z-transform
X(z)=\(\sum_{n=-\infty}^{\infty} x(n) z^{-n}\)
=>Z{x(n-k)}=\(X^1(z)=\sum_{n=-\infty}^{\infty} x(n-k) z^{-n}\)
Let n-k=l
=> X1(z)=\(\sum_{l=-\infty}^{\infty} x(l) z^{-l-k}=z^{-k}.\sum_{l=-\infty}^{\infty} x(l) z^{-l}= z^{-k}X(z)\)
5. What is the z-transform of the signal defined as x(n)=u(n)-u(n-N)?
a) \(\frac{1+z^N}{1+z^{-1}}\)
b) \(\frac{1-z^N}{1+z^{-1}}\)
c) \(\frac{1+z^{-N}}{1+z^{-1}}\)
d) \(\frac{1-z^{-N}}{1-z^{-1}}\)
Explanation:
We know that \(Z{u(n)}=\frac{1}{1-z^{-1}}\)
And by the time shifting property, we have Z{x(n-k)}=z-k.Z{x(n)}
=>Z{u(n-N)}=\(z^{-N}.\frac{1}{1-z^{-1}}\)
=>Z{u(n)-u(n-N)}=\(\frac{1-z^{-N}}{1-z^{-1}}\).
6. If X(z) is the z-transform of the signal x(n) then what is the z-transform of anx(n)?
a) X(az)
b) X(az-1)
c) X(a-1z)
d) None of the mentioned
Explanation: We know that from the definition of z-transform
Z{anx(n)}=\(\sum_{n=-\infty}^{\infty}a^n x(n) z^{-n}=\sum_{n=-\infty}^{\infty}x(n)(a^{-1}z)^{-n}=X(a^{-1}z)\).
7. If the ROC of X(z) is r1<|z|<r2, then what is the ROC of X(a-1z)?
a) |a|r1<|z|<|a|r2
b) |a|r1>|z|>|a|r2
c) |a|r1<|z|>|a|r2
d) |a|r1>|z|<|a|r2
Explanation: Given ROC of X(z) is r1<|z|<r2
Then ROC of X(a-1z) will be given by r1<|a-1z |<r2=|a|r1<|z|<|a|r2
8. If X(z) is the z-transform of the signal x(n), then what is the z-transform of the signal x(-n)?
a) X(-z)
b) X(z-1)
c) X-1(z)
d) None of the mentioned
Explanation: From the definition of z-transform, we have
Z{x(-n)}=\(\sum_{n=-\infty}^{\infty} x(-n) z^{-n}=\sum_{n=-\infty}^{\infty} x(-n) (z^{-1})^{-(-n)}=X(z^{-1})\)
9. X(z) is the z-transform of the signal x(n), then what is the z-transform of the signal nx(n)?
a) \(-z\frac{dX(z)}{dz}\)
b) \(z\frac{dX(z)}{dz}\)
c) \(-z^{-1}\frac{dX(z)}{dz}\)
d) \(z^{-1}\frac{dX(z)}{dz}\)
Explanation:
From the definition of z-transform, we have
X(z)=\(\sum_{n=-\infty}^{\infty} x(n) z^{-n}\)
On differentiating both sides, we have
\(\frac{dX(z)}{dz}=\sum_{n=-\infty}^{\infty} (-n) x(n) z^{-n-1}=-z^{-1} \sum_{n=-\infty}^{\infty}nx(n) z^{-n}=-z^{-1}Z\{nx(n)\}\)
Therefore, we get \(-z\frac{dX(z)}{dz}\) = Z{nx(n)}.
10. What is the z-transform of the signal x(n)=nanu(n)?
a) \(\frac{(az)^{-1}}{(1-(az)^{-1})^2}\)
b) \(\frac{az^{-1}}{(1-(az)^{-1})^2}\)
c) \(\frac{az^{-1}}{(1-az^{-1})^2}\)
d) \(\frac{az^{-1}}{(1+az^{-1})^2}\)
Explanation:
We know that Z{anu(n)}=\(\frac{1}{1-az^{-1}}\)=X(z) (say)
Now the z-transform of nanu(n)=\(-z\frac{dX(z)}{dz} = \frac{az^{-1}}{(1-az^{-1})^2}\)