Digital Signal Processing Multiple Choice Questions on “Properties of Z Transform-1”.
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)
Answer: B
Clarification: 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
Answer: A
Clarification: 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} sinomega_0}{1+2z^{-1} cosomega_0+z^{-2}})
B. (frac{z^{-1} sinomega_0}{1-2z^{-1} cosomega_0-z^{-2}})
C. (frac{z^{-1} cosomega_0}{1-2z^{-1} cosomega_0+z^{-2}})
D. (frac{z^{-1} sinomega_0}{1-2z^{-1} cosomega_0+z^{-2}})
Answer: D
Clarification: 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} sinomega_0}{1-2z^{-1} cosomega_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)
Answer: B
Clarification: 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}})
Answer: D
Clarification:
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
Answer: C
Clarification: 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|
A. |a|r1<|z|<|a|r2
B. |a|r1>|z|>|a|r2
C. |a|r1<|z|>|a|r2
D. |a|r1>|z|<|a|r2
Answer: A
Clarification: Given ROC of X(z) is r1<|z|
Then ROC of X(a-1z) will be given by r1<|a-1z |
8. What is the z-transform of the signal x(n)=an(sinω0n)u(n)?
A.(frac{az^{-1} sinomega_0}{1+2 az^{-1} cosomega_0+a^2 z^{-2}})
B.(frac{az^{-1} sinomega_0}{1-2 az^{-1} cosomega_0- a^2 z^{-2}})
C.(frac{(az)^{-1} cosomega_0}{1-2 az^{-1} cosomega_0+a^2 z^{-2}})
D.(frac{az^{-1} sinomega_0}{1-2 az^{-1} cosomega_0+a^2 z^{-2}})
Answer: D
Clarification:
we know that by the linearity property of z-transform of sin(ω0n)u(n) is X(z)=(frac{z^{-1} sinomega_0}{1-2z^{-1} cosomega_0+z^{-2}})
Now by the scaling in the z-domain property, we have z-transform of an (sin(ω0n))u(n) as
X(az)=(frac{az^{-1} sinomega_0}{1-2az^{-1} cosω_0+a^2 z^{-2}})
9. 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
Answer: B
Clarification: 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}))
10. X(z) is the z-transform of the signal x(n), then what is the z-transform of the signal nx(n)?
A. (-zfrac{dX(z)}{dz})
B. (zfrac{dX(z)}{dz})
C. (-z^{-1}frac{dX(z)}{dz})
D. (z^{-1}frac{dX(z)}{dz})