# Fourier Analysis Questions and Answers – Linear Difference Equations and Z – Transforms

«
»

This set of Fourier Analysis Interview Questions and Answers focuses on “Linear Difference Equations and Z – Transforms”.

1. Find the Z-Transform of $$^nC_p$$.
a) (1-z-1)n
b) (1+z-1)n
c) (1-z-1)-n
d) (1+z-1)-n

Explanation: Using the Z-Transform formula, it can be written as
$$Z(^nC_p) = 1+ ^nC_1 z^{-1}+ ^nC_2 z^{-2} + ^nC_3 z^{-3}$$+……… which can be further equated to (1+z-1)n.

2. Find the function whose Z – Transform is $$\frac{1}{z}$$.
a) δ(n)
b) δ(n+1)
c) U(n)
d) U(n+1)

Explanation: δ(n) exists only at n=0 and δ(n+1) exists only at n=1. Therefore while substituting this function, the Z – Transform at every other place becomes zero except at n=1. Therefore the Z-Transform of δ(n+1) is $$\frac{1}{z}.$$

3. Find the function whose Z transform is $$e^{\frac{1}{z}}$$.
a) log(n)
b) $$\frac{1}{n}$$
c) $$\frac{1}{n!}$$
d) $$\frac{1}{(n+1)!}$$

Explanation: Using the definition of Z- Transform we have $$∑_{n=0}^∞ (\frac{1}{n!} z^{-n})$$. Now, expanding this we get $$1+\frac{z^{-1}}{1} + \frac{z^{-2}}{2} + \frac{z^{-3}}{3}$$+ …………. This is nothing but the expansion of $$e^{\frac{1}{z}}$$, hence the answer is $$\frac{1}{n!}.$$
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

4. Find the inverse Z- Transform of $$(\frac{z}{z-a})^3$$.
a) $$\frac{1}{2} . (n+1) (n-2) a^{n-2} U(n)$$
b) $$\frac{1}{2} . (n-1) (n-2) a^{n-3} U(n)$$
c) $$\frac{1}{2} . (n-1) (n+2) a^{n-1} U(n)$$
d) $$\frac{1}{2} . (n+1) (n+2) a^n U(n)$$

Explanation: The inverse Z-Transform of $$\frac{z}{z-a}$$ is $$a^n$$. The inverse Z-Transform of $$(\frac{z}{z-a})^2$$ is the convolution of an and an. Now, the inverse Z-Transform of $$(\frac{z}{z-a})^3$$ is the convolution of the result of the previous step with an an. Thus we get the answer $$\frac{1}{2} . (n+1) (n+2) a^n U(n)$$

5. Find the inverse Z – Transform of $$log \frac{z}{z+1}$$.
a) $$\frac{(-1)^n}{n}$$
b) $$\frac{(-1)^{n+1}}{n}$$
c) $$\frac{1}{n}$$
d) $$\frac{(-1)^n}{n+1}$$

Explanation: First, substitute z as $$(\frac{1}{y})$$ and then expand the got result. This is in the format of the Z-Transform expansion. Thus we get the required results.

6. Find the Z – Transform of sinh ⁡nθ.
a) $$\frac{sinh⁡θ}{z^2-2z cosh⁡θ+1}$$
b) $$\frac{1}{2} \frac{sinh⁡θ}{z^2-2z cosh⁡θ+1}$$
c) $$\frac{z sinh⁡θ)}{z^2-2z cosh⁡θ+1}$$
d) $$\frac{z(z-sinh⁡θ)}{z^2-2z cosh⁡θ+1}$$

Explanation: The first step to solve this is to expand the function sinh⁡nθ. The expansion of this function is of the form an. First we have to find the Z-Transform of 1 and then we have to use damping rule. To, get the answer, we take L.C.M.

7. Find the value of u3 if $$U(z) = \frac{3z^2+2z+10}{(z-1)^4}$$.
a) 12
b) 13
c) 14
d) 15

Explanation: Taking limz→∞ U(z), we get 0 which is u0. Now using the shifting property and again using the limit we get u1 which is 0. Again, by using the shifting property we get u2 which is 3. Now, by using shifting by 3 properties, we get the value of u3 which is 14.

8. Find the Z – Transform of np.
a) $$-z\frac{d}{dz}(Z(n^{p-1}))$$
b) $$z\frac{d}{dz}(Z(n^p))$$
c) $$-z\frac{d}{dz}(Z(n^{p+1}))$$
d) $$z\frac{d}{dz}(Z(n^{p+1}))$$

Explanation: The Z Transform of 1 can be found just by infinite G.P. sum. The Z- Transform can be found by differentiating the Z-Transform of 1 and multiplying by (-z). And the Z-Transform of n2, can also be found by differentiating the Z-Transform of n and multiplying by (-z). Hence the general form is $$-z\frac{d}{dz}(Z(n^{p-1}))$$.

9. The Z – Transform of a function is given by $$U(z) = \frac{z^3+6z^2+9z+3}{(z-1)^4}$$. Find the Z-Transform of un+2.
a) $$\frac{10z^3+3z^2+7z^1-1}{(z-1)^4}$$
b) $$\frac{10z^4+3z^3+7z^2-z}{(z-1)^4}$$
c) $$\frac{10z^4+4z^3+7z^2-2z}{(z-1)^4}$$
d) $$\frac{10z^4+3z^3-4z}{(z-1)^4}$$

Explanation: First we have to find u0, which can be found by applying limits to U(z). Now shifting by 1 and then applying limits we get u1. Now using the second shift property, we find the Z-Transform of un+2.

10. Find u2 if $$U(z) = \frac{z^3+6z^2+9z+3}{(z-1)^4}$$.
a) 8
b) 9
c) 10
d) 11

Explanation: The first step is to find the limit of the U(z), hence getting the u0. And again doing this we get u1. And again doing the shifting property, we get un+2. And doing the limits, we get the u2.

11. Find the order of the difference equation Δ3yn – Δ2yn – Δyn = 3.
a) 3
b) 4
c) 2
d) 5

Explanation: The first step is to expand the given equation by replacing every Δyn by (yn+1– yn). Order of a difference equation is given by, $$\frac{n+3-n}{1}$$ which is actually 3.

12. Find the order of the difference equation yn+3 -3 yn+1 – yn-2 = 4.
a) 3
b) 4
c) 5
d) 6

Explanation: The order of the given difference equation can be written as Order = $$\frac{n+3-n+2}{1}$$. Therefore the order is 5.

13. Find the difference equation of yn = A 3n + B 5n.
a) yn+2 -16 yn+1 + 15 yn-1 = 0
b) yn+3 -14 yn+1 + 30 yn = 0
c) 2 yn+2 -14 yn+1 + 15 yn = 0
d) 2 yn+2 -16 yn+1 + 30 yn = 0

Explanation: This can be solved using the determinant.
$$\begin{bmatrix} y_n & 1 & 1\\ y_{n+1} & 3 & 5\\ y_{n+2} & 9 & 25\\ \end{bmatrix}$$ = 0. Now, by solving the determinant, we get the required difference equation.

14. Find the difference equation of y = ax + b.
a) Δ2y = 0
b) Δ2y = 1
c) Δ2y + 3Δy = 2
d) Δ2y + 4Δy = 5

Explanation: First step is to take Δ operator on both sides of the given equation. Now, since here we have 2 unknown variables, we have take the Δ operator twice on both the sides, hence getting the required results.

15. Solve un+2 + 10 un+1 + 9 un = 2n.
a) $$u_n = \frac{2^{n+1}}{33}+\frac{(-9)^{n+1}}{88}+\frac{(-1)^{n+1}}{24}$$
b) $$u_n = \frac{2^n}{33}+\frac{(-9)^n}{88}+\frac{(-1)^{n-1}}{24}$$
c) $$u_n = \frac{2^{n+1}}{11}+\frac{(-9)^{n+1}}{88}+\frac{(-1)^n}{24}$$
d) $$u_n = \frac{2^n}{11}+\frac{(-9)^n}{88}+\frac{(-1)^{n-1}}{24}$$

Explanation: Take Z – Transformation on both sides. Now keep U(z) on one side and take everything else to other side. Now by partial fractions method, we get it in the format of an. Now, taking inverse Z-Transform on both sides, we get the required results.

Sanfoundry Global Education & Learning Series – Fourier Analysis.

To practice all areas of Fourier Analysis for Interviews, here is complete set of 1000+ Multiple Choice Questions and Answers. 