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}

View Answer

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)

View Answer

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) \(\frac{1}{(n+1)!} \)

View Answer

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 \(\frac{1}{n!}.\)

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) \(\frac{1}{2} . (n+1) (n+2) a^n U(n) \)

View Answer

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 a

^{n}and a

^{n}. Now, the inverse Z-Transform of \((\frac{z}{z-a})^3\) is the convolution of the result of the previous step with a

^{n}a

^{n}. 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) \(\frac{(-1)^n}{n+1} \)

View Answer

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) \(\frac{sinhθ}{z^2-2z coshθ+1} \)

b) \(\frac{1}{2} \frac{sinhθ}{z^2-2z coshθ+1} \)

c) \(\frac{z sinhθ)}{z^2-2z coshθ+1} \)

d) \(\frac{z(z-sinhθ)}{z^2-2z coshθ+1} \)

View Answer

Explanation: The first step to solve this is to expand the function sinhnθ. The expansion of this function is of the form a

^{n}. 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 u_{3} if \(U(z) = \frac{3z^2+2z+10}{(z-1)^4} \).

a) 12

b) 13

c) 14

d) 15

View Answer

Explanation: Taking lim

_{z→∞}U(z), we get 0 which is u

_{0}. Now using the shifting property and again using the limit we get u

_{1}which is 0. Again, by using the shifting property we get u

_{2}which is 3. Now, by using shifting by 3 properties, we get the value of u

_{3}which is 14.

8. Find the Z – Transform of n^{p}.

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) \(z\frac{d}{dz}(Z(n^{p+1})) \)

View Answer

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 n

^{2}, 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 u_{n+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) \(\frac{10z^4+3z^3-4z}{(z-1)^4} \)

View Answer

Explanation: First we have to find u

_{0}, which can be found by applying limits to U(z). Now shifting by 1 and then applying limits we get u

_{1}. Now using the second shift property, we find the Z-Transform of u

_{n+2}.

10. Find u_{2} if \(U(z) = \frac{z^3+6z^2+9z+3}{(z-1)^4}\).

a) 8

b) 9

c) 10

d) 11

View Answer

Explanation: The first step is to find the limit of the U(z), hence getting the u

_{0}. And again doing this we get u

_{1}. And again doing the shifting property, we get u

_{n+2}. And doing the limits, we get the u

_{2}.

11. Find the order of the difference equation Δ^{3}y_{n} – Δ^{2}y_{n} – Δy_{n} = 3.

a) 3

b) 4

c) 2

d) 5

View Answer

Explanation: The first step is to expand the given equation by replacing every Δy

_{n}by (y

_{n+1}– y

_{n}). 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 y_{n+3} -3 y_{n+1} – y_{n-2} = 4.

a) 3

b) 4

c) 5

d) 6

View Answer

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 y_{n} = A 3^{n} + B 5^{n}.

a) y_{n+2} -16 y_{n+1} + 15 y_{n-1} = 0

b) y_{n+3} -14 y_{n+1} + 30 y_{n} = 0

c) 2 y_{n+2} -14 y_{n+1} + 15 y_{n} = 0

d) 2 y_{n+2} -16 y_{n+1} + 30 y_{n} = 0

View Answer

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) Δ^{2}y = 0

b) Δ^{2}y = 1

c) Δ^{2}y + 3Δy = 2

d) Δ^{2}y + 4Δy = 5

View Answer

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 u_{n+2} + 10 u_{n+1} + 9 u_{n} = 2^{n}.

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) \(u_n = \frac{2^n}{11}+\frac{(-9)^n}{88}+\frac{(-1)^{n-1}}{24} \)

View Answer

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 a

^{n}. Now, taking inverse Z-Transform on both sides, we get the required results.

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