Z Transform From Difference Equation MCQ Quiz in मल्याळम - Objective Question with Answer for Z Transform From Difference Equation - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 19, 2025
Latest Z Transform From Difference Equation MCQ Objective Questions
Top Z Transform From Difference Equation MCQ Objective Questions
Z Transform From Difference Equation Question 1:
Consider two discrete time signals: x[n] = {1, 1, 0} and y [n] = {1, 0, 1} for n = 0, 1, 2. The convoluted sequence x[n] * y[n] is
Answer (Detailed Solution Below)
Z Transform From Difference Equation Question 1 Detailed Solution
Taking z- transform of x[z] and y[z] we get
X[z] = 1 + Z-1 y(z) = 1 + Z-2
X[z] × y[z] = 1 + Z-1 + Z-2 + Z-3
The sequence x[n] * y[n] is {1, 1, 1, 1} for n = 0, 1, 2, 3Z Transform From Difference Equation Question 2:
Difference equation of a system is given by
\(10y\left( {n + 2} \right) + 60y\left( {n + 1} \right) + 90y\left( n \right) =2 x\left( {n + 1} \right) + 6x\left( n \right)\)
Then transfer function will be
Answer (Detailed Solution Below)
Z Transform From Difference Equation Question 2 Detailed Solution
\(10y\left( {n + 2} \right) + 60y\left( {n + 1} \right) + 90y\left( n \right) =2 x\left( {n + 1} \right) + 6x\left( n \right)\)
Taking Z transfer of the difference equation we get
\(\eqalign{ & 10{z^2}Y\left( z \right) + 60zY\left( z \right) + 90Y\left( z \right) = 2zX\left( z \right) + 6X\left( z \right) \cr & \Rightarrow \frac{{Y\left( z \right)}}{{X\left( z \right)}} = H\left( z \right) = \frac{{2z + 6}}{{10{z^2} + 60z + 90}} = \frac{{2z + 6}}{{10\left( {{z^2} + 6z + 9} \right)}} \cr & = \frac{{z + 3}}{{5{{\left( {z + 3} \right)}^2}}} = \frac{1}{{5\left( {z + 3} \right)}} \cr}\)
Z Transform From Difference Equation Question 3:
A system has input \(\rm x[n]\) and output \(\rm y[n]\) related by
\(\rm y\left[ {n - 1} \right] + 2y\left[ n \right] = x\left[ n \right]\)
The output of the system for \(\rm n≥0\) when \(\rm x[n]=0\) and \(\rm y[-1]=2\) is:
Answer (Detailed Solution Below)
\(\rm \left[ { - {{\left( { - \frac{1}{2}} \right)}^n}} \right]u\left[ n \right]\)
Z Transform From Difference Equation Question 3 Detailed Solution
Taking z- transform we have
\(\rm {z^{ - 1}}Y\left( z \right) + y\left[ { - 1} \right] + 2Y\left( z \right) = X\left( z \right)\)
For zero input response we put \(\rm X\left( z \right) = 0\)
\(\rm Y\left( z \right) = \frac{{ - 1}}{{1 + \frac{1}{2}{z^{ - 1}}}}\)
\(\rm \Rightarrow y\left[ n \right] = - {\left( { - \frac{1}{2}} \right)^n}u\left[ n \right]\)
Z Transform From Difference Equation Question 4:
Consider a causal LTI system described by
\({\rm{y}}\left[ {\rm{n}} \right] + 3{\rm{y}}\left[ {{\rm{n}} - 1} \right] = {\rm{x}}\left[ {\rm{n}} \right]\)
with \({\rm{x}}\left[ {\rm{n}} \right] = {\rm{\alpha u}}\left[ {\rm{n}} \right]{\rm{\ and\ y}}\left[ { - 1} \right] = {\rm{\beta }}\)
we have \({\rm{y}}\left[ {\rm{n}} \right] = \left[ {3{{\left( { - 3} \right)}^n} + 2} \right]{\rm{u}}\left[ {\rm{n}} \right]{\rm{}}\) for \({\rm{n}} \ge 0\). The value of \({\rm{\alpha }} + {\rm{\beta }}\) is _____
Answer (Detailed Solution Below) 9
Z Transform From Difference Equation Question 4 Detailed Solution
Given,
\({\rm{y}}\left[ {\rm{n}} \right] + 3{\rm{y}}\left[ {{\rm{n}} - 1} \right] = {\rm{x}}\left[ {\rm{n}} \right]\)
Taking unilateral Z transform we have
\(\begin{array}{l} {\rm{Y}}\left( {\rm{z}} \right) + 3\left( {{z^{ - 1}}{\rm{Y}}\left( {\rm{z}} \right) + {\rm{y}}\left[ { - 1} \right]} \right) = {\rm{X}}\left( {\rm{z}} \right)\\ \Rightarrow Y\left( z \right) = \frac{\alpha }{{\left( {1 + 3{z^{ - 1}}} \right)\left( {1 - {z^{ - 1}}} \right)}} - \frac{{3\beta }}{{1 + 3{z^{ - 1}}}}\\ \left( {\alpha u\left[ n \right] \leftrightarrow \frac{\alpha }{{1 - {z^{ - 1}}}}} \right) \end{array}\)
\( \Rightarrow Y\left( z \right) = \frac{{\alpha - 3\beta + 3\beta {z^{ - 1}}}}{{\left( {1 + 3{z^{ - 1}}} \right)\left( {1 - {z^{ - 1}}} \right)}}\) ______ (i)
Now from \({\rm{y}}\left[ {\rm{n}} \right] = \left[ {3{{\left( { - 3} \right)}^n} + 2} \right]{\rm{u}}\left[ {\rm{n}} \right]\), we have
Taking unilateral z transform, \(Y\left( z \right) = \frac{3}{{1 + 3{z^{ - 1}}}} + \frac{2}{{1 - {z^{ - 1}}}}\)
\(= \frac{{5 + 3{z^{ - 1}}}}{{\left( {1 + 3{z^{ - 1}}} \right)\left( {1 - {z^{ - 1}}} \right)}}\) _____(ii)
Comparing (i) and (ii)
We have β = 1
\(\begin{array}{l} {\rm{\alpha }}--3{\rm{\beta }} = 5\\ \Rightarrow {\rm{\alpha }} = 8 \end{array}\)
Thus, \({\rm{\alpha }} + {\rm{\beta }} = 8 + 1 = 9\)
Z Transform From Difference Equation Question 5:
The difference equation representation for a system is:\(y\left( n \right) - \frac{1}{3}y\left( {n - 1} \right) = 2x\left( n \right)\)\(h(1)=2\)The natural response of system h(n) is:
Answer (Detailed Solution Below)
\(\ 2{\left( {\frac{1}{3}} \right)^n}u(n)\)
Z Transform From Difference Equation Question 5 Detailed Solution
Taking z-transform
\(Y\left( z \right) - \frac{1}{3}{z^{ - 1}}Y\left( z \right) = 2X\left( z \right)\)
\( \Rightarrow H\left( z \right) = \frac{{Y\left( z \right)}}{{X\left( z \right)}} = \frac{2}{{1 - \left( {\frac{1}{3}} \right){Z^{ - 1}}}}\;\)
Taking inverse z-transform
\(H\left( n \right) = 2{\left( {\frac{1}{3}} \right)^n}u\left( n \right)\)
it satisfied H(1) = 2