Discrete Time Systems Implementation MCQ Quiz – Objective Question with Answer for Discrete Time Systems Implementation

101. We use y{‘}(nT)=-ay(nT)+bx(nT) to substitute for the derivative in y(nT) = \(\frac{T}{2} [y^{‘} (nT)+y^{‘} (nT-T)]+y(nT-T)\) and thus obtain a difference equation for the equivalent discrete-time system. With y(n) = y(nT) and x(n) = x(nT), we obtain the result as of the following?

A. \((1+\frac{aT}{2})Y(z)-(1-\frac{aT}{2})y(n-1)=\frac{bT}{2} [x(n)+x(n-1)]\)

B. \((1+\frac{aT}{n})Y(z)-(1-\frac{aT}{n})y(n-1)=\frac{bT}{n} [x(n)+x(n-1)]\)

C. \((1+\frac{aT}{2})Y(z)+(1-\frac{aT}{2})y(n-1)=\frac{bT}{2} (x(n)-x(n-1))\)

D. \((1+\frac{aT}{2})Y(z)+(1-\frac{aT}{2})y(n-1)=\frac{bT}{2} (x(n)+x(n+1))\)

Answer: A

When we substitute the given equation in the derivative of another we get the resultant required equation.

 

102. The z-transform of below difference equation is?

\((1+\frac{aT}{2})Y(z)-(1-\frac{aT}{2})y(n-1)=\frac{bT}{2} [x(n)+ x(n-1)]\)

A. \((1+\frac{aT}{2})Y(z)-(1-\frac{aT}{2}) z^{-1} Y(z)=\frac{bT}{2} (1+z^{-1})X(z)\)

B. \((1+\frac{aT}{n})Y(z)-(1-\frac{aT}{2}) z^{-1} Y(z)=\frac{bT}{n} (1+z^{-1})X(z)\)

C. \((1+\frac{aT}{2})Y(z)+(1-\frac{aT}{n}) z^{-1} Y(z)=\frac{bT}{2} (1+z^{-1})X(z)\)

D. \((1+\frac{aT}{2})Y(z)-(1+\frac{aT}{2}) z^{-1} Y(z)=\frac{bT}{2} (1+z^{-1})X(z)\)

Answer: A

By performing the z-transform of the given equation, we get the required output/equation.

 

103. What is the system function of the equivalent digital filter? H(z) = Y(z)/X(z) = ?

A. \(\frac{(\frac{bT}{2})(1+z^{-1})}{1+\frac{aT}{2}-(1-\frac{aT}{2}) z^{-1}}\)

B. \(\frac{(\frac{bT}{2})(1-z^{-1})}{1+\frac{aT}{2}-(1+\frac{aT}{2}) z^{-1}}\)

C. \(\frac{b}{\frac{2}{T}(\frac{1-z^{-1}}{1+z^{-1}}+A.}\)

D. \(\frac{(\frac{bT}{2})(1-z^{-1})}{1+\frac{aT}{2}-(1+\frac{aT}{2}) z^{-1}}\) & \(\frac{b}{\frac{2}{T}(\frac{1-z^{-1}}{1+z^{-1}}+A.}\)

Answer: D

As we considered analog linear filter with system function H(s) = b/s+a
Hence, we got an equivalent system function

where, s = \(\frac{2}{T}(\frac{1-z^{-1}}{1+z^{-1}})\).

 

104. In the Bilinear Transformation mapping, which of the following are correct?

A. All points in the LHP of s are mapped inside the unit circle in the z-plane
B. All points in the RHP of s are mapped outside the unit circle in the z-plane
C. All points in the LHP & RHP of s are mapped inside & outside the unit circle in the z-plane
D. None of the mentioned

Answer: C

The bilinear transformation is a conformal mapping that transforms the jΩ-axis into the unit circle in the z-plane and all the points are linked as mentioned above.

 

105. In Nth order differential equation, the characteristics of bilinear transformation, let z=rejw,s=o+jΩ Then for s = \(\frac{2}{T}(\frac{1-z^{-1}}{1+z^{-1}})\), the values of Ω, ℴ are

A. ℴ = \(\frac{2}{T}(\frac{r^2-1}{1+r^2+2rcosω})\), Ω = \(\frac{2}{T}(\frac{2rsinω}{1+r^2+2rcosω})\)

B. Ω = \(\frac{2}{T}(\frac{r^2-1}{1+r^2+2rcosω})\), ℴ = \(\frac{2}{T}(\frac{2rsinω}{1+r^2+2rcosω})\)

C. Ω=0, ℴ=0

D. None

Answer: A

s = \(\frac{2}{T}(\frac{z-1}{z+1}) \)

= \(\frac{2}{T}(\frac{re^jw-1}{re^jw+1})\)

= \(\frac{2}{T}(\frac{r^2-1}{1+r^2+2rcosω}+j \frac{2rsinω}{1+r^2+2rcosω})(s = ℴ+jΩ)\)

 

106. In equation ℴ = \(\frac{2}{T}(\frac{r^2-1}{1+r^2+2rcosω})\) if r < 1 then ℴ < 0 and then mapping from s-plane to z-plane occurs in which of the following order?

A. LHP in s-plane maps into the inside of the unit circle in the z-plane

B. RHP in s-plane maps into the outside of the unit circle in the z-plane

C. All of the mentioned

D. None of the mentioned

Answer: A

In the above equation, if we substitute the values of r, ℴ then we get mapping in the required way

In equation ℴ = \(\frac{2}{T}(\frac{r^2-1}{1+r^2+2rcosω})\) if r < 1 then ℴ < 0 and then mapping from s-plane to z-plane occurs in which LHP in s-plane maps into the inside of the unit circle in the z-plane.

 

107. In equation ℴ = \(\frac{2}{T}(\frac{r^2-1}{1+r^2+2rcosω})\), if r > 1 then ℴ > 0 and then mapping from s-plane to z-plane occurs in which of the following order?

A. LHP in s-plane maps into the inside of the unit circle in the z-plane
B. RHP in s-plane maps into the outside of the unit circle in the z-plane
C. All of the mentioned
D. None of the mentioned

Answer: B

In equation ℴ = \(\frac{2}{T}(\frac{r^2-1}{1+r^2+2rcosω})\), if r > 1 then ℴ > 0 and then mapping from s-plane to z-plane occurs in which RHP in s-plane maps into the outside of the unit circle in the z-plane

In the above equation, if we substitute the values of r, ℴ then we get mapping in the required way.

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