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?

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

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

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?

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

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

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

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

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

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

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

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

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

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

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

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

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

y(nT) = \(\frac{T}{2} [y^{‘} (nT)+y^{‘} (T-nT)]+y(nT-T)\)

y(nT) = \(\frac{T}{2} [y^{‘} (nT)+y^{‘} (nT-T)]+y(nT-T)\)

y(nT) = \(\frac{T}{2} [y^{‘} (nT)+y^{‘} (T-nT)]+y(T-nT)\)

y(nT) = \(\frac{T}{2} [y^{‘} (nT)+y^{‘} (nT-T)]+y(T-nT)\)

1.	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» C. All of the mentioned

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

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

4.	The z-transform of below difference equation is?
A.	\((1+\frac{aT}{2})Y(z)-(1-\frac{aT}{2})y(n-1)=\frac{bT}{2} [x(n)+ x(n-1)]\)
B.	\((1+\frac{aT}{2})Y(z)-(1-\frac{aT}{2}) z^{-1} Y(z)=\frac{bT}{2} (1+z^{-1})X(z)\)
C.	\((1+\frac{aT}{n})Y(z)-(1-\frac{aT}{2}) z^{-1} Y(z)=\frac{bT}{n} (1+z^{-1})X(z)\)
D.	\((1+\frac{aT}{2})Y(z)+(1-\frac{aT}{n}) z^{-1} Y(z)=\frac{bT}{2} (1+z^{-1})X(z)\)
E.	\((1+\frac{aT}{2})Y(z)-(1+\frac{aT}{2}) z^{-1} Y(z)=\frac{bT}{2} (1+z^{-1})X(z)\)
Answer» B. \((1+\frac{aT}{2})Y(z)-(1-\frac{aT}{2}) z^{-1} Y(z)=\frac{bT}{2} (1+z^{-1})X(z)\)

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

6.	The approximation of the integral in y(t) = \(\int_{t_0}^t y'(τ)dt+y(t_0)\) by the Trapezoidal formula at t = nT and t0=nT-T yields equation?
A.	y(nT) = \(\frac{T}{2} [y^{‘} (nT)+y^{‘} (T-nT)]+y(nT-T)\)
B.	y(nT) = \(\frac{T}{2} [y^{‘} (nT)+y^{‘} (nT-T)]+y(nT-T)\)
C.	y(nT) = \(\frac{T}{2} [y^{‘} (nT)+y^{‘} (T-nT)]+y(T-nT)\)
D.	y(nT) = \(\frac{T}{2} [y^{‘} (nT)+y^{‘} (nT-T)]+y(T-nT)\)
Answer» C. y(nT) = \(\frac{T}{2} [y^{‘} (nT)+y^{‘} (T-nT)]+y(T-nT)\)