Digital Signal Processing

IIR Filter Design by the Bilinear Transformation

Question 1
Marks : +2 | -2
Pass Ratio : 100%
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
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})\\)
Ω=0, ℴ=0
None
Explanation:
s = \\(\\frac{2}{T}(\\frac{z-1}{z+1}) \\)
Question 2
Marks : +2 | -2
Pass Ratio : 100%
In Bilinear Transformation, aliasing of frequency components is been avoided.
True
False
Explanation:
The bilinear transformation is a conformal mapping that transforms the jΩ-axis into the unit circle in the z-plane only once, thus avoiding the aliasing.
Question 3
Marks : +2 | -2
Pass Ratio : 100%
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
All of the mentioned
None of the mentioned
Explanation:
In the above equation, if we substitute the values of r, â„´ then we get mapping in the required way
Question 4
Marks : +2 | -2
Pass Ratio : 100%
What is the system function of the equivalent digital filter? H(z) = Y(z)/X(z) = ?
\\(\\frac{(\\frac{bT}{2})(1+z^{-1})}{1+\\frac{aT}{2}-(1-\\frac{aT}{2}) z^{-1}}\\)
\\(\\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)}\\)
\\(\\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)}\\)
Explanation:
As we considered analog linear filter with system function H(s) = b/s+a
Question 5
Marks : +2 | -2
Pass Ratio : 100%
In IIR Filter design by the Bilinear Transformation, the Bilinear Transformation is a mapping from
Z-plane to S-plane
S-plane to Z-plane
S-plane to J-plane
J-plane to Z-plane
Explanation:
From the equation,
Question 6
Marks : +2 | -2
Pass Ratio : 100%
In the Bilinear Transformation mapping, which of the following are correct?
All points in the LHP of s are mapped inside the unit circle in the z-plane
All points in the RHP of s are mapped outside the unit circle in the z-plane
All points in the LHP & RHP of s are mapped inside & outside the unit circle in the z-plane
None of the mentioned
Explanation:
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.
Question 7
Marks : +2 | -2
Pass Ratio : 100%
Is IIR Filter design by Bilinear Transformation is the advanced technique when compared to other design techniques?
True
False
Explanation:
Because in other techniques, only lowpass filters and limited class of bandpass filters are been supported. But this technique overcomes the limitations of other techniques and supports more.
Question 8
Marks : +2 | -2
Pass Ratio : 100%
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?
\\((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))\\)
Explanation:
When we substitute the given equation in the derivative of other we get the resultant required equation.
Question 9
Marks : +2 | -2
Pass Ratio : 100%
The approximation of the integral in y(t) = \\(\\int_{t_0}^t y\'(Ï„)dt+y(t_0)\\) by the Trapezoidal formula at t = nT and t0=nT-T yields equation?
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)\\)
Explanation:
By integrating the equation,
Question 10
Marks : +2 | -2
Pass Ratio : 100%
The z-transform of below difference equation is?
\\((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)\\)
Explanation:
By performing the z-transform of the given equation, we get the required output/equation.