This set of tricky Digital Signal Processing Questions & Answers focuses on “IIR Filter Design by the Bilinear Transformation”.

1. In IIR Filter design by the Bilinear Transformation, the Bilinear Transformation is a mapping from

a) Z-plane to S-plane

b) S-plane to Z-plane

c) S-plane to J-plane

d) J-plane to Z-plane

View Answer

Explanation: From the equation,

S=\(\frac{2}{T}(\frac{1-z^{-1}}{1+z^{-1}})\) it is clear that transformation occurs from s-plane to z-plane

2. In Bilinear Transformation, aliasing of frequency components is been avoided.

a) True

b) False

View Answer

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.

3. Is IIR Filter design by Bilinear Transformation is the advanced technique when compared to other design techniques?

a) True

b) False

View Answer

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.

4. 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 t_{0}=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) = \(\frac{T}{2} [y^{‘} (nT)+y^{‘} (nT-T)]+y(T-nT)\)

View Answer

Explanation: By integrating the equation,

y(t) = \(\int_{t_0}^t y^{‘} (τ)dt+y(t_0)\) at t=nT and t

_{0}=nT-T we get equation,

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

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

View Answer

Explanation: When we substitute the given equation in the derivative of other we get the resultant required equation.

6. 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) \((1+\frac{aT}{2})Y(z)-(1+\frac{aT}{2}) z^{-1} Y(z)=\frac{bT}{2} (1+z^{-1})X(z)\)

View Answer

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

7. 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) Both a and c

View Answer

Explanation: 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}}).

8. 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) Both a & b

d) None of the mentioned

View Answer

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.

9. In Nth order differential equation, the characteristics of bilinear transformation, let z=re^{jw},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

View Answer

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

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

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

10. 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

View Answer

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

11. 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

View Answer

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

**Sanfoundry Global Education & Learning Series – Digital Signal Processing.**

To practice tricky questions on all areas of Digital Signal Processing, __here is complete set of 1000+ Multiple Choice Questions and Answers__.