Digital Signal Processing Multiple Choice Questions on “Analysis of LTI System in Z Domain”.
1. What is the unit step response of the system described by the difference equation?
y(n)=0.9y(n-1)-0.81y(n-2)+x(n) under the initial conditions y(-1)=y(-2)=0?
A. [1.099+1.088(0.9)n.(cos(frac{pi}{3} n+5.2^o))]u(n)
B. [1.099+1.088(0.9)n.(cos(frac{pi}{3} n-5.2^o))]u(n)
C. [1.099+1.088(0.9)n.(cos(frac{pi}{3} n-5.2^o))]
D. None of the mentioned
Answer: B
Clarification: The system function is H(z)=(frac{1}{1-0.9z^{-1}+0.81z^{-2}})
The system has two complex-conjugate poles at p1=0.9ejπ/3 and p2=0.9e -jπ/3
The z-transform of the unit step sequence is
X(z)=(frac{1}{1-z^{-1}})
Therefore,
Yzs(z) = (frac{1}{(1-0.9e^{jπ/3} z^{-1})(1-0.9e{-jπ/3} z^{-1})(1-z^{-1})})
=(frac{0.542-j0.049}{(1-0.9e^{jπ/3} z^{-1})} + frac{0.542-j0.049}{(1-0.9e^{jπ/3} z^{-1})} + frac{1.099}{1-z^{-1}})
and hence the zero state response is yzs(n)= [1.099+1.088(0.9)n.(cos(frac{pi}{3} n-5.2^o))]u(n)
Since the initial conditions are zero in this case, we can conclude that y(n)=yzs(n).
2. If all the poles of H(z) are outside the unit circle, then the system is said to be _____________
A. Only causal
B. Only BIBO stable
C. BIBO stable and causal
D. None of the mentioned
Answer: D
Clarification: If all the poles of H(z) are outside an unit circle, it means that the system is neither causal nor BIBO stable.
3. If pk, k=1,2,…N are the poles of the system and |pk| < 1 for all k, then the natural response of such a system is called as Transient response.
A. True
B. False
Answer: A
Clarification: If |pk| < 1 for all k, then ynr(n) decays to 0 as n approaches infinity. In such a case we refer to the natural response of the system as the transient response.
4. If all the poles have small magnitudes, then the rate of decay of signal is __________
A. Slow
B. Constant
C. Rapid
D. None of the mentioned
Answer: C
Clarification: If the magnitudes of the poles of the response of any system is very small i.e., almost equal to zero, then the system decays very rapidly.
5. If one or more poles are located near the unit circle, then the rate of decay of signal is _________
A. Slow
B. Constant
C. Rapid
D. None of the mentioned
Answer: A
Clarification: If the magnitudes of the poles of the response of any system is almost equal to one, then the system decays very slowly or the transient will persist for a relatively long time.
6. What is the transient response of the system described by the difference equation y(n)=0.5y(n-1)+x(n) when the input signal is x(n)= 10cos(πn/4)u(n) and the system is initially at rest?
A. (0.5)nu(n)
B. 0.5(6.3)nu(n)
C. 6.3(0.5)n
D. 6.3(0.5)nu(n)
Answer: D
Clarification: The system function for the system is
H(z)=(frac{1}{1-0.5z^{-1}})
and therefore the system has a pole at z=0.5. The z-transform of the input signal is
Y(z)=H(z)X(z)
=(frac{10(1-(frac{1}{sqrt{2}}) z^{-1})}{(1-0.5z^{-1})(1-e^{jπ/4} z^{-1})(1-e^{-jπ}/4 z^{-1})})
=(frac{6.3}{1-0.5z^{-1}} + frac{6.78e^{-j28.7}}{(1-e^{jπ/4} z^{-1})} + frac{6.78e^{j28.7}}{(1-e^{-jπ/4} z^{-1})})
The natural or transient response is
ynr(n)= 6.3(0.5)nu(n)
7. What is the steady-state response of the system described by the difference equation y(n)=0.5y(n-1)+x(n) when the input signal is x(n)= 10cos(πn/4)u(n) and the system is initially at rest?
A. 13.56cos(πn/4 -28.7o)
B. 13.56cos(πn/4 +28.7o)u(n)
C. 13.56cos(πn/4 -28.7o)u(n)
D. None of the mentioned
Answer: C
Clarification: The system function for the system is
H(z)=(frac{1}{1-0.5z^{-1}})
and therefore the system has a pole at z=0.5. The z-transform of the input signal is
Y(z)=H(z)X(z)
=(frac{10(1-(frac{1}{sqrt{2}}) z^{-1})}{(1-0.5z^{-1})(1-e^{jπ/4} z^{-1})(1-e^{-jπ}/4 z^{-1})})
=(frac{6.3}{1-0.5z^{-1}} + frac{6.78e^{-j28.7}}{(1-e^{jπ/4} z^{-1})} + frac{6.78e^{j28.7}}{(1-e^{-jπ/4} z^{-1})})
The forced state response or steady-state response is
yfr(n)=13.56cos(πn/4 -28.70)u(n)