1. If an LTI system is described by the difference equation y(n)=ay(n-1)+bx(n), 0<a<1, then what is the output of the system when input x(n)=\(5+12sin\frac{π}{2}n-20cos(πn+\frac{π}{4})\)?(Given a=0.9 and b=0.1)
a) \(5+0.888sin(\frac{π}{2}n-420)-1.06cos(πn-\frac{π}{4})\)
b) \(5+0.888sin(\frac{π}{2}n-420)+1.06cos(πn+\frac{π}{4})\)
c) \(5+0.888sin(\frac{π}{2}n-420)-1.06cos(πn+\frac{π}{4})\)
d) \(5+0.888sin(\frac{π}{2}n+420)-1.06cos(πn+\frac{π}{4})\)
Explanation: From the given difference equation, we obtain
|H(ω)|=\(\frac{|b|}{\sqrt{1-2acosω+a^2}}\)
We get |H(0)|=1, |H(π/2)|=0.074 and |H(π)|=0.053
θ(0)=0, θ(π/2)=-420 and θ(π)=0 and we know that y(n)=H(ω)x(n)
=>y(n)=\(5+0.888sin(\frac{π}{2}n-42^0)-1.06cos(πn+\frac{π}{4})\)
2. The output of the Linear time invariant system cannot contain the frequency components that are not contained in the input signal.
a) True
b) False
Explanation: If x(n) is the input of an LTI system, then we know that the output of the system y(n) is y(n)= H(ω)x(n) which means the frequency components are just amplified but no new frequency components are added.
3. An LTI system is characterized by its impulse response h(n)=(1/2)nu(n). What is the spectrum of the output signal when the system is excited by the signal x(n)=(1/4)nu(n)?
a) \(\frac{1}{(1-\frac{1}{2} e^{-jω})(1+\frac{1}{4} e^{-jω})}\)
b) \(\frac{1}{(1-\frac{1}{2} e^{-jω})(1-\frac{1}{4} e^{-jω})}\)
c) \(\frac{1}{(1+\frac{1}{2} e^{-jω})(1-\frac{1}{4} e^{-jω})}\)
d) \(\frac{1}{(1+\frac{1}{2} e^{-jω})(1+\frac{1}{4} e^{-jω})}\)
Explanation: The frequency response function of the system is
H(ω) = \(\sum_{n=0}^∞ (\frac{1}{2})^n e^{-jωn}\)
=\(\frac{1}{1-\frac{1}{2} e^{-jω}}\)
Similarly, the input sequence x(n) has a Fourier transform
X(ω)=\(\sum_{n=0}^∞ (\frac{1}{4})^n e^{-jωn}\)
=\(\frac{1}{1-\frac{1}{4} e^{-jω}}\)
Hence the spectrum of the signal at the output of the system is
Y(ω)=X(ω)H(ω)
=\(\frac{1}{(1-\frac{1}{2} e^{-jω})(1-\frac{1}{4} e^{-jω})}\).
4. What is the frequency response of the system described by the system function H(z)=\(\frac{1}{1-0.8z^{-1}}\)?
a) \(\frac{e^{jω}}{e^{jω}-0.8}\)
b) \(\frac{e^{jω}}{e^{jω}+0.8}\)
c) \(\frac{e^{-jω}}{e^{-jω}-0.8}\)
d) None of the mentioned
Explanation: Given H(z)=\(\frac{1}{1-0.8z^{-1}}\)=z/(z-0.8)
Clearly, H(z) has a zero at z=0 and a pole at p=0.8. Hence the frequency response of the system is given as
H(ω)=\(\frac{e^{jω}}{e^{jω}-0.8}\).
5. If a system is said to be invertible, then?
a) One-to-one correspondence between its input and output signals
b) One-to-many correspondence between its input and output signals
c) Many-to-one correspondence between its input and output signals
d) None of the mentioned
Explanation: If we know the output of a system y(n) of a system and if we can determine the input x(n) of the system uniquely, then the system is said to be invertible. That is there should be one-to-one correspondence between the input and output signals.
6. If h(n) is the impulse response of an LTI system T and h1(n) is the impulse response of the inverse system T-1, then which of the following is true?
a) [h(n)*h1(n)].x(n)=x(n)
b) [h(n).h1(n)].x(n)=x(n)
c) [h(n)*h1(n)]*x(n)=x(n)
d) [h(n).h1(n)]*x(n)=x(n)
Explanation: If h(n) is the impulse response of an LTI system T and h1(n) is the impulse response of the inverse system T-1, then we know that h(n)*h1(n)=δ(n)=>[h(n)*h1(n)]*x(n)=x(n).
7. What is the inverse of the system with impulse response h(n)=(1/2)nu(n)?
a) δ(n)+1/2 δ(n-1)
b) δ(n)-1/2 δ(n-1)
c) δ(n)-1/2 δ(n+1)
d) δ(n)+1/2 δ(n+1)
Explanation: Given impulse response is h(n)=(1/2)nu(n)
The system function corresponding to h(n) is
H(z)=\(\frac{1}{1-\frac{1}{2} z^{-1}}\) ROC:|z|>1/2
This system is both stable and causal. Since H(z) is all pole system, its inverse is FIR and is given by the system function
HI(z)=\(1-\frac{1}{2} z^{-1}\)
Hence its impulse response is δ(n)-1/2 δ(n-1).
8. What is the inverse of the system with impulse response h(n)=δ(n)-1/2δ(n-1)?
a) (1/2)nu(n)
b) -(1/2)nu(-n-1)
c) (1/2)nu(n) & -(1/2)nu(-n-1)
d) None of the mentioned
Explanation: The system function of given system is H(z)=\(1-\frac{1}{2} z^{-1}\)
The inverse of the system has a system function as H(z)=\(\frac{1}{1-\frac{1}{2} z^{-1}}\)
Thus it has a zero at origin and a pole at z=1/2.So, two possible cases are |z|>1/2 and |z|<1/2
So, h(n)= (1/2)nu(n) for causal and stable(|z|>1/2)
and h(n)= -(1/2)nu(-n-1) for anti causal and unstable for |z|<1/2.
9. What is the causal inverse of the FIR system with impulse response h(n)=δ(n)-aδ(n-1)?
a) δ(n)-aδ(n-1)
b) δ(n)+aδ(n-1)
c) a-n
d) an
Explanation: Given h(n)= δ(n)-aδ(n-1)
Since h(0)=1, h(1)=-a and h(n)=0 for n≥a, we have
hI(0)=1/h(0)=1.
and
hI(n)=-ahI(n-1) for n≥1
Consequently, hI(1)=a, hI(2)=a2,….hI(n)=an
Which corresponds to a causal IIR system as expected.
10. If the frequency response of an FIR system is given as H(z)=6+z-1-z-2, then the system is ___________
a) Minimum phase
b) Maximum phase
c) Mixed phase
d) None of the mentioned
Explanation: Given H(z)=6+z-1-z-2
By factoring the system function we find the zeros for the system.
The zeros of the given system are at z=-1/2, 1/3
So, the system is minimum phase.