Digital Signal Processing test focuses on “Design of Low Pass Butterworth Filters-2”.
1. What is the cutoff frequency of the Butterworth filter with a pass band gain KP=-1 dB at ΩP=4 rad/sec and stop band attenuation greater than or equal to 20dB at ΩS=8 rad/sec?
A. 3.5787 rad/sec
B. 1.069 rad/sec
C. 6 rad/sec
D. 4.5787 rad/sec
Answer: D
Clarification: We know that the equation for the cutoff frequency of a Butterworth filter is given as
ΩC = (frac{Omega_P}{(10^{-K_P/10}-1)^{1/2N}})
We know that KP=-1 dB, ΩP=4 rad/sec and N=5
Upon substituting the values in the above equation, we get
ΩC=4.5787 rad/sec.
2. What is the system function of the Butterworth filter with specifications as pass band gain KP=-1 dB at ΩP=4 rad/sec and stop band attenuation greater than or equal to 20dB at ΩS=8 rad/sec?
A. (frac{1}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+2012.4})
B. (frac{1}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+1})
C. (frac{2012.4}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+2012.4})
D. None of the mentioned
Answer: C
Clarification: From the given question,
KP=-1 dB, ΩP=4 rad/sec, KS=-20 dB and ΩS=8 rad/sec
We find out order as N=5 and ΩC=4.5787 rad/sec
We know that for a 5th order normalized low pass Butterworth filter, system equation is given as
H5(s)=(frac{1}{(s+1)(s^2+0.618s+1)(s^2+1.618s+1)})
The specified low pass filter is obtained by applying low pass-to-low pass transformation on the normalized low pass filter.
That is, Ha(s)=H5(s)|s→s/Ωc
=H5(s)|s→s/4.5787
upon calculating, we get
Ha(s)=({2012.4}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+2012.4})
3. If H(s)=(frac{1}{s^2+s+1}) represent the transfer function of a low pass filter (not Butterworth) with a pass band of 1 rad/sec, then what is the system function of a low pass filter with a pass band 10 rad/sec?
A. (frac{100}{s^2+10s+100})
B. (frac{s^2}{s^2+s+1})
C. (frac{s^2}{s^2+10s+100})
D. None of the mentioned
Answer: A
Clarification: The low pass-to-low pass transformation is
s→s/Ωu
Hence the required low pass filter is
Ha(s)=H(s)|s→s/10
=(frac{100}{s^2+10s+100}).
4. If H(s)=(frac{1}{s^2+s+1}) represent the transfer function of a low pass filter (not Butterworth) with a pass band of 1 rad/sec, then what is the system function of a high pass filter with a cutoff frequency of 1rad/sec?
A. (frac{100}{s^2+10s+100})
B. (frac{s^2}{s^2+s+1})
C. (frac{s^2}{s^2+10s+100})
D. None of the mentioned
Answer: B
Clarification: The low pass-to-high pass transformation is
s→Ωu/s
Hence the required high pass filter is
Ha(s)= H(s)|s→1/s
=(frac{s^2}{s^2+s+1})
5. If H(s)=(frac{1}{s^2+s+1}) represent the transfer function of a low pass filter (not Butterworth) with a pass band of 1 rad/sec, then what is the system function of a high pass filter with a cutoff frequency of 10 rad/sec?
A. (frac{100}{s^2+10s+100})
B. (frac{s^2}{s^2+s+1})
C. (frac{s^2}{s^2+10s+100})
D. None of the mentioned
Answer: C
Clarification: The low pass-to-high pass transformation is
s→Ωu/s
Hence the required low pass filter is
Ha(s)=H(s)|s→10/s
=(frac{s^2}{s^2+10s+100})
6. If H(s)=(frac{1}{s^2+s+1}) represent the transfer function of a low pass filter (not Butterworth) with a pass band of 1 rad/sec, then what is the system function of a band pass filter with a pass band of 10 rad/sec and a center frequency of 100 rad/sec?
A. (frac{s^2}{s^4+10s^3+20100s^2+10^5 s+1})
B. (frac{100s^2}{s^4+10s^3+20100s^2+10^5 s+1})
C. (frac{s^2}{s^4+10s^3+20100s^2+10^5 s+10^8})
D. (frac{100s^2}{s^4+10s^3+20100s^2+10^5 s+10^8})
Answer: D
Clarification: The low pass-to-band pass transformation is
(srightarrowfrac{s^2+Omega_u Omega_l}{s(Omega_u-Omega_l)})
Thus the required band pass filter has a transform function as
Ha(s)=(frac{100s^2}{s^4+10s^3+20100s^2+10^5 s+10^8}).
7. If H(s)=(frac{1}{s^2+s+1}) represent the transfer function of a low pass filter (not Butterworth) with a pass band of 1 rad/sec, then what is the system function of a stop band filter with a stop band of 2 rad/sec and a center frequency of 10 rad/sec?
A. (frac{(s^2+100)^2}{s^4+2s^3+204s^2+200s+10^4})
B. (frac{(s^2+10)^2}{s^4+2s^3+204s^2+200s+10^4})
C. (frac{(s^2+10)^2}{s^4+2s^3+400s^2+200s+10^4})
D. None of the mentioned